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κεῖται.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Definition'",5]],["Association",["Rule","'VertexLabel'","'D1.5'"],["Rule","'Text'","'A surface is that which has length and breadth only.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'ἐπιφάνεια δέ ἐστιν, ὃ μῆκος καὶ πλάτος μόνον 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κύκλον.'"],["Rule","'GreekTextWordCount'",29],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Definition'",18]],["Association",["Rule","'VertexLabel'","'D1.18'"],["Rule","'Text'","'A semicircle is the figure contained by the diameter and the circumference cut off by it. And the centre of the semicircle is the same as that of the circle.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'ἡμικύκλιον δέ ἐστι τὸ περιεχόμενον σχῆμα ὑπό τε τῆς διαμέτρου καὶ τῆς ἀπολαμβανομένης ὑπ᾽ αὐτῆς περιφερείας. κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό, ὃ καὶ τοῦ κύκλου ἐστίν.'"],["Rule","'GreekTextWordCount'",28],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Definition'",19]],["Association",["Rule","'VertexLabel'","'D1.19'"],["Rule","'Text'","'Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'σχήματα εὐθύγραμμά ἐστι τὰ ὑπὸ εὐθειῶν περιεχόμενα, τρίπλευρα μὲν τὰ ὑπὸ τριῶν, τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων, πολύπλευρα δὲ τὰ ὑπὸ πλειόνων á¼¢ τεσσάρων εὐθειῶν περιεχόμενα.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Definition'",20]],["Association",["Rule","'VertexLabel'","'D1.20'"],["Rule","'Text'","'Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'τῶν δὲ τριπλεύρων σχημάτων ἰσόπλευρον μὲν τρίγωνόν ἐστι τὸ τὰς τρεῖς ἴσας ἔχον πλευράς, ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔχον πλευράς, σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχον πλευράς.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Definition'",21]],["Association",["Rule","'VertexLabel'","'D1.21'"],["Rule","'Text'","'Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acuteangled triangle that which has its three angles acute.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἔτι δὲ τῶν τριπλεύρων σχημάτων ὀρθογώνιον μὲν τρίγωνόν ἐστι τὸ ἔχον ὀρθὴν γωνίαν, ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γωνίαν, ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον γωνίας.'"],["Rule","'GreekTextWordCount'",27],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Definition'",22]],["Association",["Rule","'VertexLabel'","'D1.22'"],["Rule","'Text'","'Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia.'"],["Rule","'TextWordCount'",61],["Rule","'GreekText'","'τῶν δὲ τετραπλεύρων σχημάτων τετράγωνον μέν ἐστιν, ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον, ἑτερόμηκες δέ, ὃ ὀρθογώνιον μέν, οὐκ ἰσόπλευρον δέ, ῥόμβος δέ, ὃ ἰσόπλευρον μέν, οὐκ ὀρθογώνιον δέ, ῥομβοειδὲς δὲ τὸ τὰς ἀπεναντίον πλευράς τε καὶ γωνίας ἴσας ἀλλήλαις ἔχον, ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώνιον: τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια καλείσθω.'"],["Rule","'GreekTextWordCount'",54],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Definition'",23]],["Association",["Rule","'VertexLabel'","'D1.23'"],["Rule","'Text'","'Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction.'"],["Rule","'TextWordCount'",27],["Rule","'GreekText'","'παράλληλοί εἰσιν εὐθεῖαι, αἵτινες ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι καὶ ἐκβαλλόμεναι εἰς ἄπειρον ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Definition'",1]],["Association",["Rule","'VertexLabel'","'D2.1'"],["Rule","'Text'","'Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'πᾶν παραλληλόγραμμον ὀρθογώνιον περιέχεσθαι λέγεται ὑπὸ δύο τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν εὐθειῶν.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Definition'",2]],["Association",["Rule","'VertexLabel'","'D2.2'"],["Rule","'Text'","'And in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a gnomon.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'παντὸς δὲ παραλληλογράμμου χωρίου τῶν περὶ τὴν διάμετρον αὐτοῦ παραλληλογράμμων ἓν ὁποιονοῦν σὺν τοῖς δυσὶ παραπληρώμασι γνώμων καλείσθω.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",1]],["Association",["Rule","'VertexLabel'","'D3.1'"],["Rule","'Text'","'Equal circles are those the diameters of which are equal, or the radii of which are equal.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἴσοι κύκλοι εἰσίν, ὧν αἱ διάμετροι ἴσαι εἰσίν, á¼¢ ὧν αἱ ἐκ τῶν κέντρων ἴσαι εἰσίν.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",2]],["Association",["Rule","'VertexLabel'","'D3.2'"],["Rule","'Text'","'A straight line is said to touch a circle which, meeting the circle and being produced, does not cut the circle.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'εὐθεῖα κύκλου ἐφάπτεσθαι λέγεται, ἥτις ἁπτομένη τοῦ κύκλου καὶ ἐκβαλλομένη οὐ τέμνει τὸν κύκλον.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",3]],["Association",["Rule","'VertexLabel'","'D3.3'"],["Rule","'Text'","'Circles are said to touch one another which, meeting one another, do not cut one another.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'κύκλοι ἐφάπτεσθαι ἀλλήλων λέγονται οἵτινες ἁπτόμενοι ἀλλήλων οὐ τέμνουσιν ἀλλήλους.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",4]],["Association",["Rule","'VertexLabel'","'D3.4'"],["Rule","'Text'","'In a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ἐν κύκλῳ ἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι λέγονται, ὅταν αἱ ἀπὸ τοῦ κέντρου ἐπ᾽ αὐτὰς κάθετοι ἀγόμεναι ἴσαι ὦσιν.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",5]],["Association",["Rule","'VertexLabel'","'D3.5'"],["Rule","'Text'","'And that straight line is said to be at a greater distance on which the greater perpendicular falls.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'μεῖζον δὲ ἀπέχειν λέγεται, ἐφ᾽ ἣν ἡ μείζων κάθετος πίπτει.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",6]],["Association",["Rule","'VertexLabel'","'D3.6'"],["Rule","'Text'","'A segment of a circle is the figure contained by a straight line and a circumference of a circle.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'τμῆμα κύκλου ἐστὶ τὸ περιεχόμενον σχῆμα ὑπό τε εὐθείας καὶ κύκλου περιφερείας.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",7]],["Association",["Rule","'VertexLabel'","'D3.7'"],["Rule","'Text'","'An angle of a segment is that contained by a straight line and a circumference of a circle.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'τμήματος δὲ γωνία ἐστὶν ἡ περιεχομένη ὑπό τε εὐθείας καὶ κύκλου περιφερείας.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",8]],["Association",["Rule","'VertexLabel'","'D3.8'"],["Rule","'Text'","'An angle in a segment is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the extremities of the straight line which is the base of the segment, is contained by the straight lines so joined.'"],["Rule","'TextWordCount'",49],["Rule","'GreekText'","'ἐν τμήματι δὲ γωνία ἐστίν, ὅταν ἐπὶ τῆς περιφερείας τοῦ τμήματος ληφθῇ τι σημεῖον καὶ ἀπ᾽ αὐτοῦ ἐπὶ τὰ πέρατα τῆς εὐθείας, á¼¥ ἐστι βάσις τοῦ τμήματος, ἐπιζευχθῶσιν εὐθεῖαι, ἡ περιεχομένη γωνία ὑπὸ τῶν ἐπιζευχθεισῶν εὐθειῶν.'"],["Rule","'GreekTextWordCount'",37],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",9]],["Association",["Rule","'VertexLabel'","'D3.9'"],["Rule","'Text'","'And, when the straight lines containing the angle cut off a circumference, the angle is said to stand upon that circumference.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'ὅταν δὲ αἱ περιέχουσαι τὴν γωνίαν εὐθεῖαι ἀπολαμβάνωσί τινα περιφέρειαν, ἐπ᾽ ἐκείνης λέγεται βεβηκέναι ἡ γωνία.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",10]],["Association",["Rule","'VertexLabel'","'D3.10'"],["Rule","'Text'","'A sector of a circle is the figure which, when an angle is constructed at the centre of the circle, is contained by the straight lines containing the angle and the circumference cut off by them.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'Τομεὺς δὲ κύκλου ἐστίν, ὅταν πρὸς τῷ κέντρῳ τοῦ κύκλου συσταθῇ γωνία, τὸ περιεχόμενον σχῆμα ὑπό τε τῶν τὴν γωνίαν περιεχουσῶν εὐθειῶν καὶ τῆς ἀπολαμβανομένης ὑπ᾽ αὐτῶν περιφερείας.'"],["Rule","'GreekTextWordCount'",29],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Definition'",11]],["Association",["Rule","'VertexLabel'","'D3.11'"],["Rule","'Text'","'Similar segments of circles are those which admit equal angles, or in which the angles are equal to one another.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ὅμοια τμήματα κύκλων ἐστὶ τὰ δεχόμενα γωνίας ἴσας, á¼¢ ἐν οἷς αἱ γωνίαι ἴσαι ἀλλήλαις εἰσίν.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Definition'",1]],["Association",["Rule","'VertexLabel'","'D4.1'"],["Rule","'Text'","'A rectilineal figure is said to be inscribed in a rectilineal figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'σχῆμα εὐθύγραμμον εἰς σχῆμα εὐθύγραμμον ἐγγράφεσθαι λέγεται, ὅταν ἑκάστη τῶν τοῦ ἐγγραφομένου σχήματος γωνιῶν ἑκάστης πλευρᾶς τοῦ, εἰς ὃ ἐγγράφεται, ἅπτηται.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Definition'",2]],["Association",["Rule","'VertexLabel'","'D4.2'"],["Rule","'Text'","'Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'σχῆμα δὲ ὁμοίως περὶ σχῆμα περιγράφεσθαι λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου ἑκάστης γωνίας τοῦ, περὶ ὃ περιγράφεται, ἅπτηται.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Definition'",3]],["Association",["Rule","'VertexLabel'","'D4.3'"],["Rule","'Text'","'A rectilineal figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'σχῆμα εὐθύγραμμον εἰς κύκλον ἐγγράφεσθαι λέγεται, ὅταν ἑκάστη γωνία τοῦ ἐγγραφομένου ἅπτηται τῆς τοῦ κύκλου περιφερείας.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Definition'",4]],["Association",["Rule","'VertexLabel'","'D4.4'"],["Rule","'Text'","'A rectilineal figure is said to be circumscribed about a circle, when each side of the circumscribed figure touches the circumference of the circle.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'σχῆμα δὲ εὐθύγραμμον περὶ κύκλον περιγράφεσθαι λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου ἐφάπτηται τῆς τοῦ κύκλου περιφερείας.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Definition'",5]],["Association",["Rule","'VertexLabel'","'D4.5'"],["Rule","'Text'","'Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'κύκλος δὲ εἰς σχῆμα ὁμοίως ἐγγράφεσθαι λέγεται, ὅταν ἡ τοῦ κύκλου περιφέρεια ἑκάστης πλευρᾶς τοῦ, εἰς ὃ ἐγγράφεται, ἅπτηται.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Definition'",6]],["Association",["Rule","'VertexLabel'","'D4.6'"],["Rule","'Text'","'A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'κύκλος δὲ περὶ σχῆμα περιγράφεσθαι λέγεται, ὅταν ἡ τοῦ κύκλου περιφέρεια ἑκάστης γωνίας τοῦ, περὶ ὃ περιγράφεται, ἅπτηται.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Definition'",7]],["Association",["Rule","'VertexLabel'","'D4.7'"],["Rule","'Text'","'A straight line is said to be fitted into a circle when its extremities are on the circumference of the circle.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'εὐθεῖα εἰς κύκλον ἐναρμόζεσθαι λέγεται, ὅταν τὰ πέρατα αὐτῆς ἐπὶ τῆς περιφερείας ᾖ τοῦ κύκλου.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",1]],["Association",["Rule","'VertexLabel'","'D5.1'"],["Rule","'Text'","'A magnitude is a part of a magnitude, the less of the greater, when it measures the greater.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'μέρος ἐστὶ μέγεθος μεγέθους τὸ ἔλασσον τοῦ μείζονος, ὅταν καταμετρῇ τὸ μεῖζον.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",2]],["Association",["Rule","'VertexLabel'","'D5.2'"],["Rule","'Text'","'The greater is a multiple of the less when it is measured by the less.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'πολλαπλάσιον δὲ τὸ μεῖζον τοῦ ἐλάττονος, ὅταν καταμετρῆται ὑπὸ τοῦ ἐλάττονος.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",3]],["Association",["Rule","'VertexLabel'","'D5.3'"],["Rule","'Text'","'A ratio is a sort of relation in respect of size between two magnitudes of the same kind.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'λόγος ἐστὶ δύο μεγεθῶν ὁμογενῶν ἡ κατὰ πηλικότητά ποια σχέσις.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",4]],["Association",["Rule","'VertexLabel'","'D5.4'"],["Rule","'Text'","'Magnitudes are said to have a ratio to one another which are capable, when multiplied, of exceeding one another.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'λόγον ἔχειν πρὸς ἄλληλα μεγέθη λέγεται, ἃ δύναται πολλαπλασιαζόμενα ἀλλήλων ὑπερέχειν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",5]],["Association",["Rule","'VertexLabel'","'D5.5'"],["Rule","'Text'","'Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever be taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order.'"],["Rule","'TextWordCount'",63],["Rule","'GreekText'","'ἐν τῷ αὐτῷ λόγῳ μεγέθη λέγεται εἶναι πρῶτον πρὸς δεύτερον καὶ τρίτον πρὸς τέταρτον, ὅταν τὰ τοῦ πρώτου καὶ τρίτου ἰσάκις πολλαπλάσια τῶν τοῦ δευτέρου καὶ τετάρτου ἰσάκις πολλαπλασίων καθ᾽ ὁποιονοῦν πολλαπλασιασμὸν ἑκάτερον ἑκατέρου á¼¢ ἅμα ὑπερέχῃ á¼¢ ἅμα ἴσα ᾖ á¼¢ ἅμα ἐλλείπῃ ληφθέντα κατάλληλα.'"],["Rule","'GreekTextWordCount'",47],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",6]],["Association",["Rule","'VertexLabel'","'D5.6'"],["Rule","'Text'","'Let magnitudes which have the same ratio be called proportional.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'τὰ δὲ τὸν αὐτὸν ἔχοντα λόγον μεγέθη ἀνάλογον καλείσθω.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",7]],["Association",["Rule","'VertexLabel'","'D5.7'"],["Rule","'Text'","'When, of the equimultiples, the multiple of the first magnitude exceeds the multiple of the second, but the multiple of the third does not exceed the multiple of the fourth, then the first is said to have a greater ratio to the second than the third has to the fourth.'"],["Rule","'TextWordCount'",50],["Rule","'GreekText'","'ὅταν δὲ τῶν ἰσάκις πολλαπλασίων τὸ μὲν τοῦ πρώτου πολλαπλάσιον ὑπερέχῃ τοῦ τοῦ δευτέρου πολλαπλασίου, τὸ δὲ τοῦ τρίτου πολλαπλάσιον μὴ ὑπερέχῃ τοῦ τοῦ τετάρτου πολλαπλασίου, τότε τὸ πρῶτον πρὸς τὸ δεύτερον μείζονα λόγον ἔχειν λέγεται, ἤπερ τὸ τρίτον πρὸς τὸ τέταρτον.'"],["Rule","'GreekTextWordCount'",42],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",8]],["Association",["Rule","'VertexLabel'","'D5.8'"],["Rule","'Text'","'A proportion in three terms is the least possible.'"],["Rule","'TextWordCount'",9],["Rule","'GreekText'","'ἀναλογία δὲ ἐν τρισὶν ὅροις ἐλαχίστη ἐστίν.'"],["Rule","'GreekTextWordCount'",7],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",9]],["Association",["Rule","'VertexLabel'","'D5.9'"],["Rule","'Text'","'When three magnitudes are proportional, the first is said to have to the third the duplicate ratio of that which it has to the second.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ὅταν δὲ τρία μεγέθη ἀνάλογον ᾖ, τὸ πρῶτον πρὸς τὸ τρίτον διπλασίονα λόγον ἔχειν λέγεται ἤπερ πρὸς τὸ δεύτερον.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",10]],["Association",["Rule","'VertexLabel'","'D5.10'"],["Rule","'Text'","'When four magnitudes are continuously proportional, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on continually, whatever be the proportion.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ὅταν δὲ τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ πρῶτον πρὸς τὸ τέταρτον τριπλασίονα λόγον ἔχειν λέγεται ἤπερ πρὸς τὸ δεύτερον, καὶ ἀεὶ ἑξῆς ὁμοίως, ὡς ἂν ἡ ἀναλογία ὑπάρχῃ.'"],["Rule","'GreekTextWordCount'",28],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",11]],["Association",["Rule","'VertexLabel'","'D5.11'"],["Rule","'Text'","'The term corresponding magnitudes is used of antecedents in relation to antecedents, and of consequents in relation to consequents.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ὁμόλογα μεγέθη λέγεται τὰ μὲν ἡγούμενα τοῖς ἡγουμένοις τὰ δὲ ἑπόμενα τοῖς ἑπομένοις.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",12]],["Association",["Rule","'VertexLabel'","'D5.12'"],["Rule","'Text'","'Alternate ratio means taking the antecedent in relation to the antecedent and the consequent in relation to the consequent.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἐναλλὰξ λόγος ἐστὶ λῆψις τοῦ ἡγουμένου πρὸς τὸ ἡγούμενον καὶ τοῦ ἑπομένου πρὸς τὸ ἑπόμενον.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",13]],["Association",["Rule","'VertexLabel'","'D5.13'"],["Rule","'Text'","'Inverse ratio means taking the consequent as antecedent in relation to the antecedent as consequent.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἀνάπαλιν λόγος ἐστὶ λῆψις τοῦ ἑπομένου ὡς ἡγουμένου πρὸς τὸ ἡγούμενον ὡς ἑπόμενον.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",14]],["Association",["Rule","'VertexLabel'","'D5.14'"],["Rule","'Text'","'Composition of a ratio means taking the antecedent together with the consequent as one in relation to the consequent by itself.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'σύνθεσις λόγου ἐστὶ λῆψις τοῦ ἡγουμένου μετὰ τοῦ ἑπομένου ὡς ἑνὸς πρὸς αὐτὸ τὸ ἑπόμενον.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",15]],["Association",["Rule","'VertexLabel'","'D5.15'"],["Rule","'Text'","'Separation of a ratio means taking the excess by which the antecedent exceeds the consequent in relation to the consequent by itself.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'διαίρεσις λόγου ἐστὶ λῆψις τῆς ὑπεροχῆς, ᾗ ὑπερέχει τὸ ἡγούμενον τοῦ ἑπομένου, πρὸς αὐτὸ τὸ ἑπόμενον.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",16]],["Association",["Rule","'VertexLabel'","'D5.16'"],["Rule","'Text'","'Conversion of a ratio means taking the antecedent in relation to the excess by which the antecedent exceeds the consequent.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ἀναστροφὴ λόγου ἐστὶ λῆψις τοῦ ἡγουμένου πρὸς τὴν ὑπεροχήν, ᾗ ὑπερέχει τὸ ἡγούμενον τοῦ ἑπομένου.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",17]],["Association",["Rule","'VertexLabel'","'D5.17'"],["Rule","'Text'","'A ratio ex aequali arises when, there being several magnitudes and another set equal to them in multitude which taken two and two are in the same proportion, as the first is to the last among the first magnitudes, so is the first to the last among the second magnitudes; Or, in other words, it means taking the extreme terms by virtue of the removal of the intermediate terms.'"],["Rule","'TextWordCount'",69],["Rule","'GreekText'","'δι᾽ ἴσου λόγος ἐστὶ πλειόνων ὄντων μεγεθῶν καὶ ἄλλων αὐτοῖς ἴσων τὸ πλῆθος σύνδυο λαμβανομένων καὶ ἐν τῷ αὐτῷ λόγῳ, ὅταν ᾖ ὡς ἐν τοῖς πρώτοις μεγέθεσι τὸ πρῶτον πρὸς τὸ ἔσχατον, οὕτως ἐν τοῖς δευτέροις μεγέθεσι τὸ πρῶτον πρὸς τὸ ἔσχατον: á¼¢ ἄλλως: λῆψις τῶν ἄκρων καθ᾽ ὑπεξαίρεσιν τῶν μέσων.'"],["Rule","'GreekTextWordCount'",53],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Definition'",18]],["Association",["Rule","'VertexLabel'","'D5.18'"],["Rule","'Text'","'A perturbed proportion arises when, there being three magnitudes and another set equal to them in multitude, as antecedent is to consequent among the first magnitudes, so is antecedent to consequent among the second magnitudes, while, as the consequent is to a third among the first magnitudes, so is a third to the antecedent among the second magnitudes.'"],["Rule","'TextWordCount'",58],["Rule","'GreekText'","'τεταραγμένη δὲ ἀναλογία ἐστίν, ὅταν τριῶν ὄντων μεγεθῶν καὶ ἄλλων αὐτοῖς ἴσων τὸ πλῆθος γίνηται ὡς μὲν ἐν τοῖς πρώτοις μεγέθεσιν ἡγούμενον πρὸς ἑπόμενον, οὕτως ἐν τοῖς δευτέροις μεγέθεσιν ἡγούμενον πρὸς ἑπόμενον, ὡς δὲ ἐν τοῖς πρώτοις μεγέθεσιν ἑπόμενον πρὸς ἄλλο τι, οὕτως ἐν τοῖς δευτέροις ἄλλο τι πρὸς ἡγούμενον.'"],["Rule","'GreekTextWordCount'",50],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Definition'",1]],["Association",["Rule","'VertexLabel'","'D6.1'"],["Rule","'Text'","'Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ὅμοια σχήματα εὐθύγραμμά ἐστιν, ὅσα τάς τε γωνίας ἴσας ἔχει κατὰ μίαν καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Definition'",2]],["Association",["Rule","'VertexLabel'","'D6.2'"],["Rule","'Text'","'When two sides of one figure together with two sides of another figure form antecedents and consequents in a proportion, the figures are reciprocally related.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'Ἀντιπεπονθότα δὲ σχήματά ἐστιν, ὅταν ἐν ἑκατέρῳ τῶν σχημάτων ἡγούμενοί τε καὶ ἑπόμενοι λόγοι ὦσιν.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Definition'",3]],["Association",["Rule","'VertexLabel'","'D6.3'"],["Rule","'Text'","'A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἄκρον καὶ μέσον λόγον εὐθεῖα τετμῆσθαι λέγεται, ὅταν ᾖ ὡς ἡ ὅλη πρὸς τὸ μεῖζον τμῆμα, οὕτως τὸ μεῖζον πρὸς τὸ ἔλαττον.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Definition'",4]],["Association",["Rule","'VertexLabel'","'D6.4'"],["Rule","'Text'","'The height of any figure is the perpendicular drawn from the vertex to the base.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ὕψος ἐστὶ παντὸς σχήματος ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν βάσιν κάθετος ἀγομένη.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Definition'",5]],["Association",["Rule","'VertexLabel'","'D6.5'"],["Rule","'Text'","'A ratio is said to be compounded of ratios when the sizes of the ratios multiplied together make som (?ratio, or size). A parallelogram not \"filling\" a line is a figure lesser than the line. If it has a surplus and the line is not enough, it is a figure greater than the line.'"],["Rule","'TextWordCount'",54],["Rule","'GreekText'","'Λόγος ἐκ λόγων συγκεῖσθαι λέγεται, ὅταν αἱ τῶν λόγων πηλικότητες ἐφ᾽ ἑαυτὰς πολλαπλασιασθεῖσαι ποιῶσί τινα.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",1]],["Association",["Rule","'VertexLabel'","'D7.1'"],["Rule","'Text'","'An unit is that by virtue of which each of the things that exist is called one.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'μονάς ἐστιν, καθ᾽ ἣν ἕκαστον τῶν ὄντων ἓν λέγεται.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",2]],["Association",["Rule","'VertexLabel'","'D7.2'"],["Rule","'Text'","'A number is a multitude composed of units.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'ἀριθμὸς δὲ τὸ ἐκ μονάδων συγκείμενον 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Περισσάκις ἀρτιός ἐστιν ὁ ὑπὸ περισσοῦ ἀριθμοῦ μετρούμενος κατὰ ἄρτιον ἀριθμόν.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",10]],["Association",["Rule","'VertexLabel'","'D7.10'"],["Rule","'Text'","'An odd-times odd number is that which is measured by an odd number according to an odd number.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'περισσάκις δὲ περισσὸς ἀριθμός ἐστιν ὁ ὑπὸ περισσοῦ ἀριθμοῦ μετρούμενος κατὰ περισσὸν 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μετρούμενος.'"],["Rule","'GreekTextWordCount'",7],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",12]],["Association",["Rule","'VertexLabel'","'D7.12'"],["Rule","'Text'","'Numbers prime to one another are those which are measured by an unit alone as a common measure.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'πρῶτοι πρὸς ἀλλήλους ἀριθμοί εἰσιν οἱ μονάδι μόνῃ μετρούμενοι κοινῷ μέτρῳ.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",13]],["Association",["Rule","'VertexLabel'","'D7.13'"],["Rule","'Text'","'A composite number is that which is measured by some number.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'σύνθετος ἀριθμός ἐστιν ὁ ἀριθμῷ τινι μετρούμενος.'"],["Rule","'GreekTextWordCount'",7],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",14]],["Association",["Rule","'VertexLabel'","'D7.14'"],["Rule","'Text'","'Numbers composite to one another are those which are measured by some number as a common measure.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'σύνθετοι δὲ πρὸς ἀλλήλους ἀριθμοί εἰσιν οἱ ἀριθμῷ τινι μετρούμενοι κοινῷ 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τις.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",16]],["Association",["Rule","'VertexLabel'","'D7.16'"],["Rule","'Text'","'And, when two numbers having multiplied one another make some number, the number so produced is called plane, and its sides are the numbers which have multiplied one another.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'ὅταν δὲ δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος ἐπίπεδος καλεῖται, πλευραὶ δὲ αὐτοῦ οἱ πολλαπλασιάσαντες ἀλλήλους ἀριθμοί.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",17]],["Association",["Rule","'VertexLabel'","'D7.17'"],["Rule","'Text'","'And, when three numbers having multiplied one another make some number, the number so produced is solid, and its sides are the numbers which have multiplied one another.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ὅταν δὲ τρεῖς ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος στερεός ἐστιν, πλευραὶ δὲ αὐτοῦ οἱ πολλαπλασιάσαντες ἀλλήλους 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ὦσιν.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Definition'",21]],["Association",["Rule","'VertexLabel'","'D7.21'"],["Rule","'Text'","'Similar plane and solid numbers are those which have their sides proportional.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'ὅμοιοι ἐπίπεδοι καὶ στερεοὶ ἀριθμοί εἰσιν οἱ ἀνάλογον ἔχοντες τὰς 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ὤν.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",1.1]],["Association",["Rule","'VertexLabel'","'D10.1.1'"],["Rule","'Text'","'Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'σύμμετρα μεγέθη λέγεται τὰ τῷ αὐτῷ μέτρῳ μετρούμενα, ἀσύμμετρα δέ, ὧν μηδὲν ἐνδέχεται κοινὸν μέτρον 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Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square or in square only, rational, but those which are incommensurable with it irrational.'"],["Rule","'TextWordCount'",71],["Rule","'GreekText'","'τούτων ὑποκειμένων δείκνυται, ὅτι τῇ προτεθείσῃ εὐθείᾳ ὑπάρχουσιν εὐθεῖαι πλήθει ἄπειροι σύμμετροί τε καὶ ἀσύμμετροι αἱ μὲν μήκει μόνον, αἱ δὲ καὶ δυνάμει. καλείσθω οὖν ἡ μὲν προτεθεῖσα εὐθεῖα ῥητή, καὶ αἱ ταύτῃ σύμμετροι εἴτε μήκει καὶ δυνάμει εἴτε δυνάμει μόνον ῥηταί, αἱ δὲ ταύτῃ ἀσύμμετροι ἄλογοι 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ῥητόν, καὶ τὰ τούτῳ σύμμετρα ῥητά, τὰ δὲ τούτῳ ἀσύμμετρα ἄλογα καλείσθω, καὶ αἱ δυνάμεναι αὐτὰ ἄλογοι, εἰ μὲν τετράγωνα εἴη, αὐταὶ αἱ πλευραί, εἰ δὲ ἕτερά τινα εὐθύγραμμα, αἱ ἴσα αὐτοῖς τετράγωνα ἀναγράφουσαι.'"],["Rule","'GreekTextWordCount'",42],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",2.1]],["Association",["Rule","'VertexLabel'","'D10.2.1'"],["Rule","'Text'","'Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line;'"],["Rule","'TextWordCount'",67],["Rule","'GreekText'","'ὑποκειμένης ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων διῃρημένης εἰς τὰ ὀνόματα, ἧς τὸ μεῖζον ὄνομα τοῦ ἐλάσσονος μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει, ἐὰν μὲν τὸ μεῖζον ὄνομα σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἡ ὅλη ἐκ δύο ὀνομάτων πρώτη.'"],["Rule","'GreekTextWordCount'",42],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",2.2]],["Association",["Rule","'VertexLabel'","'D10.2.2'"],["Rule","'Text'","'but if the lesser term be commensurable in length with the rational straight line set out, let the whole be called a second binomial;'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'ἐὰν δὲ τὸ ἔλασσον ὄνομα σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ δύο ὀνομάτων δευτέρα.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",2.3]],["Association",["Rule","'VertexLabel'","'D10.2.3'"],["Rule","'Text'","'and if neither of the terms be commensurable in length with the rational straight line set out, let the whole be called a third binomial.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ἐὰν δὲ μηδέτερον τῶν ὀνομάτων σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ δύο ὀνομάτων τρίτη.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",2.4]],["Association",["Rule","'VertexLabel'","'D10.2.4'"],["Rule","'Text'","'Again, if the square on the greater term be greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a fourth binomial;'"],["Rule","'TextWordCount'",53],["Rule","'GreekText'","'πάλιν δὴ ἐὰν τὸ μεῖζον ὄνομα τοῦ ἐλάσσονος μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν μὲν τὸ μεῖζον ὄνομα σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ δύο ὀνομάτων τετάρτη.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",2.5]],["Association",["Rule","'VertexLabel'","'D10.2.5'"],["Rule","'Text'","'if the lesser, a fifth binomial;'"],["Rule","'TextWordCount'",6],["Rule","'GreekText'","'ἐὰν δὲ τὸ ἔλασσον, πέμπτη.'"],["Rule","'GreekTextWordCount'",5],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",2.6]],["Association",["Rule","'VertexLabel'","'D10.2.6'"],["Rule","'Text'","'and if neither, a sixth binomial.'"],["Rule","'TextWordCount'",6],["Rule","'GreekText'","'ἐὰν δὲ μηδέτερον, ἕκτη.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",3.1]],["Association",["Rule","'VertexLabel'","'D10.3.1'"],["Rule","'Text'","'Given a rational straight line and an apotome, if the square on the whole be greater than the square on the annex by the square on a straight line commensurable in length with the whole, and the whole be commensurable in length with the rational straight line set out, let the apotome be called a first apotome.'"],["Rule","'TextWordCount'",57],["Rule","'GreekText'","'ὑποκειμένης ῥητῆς καὶ ἀποτομῆς, ἐὰν μὲν ἡ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει, καὶ ἡ ὅλη σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, καλείσθω ἀποτομὴ πρώτη.'"],["Rule","'GreekTextWordCount'",29],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",3.2]],["Association",["Rule","'VertexLabel'","'D10.3.2'"],["Rule","'Text'","'But if the annex be commensurable in length with the rational straight line set out, and the square on the whole be greater than that on the annex by the square on a straight line commensurable with the whole, let the apotome be called a second apotome.'"],["Rule","'TextWordCount'",47],["Rule","'GreekText'","'ἐὰν δὲ ἡ προσαρμόζουσα σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καλείσθω ἀποτομὴ δευτέρα.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",3.3]],["Association",["Rule","'VertexLabel'","'D10.3.3'"],["Rule","'Text'","'But if neither be commensurable in length with the rational straight line set out, and the square on the whole be greater than the square on the annex by the square on a straight line commensurable with the whole, let the apotome be called a third apotome.'"],["Rule","'TextWordCount'",47],["Rule","'GreekText'","'ἐὰν δὲ μηδετέρα σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, ἡ δὲ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καλείσθω ἀποτομὴ τρίτη.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",3.4]],["Association",["Rule","'VertexLabel'","'D10.3.4'"],["Rule","'Text'","'Again, if the square on the whole be greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole be commensurable in length with the rational straight line set out, let the apotome be called a fourth apotome;'"],["Rule","'TextWordCount'",49],["Rule","'GreekText'","'πάλιν, ἐὰν ἡ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν μὲν ἡ ὅλη σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, καλείσθω ἀποτομὴ τετάρτη.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",3.5]],["Association",["Rule","'VertexLabel'","'D10.3.5'"],["Rule","'Text'","'if the annex be so commensurable, a fifth;'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'ἐὰν δὲ ἡ προσαρμόζουσα, πέμπτη.'"],["Rule","'GreekTextWordCount'",5],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Definition'",3.6]],["Association",["Rule","'VertexLabel'","'D10.3.6'"],["Rule","'Text'","'and, if neither, a sixth.'"],["Rule","'TextWordCount'",5],["Rule","'GreekText'","'ἐὰν δὲ μηδετέρα, ἕκτη.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",1]],["Association",["Rule","'VertexLabel'","'D11.1'"],["Rule","'Text'","'A solid is that which has length, breadth, and depth.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'στερεόν ἐστι τὸ μῆκος καὶ πλάτος καὶ βάθος ἔχον.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",2]],["Association",["Rule","'VertexLabel'","'D11.2'"],["Rule","'Text'","'An extremity of a solid is a surface.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'στερεοῦ δὲ πέρας ἐπιφάνεια.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",3]],["Association",["Rule","'VertexLabel'","'D11.3'"],["Rule","'Text'","'A straight line is at right angles to a plane, when it makes right angles with all the straight lines which meet it and are in the plane.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'εὐθεῖα πρὸς ἐπίπεδον ὀρθή ἐστιν, ὅταν πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιῇ γωνίας.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",4]],["Association",["Rule","'VertexLabel'","'D11.4'"],["Rule","'Text'","'A plane is at right angles to a plane when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐπίπεδον πρὸς ἐπίπεδον ὀρθόν ἐστιν, ὅταν αἱ τῇ κοινῇ τομῇ τῶν ἐπιπέδων πρὸς ὀρθὰς ἀγόμεναι εὐθεῖαι ἐν ἑνὶ τῶν ἐπιπέδων τῷ λοιπῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",5]],["Association",["Rule","'VertexLabel'","'D11.5'"],["Rule","'Text'","'The inclination of a straight line to a plane is, assuming a perpendicular drawn from the extremity of the straight line which is elevated above the plane to the plane, and a straight line joined from the point thus arising to the extremity of the straight line which is in the plane, the angle contained by the straight line so drawn and the straight line standing up.'"],["Rule","'TextWordCount'",67],["Rule","'GreekText'","'εὐθείας πρὸς ἐπίπεδον κλίσις ἐστίν, ὅταν ἀπὸ τοῦ μετεώρου πέρατος τῆς εὐθείας ἐπὶ τὸ ἐπίπεδον κάθετος ἀχθῇ, καὶ ἀπὸ τοῦ γενομένου σημείου ἐπὶ τὸ ἐν τῷ ἐπιπέδῳ πέρας τῆς εὐθείας εὐθεῖα ἐπιζευχθῇ, ἡ περιεχομένη γωνία ὑπὸ τῆς ἀχθείσης καὶ τῆς ἐφεστώσης.'"],["Rule","'GreekTextWordCount'",41],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",6]],["Association",["Rule","'VertexLabel'","'D11.6'"],["Rule","'Text'","'The inclination of a plane to a plane is the acute angle contained by the straight lines drawn at right angles to the common section at the same point, one in each of the planes.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'ἐπιπέδου πρὸς ἐπίπεδον κλίσις ἐστὶν ἡ περιεχομένη ὀξεῖα γωνία ὑπὸ τῶν πρὸς ὀρθὰς τῇ κοινῇ τομῇ ἀγομένων πρὸς τῷ αὐτῷ σημείῳ ἐν ἑκατέρῳ τῶν ἐπιπέδων.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",7]],["Association",["Rule","'VertexLabel'","'D11.7'"],["Rule","'Text'","'A plane is said to be similarly inclined to a plane as another is to another when the said angles of the inclinations are equal to one another.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐπίπεδον πρὸς ἐπίπεδον ὁμοίως κεκλίσθαι λέγεται καὶ ἕτερον πρὸς ἕτερον, ὅταν αἱ εἰρημέναι τῶν κλίσεων γωνίαι ἴσαι ἀλλήλαις ὦσιν.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",8]],["Association",["Rule","'VertexLabel'","'D11.8'"],["Rule","'Text'","'Parallel planes are those which do not meet.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'παράλληλα ἐπίπεδά ἐστι τὰ ἀσύμπτωτα.'"],["Rule","'GreekTextWordCount'",5],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",9]],["Association",["Rule","'VertexLabel'","'D11.9'"],["Rule","'Text'","'Similar solid figures are those contained by similar planes equal in multitude.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'ὅμοια στερεὰ σχήματά ἐστι τὰ ὑπὸ ὁμοίων ἐπιπέδων περιεχόμενα ἴσων τὸ πλῆθος.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",10]],["Association",["Rule","'VertexLabel'","'D11.10'"],["Rule","'Text'","'Equal and similar solid figures are those contained by similar planes equal in multitude and in magnitude.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἴσα δὲ καὶ ὅμοια στερεὰ σχήματά ἐστι τὰ ὑπὸ ὁμοίων ἐπιπέδων περιεχόμενα ἴσων τῷ πλήθει καὶ τῷ μεγέθει.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",11]],["Association",["Rule","'VertexLabel'","'D11.11'"],["Rule","'Text'","'A solid angle is the inclination constituted by more than two lines which meet one another and are not in the same surface, towards all the lines. Otherwise: A solid angle is that which is contained by more than two plane angles which are not in the same plane and are constructed to one point.'"],["Rule","'TextWordCount'",55],["Rule","'GreekText'","'στερεὰ γωνία ἐστὶν ἡ ὑπὸ πλειόνων á¼¢ δύο γραμμῶν ἁπτομένων ἀλλήλων καὶ μὴ ἐν τῇ αὐτῇ ἐπιφανείᾳ οὐσῶν πρὸς πάσαις ταῖς γραμμαῖς κλίσις. ἄλλως: στερεὰ γωνία ἐστὶν ἡ ὑπὸ πλειόνων á¼¢ δύο γωνιῶν ἐπιπέδων περιεχομένη μὴ οὐσῶν ἐν τῷ αὐτῷ ἐπιπέδῳ πρὸς ἑνὶ σημείῳ συνισταμένων.'"],["Rule","'GreekTextWordCount'",45],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",12]],["Association",["Rule","'VertexLabel'","'D11.12'"],["Rule","'Text'","'A pyramid is a solid figure, contained by planes, which is constructed from one plane to one point.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'πυραμίς ἐστι σχῆμα στερεὸν ἐπιπέδοις περιεχόμενον ἀπὸ ἑνὸς ἐπιπέδου πρὸς ἑνὶ σημείῳ συνεστώς.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",13]],["Association",["Rule","'VertexLabel'","'D11.13'"],["Rule","'Text'","'A prism is a solid figure contained by planes two of which, namely those which are opposite, are equal, similar and parallel, while the rest are parallelograms.'"],["Rule","'TextWordCount'",27],["Rule","'GreekText'","'πρίσμα ἐστὶ σχῆμα στερεὸν ἐπιπέδοις περιεχόμενον, ὧν δύο τὰ ἀπεναντίον ἴσα τε καὶ ὅμοιά ἐστι καὶ παράλληλα, τὰ δὲ λοιπὰ παραλληλόγραμμα.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",14]],["Association",["Rule","'VertexLabel'","'D11.14'"],["Rule","'Text'","'When, the diameter of a semicircle remaining fixed, the semicircle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a sphere.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'σφαῖρά ἐστιν, ὅταν ἡμικυκλίου μενούσης τῆς διαμέτρου περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, τὸ περιληφθὲν σχῆμα.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",15]],["Association",["Rule","'VertexLabel'","'D11.15'"],["Rule","'Text'","'The axis of the sphere is the straight line which remains fixed and about which the semicircle is turned.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἄξων δὲ τῆς σφαίρας ἐστὶν ἡ μένουσα εὐθεῖα, περὶ ἣν τὸ ἡμικύκλιον στρέφεται.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",16]],["Association",["Rule","'VertexLabel'","'D11.16'"],["Rule","'Text'","'The centre of the sphere is the same as that of the semicircle.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'κέντρον δὲ τῆς σφαίρας ἐστὶ τὸ αὐτό, ὃ καὶ τοῦ ἡμικυκλίου.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",17]],["Association",["Rule","'VertexLabel'","'D11.17'"],["Rule","'Text'","'A diameter of the sphere is any straight line drawn through the centre and terminated in both directions by the surface of the sphere.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'διάμετρος δὲ τῆς σφαίρας ἐστὶν εὐθεῖά τις διὰ τοῦ κέντρου ἠγμένη καὶ περατουμένη ἐφ᾽ ἑκάτερα τὰ μέρη ὑπὸ τῆς ἐπιφανείας τῆς σφαίρας.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",18]],["Association",["Rule","'VertexLabel'","'D11.18'"],["Rule","'Text'","'When, one side of those about the right angle in a right-angled triangle remaining fixed, the triangle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cone. And, if the straight line which remains fixed be equal to the remaining side about the right angle which is carried round, the cone will be right-angled; if less, obtuse-angled; and if greater, acute-angled.'"],["Rule","'TextWordCount'",75],["Rule","'GreekText'","'κῶνός ἐστιν, ὅταν ὀρθογωνίου τριγώνου μενούσης μιᾶς πλευρᾶς τῶν περὶ τὴν ὀρθὴν γωνίαν περιενεχθὲν τὸ τρίγωνον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, τὸ περιληφθὲν σχῆμα. κἂν μὲν ἡ μένουσα εὐθεῖα ἴση ᾖ τῇ λοιπῇ τῇ περὶ τὴν ὀρθὴν περιφερομένῃ, ὀρθογώνιος ἔσται ὁ κῶνος, ἐὰν δὲ ἐλάττων, ἀμβλυγώνιος, ἐὰν δὲ μείζων, ὀξυγώνιος.'"],["Rule","'GreekTextWordCount'",53],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",19]],["Association",["Rule","'VertexLabel'","'D11.19'"],["Rule","'Text'","'The axis of the cone is the straight line which remains fixed and about which the triangle is turned.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἄξων δὲ τοῦ κώνου ἐστὶν ἡ μένουσα εὐθεῖα, περὶ ἣν τὸ τρίγωνον στρέφεται.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",20]],["Association",["Rule","'VertexLabel'","'D11.20'"],["Rule","'Text'","'And the base is the circle described by the straight line which is carried round.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'βάσις δὲ ὁ κύκλος ὁ ὑπὸ τῆς περιφερομένης εὐθείας γραφόμενος.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",21]],["Association",["Rule","'VertexLabel'","'D11.21'"],["Rule","'Text'","'When, one side of those about the right angle in a rectangular parallelogram remaining fixed, the parallelogram is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cylinder.'"],["Rule","'TextWordCount'",41],["Rule","'GreekText'","'κύλινδρός ἐστιν, ὅταν ὀρθογωνίου παραλληλογράμμου μενούσης μιᾶς πλευρᾶς τῶν περὶ τὴν ὀρθὴν γωνίαν περιενεχθὲν τὸ παραλληλόγραμμον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, τὸ περιληφθὲν σχῆμα.'"],["Rule","'GreekTextWordCount'",27],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",22]],["Association",["Rule","'VertexLabel'","'D11.22'"],["Rule","'Text'","'The axis of the cylinder is the straight line which remains fixed and about which the parallelogram is turned.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἄξων δὲ τοῦ κυλίνδρου ἐστὶν ἡ μένουσα εὐθεῖα, περὶ ἣν τὸ παραλληλόγραμμον στρέφεται.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",23]],["Association",["Rule","'VertexLabel'","'D11.23'"],["Rule","'Text'","'And the bases are the circles described by the two sides opposite to one another which are carried round.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'βάσεις δὲ οἱ κύκλοι οἱ ὑπὸ τῶν ἀπεναντίον περιαγομένων δύο πλευρῶν γραφόμενοι.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",24]],["Association",["Rule","'VertexLabel'","'D11.24'"],["Rule","'Text'","'Similar cones and cylinders are those in which the axes and the diameters of the bases are proportional.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'ὅμοιοι κῶνοι καὶ κύλινδροί εἰσιν, ὧν οἵ τε ἄξονες καὶ αἱ διάμετροι τῶν βάσεων ἀνάλογόν εἰσιν.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",25]],["Association",["Rule","'VertexLabel'","'D11.25'"],["Rule","'Text'","'A cube is a solid figure contained by six equal squares.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'κύβος ἐστὶ σχῆμα στερεὸν ὑπὸ ἓξ τετραγώνων ἴσων περιεχόμενον.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",26]],["Association",["Rule","'VertexLabel'","'D11.26'"],["Rule","'Text'","'An octahedron is a solid figure contained by eight equal and equilateral triangles.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ὀκτάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ ὀκτὼ τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",27]],["Association",["Rule","'VertexLabel'","'D11.27'"],["Rule","'Text'","'An icosahedron is a solid figure contained by twenty equal and equilateral triangles.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'εἰκοσάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ εἴκοσι τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Definition'",28]],["Association",["Rule","'VertexLabel'","'D11.28'"],["Rule","'Text'","'A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'δωδεκάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ δώδεκα πενταγώνων ἴσων καὶ ἰσοπλεύρων καὶ ἰσογωνίων περιεχόμενον.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["Missing","'NotApplicable'"]],["Rule","'Proof'",["Missing","'NotApplicable'"]],["Rule","'ProofWordCount'",["Missing","'NotApplicable'"]],["Rule","'GreekProof'",["Missing","'NotApplicable'"]],["Rule","'GreekProofWordCount'",["Missing","'NotApplicable'"]]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'1.1'"],["Rule","'Text'","'On a given finite straight line to construct an equilateral triangle.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συστήσασθαι.'"],["Rule","'GreekTextWordCount'",8],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Postulate'",1]],["List",["Rule","'Postulate'",3]],["List",["Rule","'Book'",1],["Rule","'Definition'",15]]]],["Rule","'Proof'","'Let AB be the given finite straight line. Thus it is required to constructan equilateral triangle on the straight line AB. With centre A and distance AB let the circle BCD be described; [Post. 3]again, with centre B and distance BA let the circle ACE be described; [Post. 3] and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1] Now, since the point A is the centre of the circle CDB, AC is equal to AB. [Def. 15] Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15] But CA was also proved equal to AB;therefore each of the straight lines CA, CB is equal to AB. And things which are equal to the same thing are also equal to one another; [C. N. 1] therefore CA is also equal to CB. Therefore the three straight lines CA, AB, BC areequal to one another. Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB.'"],["Rule","'ProofWordCount'",190],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ. δεῖ δὴ ἐπὶ τῆς ΑΒ εὐθείας τρίγωνον ἰσόπλευρον συστήσασθαι. κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΒ κύκλος γεγράφθω ὁ ΒΓΔ, καὶ πάλιν κέντρῳ μὲν τῷ Β διαστήματι δὲ τῷ ΒΑ κύκλος γεγράφθω ὁ ΑΓΕ, καὶ ἀπὸ τοῦ Γ σημείου, καθ᾽ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι, ἐπὶ τὰ Α, Β σημεῖα ἐπεζεύχθωσαν εὐθεῖαι αἱ ΓΑ, ΓΒ. καὶ ἐπεὶ τὸ Α σημεῖον κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου, ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ: πάλιν, ἐπεὶ τὸ Β σημεῖον κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου, ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ. ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση: ἑκατέρα ἄρα τῶν ΓΑ, ΓΒ τῇ ΑΒ ἐστὶν ἴση. τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα: καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστὶν ἴση: αἱ τρεῖς ἄρα αἱ ΓΑ, ΑΒ, ΒΓ ἴσαι ἀλλήλαις εἰσίν. ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον, καὶ συνέσταται ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τῆς ΑΒ. Ἐπὶ τῆς δοθείσης ἄρα εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συνέσταται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",167]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'1.2'"],["Rule","'Text'","'To place at a given point (as an extremity) a straight line equal to a given straight line.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'πρὸς τῷ δοθέντι σημείῳ τῇ δοθείσῃ εὐθείᾳ ἴσην εὐθεῖαν θέσθαι.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Common Notion'",3]],["List",["Rule","'Postulate'",1]],["List",["Rule","'Postulate'",2]],["List",["Rule","'Postulate'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let A be the given point, and BC the given straight line. Thus it is required to place at the point A (as an extremity)a straight line equal to the given straight line BC. From the point A to the point B let the straight line AB be joined; [Post. 1] and on it let the equilateral triangleDAB be constructed. [I. 1] Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2] with centre B and distance BC let thecircle CGH be described; [Post. 3] and again, with centre D and distance DG let the circle GKL be described. [Post. 3] Then, since the point B is the centre of the circle CGH, BC is equal to BG. Again, since the point D is the centre of the circle GKL, DL is equal to DG. And in these DA is equal to DB; therefore the remainder AL is equal to the remainder BG. [C. N. 3]  But BC was also proved equal to BG; therefore each of the straight lines AL, BC is equal to BG. And things which are equal to the same thing are also equal to one another; [C. N. 1]therefore AL is also equal to BC. Therefore at the given point A the straight line AL is placed equal to the given straight line BC.'"],["Rule","'ProofWordCount'",228],["Rule","'GreekProof'","'ἔστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ: δεῖ δὴ πρὸς τῷ Α σημείῳ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην εὐθεῖαν θέσθαι. ἐπεζεύχθω γὰρ ἀπὸ τοῦ Α σημείου ἐπὶ τὸ Β σημεῖον εὐθεῖα ἡ ΑΒ, καὶ συνεστάτω ἐπ᾽ αὐτῆς τρίγωνον ἰσόπλευρον τὸ ΔΑΒ, καὶ ἐκβεβλήσθωσαν ἐπ᾽ εὐθείας ταῖς ΔΑ, ΔΒ εὐθεῖαι αἱ ΑΕ, ΒΖ, καὶ κέντρῳ μὲν τῷ Β διαστήματι δὲ τῷ ΒΓ κύκλος γεγράφθω ὁ ΓΗΘ, καὶ πάλιν κέντρῳ τῷ Δ καὶ διαστήματι τῷ ΔΗ κύκλος γεγράφθω ὁ ΗΚΛ. ἐπεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ τοῦ ΓΗΘ κύκλου, ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ. πάλιν, ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ ΚΛΗ κύκλου, ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ, ὧν ἡ ΔΑ τῇ ΔΒ ἴση ἐστίν. λοιπὴ ἄρα ἡ ΑΛ λοιπῇ τῇ ΒΗ ἐστὶν ἴση. ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση: ἑκατέρα ἄρα τῶν ΑΛ, ΒΓ τῇ ΒΗ ἐστὶν ἴση. τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα: καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστὶν ἴση. πρὸς ἄρα τῷ δοθέντι σημείῳ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση εὐθεῖα κεῖται ἡ ΑΛ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",189]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'1.3'"],["Rule","'Text'","'Given two unequal straight lines, to cut off from the greater a straight line equal to the less.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'δύο δοθεισῶν εὐθειῶν ἀνίσων ἀπὸ τῆς μείζονος τῇ ἐλάσσονι ἴσην εὐθεῖαν ἀφελεῖν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Postulate'",3]],["List",["Rule","'Book'",1],["Rule","'Definition'",15]],["List",["Rule","'Book'",1],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let AB, C be the two given unequal straight lines, and let AB be the greater of them. Thus it is required to cut off from AB the greater a straight line equal to C the less. At the point A let AD be placed equal to the straight line C; [I. 2] and with centre A and distance AD let the circle DEF be described. [Post. 3] Now, since the point A is the centre of the circle DEF, AE is equal to AD. [Def. 15]But C is also equal to AD. Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C. N. 1] Therefore, given the two straight lines AB, C, from AB the greater AE has been cut off equal to C the less.'"],["Rule","'ProofWordCount'",139],["Rule","'GreekProof'","'ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι αἱ ΑΒ, Γ, ὧν μείζων ἔστω ἡ ΑΒ: δεῖ δὴ ἀπὸ τῆς μείζονος τῆς ΑΒ τῇ ἐλάσσονι τῇ Γ ἴσην εὐθεῖαν ἀφελεῖν. κείσθω πρὸς τῷ Α σημείῳ τῇ Γ εὐθείᾳ ἴση ἡ ΑΔ: καὶ κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΔ κύκλος γεγράφθω ὁ ΔΕΖ. καὶ ἐπεὶ τὸ Α σημεῖον κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου, ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ: ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση. ἑκατέρα ἄρα τῶν ΑΕ, Γ τῇ ΑΔ ἐστιν ἴση: ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση. δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν ΑΒ, Γ ἀπὸ τῆς μείζονος τῆς ΑΒ τῇ ἐλάσσονι τῇ Γ ἴση ἀφῄρηται ἡ ΑΕ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",117]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'1.4'"],["Rule","'Text'","'If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.'"],["Rule","'TextWordCount'",60],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν βάσιν τῇ βάσει ἴσην ἕξει, καὶ τὸ τρίγωνον τῷ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν.'"],["Rule","'GreekTextWordCount'",58],["Rule","'References'",["List",["List",["Rule","'Common Notion'",4]]]],["Rule","'Proof'","'Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF, and the angle BAC equal to theangle EDF. I say that the base BC is also equal to the base EF, the triangle ABC will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, thatis, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. For, if the triangle ABC be applied to the triangle DEF, and if the point A be placed on the point D and the straight line AB on DE, then the point B will also coincide with E, because AB is equal to DE. Again, AB coinciding with DE, the straight line AC will also coincide with DF, because the angle BAC is equal to the angle EDF; hence the point C will also coincide with the point F, because AC is again equal to DF. But B also coincided with E; hence the base BC will coincide with the base EF. [For if, when B coincides with E and C with F, the base BC does not coincide with the base EF, two straight lines will enclose a space: which is impossible. Therefore the base BC will coincide with EF]and will be equal to it. [C. N. 4] Thus the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it. And the remaining angles will also coincide with the remaining angles and will be equal to them, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.'"],["Rule","'ProofWordCount'",294],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς τὰς ΑΒ, ΑΓ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴσην. λέγω, ὅτι καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ. Ἐφαρμοζομένου γὰρ τοῦ ΑΒΓ τριγώνου ἐπὶ τὸ ΔΕΖ τρίγωνον καὶ τιθεμένου τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ, ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ: ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ διὰ τὸ ἴσην εἶναι τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ: ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ἐφαρμόσει διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ. ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει: ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ἐφαρμόσει. εἰ γὰρ τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος τοῦ δὲ Γ ἐπὶ τὸ Ζ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει, δύο εὐθεῖαι χωρίον περιέξουσιν: ὅπερ ἐστὶν ἀδύνατον. ἐφαρμόσει ἄρα ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ καὶ ἴση αὐτῇ ἔσται: ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει καὶ ἴσον αὐτῷ ἔσται, καὶ αἱ λοιπαὶ γωνίαι ἐπὶ τὰς λοιπὰς γωνίας ἐφαρμόσουσι καὶ ἴσαι αὐταῖς ἔσονται, ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ. ἐὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δύο πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν βάσιν τῇ βάσει ἴσην ἕξει, καὶ τὸ τρίγωνον τῷ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",346]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'1.5'"],["Rule","'Text'","'In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.'"],["Rule","'TextWordCount'",33],["Rule","'GreekText'","'τῶν ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ προσεκβληθεισῶν τῶν ἴσων εὐθειῶν αἱ ὑπὸ τὴν βάσιν γωνίαι ἴσαι ἀλλήλαις ἔσονται.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Postulate'",1]],["List",["Rule","'Postulate'",2]],["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let ABC be an isosceles triangle having the side AB equal to the side AC; and let the straight lines BD, CE be produced further in a straight line with AB, AC. [Post. 2] I say that the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE. Let a point F be taken at random on BD; from AE the greater let AG be cut off equal to AF the less; [I. 3] and let the straight lines FC, GB be joined. [Post. 1] Then, since AF is equal to AG and AB to AC, the two sides FA, AC are equal to the two sides GA, AB, respectively; and they contain a common angle, the angle FAG.Therefore the base FC is equal to the base GB, and the triangle AFC is equal to the triangle AGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4] And, since the whole AF is equal to the whole AG, and in these AB is equal to AC, the remainder BF is equal to the remainder CG. But FC was also proved equal to GB;therefore the two sides BF, FC are equal to the two sides CG, GB respectively; and the angle BFC is equal to the angle CGB, while the base BC is common to them; therefore the triangle BFC is also equal to the triangle CGB,and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. Accordingly, since the whole angle ABG was proved equal to the angle ACF, and in these the angle CBG is equal to the angle BCF, the remaining angle ABC is equal to the remaining angle ACB;and they are at the base of the triangle ABC. But the angle FBC was also proved equal to the angle GCB; and they are under the base.'"],["Rule","'ProofWordCount'",363],["Rule","'GreekProof'","'ἔστω τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ ἴσην ἔχον τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ, καὶ προσεκβεβλήσθωσαν ἐπ᾽ εὐθείας ταῖς ΑΒ, ΑΓ εὐθεῖαι αἱ ΒΔ, ΓΕ: λέγω, ὅτι ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ἴση ἐστίν, ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ. εἰλήφθω γὰρ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ, καὶ ἀφῃρήσθω ἀπὸ τῆς μείζονος τῆς ΑΕ τῇ ἐλάσσονι τῇ ΑΖ ἴση ἡ ΑΗ, καὶ ἐπεζεύχθωσαν αἱ ΖΓ, ΗΒ εὐθεῖαι. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ ἡ δὲ ΑΒ τῇ ΑΓ, δύο δὴ αἱ ΖΑ, ΑΓ δυσὶ ταῖς ΗΑ, ΑΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνίαν κοινὴν περιέχουσι τὴν ὑπὸ ΖΑΗ: βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ ἴση ἐστίν, καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ, ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ. καὶ ἐπεὶ ὅλη ἡ ΑΖ ὅλῃ τῇ ΑΗ ἐστιν ἴση, ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση, λοιπὴ ἄρα ἡ ΒΖ λοιπῇ τῇ ΓΗ ἐστιν ἴση. ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση: δύο δὴ αἱ ΒΖ, ΖΓ δυσὶ ταῖς ΓΗ, ΗΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΒΖΓ γωνίᾳ τῇ ὑπὸ ΓΗΒ ἴση, καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ: καὶ τὸ ΒΖΓ ἄρα τρίγωνον τῷ ΓΗΒ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ. ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ἐδείχθη ἴση, ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ἴση, λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓ λοιπῇ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση: καί εἰσι πρὸς τῇ βάσει τοῦ ΑΒΓ τριγώνου. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴση: καί εἰσιν ὑπὸ τὴν βάσιν. τῶν ἄρα ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ προσεκβληθεισῶν τῶν ἴσων εὐθειῶν αἱ ὑπὸ τὴν βάσιν γωνίαι ἴσαι ἀλλήλαις ἔσονται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",349]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'1.6'"],["Rule","'Text'","'If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ἐὰν τριγώνου αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν, καὶ αἱ ὑπὸ τὰς ἴσας γωνίαις ὑποτείνουσαι πλευραὶ ἴσαι ἀλλήλαις ἔσονται.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List"]],["Rule","'Proof'","'Let ABC be a triangle having the angle ABC equal to the angle ACB; I say that the side AB is also equal to the side AC. For, if AB is unequal to AC, one of them is greater. Let AB be greater; and from AB the greater let DB be cut off equal to AC the less; let DC be joined. Then, since DB is equal to AC, and BC is common, the two sides DB, BC are equal to the two sides AC, CB respectively; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB, the less to the greater: which is absurd. Therefore AB is not unequal to AC; it is therefore equal to it.'"],["Rule","'ProofWordCount'",139],["Rule","'GreekProof'","'ἔστω τρίγωνον τὸ ΑΒΓ ἴσην ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν τῇ ὑπὸ ΑΓΒ γωνίᾳ: λέγω, ὅτι καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ἐστιν ἴση. εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ, ἡ ἑτέρα αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ΑΒ, καὶ ἀφῃρήσθω ἀπὸ τῆς μείζονος τῆς ΑΒ τῇ ἐλάττονι τῇ ΑΓ ἴση ἡ ΔΒ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ κοινὴ δὲ ἡ ΒΓ, δύο δὴ αἱ ΔΒ, ΒΓ δύο ταῖς ΑΓ, ΓΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΔΒΓ γωνίᾳ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση: βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ ἴση ἐστίν, καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ἴσον ἔσται, τὸ ἔλασσον τῷ μείζονι: ὅπερ ἄτοπον: οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ: ἴση ἄρα. ἐὰν ἄρα τριγώνου αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν, καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι πλευραὶ ἴσαι ἀλλήλαις ἔσονται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",153]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'1.7'"],["Rule","'Text'","'Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.'"],["Rule","'TextWordCount'",62],["Rule","'GreekText'","'ἐπὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι ἴσαι ἑκατέρα ἑκατέρᾳ οὐ συσταθήσονται πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι ταῖς ἐξ ἀρχῆς εὐθείαις.'"],["Rule","'GreekTextWordCount'",33],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight linesAD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA isequal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined. Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5]therefore the angle ADC is greater than the angle DCB; therefore the angle CDB is much greater than the angle DCB. Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB. But it was also proved much greater than it: which is impossible.'"],["Rule","'ProofWordCount'",164],["Rule","'GreekProof'","'εἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ, ΓΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ ἴσαι ἑκατέρα ἑκατέρᾳ συνεστάτωσαν πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ τῷ τε Γ καὶ Δ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι, ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α, τὴν δὲ ΓΒ τῇ ΔΒ τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β, καὶ ἐπεζεύχθω ἡ ΓΔ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ: μείζων ἄρα ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ: πολλῷ ἄρα ἡ ὑπὸ ΓΔΒ μείζων ἐστὶ τῆς ὑπὸ ΔΓΒ. πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ. ἐδείχθη δὲ αὐτῆς καὶ πολλῷ μείζων: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐπὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι ἴσαι ἑκατέρα ἑκατέρᾳ συσταθήσονται πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι ταῖς ἐξ ἀρχῆς εὐθείαις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",176]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'1.8'"],["Rule","'Text'","'If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δύο πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρα, ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην, καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἕξει τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην.'"],["Rule","'GreekTextWordCount'",34],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",7]]]],["Rule","'Proof'","'Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let them have the base BC equalto the base EF; I say that the angle BAC is also equal to the angle EDF. For, if the triangle ABC be applied to the triangle DEF, and if the point B be placed onthe point E and the straight line BC on EF, the point C will also coincide with F, because BC is equal to EF. Then, BC coinciding with EF, BA, AC will also coincide with ED, DF; for, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with ED, DF but fall beside them as EG, GF, then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it. But they cannot be so constructed. [I. 7] Therefore it is not possible that, if the base BC be applied to the base EF, the sides BA, AC should not coincide with ED, DF; they will therefore coincide,  so that the angle BAC will also coincide with the angle EDF, and will be equal to it.'"],["Rule","'ProofWordCount'",255],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ: ἐχέτω δὲ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην: λέγω, ὅτι καὶ γωνία ἡ ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. Ἐφαρμοζομένου γὰρ τοῦ ΑΒΓ τριγώνου ἐπὶ τὸ ΔΕΖ τρίγωνον καὶ τιθεμένου τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ: ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ ἐφαρμόσουσι καὶ αἱ ΒΑ, ΓΑ ἐπὶ τὰς ΕΔ, ΔΖ. εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ἐφαρμόσει, αἱ δὲ ΒΑ, ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ, ΔΖ οὐκ ἐφαρμόσουσιν ἀλλὰ παραλλάξουσιν ὡς αἱ ΕΗ, ΗΖ, συσταθήσονται ἐπὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι ἴσαι ἑκατέρα ἑκατέρᾳ πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι. οὐ συνίστανται δέ: οὐκ ἄρα ἐφαρμοζομένης τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν οὐκ ἐφαρμόσουσι καὶ αἱ ΒΑ, ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ, ΔΖ. ἐφαρμόσουσιν ἄρα: ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ ἐφαρμόσει καὶ ἴση αὐτῇ ἔσται. ἐὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δύο πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν βάσιν τῇ βάσει ἴσην ἔχῃ, καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἕξει τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",248]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'1.9'"],["Rule","'Text'","'To bisect a given rectilineal angle.'"],["Rule","'TextWordCount'",6],["Rule","'GreekText'","'τὴν δοθεῖσαν γωνίαν εὐθύγραμμον δίχα τεμεῖν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let the angle BAC be the given rectilineal angle. Thus it is required to bisect it. Let a point D be taken at random on AB; let AE be cut off from AC equal to AD; [I. 3] let DE be joined, and on DE let the equilateral triangle DEF be constructed; let AF be joined. I say that the angle BAC has been bisected by the straight line AF. For, since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively. And the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF. [I. 8] Therefore the given rectilineal angle BAC has been bisected by the straight line AF.'"],["Rule","'ProofWordCount'",131],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΒΑΓ. δεῖ δὴ αὐτὴν δίχα τεμεῖν. εἰλήφθω ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ, καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ τῇ ΑΔ ἴση ἡ ΑΕ, καὶ ἐπεζεύχθω ἡ ΔΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ, καὶ ἐπεζεύχθω ἡ ΑΖ: λέγω, ὅτι ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΖ εὐθείας. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΖ, δύο δὴ αἱ ΔΑ, ΑΖ δυσὶ ταῖς ΕΑ, ΑΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ. καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ἴση ἐστίν: γωνία ἄρα ἡ ὑπὸ ΔΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖ ἴση ἐστίν. ἡ ἄρα δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΒΑΓ δίχα τέτμηται ὑπὸ τῆς ΑΖ εὐθείας: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",122]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'1.10'"],["Rule","'Text'","'To bisect a given finite straight line.'"],["Rule","'TextWordCount'",7],["Rule","'GreekText'","'τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην δίχα τεμεῖν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",1]],["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",9]]]],["Rule","'Proof'","'Let AB be the given finite straight line. Thus it is required to bisect the finite straight line AB. Let the equilateral triangle ABC be constructed on it, [I. 1] and let the angle ACB be bisected by the straight line CD; [I. 9] I say that the straight line AB has been bisected at the point D. For, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD respectively; and the angle ACD is equal to the angle BCD; therefore the base AD is equal to the base BD. [I. 4] Therefore the given finite straight line AB has been bisected at D.'"],["Rule","'ProofWordCount'",117],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ: δεῖ δὴ τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα τεμεῖν. συνεστάτω ἐπ᾽ αὐτῆς τρίγωνον ἰσόπλευρον τὸ ΑΒΓ, καὶ τετμήσθω ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ: λέγω, ὅτι ἡ ΑΒ εὐθεῖα δίχα τέτμηται κατὰ τὸ Δ σημεῖον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, κοινὴ δὲ ἡ ΓΔ, δύο δὴ αἱ ΑΓ, ΓΔ δύο ταῖς ΒΓ, ΓΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση ἐστίν: βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔ ἴση ἐστίν. ἡ ἄρα δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ δίχα τέτμηται κατὰ τὸ Δ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",104]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'1.11'"],["Rule","'Text'","'To draw a straight line at right angles to a given straight line from a given point on it.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'τῇ δοθείσῃ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",1]],["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let AB be the given straight line, and C the given point on it. Thus it is required to draw from the point C a straight line at right angles to the straight line AB. Let a point D be taken at random on AC;let CE be made equal to CD; [I. 3] on DE let the equilateral triangle FDE be constructed, [I. 1] and let FC be joined; I say that the straight line FC has been drawn at rightangles to the given straight line AB from C the given point on it. For, since DC is equal to CE, and CF is common, the two sides DC, CF are equal to the two sides EC, CF respectively; and the base DF is equal to the base FE; therefore the angle DCF is equal to the angle ECF; [I. 8]and they are adjacent angles. But, when a straight line set up on a straight line makesthe adjacent angles equal to one another, each of the equal angles is right; [Def. 10] therefore each of the angles DCF, FCE is right. Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given pointC on it.'"],["Rule","'ProofWordCount'",206],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ τὸ δὲ δοθὲν σημεῖον ἐπ᾽ αὐτῆς τὸ Γ: δεῖ δὴ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν. εἰλήφθω ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ, καὶ κείσθω τῇ ΓΔ ἴση ἡ ΓΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον τὸ ΖΔΕ, καὶ ἐπεζεύχθω ἡ ΖΓ: λέγω, ὅτι τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ, κοινὴ δὲ ἡ ΓΖ, δύο δὴ αἱ ΔΓ, ΓΖ δυσὶ ταῖς ΕΓ, ΓΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ ἴση ἐστίν: γωνία ἄρα ἡ ὑπὸ ΔΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ ἴση ἐστίν: καί εἰσιν ἐφεξῆς. ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΔΓΖ, ΖΓΕ. τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",182]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'1.12'"],["Rule","'Text'","'To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον ἀπὸ τοῦ δοθέντος σημείου, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Postulate'",1]],["List",["Rule","'Postulate'",3]],["List",["Rule","'Book'",1],["Rule","'Definition'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",10]]]],["Rule","'Proof'","'Let AB be the given infinite straight line, and C the given point which is not on it;thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line. For let a point D be takenat random on the other side of the straight line AB, and with centre C and distance CD let the circle EFG be described; [Post. 3] let the straight line EG be bisected at H, [I. 10]and let the straight lines CG, CH, CE be joined. [Post. 1] I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it. For, since GH is equal to HE, and HC is common, the two sides GH, HC are equal to the two sides EH, HC respectively; and the base CG is equal to the base CE;therefore the angle CHG is equal to the angle EHC. [I. 8]And they are adjacent angles. But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other iscalled a perpendicular to that on which it stands. [Def. 10] Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.'"],["Rule","'ProofWordCount'",245],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ τὸ δὲ δοθὲν σημεῖον, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, τὸ Γ: δεῖ δὴ ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. εἰλήφθω γὰρ ἐπὶ τὰ ἕτερα μέρη τῆς ΑΒ εὐθείας τυχὸν σημεῖον τὸ Δ, καὶ κέντρῳ μὲν τῷ Γ διαστήματι δὲ τῷ ΓΔ κύκλος γεγράφθω ὁ ΕΖΗ, καὶ τετμήσθω ἡ ΕΗ εὐθεῖα δίχα κατὰ τὸ Θ, καὶ ἐπεζεύχθωσαν αἱ ΓΗ, ΓΘ, ΓΕ εὐθεῖαι: λέγω, ὅτι ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, κάθετος ἦκται ἡ ΓΘ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ, κοινὴ δὲ ἡ ΘΓ, δύο δὴ αἱ ΗΘ, ΘΓ δύο ταῖς ΕΘ, ΘΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ἐστιν ἴση: γωνία ἄρα ἡ ὑπὸ ΓΘΗ γωνίᾳ τῇ ὑπὸ ΕΘΓ ἐστιν ἴση. καί εἰσιν ἐφεξῆς. ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν, καὶ ἡ ἐφεστηκυῖα εὐθεῖα κάθετος καλεῖται ἐφ᾽ ἣν ἐφέστηκεν. ἐπὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, κάθετος ἦκται ἡ ΓΘ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",216]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'1.13'"],["Rule","'Text'","'If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐὰν εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι δύο ὀρθὰς á¼¢ δυσὶν ὀρθαῖς ἴσας ποιήσει.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Common Notion'",2]],["List",["Rule","'Book'",1],["Rule","'Definition'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'For let any straight line AB set up on the straight lineCD make the angles CBA, ABD; I say that the angles CBA, ABD are either two right angles or equal to two right angles. Now, if the angle CBA is equal tothe angle ABD, they are two right angles. [Def. 10] But, if not, let BE be drawn from the point B at right angles to CD; [I. 11] therefore the angles CBE, EBD are two right angles. Then, since the angle CBE is equal to the two angles CBA, ABE, let the angle EBD be added to each; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [C. N. 2]  Again, since the angle DBA is equal to the two angles DBE, EBA, let the angle ABC be added to each; therefore the angles DBA. ABC are equal to the three angles DBE, EBA, ABC. [C. N. 2]  But the angles CBE, EBD were also proved equal to the same three angles; and things which are equal to the same thing are also equal to one another; [C. N. 1]therefore the angles CBE, EBD are also equal to the angles DBA, ABC. But the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are also equal to two right angles.'"],["Rule","'ProofWordCount'",221],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ ἐπ᾽ εὐθεῖαν τὴν ΓΔ σταθεῖσα γωνίας ποιείτω τὰς ὑπὸ ΓΒΑ, ΑΒΔ: λέγω, ὅτι αἱ ὑπὸ ΓΒΑ, ΑΒΔ γωνίαι ἤτοι δύο ὀρθαί εἰσιν á¼¢ δυσὶν ὀρθαῖς ἴσαι. εἰ μὲν οὖν ἴση ἐστὶν ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ, δύο ὀρθαί εἰσιν. εἰ δὲ οὔ, ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΓΔ εὐθείᾳ πρὸς ὀρθὰς ἡ ΒΕ: αἱ ἄρα ὑπὸ ΓΒΕ, ΕΒΔ δύο ὀρθαί εἰσιν: καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ δυσὶ ταῖς ὑπὸ ΓΒΑ, ΑΒΕ ἴση ἐστίν, κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ: αἱ ἄρα ὑπὸ ΓΒΕ, ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ, ΑΒΕ, ΕΒΔ ἴσαι εἰσίν. πάλιν, ἐπεὶ ἡ ὑπὸ ΔΒΑ δυσὶ ταῖς ὑπὸ ΔΒΕ, ΕΒΑ ἴση ἐστίν, κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ: αἱ ἄρα ὑπὸ ΔΒΑ, ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ, ΕΒΑ, ΑΒΓ ἴσαι εἰσίν. ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ, ΕΒΔ τρισὶ ταῖς αὐταῖς ἴσαι: τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα: καὶ αἱ ὑπὸ ΓΒΕ, ΕΒΔ ἄρα ταῖς ὑπὸ ΔΒΑ, ΑΒΓ ἴσαι εἰσίν: ἀλλὰ αἱ ὑπὸ ΓΒΕ, ΕΒΔ δύο ὀρθαί εἰσιν: καὶ αἱ ὑπὸ ΔΒΑ, ΑΒΓ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἐὰν ἄρα εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι δύο ὀρθὰς á¼¢ δυσὶν ὀρθαῖς ἴσας ποιήσει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",199]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'1.14'"],["Rule","'Text'","'If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.'"],["Rule","'TextWordCount'",42],["Rule","'GreekText'","'ἐὰν πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο εὐθεῖαι μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς ἴσας ποιῶσιν, ἐπ᾽ εὐθείας ἔσονται ἀλλήλαις αἱ εὐθεῖαι.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Common Notion'",3]],["List",["Rule","'Postulate'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'For with any straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles; I say that BD is in a straight line with CB. For, if BD is not in a straight line with BC, let BE be in a straight line with CB. Then, since the straight line AB stands on the straight line CBE,the angles ABC, ABE are equal to two right angles. [I. 13]But the angles ABC, ABD are also equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C. N. 1] Let the angle CBA be subtracted from each;therefore the remaining angle ABE is equal to the remaining angle ABD, [C. N. 3] the less to the greater: which is impossible. Therefore BE is not in a straight line with CB. Similarly we can prove that neither is any other straightline except BD. Therefore CB is in a straight line with BD.'"],["Rule","'ProofWordCount'",184],["Rule","'GreekProof'","'πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β δύο εὐθεῖαι αἱ ΒΓ, ΒΔ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΑΒΓ, ΑΒΔ δύο ὀρθαῖς ἴσας ποιείτωσαν: λέγω, ὅτι ἐπ᾽ εὐθείας ἐστὶ τῇ ΓΒ ἡ ΒΔ. εἰ γὰρ μή ἐστι τῇ ΒΓ ἐπ᾽ εὐθείας ἡ ΒΔ, ἔστω τῇ ΓΒ ἐπ᾽ εὐθείας ἡ ΒΕ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΒ ἐπ᾽ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν, αἱ ἄρα ὑπὸ ΑΒΓ, ΑΒΕ γωνίαι δύο ὀρθαῖς ἴσαι εἰσίν: εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ, ΑΒΔ δύο ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΓΒΑ, ΑΒΕ ταῖς ὑπὸ ΓΒΑ, ΑΒΔ ἴσαι εἰσίν. κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΓΒΑ: λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση, ἡ ἐλάσσων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐπ᾽ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΓΒ τῇ ΒΔ. ἐὰν ἄρα πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο εὐθεῖαι μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς ἴσας ποιῶσιν, ἐπ᾽ εὐθείας ἔσονται ἀλλήλαις αἱ εὐθεῖαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",193]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'1.15'"],["Rule","'Text'","'If two straight lines cut one another, they make the vertical angles equal to one another.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς κατὰ κορυφὴν γωνίας ἴσας ἀλλήλαις ποιοῦσιν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Common Notion'",3]],["List",["Rule","'Postulate'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'For let the straight lines AB, CD cut one another at the point E; I say that the angle AEC is equal to the angle DEB, and the angle CEB to the angle AED. For, since the straight line AE standson the straight line CD, making the angles CEA, AED, the angles CEA, AED are equal to two right angles. [I. 13]  Again, since the straight line DE stands on the straight line AB, making the angles AED, DEB, the angles AED, DEB are equal to two right angles. [I. 13]  But the angles CEA, AED were also proved equal to two right angles; therefore the angles CEA, AED are equal to the angles AED DEB. [Post. 4 and C. N. 1]Let the angle AED be subtracted from each; therefore the remaining angle CEA is equal to the remaining angle BED. [C. N. 3] Similarly it can be proved that the angles CEB, DEA are also equal.'"],["Rule","'ProofWordCount'",158],["Rule","'GreekProof'","'δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον: λέγω, ὅτι ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ, ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ. ἐπεὶ γὰρ εὐθεῖα ἡ ΑΕ ἐπ᾽ εὐθεῖαν τὴν ΓΔ ἐφέστηκε γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ, ΑΕΔ, αἱ ἄρα ὑπὸ ΓΕΑ, ΑΕΔ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. πάλιν, ἐπεὶ εὐθεῖα ἡ ΔΕ ἐπ᾽ εὐθεῖαν τὴν ΑΒ ἐφέστηκε γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ, ΔΕΒ, αἱ ἄρα ὑπὸ ΑΕΔ, ΔΕΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ, ΑΕΔ δυσὶν ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΓΕΑ, ΑΕΔ ταῖς ὑπὸ ΑΕΔ, ΔΕΒ ἴσαι εἰσίν. κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΑΕΔ: λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν: ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ ὑπὸ ΓΕΒ, ΔΕΑ ἴσαι εἰσίν. ἐὰν ἄρα δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς κατὰ κορυφὴν γωνίας ἴσας ἀλλήλαις ποιοῦσιν: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερὸν ὅτι, ἐὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς πρὸς τῇ τομῇ γωνίας τέτρασιν ὀρθαῖς ἴσας ποιήσουσιν.'"],["Rule","'GreekProofWordCount'",169]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'1.16'"],["Rule","'Text'","'In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'παντὸς τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν μείζων ἐστίν.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Common Notion'",5]],["List",["Rule","'Postulate'",1]],["List",["Rule","'Postulate'",2]],["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let ABC be a triangle, and let one side of it BC be produced to D; I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC. Let AC be bisected at E [I. 10], and let BE be joined and producedin a straight line to F; let EF be made equal to BE [I. 3], let FC be joined [Post. 1], and let AC be drawn through to G [Post. 2]. Then, since AE is equal to EC,and BE to EF, the two sides AE, EB are equal to the two sides CE, EF respectively; and the angle AEB is equal to the angle FEC, for they are vertical angles. [I. 15]Therefore the base AB is equal to the base FC, and the triangle ABE is equal to the triangle CFE, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]therefore the angle BAE is equal to the angle ECF. But the angle ECD is greater than the angle ECF; [C. N. 5]therefore the angle ACD is greater than the angle BAE. Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15], can be proved greater than the angle ABC as well.'"],["Rule","'ProofWordCount'",221],["Rule","'GreekProof'","'ἔστω τρίγωνον τὸ ΑΒΓ, καὶ προσεκβεβλήσθω αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ: λέγω, ὅτι ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ μείζων ἐστὶν ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ, ΒΑΓ γωνιῶν. τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε, καὶ ἐπιζευχθεῖσα ἡ ΒΕ ἐκβεβλήσθω ἐπ᾽ εὐθείας ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΒΕ ἴση ἡ ΕΖ, καὶ ἐπεζεύχθω ἡ ΖΓ, καὶ διήχθω ἡ ΑΓ ἐπὶ τὸ Η. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΓ, ἡ δὲ ΒΕ τῇ ΕΖ, δύο δὴ αἱ ΑΕ, ΕΒ δυσὶ ταῖς ΓΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΕΒ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν: κατὰ κορυφὴν γάρ: βάσις ἄρα ἡ ΑΒ βάσει τῇ ΖΓ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ. μείζων δέ ἐστιν ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ: μείζων ἄρα ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ. ὁμοίως δὴ τῆς ΒΓ τετμημένης δίχα δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ, τουτέστιν ἡ ὑπὸ ΑΓΔ, μείζων καὶ τῆς ὑπὸ ΑΒΓ. παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν μείζων ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",214]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'1.17'"],["Rule","'Text'","'In any triangle two angles taken together in any manner are less than two right angles.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'παντὸς τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές εἰσι πάντῃ μεταλαμβανόμεναι.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Postulate'",2]],["List",["Rule","'Book'",1],["Rule","'Theorem'",13]],["List",["Rule","'Book'",1],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'Let ABC be a triangle; I say that two angles of the triangle ABC taken together in any manner are less than two right angles. For let BC be produced to D. [Post. 2] Then, since the angle ACD is an exterior angle of the triangle ABC, it is greater than the interior and opposite angle ABC. [I. 16] Let the angle ACB be added to each; therefore the angles ACD, ACB are greater than the angles ABC, BCA. But the angles ACD, ACB are equal to two right angles. [I. 13] Therefore the angles ABC, BCA are less than two right angles. Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well.'"],["Rule","'ProofWordCount'",128],["Rule","'GreekProof'","'ἔστω τρίγωνον τὸ ΑΒΓ: λέγω, ὅτι τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάττονές εἰσι πάντῃ μεταλαμβανόμεναι. Ἐκβεβλήσθω γὰρ ἡ ΒΓ ἐπὶ τὸ Δ. καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ. κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ: αἱ ἄρα ὑπὸ ΑΓΔ, ΑΓΒ τῶν ὑπὸ ΑΒΓ, ΒΓΑ μείζονές εἰσιν. ἀλλ᾽ αἱ ὑπὸ ΑΓΔ, ΑΓΒ δύο ὀρθαῖς ἴσαι εἰσίν: αἱ ἄρα ὑπὸ ΑΒΓ, ΒΓΑ δύο ὀρθῶν ἐλάσσονές εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ ΒΑΓ, ΑΓΒ δύο ὀρθῶν ἐλάσσονές εἰσι καὶ ἔτι αἱ ὑπὸ ΓΑΒ, ΑΒΓ. παντὸς ἄρα τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές εἰσι πάντῃ μεταλαμβανόμεναι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",114]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'1.18'"],["Rule","'Text'","'In any triangle the greater side subtends the greater angle.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'παντὸς τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν ὑποτείνει.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'For let ABC be a triangle having the side AC greater than AB; I say that the angle ABC is also greater than the angle BCA. For, since AC is greater than AB, let AD be made equal to AB [I. 3], and let BD bejoined. Then, since the angle ADB is an exterior angle of the triangle BCD, it is greater than the interior and opposite angle DCB. [I. 16] But the angle ADB is equal to the angle ABD, since the side AB is equal to AD; therefore the angle ABD is also greater than the angle ACB; therefore the angle ABC is much greater than the angle ACB.'"],["Rule","'ProofWordCount'",111],["Rule","'GreekProof'","'ἔστω γὰρ τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ: λέγω, ὅτι καὶ γωνία ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ. ἐπεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ, κείσθω τῇ ΑΒ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΒΔ. καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ: ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ, ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση: μείζων ἄρα καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ: πολλῷ ἄρα ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ τῆς ὑπὸ ΑΓΒ. παντὸς ἄρα τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν ὑποτείνει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",111]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'1.19'"],["Rule","'Text'","'In any triangle the greater angle is subtended by the greater side.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'παντὸς τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let ABC be a triangle having the angle ABC greater than the angle BCA; I say that the side AC is also greater than the side AB. For, if not, AC is either equal to AB or less. Now AC is not equal to AB; for then the angle ABC would also have been equal to the angle ACB; [I. 5] but it is not; therefore AC is not equal to AB. Neither is AC less than AB, for then the angle ABC would also have been less than the angle ACB; [I. 18] but it is not; therefore AC is not less than AB. And it was proved that it is not equal either. Therefore AC is greater than AB.'"],["Rule","'ProofWordCount'",121],["Rule","'GreekProof'","'ἔστω τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν τῆς ὑπὸ ΒΓΑ: λέγω, ὅτι καὶ πλευρὰ ἡ ΑΓ πλευρᾶς τῆς ΑΒ μείζων ἐστίν. εἰ γὰρ μή, ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ á¼¢ ἐλάσσων: ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ: ἴση γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ: οὐκ ἔστι δέ: οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ. οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ: ἐλάσσων γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῆς ὑπὸ ΑΓΒ: οὐκ ἔστι δέ: οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ. ἐδείχθη δέ, ὅτι οὐδὲ ἴση ἐστίν. μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ. παντὸς ἄρα τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",126]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'1.20'"],["Rule","'Text'","'In any triangle two sides taken together in any manner are greater than the remaining one.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Common Notion'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'For let ABC be a triangle; I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely BA, AC greater than BC, AB, BC greater than AC, BC, CA greater than AB. For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined. Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5] therefore the angle BCD is greater than the angle ADC. [C. N. 5] And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I. 19]therefore DB is greater than BC. But DA is equal to AC; therefore BA, AC are greater than BC. Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.'"],["Rule","'ProofWordCount'",159],["Rule","'GreekProof'","'ἔστω γὰρ τρίγωνον τὸ ΑΒΓ: λέγω, ὅτι τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, αἱ μὲν ΒΑ, ΑΓ τῆς ΒΓ, αἱ δὲ ΑΒ, ΒΓ τῆς ΑΓ, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. διήχθω γὰρ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον, καὶ κείσθω τῇ ΓΑ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ: μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ: καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν τῆς ὑπὸ ΒΔΓ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει, ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων. ἴση δὲ ἡ ΔΑ τῇ ΑΓ: μείζονες ἄρα αἱ ΒΑ, ΑΓ τῆς ΒΓ: ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ μὲν ΑΒ, ΒΓ τῆς ΓΑ μείζονές εἰσιν, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",160]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'1.21'"],["Rule","'Text'","'If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greaterangle.'"],["Rule","'TextWordCount'",43],["Rule","'GreekText'","'ἐὰν τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν περάτων δύο εὐθεῖαι ἐντὸς συσταθῶσιν, αἱ συσταθεῖσαι τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μὲν ἔσονται, μείζονα δὲ γωνίαν περιέξουσιν.'"],["Rule","'GreekTextWordCount'",28],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",16]],["List",["Rule","'Book'",1],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'On BC, one of the sides of the triangle ABC, from its extremities B, C, let the two straight lines BD, DC be constructed meeting within the triangle; I say that BD, DC are less than the remaining two sidesof the triangle BA, AC, but contain an angle BDC greater than the angle BAC. For let BD be drawn through to E. Then, since in any triangle two sides are greater than the remainingone, [I. 20] therefore, in the triangle ABE, the two sides AB, AE are greater than BE. Let EC be added to each; therefore BA, AC are greater than BE, EC. Again, since, in the triangle CED, the two sides CE, ED are greater than CD, let DB be added to each; therefore CE, EB are greater than CD, DB. But BA, AC were proved greater than BE, EC;therefore BA, AC are much greater than BD, DC. Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I. 16] therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED. For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC. But the angle BDC was proved greater than the angle CEB; therefore the angle BDC is much greater than the angle BAC.'"],["Rule","'ProofWordCount'",226],["Rule","'GreekProof'","'τριγώνου γὰρ τοῦ ΑΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ ἀπὸ τῶν περάτων τῶν Β, Γ δύο εὐθεῖαι ἐντὸς συνεστάτωσαν αἱ ΒΔ, ΔΓ: λέγω, ὅτι αἱ ΒΔ, ΔΓ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τῶν ΒΑ, ΑΓ ἐλάσσονες μέν εἰσιν, μείζονα δὲ γωνίαν περιέχουσι τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ. διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε. καὶ ἐπεὶ παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσιν, τοῦ ΑΒΕ ἄρα τριγώνου αἱ δύο πλευραὶ αἱ ΑΒ, ΑΕ τῆς ΒΕ μείζονές εἰσιν: κοινὴ προσκείσθω ἡ ΕΓ: αἱ ἄρα ΒΑ, ΑΓ τῶν ΒΕ, ΕΓ μείζονές εἰσιν. πάλιν, ἐπεὶ τοῦ ΓΕΔ τριγώνου αἱ δύο πλευραὶ αἱ ΓΕ, ΕΔ τῆς ΓΔ μείζονές εἰσιν, κοινὴ προσκείσθω ἡ ΔΒ: αἱ ΓΕ, ΕΒ ἄρα τῶν ΓΔ, ΔΒ μείζονές εἰσιν. ἀλλὰ τῶν ΒΕ, ΕΓ μείζονες ἐδείχθησαν αἱ ΒΑ, ΑΓ: πολλῷ ἄρα αἱ ΒΑ, ΑΓ τῶν ΒΔ, ΔΓ μείζονές εἰσιν. πάλιν, ἐπεὶ παντὸς τριγώνου ἡ ἐκτὸς γωνία τῆς ἐντὸς καὶ ἀπεναντίον μείζων ἐστίν, τοῦ ΓΔΕ ἄρα τριγώνου ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ. διὰ ταὐτὰ τοίνυν καὶ τοῦ ΑΒΕ τριγώνου ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ. ἀλλὰ τῆς ὑπὸ ΓΕΒ μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ: πολλῷ ἄρα ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ. ἐὰν ἄρα τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν περάτων δύο εὐθεῖαι ἐντὸς συσταθῶσιν, αἱ συσταθεῖσαι τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μέν εἰσιν, μείζονα δὲ γωνίαν περιέχουσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",239]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'1.22'"],["Rule","'Text'","'Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one.'"],["Rule","'TextWordCount'",39],["Rule","'GreekText'","'ἐκ τριῶν εὐθειῶν, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις εὐθείαις, τρίγωνον συστήσασθαι: δεῖ δὲ τὰς δύο τῆς λοιπῆς μείζονας εἶναι πάντῃ μεταλαμβανομένας διὰ τὸ καὶ παντὸς τριγώνου τὰς δύο πλευρὰς τῆς λοιπῆς μείζονας εἶναι πάντῃ μεταλαμβανομένας.'"],["Rule","'GreekTextWordCount'",36],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, and B, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C. Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3] With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined; I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C. For, since the point F is the centre of the circle DKL, FD is equal to FK. But FD is equal to A; therefore KF is also equal to A. Again, since the point G is the centre of the circle LKH, GH is equal to GK. But GH is equal to C; therefore KG is also equal to C. And FG is also equal to B; therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C. Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constructed.'"],["Rule","'ProofWordCount'",263],["Rule","'GreekProof'","'ἔστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ, ὧν αἱ δύο τῆς λοιπῆς μείζονες ἔστωσαν πάντῃ μεταλαμβανόμεναι, αἱ μὲν Α, Β τῆς Γ, αἱ δὲ Α, Γ τῆς Β, καὶ ἔτι αἱ Β, Γ τῆς Α: δεῖ δὴ ἐκ τῶν ἴσων ταῖς Α, Β, Γ τρίγωνον συστήσασθαι. Ἐκκείσθω τις εὐθεῖα ἡ ΔΕ πεπερασμένη μὲν κατὰ τὸ Δ ἄπειρος δὲ κατὰ τὸ Ε, καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΖ, τῇ δὲ Β ἴση ἡ ΖΗ, τῇ δὲ Γ ἴση ἡ ΗΘ: καὶ κέντρῳ μὲν τῷ Ζ, διαστήματι δὲ τῷ ΖΔ κύκλος γεγράφθω ὁ ΔΚΛ: πάλιν κέντρῳ μὲν τῷ Η, διαστήματι δὲ τῷ ΗΘ κύκλος γεγράφθω ὁ ΚΛΘ, καὶ ἐπεζεύχθωσαν αἱ ΚΖ, ΚΗ: λέγω, ὅτι ἐκ τριῶν εὐθειῶν τῶν ἴσων ταῖς Α, Β, Γ τρίγωνον συνέσταται τὸ ΚΖΗ. ἐπεὶ γὰρ τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου, ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ: ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση. καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση. πάλιν, ἐπεὶ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου, ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ: ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση: καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση. ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση: αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΚΖ, ΖΗ, ΗΚ τρισὶ ταῖς Α, Β, Γ ἴσαι εἰσίν. ἐκ τριῶν ἄρα εὐθειῶν τῶν ΚΖ, ΖΗ, ΗΚ, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις εὐθείαις ταῖς Α, Β, Γ, τρίγωνον συνέσταται τὸ ΚΖΗ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",240]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'1.23'"],["Rule","'Text'","'On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'πρὸς τῇ δοθείσῃ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην γωνίαν εὐθύγραμμον συστήσασθαι.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle; thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE. On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined, and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed in such a way that CD is equal to AF, CE to AG, and further DE to FG. Then, since the two sides DC, CE are equal to the two sides FA, AG respectively, and the base DE is equal to the base FG, the angle DCE is equal to the angle FAG. [I. 8] Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE.'"],["Rule","'ProofWordCount'",175],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΔΓΕ: δεῖ δὴ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ ΔΓΕ ἴσην γωνίαν εὐθύγραμμον συστήσασθαι. εἰλήφθω ἐφ᾽ ἑκατέρας τῶν ΓΔ, ΓΕ τυχόντα σημεῖα τὰ Δ, Ε, καὶ ἐπεζεύχθω ἡ ΔΕ: καὶ ἐκ τριῶν εὐθειῶν, αἵ εἰσιν ἴσαι τρισὶ ταῖς ΓΔ, ΔΕ, ΓΕ, τρίγωνον συνεστάτω τὸ ΑΖΗ, ὥστε ἴσην εἶναι τὴν μὲν ΓΔ τῇ ΑΖ, τὴν δὲ ΓΕ τῇ ΑΗ, καὶ ἔτι τὴν ΔΕ τῇ ΖΗ. ἐπεὶ οὖν δύο αἱ ΔΓ, ΓΕ δύο ταῖς ΖΑ, ΑΗ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ἴση, γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση. πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ ΔΓΕ ἴση γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ ΖΑΗ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",163]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'1.24'"],["Rule","'Text'","'If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.'"],["Rule","'TextWordCount'",39],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δύο πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας μείζονα ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν βάσιν τῆς βάσεως μείζονα ἕξει.'"],["Rule","'GreekTextWordCount'",33],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",19]],["List",["Rule","'Book'",1],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF, and let the angle at A be greater than the angle at D; I say that the base BC is also greater than the base EF. For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG  equal to the angle BAC; [I. 23] let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined. Then, since AB is equal to DE, and AC to DG,the two sides BA, AC are equal to the two sides ED, DG, respectively; and the angle BAC is equal to the angle EDG; therefore the base BC is equal to the base EG. [I. 4] Again, since DF is equal to DG,the angle DGF is also equal to the angle DFG; [I. 5]therefore the angle DFG is greater than the angle EGF. Therefore the angle EFG is much greater than the angle EGF. And, since EFG is a triangle having the angle EFG  greater than the angle EGF, and the greater angle is subtended by the greater side, [I. 19]the side EG is also greater than EF. But EG is equal to BC. Therefore BC is also greater than EF.'"],["Rule","'ProofWordCount'",246],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ, ἡ δὲ πρὸς τῷ Α γωνία τῆς πρὸς τῷ Δ γωνίας μείζων ἔστω: λέγω, ὅτι καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ μείζων ἐστίν. ἐπεὶ γὰρ μείζων ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ γωνίας, συνεστάτω πρὸς τῇ ΔΕ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ἡ ὑπὸ ΕΔΗ, καὶ κείσθω ὁποτέρᾳ τῶν ΑΓ, ΔΖ ἴση ἡ ΔΗ, καὶ ἐπεζεύχθωσαν αἱ ΕΗ, ΖΗ. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΑΓ τῇ ΔΗ, δύο δὴ αἱ ΒΑ, ΑΓ δυσὶ ταῖς ΕΔ, ΔΗ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση: βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΗ ἐστιν ἴση. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΔΖ τῇ ΔΗ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ: μείζων ἄρα ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ: πολλῷ ἄρα μείζων ἐστὶν ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ. καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ μείζονα ἔχον τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει, μείζων ἄρα καὶ πλευρὰ ἡ ΕΗ τῆς ΕΖ. ἴση δὲ ἡ ΕΗ τῇ ΒΓ: μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ. ἐὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας μείζονα ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν βάσιν τῆς βάσεως μείζονα ἕξει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",264]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'1.25'"],["Rule","'Text'","'If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.'"],["Rule","'TextWordCount'",39],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ βάσιν τῆς βάσεως μείζονα ἔχῃ, καὶ τὴν γωνίαν τῆς γωνίας μείζονα ἕξει τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let the base BC be greater than the base EF; I say that the angle BAC is also greater than the angle EDF. For, if not, it is either equal to it or less. Now the angle BAC is not equal to the angle EDF; for then the base BC would also have been equal to the base EF, [I. 4] but it is not; therefore the angle BAC is not equal to the angle EDF. Neither again is the angle BAC less than the angle EDF; for then the base BC would also have been less than the base EF, [I. 24] but it is not; therefore the angle BAC is not less than the angle EDF. But it was proved that it is not equal either; therefore the angle BAC is greater than the angle EDF.'"],["Rule","'ProofWordCount'",167],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ, τὴν δὲ ΑΓ τῇ ΔΖ: βάσις δὲ ἡ ΒΓ βάσεως τῆς ΕΖ μείζων ἔστω: λέγω, ὅτι καὶ γωνία ἡ ὑπὸ ΒΑΓ γωνίας τῆς ὑπὸ ΕΔΖ μείζων ἐστίν: εἰ γὰρ μή, ἤτοι ἴση ἐστὶν αὐτῇ á¼¢ ἐλάσσων: ἴση μὲν οὖν οὐκ ἔστιν ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ: ἴση γὰρ ἂν ἦν καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ: οὐκ ἔστι δέ. οὐκ ἄρα ἴση ἐστὶ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ: οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ: ἐλάσσων γὰρ ἂν ἦν καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ: οὐκ ἔστι δέ: οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ. ἐδείχθη δὲ ὅτι οὐδὲ ἴση: μείζων ἄρα ἐστὶν ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ. ἐὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκάτερᾳ, τὴν δὲ βάσιν τῆς βάσεως μείζονα ἔχῃ, καὶ τὴν γωνίαν τῆς γωνίας μείζονα ἕξει τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",184]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",26]],["Association",["Rule","'VertexLabel'","'1.26'"],["Rule","'Text'","'If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal tothe remaining sides and the remaining angle to the remaining angle.'"],["Rule","'TextWordCount'",54],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις á¼¢ τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν, καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ.'"],["Rule","'GreekTextWordCount'",50],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively, namely the angle ABC to the angle DEF, and the angleBCA to the angle EFD; and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF; I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE andAC to DF, and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF. For, if AB is unequal to DE, one of them is greater. Let AB be greater, and let BG be made equal to DE; and let GC be joined. Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively; and the angle GBC is equal to the angle DEF; therefore the base GC is equal to the base DF, and the triangle GBC is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]therefore the angle GCB is equal to the angle DFE. But the angle DFE is by hypothesis equal to the angle BCA;therefore the angle BCG is equal to the angle BCA, the less to the greater: which is impossible. Therefore AB is not unequal to DE, and is therefore equal to it. But BC is also equal to EF;therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and the angle ABC is equal to the angle DEF; therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF. [I. 4]  Again, let sides subtending equal angles be equal, as AB to DE; I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, andfurther the remaining angle BAC is equal to the remaining angle EDF. For, if BC is unequal to EF, one of them is greater. Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined. Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DEF,and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle BHA is equal to the angle EFD. But the angle EFD is equal to the angle BCA; therefore, in the triangle AHC, the exterior angle BHA isequal to the interior and opposite angle BCA: which is impossible. [I. 16] Therefore BC is not unequal to EF, and is therefore equal to it. But AB is also equal to DE;therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles; therefore the base AC is equal to the base DF, the triangle ABC equal to the triangle DEF, and the remaining angle BAC equal to the remaining angleEDF. [I. 4]'"],["Rule","'ProofWordCount'",566],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ, ΒΓΑ δυσὶ ταῖς ὑπὸ ΔΕΖ, ΕΖΔ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ: ἐχέτω δὲ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην, πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΒΓ τῇ ΕΖ: λέγω, ὅτι καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ, καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ, τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ. εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΔΕ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ΑΒ, καὶ κείσθω τῇ ΔΕ ἴση ἡ ΒΗ, καὶ ἐπεζεύχθω ἡ ΗΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΒΗ τῇ ΔΕ, ἡ δὲ ΒΓ τῇ ΕΖ, δύο δὴ αἱ ΒΗ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΗΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἴση ἐστίν: βάσις ἄρα ἡ ΗΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΗΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἀλλὰ ἡ ὑπὸ ΔΖΕ τῇ ὑπὸ ΒΓΑ ὑπόκειται ἴση: καὶ ἡ ὑπὸ ΒΓΗ ἄρα τῇ ὑπὸ ΒΓΑ ἴση ἐστίν, ἡ ἐλάσσων τῇ μείζονι: ὅπερ ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΔΕ. ἴση ἄρα. ἔστι δὲ καὶ ἡ ΒΓ τῇ ΕΖ ἴση: δύο δὴ αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση: βάσις ἄρα ἡ ΑΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση ἐστίν. ἀλλὰ δὴ πάλιν ἔστωσαν αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ ὑποτείνουσαι ἴσαι, ὡς ἡ ΑΒ τῇ ΔΕ: λέγω πάλιν, ὅτι καὶ αἱ λοιπαὶ πλευραὶ ταῖς λοιπαῖς πλευραῖς ἴσαι ἔσονται, ἡ μὲν ΑΓ τῇ ΔΖ, ἡ δὲ ΒΓ τῇ ΕΖ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση ἐστίν. εἰ γὰρ ἄνισός ἐστιν ἡ ΒΓ τῇ ΕΖ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων, εἰ δυνατόν, ἡ ΒΓ, καὶ κείσθω τῇ ΕΖ ἴση ἡ ΒΘ, καὶ ἐπεζεύχθω ἡ ΑΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΒΘ τῇ ΕΖ ἡ δὲ ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΑΒ, ΒΘ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνίας ἴσας περιέχουσιν: βάσις ἄρα ἡ ΑΘ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΑΒΘ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΘΑ γωνία τῇ ὑπὸ ΕΖΔ. ἀλλὰ ἡ ὑπὸ ΕΖΔ τῇ ὑπὸ ΒΓΑ ἐστιν ἴση: τριγώνου δὴ τοῦ ΑΘΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΒΓΑ: ὅπερ ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ΒΓ τῇ ΕΖ: ἴση ἄρα. ἐστὶ δὲ καὶ ἡ ΑΒ τῇ ΔΕ ἴση. δύο δὴ αἱ ΑΒ, ΒΓ δύο ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνίας ἴσας περιέχουσι: βάσις ἄρα ἡ ΑΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση. ἐὰν ἄρα δύο τρίγωνα τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις, á¼¢ τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν, καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",594]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",27]],["Association",["Rule","'VertexLabel'","'1.27'"],["Rule","'Text'","'If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιῇ, παράλληλοι ἔσονται ἀλλήλαις αἱ εὐθεῖαι.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'For let the straight line EF falling on the two straightlines AB, CD make the alternate angles AEF, EFD equal to one another; I say that AB is parallel to CD. For, if not, AB, CD when produced will meet either in the directionof B, D or towards A, C. Let them be produced and meet, in the direction of B, D, at G. Then, in the triangle GEF, the exterior angle AEF is equal to the interior and oppositeangle EFG: which is impossible. [I. 16] Therefore AB, CD when produced will not meet in the direction of B, D. Similarly it can be proved that neither will they meettowards A, C. But straight lines which do not meet in either direction are parallel; [Def. 23] therefore AB is parallel to CD.'"],["Rule","'ProofWordCount'",132],["Rule","'GreekProof'","'εἰς γὰρ δύο εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΑΕΖ, ΕΖΔ ἴσας ἀλλήλαις ποιείτω: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. εἰ γὰρ μή, ἐκβαλλόμεναι αἱ ΑΒ, ΓΔ συμπεσοῦνται ἤτοι ἐπὶ τὰ Β, Δ μέρη á¼¢ ἐπὶ τὰ Α, Γ. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν ἐπὶ τὰ Β, Δ μέρη κατὰ τὸ Η. τριγώνου δὴ τοῦ ΗΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΕΖΗ: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα αἱ ΑΒ, ΓΔ ἐκβαλλόμεναι συμπεσοῦνται ἐπὶ τὰ Β, Δ μέρη. ὁμοίως δὴ δειχθήσεται, ὅτι οὐδὲ ἐπὶ τὰ Α, Γ: αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη συμπίπτουσαι παράλληλοί εἰσιν: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. ἐὰν ἄρα εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιῇ, παράλληλοι ἔσονται αἱ εὐθεῖαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",138]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",28]],["Association",["Rule","'VertexLabel'","'1.28'"],["Rule","'Text'","'If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.'"],["Rule","'TextWordCount'",46],["Rule","'GreekText'","'ἐὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὴν ἐκτὸς γωνίαν τῇ ἐντὸς καὶ ἀπεναντίον καὶ ἐπὶ τὰ αὐτὰ μέρη ἴσην ποιῇ á¼¢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας, παράλληλοι ἔσονται ἀλλήλαις αἱ εὐθεῖαι.'"],["Rule","'GreekTextWordCount'",36],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",13]],["List",["Rule","'Book'",1],["Rule","'Theorem'",15]],["List",["Rule","'Book'",1],["Rule","'Theorem'",27]]]],["Rule","'Proof'","'For let the straight line EF falling on the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles on the same side, namely BGH, GHD, equal to two right angles; I say that AB is parallel to CD. For, since the angle EGB is equal to the angle GHD, while the angle EGB is equal to the angle AGH, [I. 15] the angle AGH is also equal to the angle GHD; and they are alternate; therefore AB is parallel to CD. [I. 27] Again, since the angles BGH, GHD are equal to two right angles, and the angles AGH, BGH are also equal to two right angles, [I. 13] the angles AGH, BGH are equal to the angles BGH, GHD. Let the angle BGH be subtracted from each; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate; therefore AB is parallel to CD. [I. 27]'"],["Rule","'ProofWordCount'",165],["Rule","'GreekProof'","'εἰς γὰρ δύο εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ τῇ ὑπὸ ΗΘΔ ἴσην ποιείτω á¼¢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη τὰς ὑπὸ ΒΗΘ, ΗΘΔ δυσὶν ὀρθαῖς ἴσας: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ, ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ ἐστιν ἴση, καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ ἐστιν ἴση: καί εἰσιν ἐναλλάξ: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. πάλιν, ἐπεὶ αἱ ὑπὸ ΒΗΘ, ΗΘΔ δύο ὀρθαῖς ἴσαι εἰσίν, εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ, ΒΗΘ δυσὶν ὀρθαῖς ἴσαι, αἱ ἄρα ὑπὸ ΑΗΘ, ΒΗΘ ταῖς ὑπὸ ΒΗΘ, ΗΘΔ ἴσαι εἰσίν: κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΒΗΘ: λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ λοιπῇ τῇ ὑπὸ ΗΘΔ ἐστιν ἴση: καί εἰσιν ἐναλλάξ: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. ἐὰν ἄρα εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὴν ἐκτὸς γωνίαν τῇ ἐντὸς καὶ ἀπεναντίον καὶ ἐπὶ τὰ αὐτὰ μέρη ἴσην ποιῇ á¼¢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας, παράλληλοι ἔσονται αἱ εὐθεῖαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",185]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",29]],["Association",["Rule","'VertexLabel'","'1.29'"],["Rule","'Text'","'A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles.'"],["Rule","'TextWordCount'",39],["Rule","'GreekText'","'ἡ εἰς τὰς παραλλήλους εὐθείας εὐθεῖα ἐμπίπτουσα τάς τε ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιεῖ καὶ τὴν ἐκτὸς τῇ ἐντὸς καὶ ἀπεναντίον ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας.'"],["Rule","'GreekTextWordCount'",33],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Common Notion'",2]],["List",["Rule","'Postulate'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",13]],["List",["Rule","'Book'",1],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'For let the straight line EF fall on the parallel straight lines AB, CD; I say that it makes the alternate angles AGH, GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the interior angles on the sameside, namely BGH, GHD, equal to two right angles. For, if the angle AGH is unequal to the angle GHD, one of them is greater. Let the angle AGH be greater. Let the angle BGH be added to each; therefore the angles AGH, BGH are greater than the angles BGH, GHD. But the angles AGH, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are less than two right angles. But straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]therefore AB, CD, if produced indefinitely, will meet; but they do not meet, because they are by hypothesis parallel. Therefore the angle AGH is not unequal to the angle GHD, and is therefore equal to it. Again, the angle AGH is equal to the angle EGB; [I. 15] therefore the angle EGB is also equal to the angle GHD. [C. N. 1] Let the angle BGH be added to each; therefore the angles EGB, BGH are equal to the angles BGH, GHD. [C. N. 2] But the angles EGB, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are also equal to two right angles.'"],["Rule","'ProofWordCount'",243],["Rule","'GreekProof'","'εἰς γὰρ παραλλήλους εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπιπτέτω ἡ ΕΖ: λέγω, ὅτι τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΑΗΘ, ΗΘΔ ἴσας ποιεῖ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη τὰς ὑπὸ ΒΗΘ, ΗΘΔ δυσὶν ὀρθαῖς ἴσας. εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΑΗΘ: κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ: αἱ ἄρα ὑπὸ ΑΗΘ, ΒΗΘ τῶν ὑπὸ ΒΗΘ, ΗΘΔ μείζονές εἰσιν. ἀλλὰ αἱ ὑπὸ ΑΗΘ, ΒΗΘ δυσὶν ὀρθαῖς ἴσαι εἰσίν. καὶ αἱ ἄρα ὑπὸ ΒΗΘ, ΗΘΔ δύο ὀρθῶν ἐλάσσονές εἰσιν. αἱ δὲ ἀπ᾽ ἐλασσόνων á¼¢ δύο ὀρθῶν ἐκβαλλόμεναι εἰς ἄπειρον συμπίπτουσιν: αἱ ἄρα ΑΒ, ΓΔ ἐκβαλλόμεναι εἰς ἄπειρον συμπεσοῦνται: οὐ συμπίπτουσι δὲ διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι: οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ: ἴση ἄρα. ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ ἐστιν ἴση: καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ ἐστιν ἴση. κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ: αἱ ἄρα ὑπὸ ΕΗΒ, ΒΗΘ ταῖς ὑπὸ ΒΗΘ, ΗΘΔ ἴσαι εἰσίν. ἀλλὰ αἱ ὑπὸ ΕΗΒ, ΒΗΘ δύο ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΒΗΘ, ΗΘΔ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν. ἡ ἄρα εἰς τὰς παραλλήλους εὐθείας εὐθεῖα ἐμπίπτουσα τάς τε ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιεῖ καὶ τὴν ἐκτὸς τῇ ἐντὸς καὶ ἀπεναντίον ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας: ὅπερ ἔδει δεῖξαι. αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι. ἔστω ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος: λέγω, ὅτι καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος. ἐμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ. καὶ ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΑΒ, ΕΖ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ. πάλιν, ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΕΖ, ΓΔ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση. καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ ἐστιν ἴση: καί εἰσιν ἐναλλάξ. παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. Αἱ ἄρα τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",353]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",30]],["Association",["Rule","'VertexLabel'","'1.30'"],["Rule","'Text'","'Straight lines parallel to the same straight line are also parallel to one another.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]]]],["Rule","'Proof'","'Let each of the straight lines AB, CD be parallel to EF; I say that AB is also parallel to CD. For let the straight line GK fall upon them; Then, since the straight line GK has fallen on the parallel straight lines AB, EF,the angle AGK is equal to the angle GHF. [I. 29] Again, since the straight line GK has fallen on the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD. [I. 29]  But the angle AGK was also proved equal to the angle GHF; therefore the angle AGK is also equal to the angle GKD; [C. N. 1]and they are alternate. Therefore AB is parallel to CD.'"],["Rule","'ProofWordCount'",118],["Rule","'GreekProof'","'῎Εστω ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος: λέγω, ὅτι καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος.  Ἐμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ. Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΑΒ, ΕΖ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ. πάλιν, ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΕΖ, ΓΔ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση. καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ ἐστιν ἴση: καί εἰσιν ἐναλλάξ. παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. [Αἱ ἄρα τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι:] ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",108]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",31]],["Association",["Rule","'VertexLabel'","'1.31'"],["Rule","'Text'","'Through a given point to draw a straight line parallel to a given straight line.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'διὰ τοῦ δοθέντος σημείου τῇ δοθείσῃ εὐθείᾳ παράλληλον εὐθεῖαν γραμμὴν ἀγαγεῖν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",27]]]],["Rule","'Proof'","'Let A be the given point, and BC the given straight line; thus it is required to draw through the point A a straight line parallel to the straight line BC. Let a point D be taken at random on BC, and let AD be joined; on the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [I. 23]; and let the straight line AF be produced in a straight line with EA. Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another, therefore EAF is parallel to BC. [I. 27] Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC.'"],["Rule","'ProofWordCount'",138],["Rule","'GreekProof'","'ἔστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ: δεῖ δὴ διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ παράλληλον εὐθεῖαν γραμμὴν ἀγαγεῖν. εἰλήφθω ἐπὶ τῆς ΒΓ τυχὸν σημεῖον τὸ Δ, καὶ ἐπεζεύχθω ἡ ΑΔ: καὶ συνεστάτω πρὸς τῇ ΔΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ἡ ὑπὸ ΔΑΕ: καὶ ἐκβεβλήσθω ἐπ᾽ εὐθείας τῇ ΕΑ εὐθεῖα ἡ ΑΖ. καὶ ἐπεὶ εἰς δύο εὐθείας τὰς ΒΓ, ΕΖ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΕΑΔ, ΑΔΓ ἴσας ἀλλήλαις πεποίηκεν, παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ. διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",119]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",32]],["Association",["Rule","'VertexLabel'","'1.32'"],["Rule","'Text'","'In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'παντὸς τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ἴση ἐστίν, καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν.'"],["Rule","'GreekTextWordCount'",27],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",13]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]]]],["Rule","'Proof'","'Let ABC be a triangle, and let one side of it BC be produced to D; I say that the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle ABC, BCA, CAB are equal to two right angles. For let CE be drawn through the point C parallel to the straight line AB. [I. 31] Then, since AB is parallel to CE, and AC has fallen upon them, the alternate angles BAC, ACE are equal to one another. [I. 29]  Again, since AB is parallel to CE, and the straight line BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. [I. 29] But the angle ACE was also proved equal to the angle BAC; therefore the whole angle ACD is equal to the two interior and opposite angles BAC, ABC. Let the angle ACB be added to each; therefore the angles ACD, ACB are equal to the three angles ABC, BCA, CAB. But the angles ACD, ACB are equal to two right angles; [I. 13] therefore the angles ABC, BCA, CAB are also equal to two right angles.'"],["Rule","'ProofWordCount'",200],["Rule","'GreekProof'","'ἔστω τρίγωνον τὸ ΑΒΓ, καὶ προσεκβεβλήσθω αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ: λέγω, ὅτι ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ταῖς ὑπὸ ΓΑΒ, ΑΒΓ, καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι αἱ ὑπὸ ΑΒΓ, ΒΓΑ, ΓΑΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἤχθω γὰρ διὰ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ παράλληλος ἡ ΓΕ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ, ΑΓΕ ἴσαι ἀλλήλαις εἰσίν. πάλιν, ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ, καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΒΔ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ ἴση: ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ἴση ἐστὶ δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ταῖς ὑπὸ ΒΑΓ, ΑΒΓ. κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ: αἱ ἄρα ὑπὸ ΑΓΔ, ΑΓΒ τρισὶ ταῖς ὑπὸ ΑΒΓ, ΒΓΑ, ΓΑΒ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΑΓΔ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΑΓΒ, ΓΒΑ, ΓΑΒ ἄρα δυσίν ὀρθαῖς ἴσαι εἰσίν. παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ἴση ἐστίν, καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",212]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",33]],["Association",["Rule","'VertexLabel'","'1.33'"],["Rule","'Text'","'The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι εὐθεῖαι καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",27]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]]]],["Rule","'Proof'","'Let AB, CD be equal and parallel, and let the straightlines AC, BD join them (at the extremities which are) in the same directions (respectively); I say that AC, BD are also equal and parallel. Let BC be joined. Then, since AB is parallel to CD,and BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29] And, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the griangle ABC is equal to the triangle DCB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]therefore the angle ACB is equal to the angle CBD. And, since the straight line BC falling on the two straight lines AC, BD has made the alternate angles equal to one another,AC is parallel to BD. [I. 27] And it was also proved equal to it.'"],["Rule","'ProofWordCount'",187],["Rule","'GreekProof'","'ἔστωσαν ἴσαι τε καὶ παράλληλοι αἱ ΑΒ, ΓΔ, καὶ ἐπιζευγνύτωσαν αὐτὰς ἐπὶ τὰ αὐτὰ μέρη εὐθεῖαι αἱ ΑΓ, ΒΔ: λέγω, ὅτι καὶ αἱ ΑΓ, ΒΔ ἴσαι τε καὶ παράλληλοί εἰσιν. ἐπεζεύχθω ἡ ΒΓ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ, ΒΓΔ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ κοινὴ δὲ ἡ ΒΓ, δύο δὴ αἱ ΑΒ, ΒΓ δύο ταῖς ΒΓ, ΓΔ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση: βάσις ἄρα ἡ ΑΓ βάσει τῇ ΒΔ ἐστιν ἴση, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΒΓΔ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ. καὶ ἐπεὶ εἰς δύο εὐθείας τὰς ΑΓ, ΒΔ εὐθεῖα ἐμπίπτουσα ἡ ΒΓ τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις πεποίηκεν, παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ. ἐδείχθη δὲ αὐτῇ καὶ ἴση. αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι εὐθεῖαι καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",188]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",34]],["Association",["Rule","'VertexLabel'","'1.34'"],["Rule","'Text'","'In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'τῶν παραλληλογράμμων χωρίων αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Common Notion'",2]],["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]]]],["Rule","'Proof'","'Let ACDB be a parallelogrammic area, and BC its diameter;I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it. For, since AB is parallel to CD, and the straight line BC has fallenupon them, the alternate angles ABC, BCD are equal to one another. [I. 29] Again, since AC is parallel to BD, and BC has fallen upon them,the alternate angles ACB, CBD are equal to one another. [I. 29] Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively, and one side equal to one side, namely thatadjoining the equal angles and common to both of them, BC; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [I. 26]therefore the side AB is equal to CD, and AC to BD, and further the angle BAC is equal to the angle CDB. And, since the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD. [C. N. 2] And the angle BAC was also proved equal to the angle CDB. Therefore in parallelogrammic areas the opposite sides and angles are equal to one another. I say, next, that the diameter also bisects the areas. For, since AB is equal to CD,and BC is common, the two sides AB, BC are equal to the two sides DC, CB respectively; and the angle ABC is equal to the angle BCD; therefore the base AC is also equal to DB, and the triangle ABC is equal to the triangle DCB. [I. 4] Therefore the diameter BC bisects the parallelogram ACDB.'"],["Rule","'ProofWordCount'",304],["Rule","'GreekProof'","'ἔστω παραλληλόγραμμον χωρίον τὸ ΑΓΔΒ, διάμετρος δὲ αὐτοῦ ἡ ΒΓ: λέγω, ὅτι τοῦ ΑΓΔΒ παραλληλογράμμου αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει. ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ, ΒΓΔ ἴσαι ἀλλήλαις εἰσίν. πάλιν, ἐπεὶ παράλληλός ἐστιν ἡ ΑΓ τῇ ΒΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ, ΓΒΔ ἴσαι ἀλλήλαις εἰσίν. δύο δὴ τρίγωνά ἐστι τὰ ΑΒΓ, ΒΓΔ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ, ΒΓΑ δυσὶ ταῖς ὑπὸ ΒΓΔ, ΓΒΔ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις κοινὴν αὐτῶν τὴν ΒΓ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ: ἴση ἄρα ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ, ἡ δὲ ΑΓ τῇ ΒΔ, καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΓΔΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΒΓΔ, ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ, ὅλη ἄρα ἡ ὑπὸ ΑΒΔ ὅλῃ τῇ ὑπὸ ΑΓΔ ἐστιν ἴση. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση. τῶν ἄρα παραλληλογράμμων χωρίων αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν. λέγω δή, ὅτι καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ, κοινὴ δὲ ἡ ΒΓ, δύο δὴ αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΓΔ, ΒΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση. καὶ βάσις ἄρα ἡ ΑΓ τῇ ΔΒ ἴση. καὶ τὸ ΑΒΓ ἄρα τρίγωνον τῷ ΒΓΔ τριγώνῳ ἴσον ἐστίν. ἡ ἄρα ΒΓ διάμετρος δίχα τέμνει τὸ ΑΒΓΔ παραλληλόγραμμον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",287]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",35]],["Association",["Rule","'VertexLabel'","'1.35'"],["Rule","'Text'","'Parallelograms which are on the same base and in the same parallels are equal to one another.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'τὰ παραλληλόγραμμα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Common Notion'",2]],["List",["Rule","'Common Notion'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]]]],["Rule","'Proof'","'Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC;I say that ABCD is equal to the parallelogram EBCF. For, since ABCD is a parallelogram, AD is equal to BC. [I. 34]  For the same reason also EF is equal to BC, so that AD is also equal to EF; [C. N. 1]and DE is common; therefore the whole AE is equal to the whole DF. [C. N. 2] But AB is also equal to DC; [I. 34] therefore the two sides EA, AB are equal to the two sidesFD, DC respectively, and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29]therefore the base EB is equal to the base FC, and the triangle EAB will be equal to the triangle FDC. [I. 4] Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [C. N. 3] Let the triangle GBC be added to each; therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. [C. N. 2]'"],["Rule","'ProofWordCount'",187],["Rule","'GreekProof'","'ἔστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΒΓΖ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΖ, ΒΓ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ τῷ ΕΒΓΖ παραλληλογράμμῳ. ἐπεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ, ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση: ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση: καὶ κοινὴ ἡ ΔΕ: ὅλη ἄρα ἡ ΑΕ ὅλῃ τῇ ΔΖ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση: δύο δὴ αἱ ΕΑ, ΑΒ δύο ταῖς ΖΔ, ΔΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΖΔΓ γωνίᾳ τῇ ὑπὸ ΕΑΒ ἐστιν ἴση ἡ ἐκτὸς τῇ ἐντός: βάσις ἄρα ἡ ΕΒ βάσει τῇ ΖΓ ἴση ἐστίν, καὶ τὸ ΕΑΒ τρίγωνον τῷ ΔΖΓ τριγώνῳ ἴσον ἔσται: κοινὸν ἀφῃρήσθω τὸ ΔΗΕ: λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ ἐστὶν ἴσον: κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον: ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ἴσον ἐστίν. τὰ ἄρα παραλληλόγραμμα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",177]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",36]],["Association",["Rule","'VertexLabel'","'1.36'"],["Rule","'Text'","'Parallelograms which are on equal bases and in the same parallels are equal to one another.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'τὰ παραλληλόγραμμα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Book'",1],["Rule","'Theorem'",33]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",35]]]],["Rule","'Proof'","'Let ABCD, EFGH be parallelograms which are on equal bases BC, FG and in the same parallels AH, BG; I say that the parallelogram ABCD is equal to EFGH. For let BE, CH be joined. Then, since BC is equal to FG while FG is equal to EH, BC is also equal to EH. [C. N. 1] But they are also parallel. And EB, HC join them; but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [I. 33] Therefore EBCH is a parallelogram. [I. 34] And it is equal to ABCD; for it has the same base BC with it, and is in the same parallels BC, AH with it. [I. 35] For the same reason also EFGH is equal to the same EBCH; [I. 35] so that the parallelogram ABCD is also equal to EFGH. [C. N. 1]'"],["Rule","'ProofWordCount'",153],["Rule","'GreekProof'","'ἔστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΖΗΘ ἐπὶ ἴσων βάσεων ὄντα τῶν ΒΓ, ΖΗ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΘ, ΒΗ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΖΗΘ. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ, ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση, καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση. εἰσὶ δὲ καὶ παράλληλοι. καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ, ΘΓ: αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι ἴσαι τε καὶ παράλληλοί εἰσι: καὶ αἱ ΕΒ, ΘΓ ἄρα ἴσαι τέ εἰσι καὶ παράλληλοι. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ. καί ἐστιν ἴσον τῷ ΑΒΓΔ: βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει τὴν ΒΓ, καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστὶν αὐτῷ ταῖς ΒΓ, ΑΘ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ἐστιν ἴσον: ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΖΗΘ ἐστιν ἴσον. τὰ ἄρα παραλληλόγραμμα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",164]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",37]],["Association",["Rule","'VertexLabel'","'1.37'"],["Rule","'Text'","'Triangles which are on the same base and in the same parallels are equal to one another.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'τὰ τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",35]]]],["Rule","'Proof'","'Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC;I say that the triangle ABC is equal to the triangle DBC. Let AD be produced in both directions to E, F; through B let BE be drawn parallel to CA, [I. 31]and through C let CF be drawn parallel to BD. [I. 31] Then each of the figures EBCA, DBCF is a parallelogram; and they are equal, for they are on the same base BC and in the same parallels BC, EF. [I. 35] Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34] And the triangle DBC is half of the parallelogram DBCF;for the diameter DC bisects it. [I. 34] [But the halves of equal things are equal to one another.] Therefore the triangle ABC is equal to the triangle DBC.'"],["Rule","'ProofWordCount'",150],["Rule","'GreekProof'","'ἔστω τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΔ, ΒΓ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ. Ἐκβεβλήσθω ἡ ΑΔ ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε, Ζ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΕ, διὰ δὲ τοῦ Γ τῇ ΒΔ παράλληλος ἤχθω ἡ ΓΖ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΕΒΓΑ, ΔΒΓΖ: καί εἰσιν ἴσα: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΕΖ: καί ἐστι τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον: ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει: τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ἥμισυ τὸ ΔΒΓ τρίγωνον: ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει. τὰ δὲ τῶν ἴσων ἡμίση ἴσα ἀλλήλοις ἐστίν. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ. τὰ ἄρα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",158]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",38]],["Association",["Rule","'VertexLabel'","'1.38'"],["Rule","'Text'","'Triangles which are on equal bases and in the same parallels are equal to one another.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'τὰ τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD; I say that the triangle ABC is equal to the triangle DEF. For let AD be produced in both directions to G, H; through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE. Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH; for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36] Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34] And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34] [But the halves of equal things are equal to one another.] Therefore the triangle ABC is equal to the triangle DEF.'"],["Rule","'ProofWordCount'",151],["Rule","'GreekProof'","'ἔστω τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἐπὶ ἴσων βάσεων τῶν ΒΓ, ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΑΔ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. Ἐκβεβλήσθω γὰρ ἡ ΑΔ ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η, θ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΗ, διὰ δὲ τοῦ Ζ τῇ ΔΕ παράλληλος ἤχθω ἡ ΖΘ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΗΒΓΑ, ΔΕΖΘ: καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΗΘ: καί ἐστι τοῦ μὲν ΗΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον. ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει: τοῦ δὲ ΔΕΖΘ παραλληλογράμμου ἥμισυ τὸ ΖΕΔ τρίγωνον: ἡ γὰρ ΔΖ διάμετρος αὐτὸ δίχα τέμνει: τὰ δὲ τῶν ἴσων ἡμίση ἴσα ἀλλήλοις ἐστίν. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. τὰ ἄρα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",161]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",39]],["Association",["Rule","'VertexLabel'","'1.39'"],["Rule","'Text'","'Equal triangles which are on the same base and on the same side are also in the same parallels.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'τὰ ἴσα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",37]]]],["Rule","'Proof'","'Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it;[I say that they are also in the same parallels.] And [For]let AD be joined; I say that AD is parallel to BC. For, if not, let AE be drawn through the point A parallel to the straight lineBC, [I. 31] and let EC be joined. Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the sameparallels. [I. 37] But ABC is equal to DBC; therefore DBC is also equal to EBC, [C. N. 1]the greater to the less: which is impossible. Therefore AE is not parallel to BC. Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BC.'"],["Rule","'ProofWordCount'",143],["Rule","'GreekProof'","'Ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ: λέγω, ὅτι καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν.  Ἐπεζεύχθω γὰρ ἡ ΑΔ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ. Εἰ γὰρ μή, ἤχθω διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ παράλληλος ἡ ΑΕ, καὶ ἐπεζεύχθω ἡ ΕΓ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΕΒΓ τριγώνῳ: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις. ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον: καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ τὸ μεῖζον τῷ ἐλάσσονι: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλη τις πλὴν τῆς ΑΔ: ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος. Τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",152]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",40]],["Association",["Rule","'VertexLabel'","'1.40'"],["Rule","'Text'","'Equal triangles which are on equal bases and on the same side are also in the same parallels.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'τὰ ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",38]]]],["Rule","'Proof'","'Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side. I say that they are also in the same parallels. For let AD be joined; I say that AD is parallel to BE. For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined. Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38] But the triangle ABC is equal to the triangle DCE; therefore the triangle DCE is also equal to the triangle FCE, [C. N. 1]the greater to the less: which is impossible. Therefore AF is not parallel to BE. Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE.'"],["Rule","'ProofWordCount'",143],["Rule","'GreekProof'","'ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΓΔΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ, ΓΕ καὶ ἐπὶ τὰ αὐτὰ μέρη. λέγω, ὅτι καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. ἐπεζεύχθω γὰρ ἡ ΑΔ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ. εἰ γὰρ μή, ἤχθω διὰ τοῦ Α τῇ ΒΕ παράλληλος ἡ ΑΖ, καὶ ἐπεζεύχθω ἡ ΖΕ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΖΓΕ τριγώνῳ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΓΕ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΕ, ΑΖ. ἀλλὰ τὸ ΑΒΓ τρίγωνον ἴσον ἐστὶ τῷ ΔΓΕ τριγώνῳ: καὶ τὸ ΔΓΕ ἄρα τρίγωνον ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ τὸ μεῖζον τῷ ἐλάσσονι: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλη τις πλὴν τῆς ΑΔ: ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος. τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",153]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",41]],["Association",["Rule","'VertexLabel'","'1.41'"],["Rule","'Text'","'If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'ἐὰν παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστι τὸ παραλληλόγραμμον τοῦ τριγώνου.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",37]]]],["Rule","'Proof'","'For let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE; I say that the parallelogram ABCD is double of the triangle BEC. For let AC be joined. Then the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels BC, AE. [I. 37] But the parallelogram ABCD is double of the triangle ABC; for the diameter AC bisects it; [I. 34]so that the parallelogram ABCD is also double of the triangle EBC.'"],["Rule","'ProofWordCount'",100],["Rule","'GreekProof'","'παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ τριγώνῳ τῷ ΕΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ταῖς ΒΓ, ΑΕ: λέγω, ὅτι διπλάσιόν ἐστι τὸ ΑΒΓΔ παραλληλόγραμμον τοῦ ΒΕΓ τριγώνου. ἐπεζεύχθω γὰρ ἡ ΑΓ. ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΕΒΓ τριγώνῳ: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΕ. ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου: ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει: ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον. ἐὰν ἄρα παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστι τὸ παραλληλόγραμμον τοῦ τριγώνου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",114]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",42]],["Association",["Rule","'VertexLabel'","'1.42'"],["Rule","'Text'","'To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'τῷ δοθέντι τριγώνῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",38]],["List",["Rule","'Book'",1],["Rule","'Theorem'",41]]]],["Rule","'Proof'","'Let ABC be the given triangle, and D the given rectilineal angle; thus it is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC. Let BC be bisected at E, and let AE be joined; on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23] through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF. Then FECG is a parallelogram. And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC, for they are on equal bases BE, EC and in the same parallels BC, AG; [I. 38]therefore the triangle ABC is double of the triangle AEC. But the parallelogram FECG is also double of the triangle AEC, for it has the same base with it and is in the same parallels with it; [I. 41] therefore the parallelogram FECG is equal to the triangle ABC. And it has the angle CEF equal to the given angle D. Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D.'"],["Rule","'ProofWordCount'",209],["Rule","'GreekProof'","'ἔστω τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ: δεῖ δὴ τῷ ΑΒΓ τριγώνῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ. τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΑΕ, καὶ συνεστάτω πρὸς τῇ ΕΓ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε τῇ Δ γωνίᾳ ἴση ἡ ὑπὸ ΓΕΖ, καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ἤχθω ἡ ΑΗ, διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος ἤχθω ἡ ΓΗ: παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ, ἴσον ἐστὶ καὶ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΕ, ΕΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΗ: διπλάσιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου. ἔστι δὲ καὶ τὸ ΖΕΓΗ παραλληλόγραμμον διπλάσιον τοῦ ΑΕΓ τριγώνου: βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει καὶ ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις: ἴσον ἄρα ἐστὶ τὸ ΖΕΓΗ παραλληλόγραμμον τῷ ΑΒΓ τριγώνῳ. καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ἴσην τῇ δοθείσῃ τῇ Δ. τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ ἴσον παραλληλόγραμμον συνέσταται τὸ ΖΕΓΗ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ, ἥτις ἐστὶν ἴση τῇ Δ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",195]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",43]],["Association",["Rule","'VertexLabel'","'1.43'"],["Rule","'Text'","'In any parallelogram the complements of the parallelograms about the diameter are equal to one another.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'παντὸς παραλληλογράμμου τῶν περὶ τὴν διάμετρον παραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Common Notion'",2]],["List",["Rule","'Common Notion'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]]]],["Rule","'Proof'","'Let ABCD be a parallelogram, and AC its diameter; and about AC let EH, FG be parallelograms, and BK, KD  the so-called complements; I say that the complement BK is equal to the complement KD. For, since ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ACD. [I. 34] Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK. For the same reason the triangle KFC is also equal to KGC. Now, since the triangle AEK is equal to the triangle AHK, and KFC to KGC,  the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C. N. 2] And the whole triangle ABC is also equal to the whole ADC; therefore the complement BK which remains is equal to thecomplement KD which remains. [C. N. 3]'"],["Rule","'ProofWordCount'",149],["Rule","'GreekProof'","'ἔστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἡ ΑΓ, περὶ δὲ τὴν ΑΓ παραλληλόγραμμα μὲν ἔστω τὰ ΖΘ, ΖΗ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΒΚ, ΚΔ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα τῷ ΚΔ παραπληρώματι. ἐπεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἡ ΑΓ, ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ. πάλιν, ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΕΘ, διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ, ἴσον ἐστὶ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον. ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον, τὸ δὲ ΚΖΓ τῷ ΚΗΓ, τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ: ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ὅλῳ τῷ ΑΔΓ ἴσον: λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα λοιπῷ τῷ ΚΔ παραπληρώματί ἐστιν ἴσον. παντὸς ἄρα παραλληλογράμμου χωρίου τῶν περὶ τὴν διάμετρον παραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",156]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",44]],["Association",["Rule","'VertexLabel'","'1.44'"],["Rule","'Text'","'To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι τριγώνῳ ἴσον παραλληλόγραμμον παραβαλεῖν ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Postulate'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",15]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",42]],["List",["Rule","'Book'",1],["Rule","'Theorem'",43]]]],["Rule","'Proof'","'Let AB be the given straight line, C the given triangle and D the given rectilineal angle;thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C. Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42];let it be placed so that BE is in a straight line with AB; let FG be drawn through to H, and let AH be drawn through A parallel to either BG or EF. [I. 31] Let HB be joined. Then, since the straight line HF falls upon the parallelsAH, EF, the angles AHF, HFE are equal to two right angles. [I. 29]Therefore the angles BHG, GFE are less than two right angles; and straight lines produced indefinitely from angles less thantwo right angles meet; [Post. 5] therefore HB, FE, when produced, will meet. Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31] and let HA, GB be produced to the points L, M. Then HLKF is a parallelogram, HK is its diameter, and AG, ME are parallelograms. and LB, BF the so-called complements, about HK; therefore LB is equal to BF. [I. 43] But BF is equal to the triangle C;therefore LB is also equal to C. [C. N. 1] And, since the angle GBE is equal to the angle ABM, [I. 15] while the angle GBE is equal to D, the angle ABM is also equal to the angle D. Therefore the parallelogram LB equal to the given triangleC has been applied to the given straight line AB, in the angle ABM which is equal to D.'"],["Rule","'ProofWordCount'",303],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν τρίγωνον τὸ Γ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ: δεῖ δὴ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον παραλληλόγραμμον παραβαλεῖν ἐν ἴσῃ τῇ Δ γωνίᾳ. συνεστάτω τῷ Γ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΒΕΖΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ, á¼¥ ἐστιν ἴση τῇ Δ: καὶ κείσθω ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΒΕ τῇ ΑΒ, καὶ διήχθω ἡ ΖΗ ἐπὶ τὸ Θ, καὶ διὰ τοῦ Α ὁποτέρᾳ τῶν ΒΗ, ΕΖ παράλληλος ἤχθω ἡ ΑΘ, καὶ ἐπεζεύχθω ἡ ΘΒ. καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ, ΕΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ, αἱ ἄρα ὑπὸ ΑΘΖ, ΘΖΕ γωνίαι δυσὶν ὀρθαῖς εἰσιν ἴσαι. αἱ ἄρα ὑπὸ ΒΘΗ, ΗΖΕ δύο ὀρθῶν ἐλάσσονές εἰσιν: αἱ δὲ ἀπὸ ἐλασσόνων á¼¢ δύο ὀρθῶν εἰς ἄπειρον ἐκβαλλόμεναι συμπίπτουσιν: αἱ ΘΒ, ΖΕ ἄρα ἐκβαλλόμεναι συμπεσοῦνται. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Κ, καὶ διὰ τοῦ Κ σημείου ὁποτέρᾳ τῶν ΕΑ, ΖΘ παράλληλος ἤχθω ἡ ΚΛ, καὶ ἐκβεβλήσθωσαν αἱ ΘΑ, ΗΒ ἐπὶ τὰ Λ, Μ σημεῖα. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ, διάμετρος δὲ αὐτοῦ ἡ ΘΚ, περὶ δὲ τὴν ΘΚ παραλληλόγραμμα μὲν τὰ ΑΗ, ΜΕ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΛΒ, ΒΖ: ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ. ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον: καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ, ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση, καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ ἐστὶν ἴση. παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΛΒ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ, á¼¥ ἐστιν ἴση τῇ Δ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",277]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",45]],["Association",["Rule","'VertexLabel'","'1.45'"],["Rule","'Text'","'To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Common Notion'",2]],["List",["Rule","'Book'",1],["Rule","'Theorem'",14]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",30]],["List",["Rule","'Book'",1],["Rule","'Theorem'",33]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",42]],["List",["Rule","'Book'",1],["Rule","'Theorem'",44]]]],["Rule","'Proof'","'Let ABCD be the given rectilineal figure and E the given rectilineal angle;thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD. Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I. 42]let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I. 44] Then, since the angle E is equal to each of the angles HKF, GHM,the angle HKF is also equal to the angle GHM. [C. N. 1]  Let the angle KHG be added to each; therefore the angles FKH, KHG are equal to the angles KHG, GHM. But the angles FKH, KHG are equal to two right angles; [I. 29]therefore the angles KHG, GHM are also equal to two right angles. Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles;therefore KH is in a straight line with HM. [I. 14] And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I. 29] Let the angle HGL be added to each;therefore the angles MHG, HGL are equal to the angles HGF, HGL. [C. N. 2] But the angles MHG, HGL are equal to two right angles; [I. 29] therefore the angles HGF, HGL are also equal to two right angles. [C. N. 1]Therefore FG is in a straight line with GL. [I. 14] And, since FK is equal and parallel to HG, [I. 34] and HG to ML also,   KF is also equal and parallel to ML; [C. N. 1; I. 30] and the straight lines KM, FL join them (at their extremities);therefore KM, FL are also equal and parallel. [I. 33] Therefore KFLM is a parallelogram. And, since the triangle ABD is equal to the parallelogram FH, and DBC to GM,  the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E.'"],["Rule","'ProofWordCount'",396],["Rule","'GreekProof'","'ἔστω τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε: δεῖ δὴ τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε. ἐπεζεύχθω ἡ ΔΒ, καὶ συνεστάτω τῷ ΑΒΔ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΖΘ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ, á¼¥ ἐστιν ἴση τῇ Ε: καὶ παραβεβλήσθω παρὰ τὴν ΗΘ εὐθεῖαν τῷ ΔΒΓ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΗΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ, á¼¥ ἐστιν ἴση τῇ Ε. καὶ ἐπεὶ ἡ Ε γωνία ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ, ΗΘΜ ἐστιν ἴση, καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ἐστιν ἴση. κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ: αἱ ἄρα ὑπὸ ΖΚΘ, ΚΘΗ ταῖς ὑπὸ ΚΘΗ, ΗΘΜ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΖΚΘ, ΚΘΗ δυσὶν ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΚΘΗ, ΗΘΜ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν. πρὸς δή τινι εὐθείᾳ τῇ ΗΘ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ δύο εὐθεῖαι αἱ ΚΘ, ΘΜ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δύο ὀρθαῖς ἴσας ποιοῦσιν: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ: καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ, ΖΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ, ΘΗΖ ἴσαι ἀλλήλαις εἰσίν. κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ: αἱ ἄρα ὑπὸ ΜΘΗ, ΘΗΛ ταῖς ὑπὸ ΘΗΖ, ΘΗΛ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΜΘΗ, ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΘΗΖ, ΘΗΛ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ. καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ ἴση τε καὶ παράλληλός ἐστιν, ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ, καὶ ἡ ΚΖ ἄρα τῇ ΜΛ ἴση τε καὶ παράλληλός ἐστιν: καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ΚΜ, ΖΛ: καὶ αἱ ΚΜ, ΖΛ ἄρα ἴσαι τε καὶ παράλληλοί εἰσιν: παραλληλόγραμμον ἄρα ἐστὶ τὸ ΚΖΛΜ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλληλογράμμῳ, τὸ δὲ ΔΒΓ τῷ ΗΜ, ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ ἐστὶν ἴσον. τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ ἴσον παραλληλόγραμμον συνέσταται τὸ ΚΖΛΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ, á¼¥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",340]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",46]],["Association",["Rule","'VertexLabel'","'1.46'"],["Rule","'Text'","'On a given straight line to describe a square.'"],["Rule","'TextWordCount'",9],["Rule","'GreekText'","'ἀπὸ τῆς δοθείσης εὐθείας τετράγωνον ἀναγράψαι.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",11]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]]]],["Rule","'Proof'","'Let AB be the given straight line; thus it is required to describe a square on the straight line AB. Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB; through the point D let DE be drawnparallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31] Therefore ADEB is a parallelogram; therefore AB is equal to DE, and AD to BE. [I. 34] But AB is equal to AD;therefore the four straight lines BA, AD, DE, EB are equal to one another; therefore the parallelogram ADEB is equilateral. I say next that it is also right-angled. For, since the straight line AD falls upon the parallelsAB, DE, the angles BAD, ADE are equal to two right angles. [I. 29] But the angle BAD is right; therefore the angle ADE is also right. And in parallelogrammic areas the opposite sides andangles are equal to one another; [I. 34] therefore each of the opposite angles ABE, BED is also right. Therefore ADEB is right-angled. And it was also proved equilateral. Therefore it is a square; and it is described on the straight line AB.'"],["Rule","'ProofWordCount'",209],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ: δεῖ δὴ ἀπὸ τῆς ΑΒ εὐθείας τετράγωνον ἀναγράψαι. ἤχθω τῇ ΑΒ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ α πρὸς ὀρθὰς ἡ ΑΓ, καὶ κείσθω τῇ ΑΒ ἴση ἡ ΑΔ: καὶ διὰ μὲν τοῦ Δ σημείου τῇ ΑΒ παράλληλος ἤχθω ἡ ΔΕ, διὰ δὲ τοῦ Β σημείου τῇ ΑΔ παράλληλος ἤχθω ἡ ΒΕ. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒ: ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΑΔ τῇ ΒΕ. ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση: αἱ τέσσαρες ἄρα αἱ ΒΑ, ΑΔ, ΔΕ, ΕΒ ἴσαι ἀλλήλαις εἰσίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ, ΔΕ εὐθεῖα ἐνέπεσεν ἡ ΑΔ, αἱ ἄρα ὑπὸ ΒΑΔ, ΑΔΕ γωνίαι δύο ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ. τῶν δὲ παραλληλογράμμων χωρίων αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν: ὀρθὴ ἄρα καὶ ἑκατέρα τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ, ΒΕΔ γωνιῶν: ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ. ἐδείχθη δὲ καὶ ἰσόπλευρον. τετράγωνον ἄρα ἐστίν: καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ἀναγεγραμμένον: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",184]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",47]],["Association",["Rule","'VertexLabel'","'1.47'"],["Rule","'Text'","'In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'ἐν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν τετραγώνοις.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Common Notion'",2]],["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",14]],["List",["Rule","'Book'",1],["Rule","'Theorem'",41]],["List",["Rule","'Book'",1],["Rule","'Theorem'",46]]]],["Rule","'Proof'","'Let ABC be a right-angled triangle having the angleBAC right; I say that the square on BC is equal to the squares on BA, AC. For let there be described on BC the square BDEC,and on BA, AC the squares GB, HC; [I. 46] through A let AL be drawn parallel to either BD or CE, and let AD, FC be joined. Then, since each of the angles BAC, BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight linesAC, AG not lying on the same side make the adjacent angles equal to two right angles; therefore CA is in a straight line with AG. [I. 14]  For the same reason BA is also in a straight line with AH. And, since the angle DBC is equal to the angle FBA: for each is right: let the angle ABC be added to each;therefore the whole angle DBA is equal to the whole angle FBC. [C. N. 2]  And, since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively,and the angle ABD is equal to the angle FBC; therefore the base AD is equal to the base FC, and the triangle ABD is equal to the triangle FBC. [I. 4] Now the parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallelsBD, AL. [I. 41] And the square GB is double of the triangle FBC, for they again have the same base FB and are in the same parallels FB, GC. [I. 41] [But the doubles of equals are equal to one another.]Therefore the parallelogram BL is also equal to the square GB. Similarly, if AE, BK be joined, the parallelogram CL can also be proved equal to the square HC;therefore the whole square BDEC is equal to the two squares GB, HC. [C. N. 2] And the square BDEC is described on BC, and the squares GB, HC on BA, AC. Therefore the square on the side BC is equal to thesquares on the sides BA, AC.'"],["Rule","'ProofWordCount'",368],["Rule","'GreekProof'","'ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν: λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνοις. Ἀναγεγράφθω γὰρ ἀπὸ μὲν τῆς ΒΓ τετράγωνον τὸ ΒΔΕΓ, ἀπὸ δὲ τῶν ΒΑ, ΑΓ τὰ ΗΒ, ΘΓ, καὶ διὰ τοῦ α ὁποτέρᾳ τῶν ΒΔ, ΓΕ παράλληλος ἤχθω ἡ ΑΛ: καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΖΓ. καὶ ἐπεὶ ὀρθή ἐστιν ἑκατέρα τῶν ὑπὸ ΒΑΓ, ΒΑΗ γωνιῶν, πρὸς δή τινι εὐθείᾳ τῇ ΒΑ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α δύο εὐθεῖαι αἱ ΑΓ, ΑΗ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς ἴσας ποιοῦσιν: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾽ εὐθείας. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ: ὀρθὴ γὰρ ἑκατέρα: κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ: ὅλη ἄρα ἡ ὑπὸ ΔΒΑ ὅλῃ τῇ ὑπὸ ΖΒΓ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΔΒ τῇ ΒΓ, ἡ δὲ ΖΒ τῇ ΒΑ, δύο δὴ αἱ ΔΒ, ΒΑ δύο ταῖς ΖΒ, ΒΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ἴση: βάσις ἄρα ἡ ΑΔ βάσει τῇ ΖΓ ἐστιν ἴση, καὶ τὸ ΑΒΔ τρίγωνον τῷ ΖΒΓ τριγώνῳ ἐστὶν ἴσον: καὶ ἐστὶ τοῦ μὲν ΑΒΔ τριγώνου διπλάσιον τὸ ΒΛ παραλληλόγραμμον: βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ταῖς ΒΔ, ΑΛ: τοῦ δὲ ΖΒΓ τριγώνου διπλάσιον τὸ ΗΒ τετράγωνον: βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι τὴν ΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ταῖς ΖΒ, ΗΓ. τὰ δὲ τῶν ἴσων διπλάσια ἴσα ἀλλήλοις ἐστίν: ἴσον ἄρα ἐστὶ καὶ τὸ ΒΛ παραλληλόγραμμον τῷ ΗΒ τετραγώνῳ. ὁμοίως δὴ ἐπιζευγνυμένων τῶν ΑΕ, ΒΚ δειχθήσεται καὶ τὸ ΓΛ παραλληλόγραμμον ἴσον τῷ ΘΓ τετραγώνῳ: ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον δυσὶ τοῖς ΗΒ, ΘΓ τετραγώνοις ἴσον ἐστίν. καί ἐστι τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ἀναγραφέν, τὰ δὲ ΗΒ, ΘΓ ἀπὸ τῶν ΒΑ, ΑΓ. τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ πλευρῶν τετραγώνοις. ἐν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν τετραγώνοις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",368]]],["Rule",["Association",["Rule","'Book'",1],["Rule","'Theorem'",48]],["Association",["Rule","'VertexLabel'","'1.48'"],["Rule","'Text'","'If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν τριγώνου τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον ἴσον ᾖ τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ὀρθή ἐστιν.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]]]],["Rule","'Proof'","'For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA, AC; I say that the angle BAC is right. For let AD be drawn from the point A at right angles to the straight line AC, let AD be made equal to BA, and let DC be joined. Since DA is equal to AB, the square on DA is also equal to the square on AB. Let the square on AC be added to each; therefore the squares on DA, AC are equal to the squares on BA, AC. But the square on DC is equal to the squares on DA, AC, for the angle DAC is right; [I. 47] and the square on BC is equal to the squares on BA, AC, for this is the hypothesis; therefore the square on DC is equal to the square on BC, so that the side DC is also equal to BC. And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8]But the angle DAC is right; therefore the angle BAC is also right.'"],["Rule","'ProofWordCount'",220],["Rule","'GreekProof'","'τριγώνου γὰρ τοῦ ΑΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγωνον ἴσον ἔστω τοῖς ἀπὸ τῶν ΒΑ, ΑΓ πλευρῶν τετραγώνοις: λέγω, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία. ἤχθω γὰρ ἀπὸ τοῦ Α σημείου τῇ ΑΓ εὐθείᾳ πρὸς ὀρθὰς ἡ ΑΔ καὶ κείσθω τῇ ΒΑ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΑΓ τετράγωνον: τὰ ἄρα ἀπὸ τῶν ΔΑ, ΑΓ τετράγωνα ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνοις. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ, ΑΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΔΓ: ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία: τοῖς δὲ ἀπὸ τῶν ΒΑ, ΑΓ ἴσον ἐστὶ τὸ ἀπὸ ΒΓ: ὑπόκειται γάρ: τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ: ὥστε καὶ πλευρὰ ἡ ΔΓ τῇ ΒΓ ἐστιν ἴση: καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο δὴ αἱ ΔΑ, ΑΓ δύο ταῖς ΒΑ, ΑΓ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΓ βάσει τῇ ΒΓ ἴση: γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ. ἐὰν ἄρα τριγώνου τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον ἴσον ᾖ τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ὀρθή ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",237]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'2.1'"],["Rule","'Text'","'If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments.'"],["Rule","'TextWordCount'",42],["Rule","'GreekText'","'ἐὰν ὦσι δύο εὐθεῖαι, τμηθῇ δὲ ἡ ἑτέρα αὐτῶν εἰς ὁσαδηποτοῦν τμήματα, τὸ περιεχόμενον ὀρθογώνιον ὑπὸ τῶν δύο εὐθειῶν ἴσον ἐστὶ τοῖς ὑπό τε τῆς ἀτμήτου καὶ ἑκάστου τῶν τμημάτων περιεχομένοις ὀρθογωνίοις.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",11]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]]]],["Rule","'Proof'","'Let A, BC be two straight lines, and let BC be cut at random at the points D, E; I say that the rectangle contained by A, BC is equal to the rectangle contained by A, BD, that contained by A, DE and that contained by A, EC. For let BF be drawn from B at right angles to BC; [I. 11] let BG be made equal to A, [I. 3] through G let GH be drawn parallel to BC, [I. 31] and through D, E, C let DK, EL, CH be drawn parallel to BG. Then BH is equal to BK, DL, EH. Now BH is the rectangle A, BC, for it is contained by GB, BC, and BG is equal to A; BK is the rectangle A, BD, for it is contained by GB, BD, and BG is equal to A; and DL is the rectangle A, DE, for DK, that is BG [I. 34], is equal to A. Similarly also EH is the rectangle A, EC. Therefore the rectangle A, BC is equal to the rectangle A, BD, the rectangle A, DE and the rectangle A, EC.'"],["Rule","'ProofWordCount'",190],["Rule","'GreekProof'","'ἔστωσαν δύο εὐθεῖαι αἱ Α, ΒΓ, καὶ τετμήσθω ἡ ΒΓ, ὡς ἔτυχεν, κατὰ τὰ Δ, Ε σημεῖα: λέγω, ὅτι τὸ ὑπὸ τῶν Α, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν Α, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ὑπὸ τῶν Α, ΔΕ καὶ ἔτι τῷ ὑπὸ τῶν Α, ΕΓ. ἤχθω γὰρ ἀπὸ τοῦ Β τῇ ΒΓ πρὸς ὀρθὰς ἡ ΒΖ, καὶ κείσθω τῇ Α ἴση ἡ ΒΗ, καὶ διὰ μὲν τοῦ Η τῇ ΒΓ παράλληλος ἤχθω ἡ ΗΘ, διὰ δὲ τῶν Δ, Ε, Γ τῇ ΒΗ παράλληλοι ἤχθωσαν αἱ ΔΚ, ΕΛ, ΓΘ. ἴσον δή ἐστι τὸ ΒΘ τοῖς ΒΚ, ΔΛ, ΕΘ. καί ἐστι τὸ μὲν ΒΘ τὸ ὑπὸ τῶν Α, ΒΓ: περιέχεται μὲν γὰρ ὑπὸ τῶν ΗΒ, ΒΓ, ἴση δὲ ἡ ΒΗ τῇ Α: τὸ δὲ ΒΚ τὸ ὑπὸ τῶν Α, ΒΔ: περιέχεται μὲν γὰρ ὑπὸ τῶν ΗΒ, ΒΔ, ἴση δὲ ἡ ΒΗ τῇ Α. τὸ δὲ ΔΛ τὸ ὑπὸ τῶν Α, ΔΕ: ἴση γὰρ ἡ ΔΚ, τουτέστιν ἡ ΒΗ, τῇ Α. καὶ ἔτι ὁμοίως τὸ ΕΘ τὸ ὑπὸ τῶν Α, ΕΓ: τὸ ἄρα ὑπὸ τῶν Α, ΒΓ ἴσον ἐστὶ τῷ τε ὑπὸ Α, ΒΔ καὶ τῷ ὑπὸ Α, ΔΕ καὶ ἔτι τῷ ὑπὸ Α, ΕΓ. ἐὰν ἄρα ὦσι δύο εὐθεῖαι, τμηθῇ δὲ ἡ ἑτέρα αὐτῶν εἰς ὁσαδηποτοῦν τμήματα, τὸ περιεχόμενον ὀρθογώνιον ὑπὸ τῶν δύο εὐθειῶν ἴσον ἐστὶ τοῖς ὑπό τε τῆς ἀτμήτου καὶ ἑκάστου τῶν τμημάτων περιεχομένοις ὀρθογωνίοις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",232]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'2.2'"],["Rule","'Text'","'If a straight line be cut at random, the rectangle contained by the whole and both of the segments is equal to the square on the whole.'"],["Rule","'TextWordCount'",27],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης καὶ ἑκατέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ὅλης τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",46]]]],["Rule","'Proof'","'For let the straight line AB be cut at random at the point C; I say that the rectangle contained by AB, BC together with the rectangle contained by BA, AC is equal to the square on AB. For let the square ADEB be described on AB [I. 46], and let CF be drawn through C parallel to either AD or BE. [I. 31]  Then AE is equal to AF, CE. Now AE is the square on AB; AF is the rectangle contained by BA, AC, for it is contained by DA, AC, and AD is equal to AB; and CE is the rectangle AB, BC, for BE is equal to AB. Therefore the rectangle BA, AC together with the rectangle AB, BC is equal to the square on AB.'"],["Rule","'ProofWordCount'",130],["Rule","'GreekProof'","'εὐθεῖα γὰρ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ὑπὸ ΒΑ, ΑΓ περιεχομένου ὀρθογωνίου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, καὶ ἤχθω διὰ τοῦ Γ ὁποτέρᾳ τῶν ΑΔ, ΒΕ παράλληλος ἡ ΓΖ. ἴσον δή ἐστι τὸ ΑΕ τοῖς ΑΖ, ΓΕ. καί ἐστι τὸ μὲν ΑΕ τὸ ἀπὸ τῆς ΑΒ τετράγωνον, τὸ δὲ ΑΖ τὸ ὑπὸ τῶν ΒΑ, ΑΓ περιεχόμενον ὀρθογώνιον: περιέχεται μὲν γὰρ ὑπὸ τῶν ΔΑ, ΑΓ, ἴση δὲ ἡ ΑΔ τῇ ΑΒ: τὸ δὲ ΓΕ τὸ ὑπὸ τῶν ΑΒ, ΒΓ: ἴση γὰρ ἡ ΒΕ τῇ ΑΒ. τὸ ἄρα ὑπὸ τῶν ΒΑ, ΑΓ μετὰ τοῦ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης καὶ ἑκατέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ὅλης τετραγώνῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",155]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'2.3'"],["Rule","'Text'","'If a straight line be cut at random, the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ προειρημένου τμήματος τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",46]]]],["Rule","'Proof'","'For let the straight line AB be cut at random at C; I say that the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC. For let the square CDEB be described on CB; [I. 46] let ED be drawn through to F, and through A let AF be drawn parallel to either CD or BE. [I. 31]  Then AE is equal to AD, CE. Now AE is the rectangle contained by AB, BC, for it is contained by AB, BE, and BE is equal to BC;   AD is the rectangle AC, CB, for DC is equal to CB; and DB is the square on CB. Therefore the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC.'"],["Rule","'ProofWordCount'",139],["Rule","'GreekProof'","'εὐθεῖα γὰρ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΒΓ τετραγώνου. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΒ τετράγωνον τὸ ΓΔΕΒ, καὶ διήχθω ἡ ΕΔ ἐπὶ τὸ Ζ, καὶ διὰ τοῦ Α ὁποτέρᾳ τῶν ΓΔ, ΒΕ παράλληλος ἤχθω ἡ ΑΖ. ἴσον δή ἐστι τὸ ΑΕ τοῖς ΑΔ, ΓΕ: καί ἐστι τὸ μὲν ΑΕ τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον: περιέχεται μὲν γὰρ ὑπὸ τῶν ΑΒ, ΒΕ, ἴση δὲ ἡ ΒΕ τῇ ΒΓ: τὸ δὲ ΑΔ τὸ ὑπὸ τῶν ΑΓ, ΓΒ: ἴση γὰρ ἡ ΔΓ τῇ ΓΒ: τὸ δὲ ΔΒ τὸ ἀπὸ τῆς ΓΒ τετράγωνον: τὸ ἄρα ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΒΓ τετραγώνου. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ προειρημένου τμήματος τετραγώνῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",176]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'2.4'"],["Rule","'Text'","'If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν τμημάτων τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ.'"],["Rule","'GreekTextWordCount'",27],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",46]]]],["Rule","'Proof'","'For let the straight line AB be cut at random at C; I say that the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB. For let the square ADEB be described on AB, [I. 46] let BD be joined; through C let CF be drawn parallel to either AD or EB, and through G let HK be drawn parallel to either AB or DE. [I. 31] Then, since CF is parallel to AD, and BD has fallen on them, the exterior angle CGB is equal to the interior and opposite angle ADB. [I. 29]  But the angle ADB is equal to the angle ABD, since the side BA is also equal to AD; [I. 5]therefore the angle CGB is also equal to the angle GBC, so that the side BC is also equal to the side CG. [I. 6]  But CB is equal to GK, and CG to KB; [I. 34] therefore GK is also equal to KB; therefore CGKB is equilateral. I say next that it is also right-angled. For, since CG is parallel to BK, the angles KBC, GCB are equal to two right angles. [I. 29]  But the angle KBC is right; therefore the angle BCG is also right, so that the opposite angles CGK, GKB are also right. [I. 34]  Therefore CGKB is right-angled; and it was also proved equilateral; therefore it is a square; and it is described on CB. For the same reason HF is also a square; and it is described on HG, that is AC. [I. 34]  Therefore the squares HF, KC are the squares on AC, CB. Now, since AG is equal to GE, and AG is the rectangle AC, CB, for GC is equal to CB, therefore GE is also equal to the rectangle AC, CB. Therefore AG, GE are equal to twice the rectangle AC, CB. But the squares HF, CK are also the squares on AC, CB; therefore the four areas HF, CK, AG, GE are equal to the squares on AC, CB and twice the rectangle contained by AC, CB. But HF, CK, AG, GE are the whole ADEB, which is the square on AB. Therefore the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB.'"],["Rule","'ProofWordCount'",391],["Rule","'GreekProof'","'εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ. λέγω, ὅτι τὸ ἀπὸ τῆς ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, καὶ ἐπεζεύχθω ἡ ΒΔ, καὶ διὰ μὲν τοῦ Γ ὁποτέρᾳ τῶν ΑΔ, ΕΒ παράλληλος ἤχθω ἡ ΓΖ, διὰ δὲ τοῦ Η ὁποτέρᾳ τῶν ΑΒ, ΔΕ παράλληλος ἤχθω ἡ ΘΚ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΓΖ τῇ ΑΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΔ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΗΒ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΔΒ. ἀλλ᾽ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΒΑ τῇ ΑΔ ἐστιν ἴση: καὶ ἡ ὑπὸ ΓΗΒ ἄρα γωνία τῇ ὑπὸ ΗΒΓ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΒΓ πλευρᾷ τῇ ΓΗ ἐστιν ἴση: ἀλλ᾽ ἡ μὲν ΓΒ τῇ ΗΚ ἐστιν ἴση, ἡ δὲ ΓΗ τῇ ΚΒ: καὶ ἡ ΗΚ ἄρα τῇ ΚΒ ἐστιν ἴση: ἰσόπλευρον ἄρα ἐστὶ τὸ ΓΗΚΒ. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΓΗ τῇ ΒΚ καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΓΒ, αἱ ἄρα ὑπὸ ΚΒΓ, ΗΓΒ γωνίαι δύο ὀρθαῖς εἰσιν ἴσαι. ὀρθὴ δὲ ἡ ὑπὸ ΚΒΓ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΓΗ: ὥστε καὶ αἱ ἀπεναντίον αἱ ὑπὸ ΓΗΚ, ΗΚΒ ὀρθαί εἰσιν. ὀρθογώνιον ἄρα ἐστὶ τὸ ΓΗΚΒ: ἐδείχθη δὲ καὶ ἰσόπλευρον: τετράγωνον ἄρα ἐστίν: καί ἐστιν ἀπὸ τῆς ΓΒ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΘΖ τετράγωνόν ἐστιν: καί ἐστιν ἀπὸ τῆς ΘΗ, τουτέστιν ἀπὸ τῆς ΑΓ: τὰ ἄρα ΘΖ, ΚΓ τετράγωνα ἀπὸ τῶν ΑΓ, ΓΒ εἰσιν. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, καί ἐστι τὸ ΑΗ τὸ ὑπὸ τῶν ΑΓ, ΓΒ: ἴση γὰρ ἡ ΗΓ τῇ ΓΒ: καὶ τὸ ΗΕ ἄρα ἴσον ἐστὶ τῷ ὑπὸ ΑΓ, ΓΒ: τὰ ἄρα ΑΗ, ΗΕ ἴσα ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. ἔστι δὲ καὶ τὰ ΘΖ, ΓΚ τετράγωνα ἀπὸ τῶν ΑΓ, ΓΒ: τὰ ἄρα τέσσαρα τὰ ΘΖ, ΓΚ, ΑΗ, ΗΕ ἴσα ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ. ἀλλὰ τὰ ΘΖ, ΓΚ, ΑΗ, ΗΕ ὅλον ἐστὶ τὸ ΑΔΕΒ, ὅ ἐστιν ἀπὸ τῆς ΑΒ τετράγωνον: τὸ ἄρα ἀπὸ τῆς ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν τμημάτων τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ: ὅπερ ἔδει δεῖξαι. πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐν τοῖς τετραγώνοις χωρίοις τὰ περὶ τὴν διάμετρον παραλληλόγραμμα τετράγωνά ἐστιν.'"],["Rule","'GreekProofWordCount'",435]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'2.5'"],["Rule","'Text'","'If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half.'"],["Rule","'TextWordCount'",42],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὸ ὑπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ἡμισείας τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",36]],["List",["Rule","'Book'",1],["Rule","'Theorem'",43]],["List",["Rule","'Book'",1],["Rule","'Theorem'",46]]]],["Rule","'Proof'","'For let a straight line AB be cut into equal segments at C and into unequal segments at D; I say that the rectangle contained by AD, DB together with the square on CD is equal to the square on CB. For let the square CEFB be described on CB, [I. 46] and let BE be joined; through D let DG be drawn parallel to either CE or BF, through H again let KM be drawn parallel to either AB or EF, and again through A let AK be drawn parallel to either CL or BM. [I. 31]  Then, since the complement CH is equal to the complement HF, [I. 43] let DM be added to each; therefore the whole CM is equal to the whole DF. But CM is equal to AL, since AC is also equal to CB; [I. 36]therefore AL is also equal to DF. Let CH be added to each; therefore the whole AH is equal to the gnomon NOP. But AH is the rectangle AD, DB, for DH is equal to DB, therefore the gnomon NOP is also equal to the rectangle AD, DB. Let LG, which is equal to the square on CD, be added to each; therefore the gnomon NOP and LG are equal to the rectangle contained by AD, DB and the square on CD. But the gnomon NOP and LG are the whole square CEFB, which is described on CB; therefore the rectangle contained by AD, DB together with the square on CD is equal to the square on CB.'"],["Rule","'ProofWordCount'",260],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω εἰς μὲν ἴσα κατὰ τὸ Γ, εἰς δὲ ἄνισα κατὰ τὸ Δ: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΔ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΒ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΒ τετράγωνον τὸ ΓΕΖΒ, καὶ ἐπεζεύχθω ἡ ΒΕ, καὶ διὰ μὲν τοῦ Δ ὁποτέρᾳ τῶν ΓΕ, ΒΖ παράλληλος ἤχθω ἡ ΔΗ, διὰ δὲ τοῦ Θ ὁποτέρᾳ τῶν ΑΒ, ΕΖ παράλληλος πάλιν ἤχθω ἡ ΚΜ, καὶ πάλιν διὰ τοῦ Α ὁποτέρᾳ τῶν ΓΛ, ΒΜ παράλληλος ἤχθω ἡ ΑΚ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΓΘ παραπλήρωμα τῷ ΘΖ παραπληρώματι, κοινὸν προσκείσθω τὸ ΔΜ: ὅλον ἄρα τὸ ΓΜ ὅλῳ τῷ ΔΖ ἴσον ἐστίν. ἀλλὰ τὸ ΓΜ τῷ ΑΛ ἴσον ἐστίν, ἐπεὶ καὶ ἡ ΑΓ τῇ ΓΒ ἐστιν ἴση: καὶ τὸ ΑΛ ἄρα τῷ ΔΖ ἴσον ἐστίν. κοινὸν προσκείσθω τὸ ΓΘ: ὅλον ἄρα τὸ ΑΘ τῷ ΜΝΞ γνώμονι ἴσον ἐστίν. ἀλλὰ τὸ ΑΘ τὸ ὑπὸ τῶν ΑΔ, ΔΒ ἐστιν: ἴση γὰρ ἡ ΔΘ τῇ ΔΒ: καὶ ὁ ΜΝΞ ἄρα γνώμων ἴσος ἐστὶ τῷ ὑπὸ ΑΔ, ΔΒ. κοινὸν προσκείσθω τὸ ΛΗ, ὅ ἐστιν ἴσον τῷ ἀπὸ τῆς ΓΔ: ὁ ἄρα ΜΝΞ γνώμων καὶ τὸ ΛΗ ἴσα ἐστὶ τῷ ὑπὸ τῶν ΑΔ, ΔΒ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΓΔ τετραγώνῳ. ἀλλὰ ὁ ΜΝΞ γνώμων καὶ τὸ ΛΗ ὅλον ἐστὶ τὸ ΓΕΖΒ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς ΓΒ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΔ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΒ τετραγώνῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὸ ὑπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ἡμισείας τετραγώνῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",283]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'2.6'"],["Rule","'Text'","'If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.'"],["Rule","'TextWordCount'",60],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας, τὸ ὑπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ τῆς προσκειμένης περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",45],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",36]],["List",["Rule","'Book'",1],["Rule","'Theorem'",43]],["List",["Rule","'Book'",1],["Rule","'Theorem'",46]]]],["Rule","'Proof'","'For let a straight line AB be bisected at the point C, and let a straight line BD be added to it in a straight line;  I say that the rectangle contained by AD, DB together with the square on CB is equal to the square on CD. For let the square CEFD be described on CD, [I. 46] and let DE be joined; through the point B let BG be drawn parallel to either EC or DF, through the point H let KM be drawn parallel to either AB or EF, and further through A let AK be drawn parallel to either CL or DM. [I. 31] Then, since AC is equal to CB, AL is also equal to CH. [I. 36]But CH is equal to HF. [I. 43] Therefore AL is also equal to HF. Let CM be added to each; therefore the whole AM is equal to the gnomon NOP. But AM is the rectangle AD, DB, for DM is equal to DB; therefore the gnomon NOP is also equal to the rectangle AD, DB. Let LG, which is equal to the square on BC, be added to each; therefore the rectangle contained by AD, DB together with the square on CB is equal to the gnomon NOP and LG. But the gnomon NOP and LG are the whole square CEFD, which is described on CD; therefore the rectangle contained by AD, DB together with the square on CB is equal to the square on CD.'"],["Rule","'ProofWordCount'",250],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω δίχα κατὰ τὸ Γ σημεῖον, προσκείσθω δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας ἡ ΒΔ: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΔ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΔ τετράγωνον τὸ ΓΕΖΔ, καὶ ἐπεζεύχθω ἡ ΔΕ, καὶ διὰ μὲν τοῦ Β σημείου ὁποτέρᾳ τῶν ΕΓ, ΔΖ παράλληλος ἤχθω ἡ ΒΗ, διὰ δὲ τοῦ Θ σημείου ὁποτέρᾳ τῶν ΑΒ, ΕΖ παράλληλος ἤχθω ἡ ΚΜ, καὶ ἔτι διὰ τοῦ Α ὁποτέρᾳ τῶν ΓΛ, ΔΜ παράλληλος ἤχθω ἡ ΑΚ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, ἴσον ἐστὶ καὶ τὸ ΑΛ τῷ ΓΘ. ἀλλὰ τὸ ΓΘ τῷ ΘΖ ἴσον ἐστίν. καὶ τὸ ΑΛ ἄρα τῷ ΘΖ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΓΜ: ὅλον ἄρα τὸ ΑΜ τῷ ΝΞΟ γνώμονί ἐστιν ἴσον. ἀλλὰ τὸ ΑΜ ἐστι τὸ ὑπὸ τῶν ΑΔ, ΔΒ: ἴση γάρ ἐστιν ἡ ΔΜ τῇ ΔΒ: καὶ ὁ ΝΞΟ ἄρα γνώμων ἴσος ἐστὶ τῷ ὑπὸ τῶν ΑΔ, ΔΒ περιεχομένῳ ὀρθογωνίῳ. κοινὸν προσκείσθω τὸ ΛΗ, ὅ ἐστιν ἴσον τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ τῷ ΝΞΟ γνώμονι καὶ τῷ ΛΗ. ἀλλὰ ὁ ΝΞΟ γνώμων καὶ τὸ ΛΗ ὅλον ἐστὶ τὸ ΓΕΖΔ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς ΓΔ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΔ τετραγώνῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας, τὸ ὑπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ τῆς προσκειμένης περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης τετραγώνῳ: ὅπερ ἔδει δεῖξαι. ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ᾽ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ. εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ: καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, κοινὸν προσκείσθω τὸ ΓΖ: ὅλον ἄρα τὸ ΑΖ ὅλῳ τῷ ΓΕ ἴσον ἐστίν: τὰ ἄρα ΑΖ, ΓΕ διπλάσιά ἐστι τοῦ ΑΖ. ἀλλὰ τὰ ΑΖ, ΓΕ ὁ ΚΛΜ ἐστι γνώμων καὶ τὸ ΓΖ τετράγωνον: ὁ ΚΛΜ ἄρα γνώμων καὶ τὸ ΓΖ διπλάσιά ἐστι τοῦ ΑΖ. ἔστι δὲ τοῦ ΑΖ διπλάσιον καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἴση γὰρ ἡ ΒΖ τῇ ΒΓ: ὁ ἄρα ΚΛΜ γνώμων καὶ τὸ ΓΖ τετράγωνον ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. κοινὸν προσκείσθω τὸ ΔΗ, ὅ ἐστιν ἀπὸ τῆς ΑΓ τετράγωνον: ὁ ἄρα ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ. ἀλλὰ ὁ ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ὅλον ἐστὶ τὸ ΑΔΕΒ καὶ τὸ ΓΖ, ἅ ἐστιν ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα: τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ᾽ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",593]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'2.7'"],["Rule","'Text'","'If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment and the square on the remaining segment.'"],["Rule","'TextWordCount'",43],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ᾽ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",41],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",43]],["List",["Rule","'Book'",1],["Rule","'Theorem'",46]]]],["Rule","'Proof'","'For let a straight line AB be cut at random at the point C;  I say that the squares on AB, BC are equal to twice the rectangle contained by AB, BC and the square on CA. For let the square ADEB be described on AB, [I. 46] and let the figure be drawn. Then, since AG is equal to GE, [I. 43] let CF be added to each; therefore the whole AF is equal to the whole CE. Therefore AF, CE are double of AF. But AF, CE are the gnomon KLM and the square CF; therefore the gnomon KLM and the square CF are double of AF. But twice the rectangle AB, BC is also double of AF; for BF is equal to BC; therefore the gnomon KLM and the square CF are equal to twice the rectangle AB, BC. Let DG, which is the square on AC, be added to each; therefore the gnomon KLM and the squares BG, GD are equal to twice the rectangle contained by AB, BC and the square on AC. But the gnomon KLM and the squares BG, GD are the whole ADEB and CF, which are squares described on AB, BC; therefore the squares on AB, BC are equal to twice the rectangle contained by AB, BC together with the square on AC.'"],["Rule","'ProofWordCount'",222],["Rule","'GreekProof'","'Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ: καὶ καταγεγράφθω τὸ σχῆμα.  Ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, κοινὸν προσκείσθω τὸ ΓΖ: ὅλον ἄρα τὸ ΑΖ ὅλῳ τῷ ΓΕ ἴσον ἐστίν: τὰ ἄρα ΑΖ, ΓΕ διπλάσιά ἐστι τοῦ ΑΖ. ἀλλὰ τὰ ΑΖ, ΓΕ ὁ ΚΛΜ ἐστι γνώμων καὶ τὸ ΓΖ τετράγωνον: ὁ ΚΛΜ ἄρα γνώμων καὶ τὸ ΓΖ διπλάσιά ἐστι τοῦ ΑΖ. ἔστι δὲ τοῦ ΑΖ διπλάσιον καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἴση γὰρ ἡ ΒΖ τῇ ΒΓ: ὁ ἄρα ΚΛΜ γνώμων καὶ τὸ ΓΖ τετράγωνον ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. κοινὸν προσκείσθω τὸ ΔΗ, ὅ ἐστιν ἀπὸ τῆς ΑΓ τετράγωνον: ὁ ἄρα ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ. ἀλλὰ ὁ ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ὅλον ἐστὶ τὸ ΑΔΕΒ καὶ τὸ ΓΖ, ἅ ἐστιν ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα: τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ [τε] δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. Ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ᾽ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",263]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'2.8'"],["Rule","'Text'","'If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line.'"],["Rule","'TextWordCount'",47],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ τετράκις ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνου ἴσον ἐστὶ τῷ ἀπό τε τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",40],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",36]],["List",["Rule","'Book'",1],["Rule","'Theorem'",43]]]],["Rule","'Proof'","'For let a straight line AB be cut at random at the point C;  I say that four times the rectangle contained by AB, BC together with the square on AC is equal to the square described on AB, BC as on one straight line. For let [the straight line]BD be produced in a straight line [with AB], and let BD be made equal to CB; let the square AEFD be described on AD, and let the figure be drawn double. Then, since CB is equal to BD, while CB is equal to GK, and BD to KN, therefore GK is also equal to KN. For the same reason QR is also equal to RP. And, since BC is equal to BD, and GK to KN, therefore CK is also equal to KD, and GR to RN. [I. 36] But CK is equal to RN, for they are complements of the parallelogram CP; [I. 43] therefore KD is also equal to GR; therefore the four areas DK, CK, GR, RN are equal to one another. Therefore the four are quadruple of CK. Again, since CB is equal to BD, while BD is equal to BK, that is CG, and CB is equal to GK, that is GQ, therefore CG is also equal to GQ. And, since CG is equal to GQ, and QR to RP, AG is also equal to MQ, and QL to RF. [I. 36] But MQ is equal to QL, for they are complements of the parallelogram ML; [I. 43] therefore AG is also equal to RF; therefore the four areas AG, MQ, QL, RF are equal to one another. Therefore the four are quadruple of AG. But the four areas CK, KD, GR, RN were proved to be quadruple of CK; therefore the eight areas, which contain the gnomon STU, are quadruple of AK. Now, since AK is the rectangle AB, BD, for BK is equal to BD, therefore four times the rectangle AB, BD is quadruple of AK. But the gnomon STU was also proved to be quadruple of AK; therefore four times the rectangle AB, BD is equal to the gnomon STU. Let OH, which is equal to the square on AC, be added to each; therefore four times the rectangle AB, BD together with the square on AC is equal to the gnomon STU and OH. But the gnomon STU and OH are the whole square AEFD, which is described on AD. therefore four times the rectangle AB, BD together with the square on AC is equal to the square on AD  But BD is equal to BC; therefore four times the rectangle contained by AB, BC together with the square on AC is equal to the square on AD, that is to the square described on AB and BC as on one straight line.'"],["Rule","'ProofWordCount'",473],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὸ τετράκις ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ, ΒΓ ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ. Ἐκβεβλήσθω γὰρ ἐπ᾽ εὐθείας τῇ ΑΒ εὐθεῖα ἡ ΒΔ, καὶ κείσθω τῇ ΓΒ ἴση ἡ ΒΔ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΑΔ τετράγωνον τὸ ΑΕΖΔ, καὶ καταγεγράφθω διπλοῦν τὸ σχῆμα. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΓΒ τῇ ΒΔ, ἀλλὰ ἡ μὲν ΓΒ τῇ ΗΚ ἐστιν ἴση, ἡ δὲ ΒΔ τῇ ΚΝ, καὶ ἡ ΗΚ ἄρα τῇ ΚΝ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΠΡ τῇ ΡΟ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΔ, ἡ δὲ ΗΚ τῇ ΚΝ, ἴσον ἄρα ἐστὶ καὶ τὸ μὲν ΓΚ τῷ ΚΔ, τὸ δὲ ΗΡ τῷ ΡΝ. ἀλλὰ τὸ ΓΚ τῷ ΡΝ ἐστιν ἴσον: παραπληρώματα γὰρ τοῦ ΓΟ παραλληλογράμμου: καὶ τὸ ΚΔ ἄρα τῷ ΗΡ ἴσον ἐστίν: τὰ τέσσαρα ἄρα τὰ ΔΚ, ΓΚ, ΗΡ, ΡΝ ἴσα ἀλλήλοις ἐστίν. τὰ τέσσαρα ἄρα τετραπλάσιά ἐστι τοῦ ΓΚ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΒΔ, ἀλλὰ ἡ μὲν ΒΔ τῇ ΒΚ, τουτέστι τῇ ΓΗ ἴση, ἡ δὲ ΓΒ τῇ ΗΚ, τουτέστι τῇ ΗΠ, ἐστιν ἴση, καὶ ἡ ΓΗ ἄρα τῇ ΗΠ ἴση ἐστίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΓΗ τῇ ΗΠ, ἡ δὲ ΠΡ τῇ ΡΟ, ἴσον ἐστὶ καὶ τὸ μὲν ΑΗ τῷ ΜΠ, τὸ δὲ ΠΛ τῷ ΡΖ. ἀλλὰ τὸ ΜΠ τῷ ΠΛ ἐστιν ἴσον: παραπληρώματα γὰρ τοῦ ΜΛ παραλληλογράμμου: καὶ τὸ ΑΗ ἄρα τῷ ΡΖ ἴσον ἐστίν: τὰ τέσσαρα ἄρα τὰ ΑΗ, ΜΠ, ΠΛ, ΡΖ ἴσα ἀλλήλοις ἐστίν: τὰ τέσσαρα ἄρα τοῦ ΑΗ ἐστι τετραπλάσια. ἐδείχθη δὲ καὶ τὰ τέσσαρα τὰ ΓΚ, ΚΔ, ΗΡ, ΡΝ τοῦ ΓΚ τετραπλάσια: τὰ ἄρα ὀκτώ, ἃ περιέχει τὸν ΣΤΥ γνώμονα, τετραπλάσιά ἐστι τοῦ ΑΚ. καὶ ἐπεὶ τὸ ΑΚ τὸ ὑπὸ τῶν ΑΒ, ΒΔ ἐστιν: ἴση γὰρ ἡ ΒΚ τῇ ΒΔ: τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ τετραπλάσιόν ἐστι τοῦ ΑΚ. ἐδείχθη δὲ τοῦ ΑΚ τετραπλάσιος καὶ ὁ ΣΤΥ γνώμων: τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ ἴσον ἐστὶ τῷ ΣΤΥ γνώμονι. κοινὸν προσκείσθω τὸ ΞΘ, ὅ ἐστιν ἴσον τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ: τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ ΑΓ τετραγώνου ἴσον ἐστὶ τῷ ΣΤΥ γνώμονι καὶ τῷ ΞΘ. ἀλλὰ ὁ ΣΤΥ γνώμων καὶ τὸ ΞΘ ὅλον ἐστὶ τὸ ΑΕΖΔ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς ΑΔ: τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ μετὰ τοῦ ἀπὸ ΑΓ ἴσον ἐστὶ τῷ ἀπὸ ΑΔ τετραγώνῳ: ἴση δὲ ἡ ΒΔ τῇ ΒΓ. τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ ΑΓ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΔ, τουτέστι τῷ ἀπὸ τῆς ΑΒ καὶ ΒΓ ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ τετράκις ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνου ἴσον ἐστὶ τῷ ἀπό τε τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",499]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'2.9'"],["Rule","'Text'","'If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section.'"],["Rule","'TextWordCount'",41],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὰ ἀπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]]]],["Rule","'Proof'","'For let a straight line AB be cut into equal segments at C, and into unequal segments at D; I say that the squares on AD, DB are double of the squares on AC, CD. For let CE be drawn from C at right angles to AB, and let it be made equal to either AC or CB; let EA, EB be joined, let DF be drawn through D parallel to EC, and FG through F parallel to AB, and let AF be joined. Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC. And, since the angle at C is right, the remaining angles EAC, AEC are equal to one right angle. [I. 32] And they are equal; therefore each of the angles CEA, CAE is half a right angle. For the same reason each of the angles CEB, EBC is also half a right angle; therefore the whole angle AEB is right. And, since the angle GEF is half a right angle, and the angle EGF is right, for it is equal to the interior and opposite angle ECB, [I. 29] the remaining angle EFG is half a right angle; [I. 32] therefore the angle GEF is equal to the angle EFG, so that the side EG is also equal to GF. [I. 6] Again, since the angle at B is half a right angle, and the angle FDB is right, for it is again equal to the interior and opposite angle ECB, [I. 29] the remaining angle BFD is half a right angle; [I. 32]therefore the angle at B is equal to the angle DFB, so that the side FD is also equal to the side DB. [I. 6]  Now, since AC is equal to CE, the square on AC is also equal to the square on CE; therefore the squares on AC, CE are double of the square on AC. But the square on EA is equal to the squares on AC, CE, for the angle ACE is right; [I. 47] therefore the square on EA is double of the square on AC. Again, since EG is equal to GF, the square on EG is also equal to the square on GF; therefore the squares on EG, GF are double of the square on GF. But the square on EF is equal to the squares on EG, GF; therefore the square on EF is double of the square on GF. But GF is equal to CD; [I. 34] therefore the square on EF is double of the square on CD. But the square on EA is also double of the square on AC; therefore the squares on AE, EF are double of the squares on AC, CD. And the square on AF is equal to the squares on AE, EF, for the angle AEF is right; [I. 47] therefore the square on AF is double of the squares on AC, CD. But the squares on AD, DF are equal to the square on AF, for the angle at D is right; [I. 47] therefore the squares on AD, DF are double of the squares on AC, CD. And DF is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD.'"],["Rule","'ProofWordCount'",549],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω εἰς μὲν ἴσα κατὰ τὸ Γ, εἰς δὲ ἄνισα κατὰ τὸ Δ: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἤχθω γὰρ ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΕ, καὶ κείσθω ἴση ἑκατέρᾳ τῶν ΑΓ, ΓΒ, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ, καὶ διὰ μὲν τοῦ Δ τῇ ΕΓ παράλληλος ἤχθω ἡ ΔΖ, διὰ δὲ τοῦ Ζ τῇ ΑΒ ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΑΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΕΑΓ γωνία τῇ ὑπὸ ΑΕΓ. καὶ ἐπεὶ ὀρθή ἐστιν ἡ πρὸς τῷ Γ, λοιπαὶ ἄρα αἱ ὑπὸ ΕΑΓ, ΑΕΓ μιᾷ ὀρθῇ ἴσαι εἰσίν: καί εἰσιν ἴσαι: ἡμίσεια ἄρα ὀρθῆς ἐστιν ἑκατέρα τῶν ὑπὸ ΓΕΑ, ΓΑΕ. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΕΒ, ΕΒΓ ἡμίσειά ἐστιν ὀρθῆς: ὅλη ἄρα ἡ ὑπὸ ΑΕΒ ὀρθή ἐστιν. καὶ ἐπεὶ ἡ ὑπὸ ΗΕΖ ἡμίσειά ἐστιν ὀρθῆς, ὀρθὴ δὲ ἡ ὑπὸ ΕΗΖ: ἴση γάρ ἐστι τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΕΓΒ: λοιπὴ ἄρα ἡ ὑπὸ ΕΖΗ ἡμίσειά ἐστιν ὀρθῆς: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΗΕΖ γωνία τῇ ὑπὸ ΕΖΗ: ὥστε καὶ πλευρὰ ἡ ΕΗ τῇ ΗΖ ἐστιν ἴση. πάλιν ἐπεὶ ἡ πρὸς τῷ β γωνία ἡμίσειά ἐστιν ὀρθῆς, ὀρθὴ δὲ ἡ ὑπὸ ΖΔΒ: ἴση γὰρ πάλιν ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΕΓΒ: λοιπὴ ἄρα ἡ ὑπὸ ΒΖΔ ἡμίσειά ἐστιν ὀρθῆς: ἴση ἄρα ἡ πρὸς τῷ Β γωνία τῇ ὑπὸ ΔΖΒ: ὥστε καὶ πλευρὰ ἡ ΖΔ πλευρᾷ τῇ ΔΒ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴσον ἐστὶ καὶ τὸ ἀπὸ ΑΓ τῷ ἀπὸ ΓΕ: τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΕ τετράγωνα διπλάσιά ἐστι τοῦ ἀπὸ ΑΓ. τοῖς δὲ ἀπὸ τῶν ΑΓ, ΓΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΑ τετράγωνον: ὀρθὴ γὰρ ἡ ὑπὸ ΑΓΕ γωνία: τὸ ἄρα ἀπὸ τῆς ΕΑ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΑΓ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΕΗ τῇ ΗΖ, ἴσον καὶ τὸ ἀπὸ τῆς ΕΗ τῷ ἀπὸ τῆς ΗΖ: τὰ ἄρα ἀπὸ τῶν ΕΗ, ΗΖ τετράγωνα διπλάσιά ἐστι τοῦ ἀπὸ τῆς ΗΖ τετραγώνου. τοῖς δὲ ἀπὸ τῶν ΕΗ, ΗΖ τετραγώνοις ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΖ τετράγωνον: τὸ ἄρα ἀπὸ τῆς ΕΖ τετράγωνον διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΗΖ. ἴση δὲ ἡ ΗΖ τῇ ΓΔ: τὸ ἄρα ἀπὸ τῆς ΕΖ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΓΔ. ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΕΑ διπλάσιον τοῦ ἀπὸ τῆς ΑΓ: τὰ ἄρα ἀπὸ τῶν ΑΕ, ΕΖ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. τοῖς δὲ ἀπὸ τῶν ΑΕ, ΕΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΑΖ τετράγωνον: ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΑΕΖ γωνία: τὸ ἄρα ἀπὸ τῆς ΑΖ τετράγωνον διπλάσιόν ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ. τῷ δὲ ἀπὸ τῆς ΑΖ ἴσα τὰ ἀπὸ τῶν ΑΔ, ΔΖ: ὀρθὴ γὰρ ἡ πρὸς τῷ Δ γωνία: τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΖ διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἴση δὲ ἡ ΔΖ τῇ ΔΒ: τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὰ ἀπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",534]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'2.10'"],["Rule","'Text'","'If a straight line be bisected, and a straight line be added to it in a straight line, the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and of the square described on the straight line made up of the half and the added straight line as on one straight line.'"],["Rule","'TextWordCount'",70],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας, τὸ ἀπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ τὸ ἀπὸ τῆς προσκειμένης τὰ συναμφότερα τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης ὡς ἀπὸ μιᾶς ἀναγραφέντος τετραγώνου.'"],["Rule","'GreekTextWordCount'",52],["Rule","'References'",["List",["List",["Rule","'Common Notion'",1]],["List",["Rule","'Postulate'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",1],["Rule","'Theorem'",11]],["List",["Rule","'Book'",1],["Rule","'Theorem'",15]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]]]],["Rule","'Proof'","'For let a straight line AB be bisected at C, and let a straight line BD be added to it in a straight line; I say that the squares on AD, DB are double of the squares on AC, CD. For let CE be drawn from the point C at right angles to AB [I. 11], and let it be made equal to either AC or CB [I. 3]; let EA, EB be joined; through E let EF be drawn parallel to AD, and through D let FD be drawn parallel to CE. [I. 31] Then, since a straight line EF falls on the parallel straight lines EC, FD, the angles CEF, EFD are equal to two right angles; [I. 29]therefore the angles FEB, EFD are less than two right angles. But straight lines produced from angles less than two right angles meet; [I. Post. 5] therefore EB, FD, if produced in the direction B, D, will meet. Let them be produced and meet at G, and let AG be joined. Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC; [I. 5] and the angle at C is right; therefore each of the angles EAC, AEC is half a right angle. [I. 32] For the same reason each of the angles CEB, EBC is also half a right angle; therefore the angle AEB is right. And, since the angle EBC is half a right angle, the angle DBG is also half a right angle. [I. 15] But the angle BDG is also right, for it is equal to the angle DCE, they being alternate; [I. 29] therefore the remaining angle DGB is half a right angle; [I. 32]therefore the angle DGB is equal to the angle DBG, so that the side BD is also equal to the side GD. [I. 6]  Again, since the angle EGF is half a right angle, and the angle at F is right, for it is equal to the opposite angle, the angle at C, [I. 34] the remaining angle FEG is half a right angle; [I. 32] therefore the angle EGF is equal to the angle FEG, so that the side GF is also equal to the side EF. [I. 6] Now, since the square on EC is equal to the square on CA, the squares on EC, CA are double of the square on CA. But the square on EA is equal to the squares on EC, CA; [I. 47] therefore the square on EA is double of the square on AC. [C. N. 1]  Again, since FG is equal to EF, the square on FG is also equal to the square on FE; therefore the squares on GF, FE are double of the square on EF. But the square on EG is equal to the squares on GF, FE; [I. 47] therefore the square on EG is double of the square on EF. And EF is equal to CD; [I. 34] therefore the square on EG is double of the square on CD. But the square on EA was also proved double of the square on AC; therefore the squares on AE, EG are double of the squares on AC, CD. And the square on AG is equal to the squares on AE, EG; [I. 47] therefore the square on AG is double of the squares on AC, CD. But the squares on AD, DG are equal to the square on AG; [I. 47] therefore the squares on AD, DG are double of the squares on AC, CD. And DG is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD.'"],["Rule","'ProofWordCount'",616],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω δίχα κατὰ τὸ Γ, προσκείσθω δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας ἡ ΒΔ: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἤχθω γὰρ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΕ, καὶ κείσθω ἴση ἑκατέρᾳ, τῶν ΑΓ, ΓΒ, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ: καὶ διὰ μὲν τοῦ Ε τῇ ΑΔ παράλληλος ἤχθω ἡ ΕΖ, διὰ δὲ τοῦ Δ τῇ ΓΕ παράλληλος ἤχθω ἡ ΖΔ. καὶ ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΕΓ, ΖΔ εὐθεῖά τις ἐνέπεσεν ἡ ΕΖ, αἱ ὑπὸ ΓΕΖ, ΕΖΔ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν: αἱ ἄρα ὑπὸ ΖΕΒ, ΕΖΔ δύο ὀρθῶν ἐλάσσονές εἰσιν: αἱ δὲ ἀπ᾽ ἐλασσόνων á¼¢ δύο ὀρθῶν ἐκβαλλόμεναι συμπίπτουσιν: αἱ ἄρα ΕΒ, ΖΔ ἐκβαλλόμεναι ἐπὶ τὰ Β, Δ μέρη συμπεσοῦνται. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Η, καὶ ἐπεζεύχθω ἡ ΑΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΕΑΓ τῇ ὑπὸ ΑΕΓ: καὶ ὀρθὴ ἡ πρὸς τῷ Γ: ἡμίσεια ἄρα ὀρθῆς ἐστιν ἑκατέρα τῶν ὑπὸ ΕΑΓ, ΑΕΓ. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΕΒ, ΕΒΓ ἡμίσειά ἐστιν ὀρθῆς: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ. καὶ ἐπεὶ ἡμίσεια ὀρθῆς ἐστιν ἡ ὑπὸ ΕΒΓ, ἡμίσεια ἄρα ὀρθῆς καὶ ἡ ὑπὸ ΔΒΗ. ἔστι δὲ καὶ ἡ ὑπὸ ΒΔΗ ὀρθή: ἴση γάρ ἐστι τῇ ὑπὸ ΔΓΕ: ἐναλλὰξ γάρ: λοιπὴ ἄρα ἡ ὑπὸ ΔΗΒ ἡμίσειά ἐστιν ὀρθῆς: ἡ ἄρα ὑπὸ ΔΗΒ τῇ ὑπὸ ΔΒΗ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΒΔ πλευρᾷ τῇ ΗΔ ἐστιν ἴση. πάλιν, ἐπεὶ ἡ ὑπὸ ΕΗΖ ἡμίσειά ἐστιν ὀρθῆς, ὀρθὴ δὲ ἡ πρὸς τῷ Ζ: ἴση γάρ ἐστι τῇ ἀπεναντίον τῇ πρὸς τῷ Γ: λοιπὴ ἄρα ἡ ὑπὸ ΖΕΗ ἡμίσειά ἐστιν ὀρθῆς: ἴση ἄρα ἡ ὑπὸ ΕΗΖ γωνία τῇ ὑπὸ ΖΕΗ: ὥστε καὶ πλευρὰ ἡ ΗΖ πλευρᾷ τῇ ΕΖ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΕΓ τῇ ΓΑ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΕΓ τετράγωνον τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ: τὰ ἄρα ἀπὸ τῶν ΕΓ, ΓΑ τετράγωνα διπλάσιά ἐστι τοῦ ἀπὸ τῆς ΓΑ τετραγώνου. τοῖς δὲ ἀπὸ τῶν ΕΓ, ΓΑ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΑ: τὸ ἄρα ἀπὸ τῆς ΕΑ τετράγωνον διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΖΗ τῇ ΕΖ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΖΕ: τὰ ἄρα ἀπὸ τῶν ΗΖ, ΖΕ διπλάσιά ἐστι τοῦ ἀπὸ τῆς ΕΖ. τοῖς δὲ ἀπὸ τῶν ΗΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΗ: τὸ ἄρα ἀπὸ τῆς ΕΗ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΖ. ἴση δὲ ἡ ΕΖ τῇ ΓΔ: τὸ ἄρα ἀπὸ τῆς ΕΗ τετράγωνον διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΓΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΕΑ διπλάσιον τοῦ ἀπὸ τῆς ΑΓ: τὰ ἄρα ἀπὸ τῶν ΑΕ, ΕΗ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. τοῖς δὲ ἀπὸ τῶν ΑΕ, ΕΗ τετραγώνοις ἴσον ἐστὶ τὸ ἀπὸ τῆς ΑΗ τετράγωνον: τὸ ἄρα ἀπὸ τῆς ΑΗ διπλάσιόν ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ. τῷ δὲ ἀπὸ τῆς ΑΗ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΔ, ΔΗ: τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΗ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἴση δὲ ἡ ΔΗ τῇ ΔΒ: τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας, τὸ ἀπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ τὸ ἀπὸ τῆς προσκειμένης τὰ συναμφότερα τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης ὡς ἀπὸ μιᾶς ἀναγραφέντος τετραγώνου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",591]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'2.11'"],["Rule","'Text'","'To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'τὴν δοθεῖσαν εὐθεῖαν τεμεῖν ὥστε τὸ ὑπὸ τῆς ὅλης καὶ τοῦ ἑτέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",46]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",2],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'Let AB be the given straight line; thus it is required to cut AB so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment. For let the square ABDC be described on AB; [I. 46] let AC be bisected at the point E, and let BE be joined; let CA be drawn through to F, and let EF be made equal to BE; let the square FH be described on AF, and let GH be drawn through to K. I say that AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on AH. For, since the straight line AC has been bisected at E, and FA is added to it, the rectangle contained by CF, FA together with the square on AE is equal to the square on EF. [II. 6]  But EF is equal to EB; therefore the rectangle CF, FA together with the square on AE is equal to the square on EB. But the squares on BA, AE are equal to the square on EB, for the angle at A is right; [I. 47] therefore the rectangle CF, FA together with the square on AE is equal to the squares on BA, AE. Let the square on AE be subtracted from each; therefore the rectangle CF, FA which remains is equal to the square on AB. Now the rectangle CF, FA is FK, for AF is equal to FG; and the square on AB is AD; therefore FK is equal to AD. Let AK be subtracted from each; therefore FH which remains is equal to HD. And HD is the rectangle AB, BH, for AB is equal to BD; and FH is the square on AH; therefore the rectangle contained by AB, BH is equal to the square on HA. therefore the given straight line AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on HA. Q. E. F.'"],["Rule","'ProofWordCount'",346],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ: δεῖ δὴ τὴν ΑΒ τεμεῖν ὥστε τὸ ὑπὸ τῆς ὅλης καὶ τοῦ ἑτέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΒΔΓ, καὶ τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε σημεῖον, καὶ ἐπεζεύχθω ἡ ΒΕ, καὶ διήχθω ἡ ΓΑ ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΒΕ ἴση ἡ ΕΖ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΑΖ τετράγωνον τὸ ΖΘ, καὶ διήχθω ἡ ΗΘ ἐπὶ τὸ Κ: λέγω, ὅτι ἡ ΑΒ τέτμηται κατὰ τὸ Θ, ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον ποιεῖν τῷ ἀπὸ τῆς ΑΘ τετραγώνῳ. ἐπεὶ γὰρ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ε, πρόσκειται δὲ αὐτῇ ἡ ΖΑ, τὸ ἄρα ὑπὸ τῶν ΓΖ, ΖΑ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΑΕ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΖ τετραγώνῳ. ἴση δὲ ἡ ΕΖ τῇ ΕΒ: τὸ ἄρα ὑπὸ τῶν ΓΖ, ΖΑ μετὰ τοῦ ἀπὸ τῆς ΑΕ ἴσον ἐστὶ τῷ ἀπὸ ΕΒ. ἀλλὰ τῷ ἀπὸ ΕΒ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΒΑ, ΑΕ: ὀρθὴ γὰρ ἡ πρὸς τῷ Α γωνία: τὸ ἄρα ὑπὸ τῶν ΓΖ, ΖΑ μετὰ τοῦ ἀπὸ τῆς ΑΕ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΕ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΑΕ: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΓΖ, ΖΑ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΓΖ, ΖΑ τὸ ΖΚ: ἴση γὰρ ἡ ΑΖ τῇ ΖΗ: τὸ δὲ ἀπὸ τῆς ΑΒ τὸ ΑΔ: τὸ ἄρα ΖΚ ἴσον ἐστὶ τῷ ΑΔ. κοινὸν ἀφῃρήσθω τὸ ΑΚ: λοιπὸν ἄρα τὸ ΖΘ τῷ ΘΔ ἴσον ἐστίν. καί ἐστι τὸ μὲν ΘΔ τὸ ὑπὸ τῶν ΑΒ, ΒΘ: ἴση γὰρ ἡ ΑΒ τῇ ΒΔ: τὸ δὲ ΖΘ τὸ ἀπὸ τῆς ΑΘ: τὸ ἄρα ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ ΘΑ τετραγώνῳ. ἡ ἄρα δοθεῖσα εὐθεῖα ἡ ΑΒ τέτμηται κατὰ τὸ Θ ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον ποιεῖν τῷ ἀπὸ τῆς ΘΑ τετραγώνῳ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",327]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'2.12'"],["Rule","'Text'","'In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the   perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.'"],["Rule","'TextWordCount'",59],["Rule","'GreekText'","'ἐν τοῖς ἀμβλυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ἀμβλεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον μεῖζόν ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ἀμβλεῖαν γωνίαν, ἐφ᾽ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐκτὸς ὑπὸ τῆς καθέτου πρὸς τῇ ἀμβλείᾳ γωνίᾳ.'"],["Rule","'GreekTextWordCount'",52],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",2],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and let BD be drawn from the point B perpendicular to CA produced;  I say that the square on BC is greater than the squares on BA, AC by twice the rectangle contained by CA, AD. For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD. [II. 4] Let the square on DB be added to each; therefore the squares on CD, DB are equal to the squares on CA, AD, DB and twice the rectangle CA, AD. But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47] and the square on AB is equal to the squares on AD, DB; [I. 47]therefore the square on CB is equal to the squares on CA, AB and twice the rectangle contained by CA, AD; so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD.'"],["Rule","'ProofWordCount'",194],["Rule","'GreekProof'","'ἔστω ἀμβλυγώνιον τρίγωνον τὸ ΑΒΓ ἀμβλεῖαν ἔχον τὴν ὑπὸ ΒΑΓ, καὶ ἤχθω ἀπὸ τοῦ Β σημείου ἐπὶ τὴν ΓΑ ἐκβληθεῖσαν κάθετος ἡ ΒΔ. λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ τετράγωνον μεῖζόν ἐστι τῶν ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. ἐπεὶ γὰρ εὐθεῖα ἡ ΓΑ τέτμηται, ὡς ἔτυχεν, κατὰ τὸ Α σημεῖον, τὸ ἄρα ἀπὸ τῆς ΔΓ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΓΑ, ΑΔ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΔΒ: τὰ ἄρα ἀπὸ τῶν ΓΔ, ΔΒ ἴσα ἐστὶ τοῖς τε ἀπὸ τῶν ΓΑ, ΑΔ, ΔΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΓΔ, ΔΒ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΓΒ: ὀρθὴ γὰρ ἡ πρὸς τῷ Δ γωνία: τοῖς δὲ ἀπὸ τῶν ΑΔ, ΔΒ ἴσον τὸ ἀπὸ τῆς ΑΒ: τὸ ἄρα ἀπὸ τῆς ΓΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΓΑ, ΑΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ: ὥστε τὸ ἀπὸ τῆς ΓΒ τετράγωνον τῶν ἀπὸ τῶν ΓΑ, ΑΒ τετραγώνων μεῖζόν ἐστι τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. ἐν ἄρα τοῖς ἀμβλυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ἀμβλεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον μεῖζόν ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ἀμβλεῖαν γωνίαν, ἐφ᾽ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐκτὸς ὑπὸ τῆς καθέτου πρὸς τῇ ἀμβλείᾳ γωνίᾳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",244]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'2.13'"],["Rule","'Text'","'In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle.'"],["Rule","'TextWordCount'",59],["Rule","'GreekText'","'ἐν τοῖς ὀξυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀξεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν τὴν ὀξεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ὀξεῖαν γωνίαν, ἐφ᾽ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐντὸς ὑπὸ τῆς καθέτου πρὸς τῇ ὀξείᾳ γωνίᾳ.'"],["Rule","'GreekTextWordCount'",52],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",2],["Rule","'Theorem'",7]]]],["Rule","'Proof'","'Let ABC be an acute-angled triangle having the angle at B acute, and let AD be drawn from the point A perpendicular to BC;  I say that the square on AC is less than the squares on CB, BA by twice the rectangle contained by CB, BD. For, since the straight line CB has been cut at random at D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on DC. [II. 7] Let the square on DA be added to each; therefore the squares on CB, BD, DA are equal to twice the rectangle contained by CB, BD and the squares on AD, DC. But the square on AB is equal to the squares on BD, DA, for the angle at D is right; [I. 47] and the square on AC is equal to the squares on AD, DC; therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD,  so that the square on AC alone is less than the squares on CB, BA by twice the rectangle contained by CB, BD.'"],["Rule","'ProofWordCount'",191],["Rule","'GreekProof'","'ἔστω ὀξυγώνιον τρίγωνον τὸ ΑΒΓ ὀξεῖαν ἔχον τὴν πρὸς τῷ Β γωνίαν, καὶ ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ: λέγω, ὅτι τὸ ἀπὸ τῆς ΑΓ τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν ΓΒ, ΒΑ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ. ἐπεὶ γὰρ εὐθεῖα ἡ ΓΒ τέτμηται, ὡς ἔτυχεν, κατὰ τὸ Δ, τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΔ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΔΓ τετραγώνῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΔΑ τετράγωνον: τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΔ, ΔΑ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τοῖς ἀπὸ τῶν ΑΔ, ΔΓ τετραγώνοις. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΒΔ, ΔΑ ἴσον τὸ ἀπὸ τῆς ΑΒ: ὀρθὴ γὰρ ἡ πρὸς τῷ Δ γωνίᾳ: τοῖς δὲ ἀπὸ τῶν ΑΔ, ΔΓ ἴσον τὸ ἀπὸ τῆς ΑΓ: τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΑ ἴσα ἐστὶ τῷ τε ἀπὸ τῆς ΑΓ καὶ τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ: ὥστε μόνον τὸ ἀπὸ τῆς ΑΓ ἔλαττόν ἐστι τῶν ἀπὸ τῶν ΓΒ, ΒΑ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ. ἐν ἄρα τοῖς ὀξυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀξεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν τὴν ὀξεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ὀξεῖαν γωνίαν, ἐφ᾽ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐντὸς ὑπὸ τῆς καθέτου πρὸς τῇ ὀξείᾳ γωνίᾳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",243]]],["Rule",["Association",["Rule","'Book'",2],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'2.14'"],["Rule","'Text'","'To construct a square equal to a given rectilineal figure.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'τῷ δοθέντι εὐθυγράμμῳ ἴσον τετράγωνον συστήσασθαι.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",45]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",2],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'Let A be the given rectilineal figure; thus it is required to construct a square equal to the rectilineal figure A. For let there be constructed the rectangular parallelogram BD equal to the rectilineal figure A. [I. 45]  Then, if BE is equal to ED, that which was enjoined will have been done; for a square BD has been constructed equal to the rectilineal figure A. But, if not, one of the straight lines BE, ED is greater. Let BE be greater, and let it be produced to F; let EF be made equal to ED, and let BF be bisected at G. With centre G and distance one of the straight lines GB, GF let the semicircle BHF be described; let DE be produced to H, and let GH be joined. Then, since the straight line BF has been cut into equal segments at G, and into unequal segments at E, the rectangle contained by BE, EF together with the square on EG is equal to the square on GF. [II. 5] But GF is equal to GH; therefore the rectangle BE, EF together with the square on GE is equal to the square on GH. But the squares on HE, EG are equal to the square on GH; [I. 47] therefore the rectangle BE, EF together with the square on GE is equal to the squares on HE, EG. Let the square on GE be subtracted from each; therefore the rectangle contained by BE, EF which remains is equal to the square on EH. But the rectangle BE, EF is BD, for EF is equal to ED; therefore the parallelogram BD is equal to the square on HE. And BD is equal to the rectilineal figure A. Therefore the rectilineal figure A is also equal to the square which can be described on EH. Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilineal figure A.'"],["Rule","'ProofWordCount'",327],["Rule","'GreekProof'","'ἔστω τὸ δοθὲν εὐθύγραμμον τὸ Α: δεῖ δὴ τῷ Α εὐθυγράμμῳ ἴσον τετράγωνον συστήσασθαι. συνεστάτω γὰρ τῷ Α εὐθυγράμμῳ ἴσον παραλληλόγραμμον ὀρθογώνιον τὸ ΒΔ: εἰ μὲν οὖν ἴση ἐστὶν ἡ ΒΕ τῇ ΕΔ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. συνέσταται γὰρ τῷ Α εὐθυγράμμῳ ἴσον τετράγωνον τὸ ΒΔ: εἰ δὲ οὔ, μία τῶν ΒΕ, ΕΔ μείζων ἐστίν. ἔστω μείζων ἡ ΒΕ, καὶ ἐκβεβλήσθω ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΕΔ ἴση ἡ ΕΖ, καὶ τετμήσθω ἡ ΒΖ δίχα κατὰ τὸ Η, καὶ κέντρῳ τῷ Η, διαστήματι δὲ ἑνὶ τῶν ΗΒ, ΗΖ ἡμικύκλιον γεγράφθω τὸ ΒΘΖ, καὶ ἐκβεβλήσθω ἡ ΔΕ ἐπὶ τὸ Θ, καὶ ἐπεζεύχθω ἡ ΗΘ. ἐπεὶ οὖν εὐθεῖα ἡ ΒΖ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς δὲ ἄνισα κατὰ τὸ Ε, τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΖ τετραγώνῳ. ἴση δὲ ἡ ΗΖ τῇ ΗΘ: τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ μετὰ τοῦ ἀπὸ τῆς ΗΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΘ. τῷ δὲ ἀπὸ τῆς ΗΘ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΘΕ, ΕΗ τετράγωνα: τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ μετὰ τοῦ ἀπὸ ΗΕ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΘΕ, ΕΗ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΗΕ τετράγωνον: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΒΕ, ΕΖ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ τετραγώνῳ. ἀλλὰ τὸ ὑπὸ τῶν ΒΕ, ΕΖ τὸ ΒΔ ἐστιν: ἴση γὰρ ἡ ΕΖ τῇ ΕΔ: τὸ ἄρα ΒΔ παραλληλόγραμμον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΘΕ τετραγώνῳ. ἴσον δὲ τὸ ΒΔ τῷ Α εὐθυγράμμῳ. καὶ τὸ Α ἄρα εὐθύγραμμον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ ἀναγραφησομένῳ τετραγώνῳ. τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ Α ἴσον τετράγωνον συνέσταται τὸ ἀπὸ τῆς ΕΘ ἀναγραφησόμενον: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",283]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'3.1'"],["Rule","'Text'","'To find the centre of a given circle.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'τοῦ δοθέντος κύκλου τὸ κέντρον εὑρεῖν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let ABC be the given circle; thus it is required to find the centre of the circle ABC. Let a straight line AB be drawnthrough it at random, and let it be bisected at the point D; from D let DC be drawn at right angles to AB and let it be drawn through to E; let CE be bisected at F;I say that F is the centre of the circle ABC. For suppose it is not, but, if possible, let G be the centre, and let GA, GD, GB be joined. Then, since AD is equal to DB, and DG is common, the two sides AD, DG are equal to the two sides BD, DG respectively; and the base GA is equal to the base GB, for they areradii; therefore the angle ADG is equal to the angle GDB. [I. 8]  But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10]therefore the angle GDB is right. But the angle FDB is also right; therefore the angle FDB is equal to the angle GDB, the greater to the less: which is impossible. Therefore G is not the centre of the circle ABC. Similarly we can prove that neither is any other point except F. Therefore the point F is the centre of the circle ABC.'"],["Rule","'ProofWordCount'",236],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ: δεῖ δὴ τοῦ ΑΒΓ κύκλου τὸ κέντρον εὑρεῖν. διήχθω τις εἰς αὐτόν, ὡς ἔτυχεν, εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω δίχα κατὰ τὸ Δ σημεῖον, καὶ ἀπὸ τοῦ Δ τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΔΓ καὶ διήχθω ἐπὶ τὸ Ε, καὶ τετμήσθω ἡ ΓΕ δίχα κατὰ τὸ Ζ: λέγω, ὅτι τὸ Ζ κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἔστω τὸ Η, καὶ ἐπεζεύχθωσαν αἱ ΗΑ, ΗΔ, ΗΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ ἡ ΔΗ, δύο δὴ αἱ ΑΔ, ΔΗ δύο ταῖς ΗΔ, ΔΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΗΑ βάσει τῇ ΗΒ ἐστιν ἴση: ἐκ κέντρου γάρ: γωνία ἄρα ἡ ὑπὸ ΑΔΗ γωνίᾳ τῇ ὑπὸ ΗΔΒ ἴση ἐστίν. ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΗΔΒ. ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΔΒ ὀρθή: ἴση ἄρα ἡ ὑπὸ ΖΔΒ τῇ ὑπὸ ΗΔΒ, ἡ μείζων τῇ ἐλάττονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Η κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλο τι πλὴν τοῦ Ζ. τὸ Ζ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἐν κύκλῳ εὐθεῖά τις εὐθεῖάν τινα δίχα καὶ πρὸς ὀρθὰς τέμνῃ, ἐπὶ τῆς τεμνούσης ἐστὶ τὸ κέντρον τοῦ κύκλου: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",231]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'3.2'"],["Rule","'Text'","'If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'ἐὰν κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",16]],["List",["Rule","'Book'",1],["Rule","'Theorem'",19]],["List",["Rule","'Book'",3],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let ABC be a circle, and let two points A, B be taken at random on its circumference; I say that the straight line joined from A to B will fall within the circle. For suppose it does not, but, if possible, let it fall outside, as AEB; let the centre of the circle ABC be taken [III. 1], and let it be D; let DA, DB be joined, and let DFE be drawn through. Then, since DA is equal to DB, the angle DAE is also equal to the angle DBE. [I. 5]And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. [I. 16]  But the angle DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE. And the greater angle is subtended by the greater side; [I. 19] therefore DB is greater than DE. But DB is equal to DF;  therefore DF is greater than DE,   the less than the greater: which is impossible. Therefore the straight line joined from A to B will not fall outside the circle. Similarly we can prove that neither will it fall on the circumference itself; therefore it will fall within.'"],["Rule","'ProofWordCount'",207],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓ, καὶ ἐπὶ τῆς περιφερείας αὐτοῦ εἰλήφθω δύο τυχόντα σημεῖα τὰ Α, Β: λέγω, ὅτι ἡ ἀπὸ τοῦ Α ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, πιπτέτω ἐκτὸς ὡς ἡ ΑΕΒ, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ δ, καὶ ἐπεζεύχθωσαν αἱ ΔΑ, ΔΒ, καὶ διήχθω ἡ ΔΖΕ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΔΒ, ἴση ἄρα καὶ γωνία ἡ ὑπὸ ΔΑΕ τῇ ὑπὸ ΔΒΕ: καὶ ἐπεὶ τριγώνου τοῦ ΔΑΕ μία πλευρὰ προσεκβέβληται ἡ ΑΕΒ, μείζων ἄρα ἡ ὑπὸ ΔΕΒ γωνία τῆς ὑπὸ ΔΑΕ. ἴση δὲ ἡ ὑπὸ ΔΑΕ τῇ ὑπὸ ΔΒΕ: μείζων ἄρα ἡ ὑπὸ ΔΕΒ τῆς ὑπὸ ΔΒΕ. ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει: μείζων ἄρα ἡ ΔΒ τῆς ΔΕ. ἴση δὲ ἡ ΔΒ τῇ ΔΖ. μείζων ἄρα ἡ ΔΖ τῆς ΔΕ ἡ ἐλάττων τῆς μείζονος: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Α ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἐκτὸς πεσεῖται τοῦ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἐπ᾽ αὐτῆς τῆς περιφερείας: ἐντὸς ἄρα. ἐὰν ἄρα κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",199]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'3.3'"],["Rule","'Text'","'If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]]]],["Rule","'Proof'","'Let ABC be a circle, and in it let a straight line CD  through the centre bisect a straight line AB not through the centre at the point F; I say that it also cuts it at right angles. For let the centre of the circle ABC  be taken, and let it be E; let EA, EB be joined. Then, since AF is equal to FB, and FE is common, two sides are equal to two sides; and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. [I. 8]  But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10]therefore each of the angles AFE, BFE is right. Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles. Again, let CD cut AB at right angles;I say that it also bisects it. that is, that AF is equal to FB. For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5] But the right angle AFE is equal to the right angle BFE,therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26]therefore AF is equal to FB.'"],["Rule","'ProofWordCount'",272],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓ, καὶ ἐν αὐτῷ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΓΔ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΒ δίχα τεμνέτω κατὰ τὸ Ζ σημεῖον: λέγω, ὅτι καὶ πρὸς ὀρθὰς αὐτὴν τέμνει. εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Ε, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΕ, δύο δυσὶν ἴσαι εἰσίν. καὶ βάσις ἡ ΕΑ βάσει τῇ ΕΒ ἴση: γωνία ἄρα ἡ ὑπὸ ΑΖΕ γωνίᾳ τῇ ὑπὸ ΒΖΕ ἴση ἐστίν. ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν: ἑκατέρα ἄρα τῶν ὑπὸ ΑΖΕ, ΒΖΕ ὀρθή ἐστιν. ἡ ΓΔ ἄρα διὰ τοῦ κέντρου οὖσα τὴν ΑΒ μὴ διὰ τοῦ κέντρου οὖσαν δίχα τέμνουσα καὶ πρὸς ὀρθὰς τέμνει. ἀλλὰ δὴ ἡ ΓΔ τὴν ΑΒ πρὸς ὀρθὰς τεμνέτω: λέγω, ὅτι καὶ δίχα αὐτὴν τέμνει, τουτέστιν, ὅτι ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴση ἐστὶν ἡ ΕΑ τῇ ΕΒ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΕΑΖ τῇ ὑπὸ ΕΒΖ. ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΖΕ ὀρθῇ τῇ ὑπὸ ΒΖΕ ἴση: δύο ἄρα τρίγωνά ἐστι τὰ ΕΑΖ, ΕΖΒ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΕΖ ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει: ἴση ἄρα ἡ ΑΖ τῇ ΖΒ. ἐὰν ἄρα ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",270]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'3.4'"],["Rule","'Text'","'If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ διὰ τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Theorem'",1]],["List",["Rule","'Book'",3],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'Let ABCD be a circle, and in it let the two straight lines AC, BD, which are not through the centre, cut one another at E; I say that they do not bisect one another. For, if possible, let them bisect one another, so that AE is equal to EC, and BE to ED; let the centre of the circle ABCD be taken [III. 1], and let it be F; let FE be joined. Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [III. 3]therefore the angle FEA is right. Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles; [III. 3]therefore the angle FEB is right. But the angle FEA was also proved right; therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible. Therefore AC, BD do not bisect one another.'"],["Rule","'ProofWordCount'",167],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε μὴ διὰ τοῦ κέντρου οὖσαι: λέγω, ὅτι οὐ τέμνουσιν ἀλλήλας δίχα. εἰ γὰρ δυνατόν, τεμνέτωσαν ἀλλήλας δίχα ὥστε ἴσην εἶναι τὴν μὲν ΑΕ τῇ ΕΓ, τὴν δὲ ΒΕ τῇ ΕΔ: καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΖΕ. ἐπεὶ οὖν εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΖΕ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ δίχα τέμνει, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΕΑ: πάλιν, ἐπεὶ εὐθεῖά τις ἡ ΖΕ εὐθεῖάν τινα τὴν ΒΔ δίχα τέμνει, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: ὀρθὴ ἄρα ἡ ὑπὸ ΖΕΒ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΖΕΑ ὀρθή: ἴση ἄρα ἡ ὑπὸ ΖΕΑ τῇ ὑπὸ ΖΕΒ ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα αἱ ΑΓ, ΒΔ τέμνουσιν ἀλλήλας δίχα. ἐὰν ἄρα ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ διὰ τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",164]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'3.5'"],["Rule","'Text'","'If two circles cut one another, they will not have the same centre.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἐὰν δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",15]]]],["Rule","'Proof'","'For let the circles ABC, CDG cut one another at the points B, C; I say that they will not have the same centre. For, if possible, let it be E; let EC be joined, and let EFG be drawn through at random. Then, since the point E is the centre of the circle ABC, EC is equal to EF. [I. Def. 15] Again, since the point E is the centre of the circle CDG, EC is equal to EG. But EC was proved equal to EF also; therefore EF is also equal to EG, the less to the greater: which is impossible. Therefore the point E is not the centre of the circles ABC, CDG.'"],["Rule","'ProofWordCount'",116],["Rule","'GreekProof'","'δύο γὰρ κύκλοι οἱ ΑΒΓ, ΓΔΗ τεμνέτωσαν ἀλλήλους κατὰ τὰ Β, Γ σημεῖα. λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον. εἰ γὰρ δυνατόν, ἔστω τὸ Ε, καὶ ἐπεζεύχθω ἡ ΕΓ, καὶ διήχθω ἡ ΕΖΗ, ὡς ἔτυχεν. καὶ ἐπεὶ τὸ Ε σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΕΓ τῇ ΕΖ. πάλιν, ἐπεὶ τὸ Ε σημεῖον κέντρον ἐστὶ τοῦ ΓΔΗ κύκλου, ἴση ἐστὶν ἡ ΕΓ τῇ ΕΗ: ἐδείχθη δὲ ἡ ΕΓ καὶ τῇ ΕΖ ἴση: καὶ ἡ ΕΖ ἄρα τῇ ΕΗ ἐστιν ἴση ἡ ἐλάσσων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Ε σημεῖον κέντρον ἐστὶ τῶν ΑΒΓ, ΓΔΗ κύκλων. ἐὰν ἄρα δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔστιν αὐτῶν τὸ αὐτὸ κέντρον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",118]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'3.6'"],["Rule","'Text'","'If two circles touch one another, they will not have the same centre.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἐὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List"]],["Rule","'Proof'","'For let the two circles ABC, CDE touch one another at the point C; I say that they will not have the same centre. For, if possible, let it be F; let FC be joined, and let FEB be drawn through at random. Then, since the point F is the centre of the circle ABC, FC is equal to FB. Again, since the point F is the centre of the circle CDE, FC is equal to FE. But FC was proved equal to FB; therefore FE is also equal to FB, the less to the greater: which is impossible. Therefore F is not the centre of the circles ABC, CDE.'"],["Rule","'ProofWordCount'",110],["Rule","'GreekProof'","'δύο γὰρ κύκλοι οἱ ΑΒΓ, ΓΔΕ ἐφαπτέσθωσαν ἀλλήλων κατὰ τὸ Γ σημεῖον: λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον. εἰ γὰρ δυνατόν, ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΖΓ, καὶ διήχθω, ὡς ἔτυχεν, ἡ ΖΕΒ. ἐπεὶ οὖν τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΖΓ τῇ ΖΒ. πάλιν, ἐπεὶ τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΓΔΕ κύκλου, ἴση ἐστὶν ἡ ΖΓ τῇ ΖΕ. ἐδείχθη δὲ ἡ ΖΓ τῇ ΖΒ ἴση: καὶ ἡ ΖΕ ἄρα τῇ ΖΒ ἐστιν ἴση, ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Ζ σημεῖον κέντρον ἐστὶ τῶν ΑΒΓ, ΓΔΕ κύκλων. ἐὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",116]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'3.7'"],["Rule","'Text'","'If on the diameter of a circle a point be taken which is not the centre of the circle, and from the point straight lines fall upon the circle, that will be greatest on which the centre is, the remainder of the same diameter will be least, and of the rest  the nearer to the straight line through the centre is always greater than the more remote, and only two equal straight lines will fall from the point on the circle, one on each side of the least straight line.'"],["Rule","'TextWordCount'",90],["Rule","'GreekText'","'ἐὰν κύκλου ἐπὶ τῆς διαμέτρου ληφθῇ τι σημεῖον, ὃ μή ἐστι κέντρον τοῦ κύκλου, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσιν εὐθεῖαί τινες, μεγίστη μὲν ἔσται, ἐφ᾽ ἧς τὸ κέντρον, ἐλαχίστη δὲ ἡ λοιπή, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ἐλαχίστης.'"],["Rule","'GreekTextWordCount'",66],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",20]],["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'Let ABCD be a circle, and let AD be a diameter of it;on AD let a point F be taken which is not the centre of the circle, let E be the centre of the circle, and from F let straight lines FB, FC, FG fall upon the circle ABCD; I say that FA is greatest, FD is least, and of the rest FB isgreater than FC, and FC than FG. For let BE, CE, GE be joined. Then, since in any triangle two sides are greater than the remaining one, [I. 20]EB, EF are greater than BF. But AE is equal to BE; therefore AF is greater than BF. Again, since BE is equal to CE, and FE is common,the two sides BE, EF are equal to the two sides CE, EF. But the angle BEF is also greater than the angle CEF; therefore the base BF is greater than the base CF. [I. 24] For the same reason CF is also greater than FG. Again, since GF, FE are greater than EG, and EG is equal to ED, GF, FE are greater than ED. Let EF be subtracted from each; therefore the remainder GF is greater than the remainder FD.  Therefore FA is greatest, FD is least, and FB is greater than FC, and FC than FG. I say also that from the point F only two equal straight lines will fall on the circle ABCD, one on each side of theleast FD. For on the straight line EF, and at the point E on it, let the angle FEH be constructed equal to the angle GEF [I. 23], and let FH be joined. Then, since GE is equal to EH,and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the angle GEF is equal to the angle HEF;  therefore the base FG is equal to the base FH. [I. 4]  I say again that another straight line equal to FG will notfall on the circle from the point F. For, if possible, let FK so fall. Then, since FK is equal to FG, and FH to FG, FK is also equal to FH, the nearer to the straight line through the centre being thus equal to the more remote: which is impossible. Therefore another straight line equal to GF will not fall from the point F upon the circle; therefore only one straight line will so fall.'"],["Rule","'ProofWordCount'",412],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΑΔ, καὶ ἐπὶ τῆς ΑΔ εἰλήφθω τι σημεῖον τὸ Ζ, ὃ μή ἐστι κέντρον τοῦ κύκλου, κέντρον δὲ τοῦ κύκλου ἔστω τὸ Ε, καὶ ἀπὸ τοῦ Ζ πρὸς τὸν ΑΒΓΔ κύκλον προσπιπτέτωσαν εὐθεῖαί τινες αἱ ΖΒ, ΖΓ, ΖΗ: λέγω, ὅτι μεγίστη μέν ἐστιν ἡ ΖΑ, ἐλαχίστη δὲ ἡ ΖΔ, τῶν δὲ ἄλλων ἡ μὲν ΖΒ τῆς ΖΓ μείζων, ἡ δὲ ΖΓ τῆς ΖΗ. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΕ, ΗΕ. καὶ ἐπεὶ παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσιν, αἱ ἄρα ΕΒ, ΕΖ τῆς ΒΖ μείζονές εἰσιν. ἴση δὲ ἡ ΑΕ τῇ ΒΕ αἱ ἄρα ΒΕ, ΕΖ ἴσαι εἰσὶ τῇ ΑΖ: μείζων ἄρα ἡ ΑΖ τῆς ΒΖ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΓΕ, κοινὴ δὲ ἡ ΖΕ, δύο δὴ αἱ ΒΕ, ΕΖ δυσὶ ταῖς ΓΕ, ΕΖ ἴσαι εἰσίν. ἀλλὰ καὶ γωνία ἡ ὑπὸ ΒΕΖ γωνίας τῆς ὑπὸ ΓΕΖ μείζων. βάσις ἄρα ἡ ΒΖ βάσεως τῆς ΓΖ μείζων ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΖ τῆς ΖΗ μείζων ἐστίν. πάλιν, ἐπεὶ αἱ ΗΖ, ΖΕ τῆς ΕΗ μείζονές εἰσιν, ἴση δὲ ἡ ΕΗ τῇ ΕΔ, αἱ ἄρα ΗΖ, ΖΕ τῆς ΕΔ μείζονές εἰσιν. κοινὴ ἀφῃρήσθω ἡ ΕΖ: λοιπὴ ἄρα ἡ ΗΖ λοιπῆς τῆς ΖΔ μείζων ἐστίν. μεγίστη μὲν ἄρα ἡ ΖΑ, ἐλαχίστη δὲ ἡ ΖΔ, μείζων δὲ ἡ μὲν ΖΒ τῆς ΖΓ, ἡ δὲ ΖΓ τῆς ΖΗ. λέγω, ὅτι καὶ ἀπὸ τοῦ Ζ σημείου δύο μόνον ἴσαι προσπεσοῦνται πρὸς τὸν ΑΒΓΔ κύκλον ἐφ᾽ ἑκάτερα τῆς ΖΔ ἐλαχίστης. συνεστάτω γὰρ πρὸς τῇ ΕΖ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε τῇ ὑπὸ ΗΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΖΕΘ, καὶ ἐπεζεύχθω ἡ ΖΘ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΗΕ τῇ ΕΘ, κοινὴ δὲ ἡ ΕΖ, δύο δὴ αἱ ΗΕ, ΕΖ δυσὶ ταῖς ΘΕ, ΕΖ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΗΕΖ γωνίᾳ τῇ ὑπὸ ΘΕΖ ἴση: βάσις ἄρα ἡ ΖΗ βάσει τῇ ΖΘ ἴση ἐστίν. λέγω δή, ὅτι τῇ ΖΗ ἄλλη ἴση οὐ προσπεσεῖται πρὸς τὸν κύκλον ἀπὸ τοῦ Ζ σημείου. εἰ γὰρ δυνατόν, προσπιπτέτω ἡ ΖΚ. καὶ ἐπεὶ ἡ ΖΚ τῇ ΖΗ ἴση ἐστίν, ἀλλὰ ἡ ΖΘ τῇ ΖΗ ἴση ἐστίν, καὶ ἡ ΖΚ ἄρα τῇ ΖΘ ἐστιν ἴση, ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῇ ἀπώτερον ἴση: ὅπερ ἀδύνατον. οὐκ ἄρα ἀπὸ τοῦ Ζ σημείου ἑτέρα τις προσπεσεῖται πρὸς τὸν κύκλον ἴση τῇ ΗΖ: μία ἄρα μόνη. ἐὰν ἄρα κύκλου ἐπὶ τῆς διαμέτρου ληφθῇ τι σημεῖον, ὃ μή ἐστι κέντρον τοῦ κύκλου, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσιν εὐθεῖαί τινες, μεγίστη μὲν ἔσται, ἐφ᾽ ἧς τὸ κέντρον, ἐλαχίστη δὲ ἡ λοιπή, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ αὐτοῦ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ἐλαχίστης: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",461]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'3.8'"],["Rule","'Text'","'If a point be taken outside a circle and from the point straight lines be drawn through to the circle, one of which is through the centre and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the centre is greatest, while of the rest   the nearer to that through the centre is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote, and only two equal straight lines will fall on the circle from the point, one on each side of the least.'"],["Rule","'TextWordCount'",125],["Rule","'GreekText'","'ἐὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον διαχθῶσιν εὐθεῖαί τινες, ὧν μία μὲν διὰ τοῦ κέντρου, αἱ δὲ λοιπαί, ὡς ἔτυχεν, τῶν μὲν πρὸς τὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη μέν ἐστιν ἡ διὰ τοῦ κέντρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, τῶν δὲ πρὸς τὴν κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη μέν ἐστιν ἡ μεταξὺ τοῦ τε σημείου καὶ τῆς διαμέτρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς ἐλαχίστης τῆς ἀπώτερόν ἐστιν ἐλάττων, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ἐλαχίστης.'"],["Rule","'GreekTextWordCount'",103],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",20]],["List",["Rule","'Book'",1],["Rule","'Theorem'",21]],["List",["Rule","'Book'",1],["Rule","'Theorem'",24]],["List",["Rule","'Book'",3],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let ABC be a circle, and let a point D be taken outside ABC; let there be drawn through from it straight lines DA, DE, DF, DC, and let DA be through the centre; I say that, of the straight lines falling on the concave circumference AEFC, the straight line DA through the centre is greatest, while DE is greater than DF and DF than DC.; but, of the straight lines falling on the convex circumference HLKG, the straight line DG between the point and the diameter AG is least; and the nearer to the least DG is always less than the more remote, namely DK than DL, and DL than DH. For let the centre of the circle ABC be taken [III. 1], and let it be M; let ME, MF, MC, MK, ML, MH be joined. Then, since AM is equal to EM, let MD be added to each; therefore AD is equal to EM, MD. But EM, MD are greater than ED; [I. 20] therefore AD is also greater than ED. Again, since ME is equal to MF, and MD is common, therefore EM, MD are equal to FM, MD; and the angle EMD is greater than the angle FMD;  therefore the base ED is greater than the base FD. [I. 24]  Similarly we can prove that FD is greater than CD; therefore DA is greatest, while DE is greater than DF, and DF than DC. Next, since MK, KD are greater than MD, [I. 20] and MG is equal to MK, therefore the remainder KD is greater than the remainder GD, so that GD is less than KD. And, since on MD, one of the sides of the triangle MLD, two straight lines MK, KD were constructed meeting within the triangle, therefore MK, KD are less than ML, LD; [I. 21] and MK is equal to ML; therefore the remainder DK is less than the remainder DL. Similarly we can prove that DL is also less than DH; therefore DG is least, while DK is less than DL, and DL than DH. I say also that only two equal straight lines will fall from the point D on the circle, one on each side of the least DG. On the straight line MD, and at the point M on it, let the angle DMB be constructed equal to the angle KMD, and let DB be joined. Then, since MK is equal to MB, and MD is common, the two sides KM, MD are equal to the two sides BM, MD respectively; and the angle KMD is equal to the angle BMD; therefore the base DK is equal to the base DB. [I. 4]  I say that no other straight line equal to the straight line DK will fall on the circle from the point D. For, if possible, let a straight line so fall, and let it be DN. Then, since DK is equal to DN,   while DK is equal to DB, DB is also equal to DN, that is, the nearer to the least DG equal to the more remote: which was proved impossible. Therefore no more than two equal straight lines will fall on the circle ABC from the point D, one on each side of DG the least.'"],["Rule","'ProofWordCount'",545],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓ, καὶ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, καὶ ἀπ᾽ αὐτοῦ διήχθωσαν εὐθεῖαί τινες αἱ ΔΑ, ΔΕ, ΔΖ, ΔΓ, ἔστω δὲ ἡ ΔΑ διὰ τοῦ κέντρου. λέγω, ὅτι τῶν μὲν πρὸς τὴν ΑΕΖΓ κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη μέν ἐστιν ἡ διὰ τοῦ κέντρου ἡ ΔΑ, μείζων δὲ ἡ μὲν ΔΕ τῆς ΔΖ ἡ δὲ ΔΖ τῆς ΔΓ, τῶν δὲ πρὸς τὴν ΘΛΚΗ κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη μέν ἐστιν ἡ ΔΗ ἡ μεταξὺ τοῦ σημείου καὶ τῆς διαμέτρου τῆς ΑΗ, ἀεὶ δὲ ἡ ἔγγιον τῆς ΔΗ ἐλαχίστης ἐλάττων ἐστὶ τῆς ἀπώτερον, ἡ μὲν ΔΚ τῆς ΔΛ, ἡ δὲ ΔΛ τῆς ΔΘ. εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου καὶ ἔστω τὸ Μ: καὶ ἐπεζεύχθωσαν αἱ ΜΕ, ΜΖ, ΜΓ, ΜΚ, ΜΛ, ΜΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΜ τῇ ΕΜ, κοινὴ προσκείσθω ἡ ΜΔ: ἡ ἄρα ΑΔ ἴση ἐστὶ ταῖς ΕΜ, ΜΔ. ἀλλ᾽ αἱ ΕΜ, ΜΔ τῆς ΕΔ μείζονές εἰσιν: καὶ ἡ ΑΔ ἄρα τῆς ΕΔ μείζων ἐστίν. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΜΕ τῇ ΜΖ, κοινὴ δὲ ἡ ΜΔ, αἱ ΕΜ, ΜΔ ἄρα ταῖς ΖΜ, ΜΔ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΕΜΔ γωνίας τῆς ὑπὸ ΖΜΔ μείζων ἐστίν. βάσις ἄρα ἡ ΕΔ βάσεως τῆς ΖΔ μείζων ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΖΔ τῆς ΓΔ μείζων ἐστίν: μεγίστη μὲν ἄρα ἡ ΔΑ, μείζων δὲ ἡ μὲν ΔΕ τῆς ΔΖ, ἡ δὲ ΔΖ τῆς ΔΓ. καὶ ἐπεὶ αἱ ΜΚ, ΚΔ τῆς ΜΔ μείζονές εἰσιν, ἴση δὲ ἡ ΜΗ τῇ ΜΚ, λοιπὴ ἄρα ἡ ΚΔ λοιπῆς τῆς ΗΔ μείζων ἐστίν: ὥστε ἡ ΗΔ τῆς ΚΔ ἐλάττων ἐστίν: καὶ ἐπεὶ τριγώνου τοῦ ΜΛΔ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΜΔ δύο εὐθεῖαι ἐντὸς συνεστάθησαν αἱ ΜΚ, ΚΔ, αἱ ἄρα ΜΚ, ΚΔ τῶν ΜΛ, ΛΔ ἐλάττονές εἰσιν: ἴση δὲ ἡ ΜΚ τῇ ΜΛ: λοιπὴ ἄρα ἡ ΔΚ λοιπῆς τῆς ΔΛ ἐλάττων ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΔΛ τῆς ΔΘ ἐλάττων ἐστίν: ἐλαχίστη μὲν ἄρα ἡ ΔΗ, ἐλάττων δὲ ἡ μὲν ΔΚ τῆς ΔΛ ἡ δὲ ΔΛ τῆς ΔΘ. λέγω, ὅτι καὶ δύο μόνον ἴσαι ἀπὸ τοῦ Δ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ΔΗ ἐλαχίστης: συνεστάτω πρὸς τῇ ΜΔ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Μ τῇ ὑπὸ ΚΜΔ γωνίᾳ ἴση γωνία ἡ ὑπὸ ΔΜΒ καὶ ἐπεζεύχθω ἡ ΔΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΜΚ τῇ ΜΒ, κοινὴ δὲ ἡ ΜΔ, δύο δὴ αἱ ΚΜ, ΜΔ δύο ταῖς ΒΜ, ΜΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΚΜΔ γωνίᾳ τῇ ὑπὸ ΒΜΔ ἴση: βάσις ἄρα ἡ ΔΚ βάσει τῇ ΔΒ ἴση ἐστίν. λέγω δή, ὅτι τῇ ΔΚ εὐθείᾳ ἄλλη ἴση οὐ προσπεσεῖται πρὸς τὸν κύκλον ἀπὸ τοῦ Δ σημείου. εἰ γὰρ δυνατόν, προσπιπτέτω καὶ ἔστω ἡ ΔΝ. ἐπεὶ οὖν ἡ ΔΚ τῇ ΔΝ ἐστιν ἴση, ἀλλ᾽ ἡ ΔΚ τῇ ΔΒ ἐστιν ἴση, καὶ ἡ ΔΒ ἄρα τῇ ΔΝ ἐστιν ἴση, ἡ ἔγγιον τῆς ΔΗ ἐλαχίστης τῇ ἀπώτερον ἐστιν ἴση: ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα πλείους á¼¢ δύο ἴσαι πρὸς τὸν ΑΒΓ κύκλον ἀπὸ τοῦ Δ σημείου ἐφ᾽ ἑκάτερα τῆς ΔΗ ἐλαχίστης προσπεσοῦνται. ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον διαχθῶσιν εὐθεῖαί τινες, ὧν μία μὲν διὰ τοῦ κέντρου αἱ δὲ λοιπαί, ὡς ἔτυχεν, τῶν μὲν πρὸς τὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη μέν ἐστιν ἡ διὰ τοῦ κέντρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, τῶν δὲ πρὸς τὴν κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη μέν ἐστιν ἡ μεταξὺ τοῦ τε σημείου καὶ τῆς διαμέτρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς ἐλαχίστης τῆς ἀπώτερόν ἐστιν ἐλάττων, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ἐλαχίστης: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",613]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'3.9'"],["Rule","'Text'","'If a point be taken within a circle, and more than two equal straight lines fall from the point on the circle, the point taken is the centre of the circle.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους á¼¢ δύο ἴσαι εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",3],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let ABC be a circle and D a point within it, and from D let more than two equal straight lines, namely DA, DB, DC, fall on the circle ABC; I say that the point D is the centre of the circle ABC. For let AB, BC be joined and bisected at the points E, F, and let ED, FD be joined and drawn through to the points G, K, H, L. Then, since AE is equal to EB, and ED is common, the two sides AE, ED are equal to the two sides BE, ED; and the base DA is equal to the base DB; therefore the angle AED is equal to the angle BED. [I. 8]  Therefore each of the angles AED, BED is right; [I. Def. 10] therefore GK cuts AB into two equal parts and at right angles. And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [III. 1] the centre of the circle is on GK. For the same reason the centre of the circle ABC is also on HL. And the straight lines GK, HL have no other point common but the point D; therefore the point D is the centre of the circle ABC.'"],["Rule","'ProofWordCount'",224],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓ, ἐντὸς δὲ αὐτοῦ σημεῖον τὸ Δ, καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν πλείους á¼¢ δύο ἴσαι εὐθεῖαι αἱ ΔΑ, ΔΒ, ΔΓ: λέγω, ὅτι τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ἐπεζεύχθωσαν γὰρ αἱ ΑΒ, ΒΓ καὶ τετμήσθωσαν δίχα κατὰ τὰ Ε, Ζ σημεῖα, καὶ ἐπιζευχθεῖσαι αἱ ΕΔ, ΖΔ διήχθωσαν ἐπὶ τὰ Η, Κ, Θ, Λ σημεῖα. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, κοινὴ δὲ ἡ ΕΔ, δύο δὴ αἱ ΑΕ, ΕΔ δύο ταῖς ΒΕ, ΕΔ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΑ βάσει τῇ ΔΒ ἴση: γωνία ἄρα ἡ ὑπὸ ΑΕΔ γωνίᾳ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν: ὀρθὴ ἄρα ἑκατέρα τῶν ὑπὸ ΑΕΔ, ΒΕΔ γωνιῶν: ἡ ΗΚ ἄρα τὴν ΑΒ τέμνει δίχα καὶ πρὸς ὀρθάς. καὶ ἐπεί, ἐὰν ἐν κύκλῳ εὐθεῖά τις εὐθεῖάν τινα δίχα τε καὶ πρὸς ὀρθὰς τέμνῃ, ἐπὶ τῆς τεμνούσης ἐστὶ τὸ κέντρον τοῦ κύκλου, ἐπὶ τῆς ΗΚ ἄρα ἐστὶ τὸ κέντρον τοῦ κύκλου. διὰ τὰ αὐτὰ δὴ καὶ ἐπὶ τῆς ΘΛ ἐστι τὸ κέντρον τοῦ ΑΒΓ κύκλου. καὶ οὐδὲν ἕτερον κοινὸν ἔχουσιν αἱ ΗΚ, ΘΛ εὐθεῖαι á¼¢ τὸ Δ σημεῖον: τὸ Δ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους á¼¢ δύο ἴσαι εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",222]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'3.10'"],["Rule","'Text'","'A circle does not cut a circle at more points than two.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'κύκλος κύκλον οὐ τέμνει κατὰ πλείονα σημεῖα á¼¢ δύο.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Theorem'",1]],["List",["Rule","'Book'",3],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, C, F, H; let BH, BG be joined and bisected at the points K, L, and from K, L let KC, LM be drawn at right angles to BH, BG and carried through to the points A, E. Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles, the centre of the circle ABC is on AC. [III. 1]  Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles, the centre of the circle ABC is on NO. But it was also proved to be on AC, and the straight lines AC, NO meet at no point except at P; therefore the point P is the centre of the circle ABC. Similarly we can prove that P is also the centre of the circle DEF; therefore the two circles ABC, DEF which cut one another have the same centre P: which is impossible. [III. 5]'"],["Rule","'ProofWordCount'",190],["Rule","'GreekProof'","'εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓ κύκλον τὸν ΔΕΖ τεμνέτω κατὰ πλείονα σημεῖα á¼¢ δύο τὰ Β, Η, Ζ, Θ, καὶ ἐπιζευχθεῖσαι αἱ ΒΘ, ΒΗ δίχα τεμνέσθωσαν κατὰ τὰ κ, Λ σημεῖα: καὶ ἀπὸ τῶν Κ, Λ ταῖς ΒΘ, ΒΗ πρὸς ὀρθὰς ἀχθεῖσαι αἱ ΚΓ, ΛΜ διήχθωσαν ἐπὶ τὰ Α, Ε σημεῖα. ἐπεὶ οὖν ἐν κύκλῳ τῷ ΑΒΓ εὐθεῖά τις ἡ ΑΓ εὐθεῖάν τινα τὴν ΒΘ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΑΓ ἄρα ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου. πάλιν, ἐπεὶ ἐν κύκλῳ τῷ αὐτῷ τῷ ΑΒΓ εὐθεῖά τις ἡ ΝΞ εὐθεῖάν τινα τὴν ΒΗ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΝΞ ἄρα ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου. ἐδείχθη δὲ καὶ ἐπὶ τῆς ΑΓ, καὶ κατ᾽ οὐδὲν συμβάλλουσιν αἱ ΑΓ, ΝΞ εὐθεῖαι á¼¢ κατὰ τὸ Ο: τὸ Ο ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι καὶ τοῦ ΔΕΖ κύκλου κέντρον ἐστὶ τὸ Ο: δύο ἄρα κύκλων τεμνόντων ἀλλήλους τῶν ΑΒΓ, ΔΕΖ τὸ αὐτό ἐστι κέντρον τὸ Ο: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα κύκλος κύκλον τέμνει κατὰ πλείονα σημεῖα á¼¢ δύο: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",182]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'3.11'"],["Rule","'Text'","'If two circles touch one another internally, and their centres be taken, the straight line joining their centres, if it be also produced, will fall on the point of contact of the circles.'"],["Rule","'TextWordCount'",33],["Rule","'GreekText'","'ἐὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐντός, καὶ ληφθῇ αὐτῶν τὰ κέντρα, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη εὐθεῖα καὶ ἐκβαλλομένη ἐπὶ τὴν συναφὴν πεσεῖται τῶν κύκλων.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List"]],["Rule","'Proof'","'For let the two circles ABC, ADE touch one another internally at the point A, and let the centre F of the circle ABC, and the centre G of ADE, be taken; I say that the straight line joined from G to F and produced will fall on A. For suppose it does not, but, if possible, let it fall as FGH, and let AF, AG be joined. Then, since AG, GF are greater than FA, that is, than FH, let FG be subtracted from each; therefore the remainder AG is greater than the remainder GH. But AG is equal to GD; therefore GD is also greater than GH, the less than the greater: which is impossible. Therefore the straight line joined from F to G will not fall outside; therefore it will fall at A on the point of contact.'"],["Rule","'ProofWordCount'",141],["Rule","'GreekProof'","'δύο γὰρ κύκλοι οἱ ΑΒΓ, ΑΔΕ ἐφαπτέσθωσαν ἀλλήλων ἐντὸς κατὰ τὸ Α σημεῖον, καὶ εἰλήφθω τοῦ μὲν ΑΒΓ κύκλου κέντρον τὸ Ζ, τοῦ δὲ ΑΔΕ τὸ Η: λέγω, ὅτι ἡ ἀπὸ τοῦ Η ἐπὶ τὸ Ζ ἐπιζευγνυμένη εὐθεῖα ἐκβαλλομένη ἐπὶ τὸ Α πεσεῖται. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, πιπτέτω ὡς ἡ ΖΗΘ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΑΗ. ἐπεὶ οὖν αἱ ΑΗ, ΗΖ τῆς ΖΑ, τουτέστι τῆς ΖΘ, μείζονές εἰσιν, κοινὴ ἀφῃρήσθω ἡ ΖΗ: λοιπὴ ἄρα ἡ ΑΗ λοιπῆς τῆς ΗΘ μείζων ἐστίν. ἴση δὲ ἡ ΑΗ τῇ ΗΔ: καὶ ἡ ΗΔ ἄρα τῆς ΗΘ μείζων ἐστὶν ἡ ἐλάττων τῆς μείζονος: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα ἐκτὸς πεσεῖται: κατὰ τὸ Α ἄρα ἐπὶ τῆς συναφῆς πεσεῖται. ἐὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐντός, καὶ ληφθῇ αὐτῶν τὰ κέντρα, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη εὐθεῖα καὶ ἐκβαλλομένη ἐπὶ τὴν συναφὴν πεσεῖται τῶν κύκλων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",155]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'3.12'"],["Rule","'Text'","'If two circles touch one another externally, the straight line joining their centres will pass through the point of contact.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ἐὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη διὰ τῆς ἐπαφῆς ἐλεύσεται.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'For let the two circles ABC, ADE touch one anotherexternally at the point A, and let the centre F of ABC, and the centre G of ADE, be taken; I say that the straight line joined from F to G will pass through the point of contact at A. For suppose it does not,but, if possible, let it pass as FCDG, and let AF, AG be joined. Then, since the point F is the centre of the circle ABC,FA is equal to FC. Again, since the point G is the centre of the circle ADE, GA is equal to GD. But FA was also proved equal to FC;therefore FA, AG are equal to FC, GD, so that the whole FG is greater than FA, AG; but it is also less [I. 20]: which is impossible. Therefore the straight line joined from F to G will not fail to pass through the point of contact at A; therefore it will pass through it.'"],["Rule","'ProofWordCount'",165],["Rule","'GreekProof'","'δύο γὰρ κύκλοι οἱ ΑΒΓ, ΑΔΕ ἐφαπτέσθωσαν ἀλλήλων ἐκτὸς κατὰ τὸ Α σημεῖον, καὶ εἰλήφθω τοῦ μὲν ΑΒΓ κέντρον τὸ Ζ, τοῦ δὲ ΑΔΕ τὸ Η: λέγω, ὅτι ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ Α ἐπαφῆς ἐλεύσεται. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἐρχέσθω ὡς ἡ ΖΓΔΗ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΑΗ. ἐπεὶ οὖν τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΖΑ τῇ ΖΓ. πάλιν, ἐπεὶ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΔΕ κύκλου, ἴση ἐστὶν ἡ ΗΑ τῇ ΗΔ. ἐδείχθη δὲ καὶ ἡ ΖΑ τῇ ΖΓ ἴση: αἱ ἄρα ΖΑ, ΑΗ ταῖς ΖΓ, ΗΔ ἴσαι εἰσίν: ὥστε ὅλη ἡ ΖΗ τῶν ΖΑ, ΑΗ μείζων ἐστίν: ἀλλὰ καὶ ἐλάττων: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ Α ἐπαφῆς οὐκ ἐλεύσεται: δι᾽ αὐτῆς ἄρα. ἐὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη εὐθεῖα διὰ τῆς ἐπαφῆς ἐλεύσεται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",167]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'3.13'"],["Rule","'Text'","'A circle does not touch a circle at more points than one, whether it touch it internally or externally.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'κύκλος κύκλου οὐκ ἐφάπτεται κατὰ πλείονα σημεῖα á¼¢ καθ᾽ ἕν, ἐάν τε ἐντὸς ἐάν τε ἐκτὸς ἐφάπτηται.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Definition'",3]],["List",["Rule","'Book'",3],["Rule","'Theorem'",2]],["List",["Rule","'Book'",3],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'For, if possible, let the circle ABDC touch the circle EBFD, first internally, at morepoints than one, namely D, B. Let the centre G of the circle ABDC, and the centre H of EBFD, be taken. Therefore the straight linejoined from G to H will fall on B, D. [III. 11] Let it so fall, as BGHD. Then, since the point G is the centre of the circle ABCD,BG is equal to GD;  therefore BG is greater than HD; therefore BH is much greater than HD. Again, since the point H is the centre of the circle EBFD,BH is equal to HD;  but it was also proved much greater than it: which is impossible. Therefore a circle does not touch a circle internally at more points than one. I say further that neither does it so touch it externally. For, if possible, let the circle ACK touch the circle ABDC at more points than one, namely A, C, and let AC be joined. Then, since on the circumference of each of the circlesABDC, ACK two points A, C have been taken at random, the straight line joining the points will fall within each circle; [III. 2] but it fell within the circle ABCD and outside ACK [III. Def. 3]: which is absurd. Therefore a circle does not touch a circle externally at more points than one. And it was proved that neither does it so touch it internally.'"],["Rule","'ProofWordCount'",240],["Rule","'GreekProof'","'εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓΔ κύκλου τοῦ ΕΒΖΔ ἐφαπτέσθω πρότερον ἐντὸς κατὰ πλείονα σημεῖα á¼¢ ἓν τὰ δ, Β. καὶ εἰλήφθω τοῦ μὲν ΑΒΓΔ κύκλου κέντρον τὸ Η, τοῦ δὲ ΕΒΖΔ τὸ Θ. ἡ ἄρα ἀπὸ τοῦ Η ἐπὶ τὸ Θ ἐπιζευγνυμένη ἐπὶ τὰ Β, Δ πεσεῖται. πιπτέτω ὡς ἡ ΒΗΘΔ. καὶ ἐπεὶ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓΔ κύκλου, ἴση ἐστὶν ἡ ΒΗ τῇ ΗΔ: μείζων ἄρα ἡ ΒΗ τῆς ΘΔ: πολλῷ ἄρα μείζων ἡ ΒΘ τῆς ΘΔ. πάλιν, ἐπεὶ τὸ Θ σημεῖον κέντρον ἐστὶ τοῦ ΕΒΖΔ κύκλου, ἴση ἐστὶν ἡ ΒΘ τῇ ΘΔ: ἐδείχθη δὲ αὐτῆς καὶ πολλῷ μείζων: ὅπερ ἀδύνατον: οὐκ ἄρα κύκλος κύκλου ἐφάπτεται ἐντὸς κατὰ πλείονα σημεῖα á¼¢ ἕν. λέγω δή, ὅτι οὐδὲ ἐκτός. εἰ γὰρ δυνατόν, κύκλος ὁ ΑΓΚ κύκλου τοῦ ΑΒΓΔ ἐφαπτέσθω ἐκτὸς κατὰ πλείονα σημεῖα á¼¢ ἓν τὰ Α, Γ, καὶ ἐπεζεύχθω ἡ ΑΓ. ἐπεὶ οὖν κύκλων τῶν ΑΒΓΔ, ΑΓΚ εἴληπται ἐπὶ τῆς περιφερείας ἑκατέρου δύο τυχόντα σημεῖα τὰ Α, Γ, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς ἑκατέρου πεσεῖται: ἀλλὰ τοῦ μὲν ΑΒΓΔ ἐντὸς ἔπεσεν, τοῦ δὲ ΑΓΚ ἐκτός: ὅπερ ἄτοπον: οὐκ ἄρα κύκλος κύκλου ἐφάπτεται ἐκτὸς κατὰ πλείονα σημεῖα á¼¢ ἕν. ἐδείχθη δέ, ὅτι οὐδὲ ἐντός. κύκλος ἄρα κύκλου οὐκ ἐφάπτεται κατὰ πλείονα σημεῖα á¼¢ καθ᾽ ἕν, ἐάν τε ἐντὸς ἐάν τε ἐκτὸς ἐφάπτηται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",220]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'3.14'"],["Rule","'Text'","'In a circle equal straight lines are equally distant from the centre, and those which are equally distant from the centre are equal to one another.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐν κύκλῳ αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου, καὶ αἱ ἴσον ἀπέχουσαι ἀπὸ τοῦ κέντρου ἴσαι ἀλλήλαις εἰσίν.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Definition'",4]],["List",["Rule","'Book'",3],["Rule","'Theorem'",1]],["List",["Rule","'Book'",3],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'Let ABDC be a circle, and let AB, CD be equal straight lines in it; I say that AB, CD are equally distant from the centre. For let the centre of the circle ABDC be taken [III. 1], and let it be E; from E let EF, EG be drawn perpendicular to AB, CD, and let AE, EC be joined. Then, since a straight line EF through the centre cuts a straight line AB not through the centre at right angles, it also bisects it. [III. 3] Therefore AF is equal to FB; therefore AB is double of AF. For the same reason CD is also double of CG; and AB is equal to CD; therefore AF is also equal to CG. And, since AE is equal to EC, the square on AE is also equal to the square on EC. But the squares on AF, EF are equal to the square on AE, for the angle at F is right; and the squares on EG, GC are equal to the square on EC, for the angle at G is right; [I. 47] therefore the squares on AF, FE are equal to the squares on CG, GE, of which the square on AF is equal to the square on CG, for AF is equal to CG; therefore the square on FE which remains is equal to the square on EG, therefore EF is equal to EG. But in a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal; [III. Def. 4] therefore AB, CD are equally distant from the centre. Next, let the straight lines AB, CD be equally distant from the centre; that is, let EF be equal to EG. I say that AB is also equal to CD. For, with the same construction, we can prove, similarly, that AB is double of AF, and CD of CG. And, since AE is equal to CE, the square on AE is equal to the square on CE. But the squares on EF, FA are equal to the square on AE, and the squares on EG, GC equal to the square on CE. [I. 47] Therefore the squares on EF, FA are equal to the squares on EG, GC, of which the square on EF is equal to the square on EG, for EF is equal to EG; therefore the square on AF which remains is equal to the square on CG; therefore AF is equal to CG. And AB is double of AF, and CD double of CG; therefore AB is equal to CD.'"],["Rule","'ProofWordCount'",438],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ ἴσαι εὐθεῖαι ἔστωσαν αἱ ΑΒ, ΓΔ: λέγω, ὅτι αἱ ΑΒ, ΓΔ ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου. εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓΔ κύκλου καὶ ἔστω τὸ Ε, καὶ ἀπὸ τοῦ Ε ἐπὶ τὰς ΑΒ, ΓΔ κάθετοι ἤχθωσαν αἱ ΕΖ, ΕΗ, καὶ ἐπεζεύχθωσαν αἱ ΑΕ, ΕΓ. ἐπεὶ οὖν εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΕΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΒ πρὸς ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει. ἴση ἄρα ἡ ΑΖ τῇ ΖΒ: διπλῆ ἄρα ἡ ΑΒ τῆς ΑΖ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΔ τῆς ΓΗ ἐστι διπλῆ: καί ἐστιν ἴση ἡ ΑΒ τῇ ΓΔ: ἴση ἄρα καὶ ἡ ΑΖ τῇ ΓΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΕΓ, ἴσον καὶ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς ΕΓ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΕ ἴσα τὰ ἀπὸ τῶν ΑΖ, ΕΖ: ὀρθὴ γὰρ ἡ πρὸς τῷ Ζ γωνία: τῷ δὲ ἀπὸ τῆς ΕΓ ἴσα τὰ ἀπὸ τῶν ΕΗ, ΗΓ: ὀρθὴ γὰρ ἡ πρὸς τῷ Η γωνία: τὰ ἄρα ἀπὸ τῶν ΑΖ, ΖΕ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΓΗ, ΗΕ, ὧν τὸ ἀπὸ τῆς ΑΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΗ: ἴση γάρ ἐστιν ἡ ΑΖ τῇ ΓΗ: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΖΕ τῷ ἀπὸ τῆς ΕΗ ἴσον ἐστίν: ἴση ἄρα ἡ ΕΖ τῇ ΕΗ. ἐν δὲ κύκλῳ ἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι λέγονται, ὅταν αἱ ἀπὸ τοῦ κέντρου ἐπ᾽ αὐτὰς κάθετοι ἀγόμεναι ἴσαι ὦσιν: αἱ ἄρα ΑΒ, ΓΔ ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου. ἀλλὰ δὴ αἱ ΑΒ, ΓΔ εὐθεῖαι ἴσον ἀπεχέτωσαν ἀπὸ τοῦ κέντρου, τουτέστιν ἴση ἔστω ἡ ΕΖ τῇ ΕΗ. λέγω, ὅτι ἴση ἐστὶ καὶ ἡ ΑΒ τῇ ΓΔ. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι διπλῆ ἐστιν ἡ μὲν ΑΒ τῆς ΑΖ, ἡ δὲ ΓΔ τῆς ΓΗ: καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΓΕ, ἴσον ἐστὶ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς ΓΕ: ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΕ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΕΖ, ΖΑ, τῷ δὲ ἀπὸ τῆς ΓΕ ἴσα τὰ ἀπὸ τῶν ΕΗ, ΗΓ. τὰ ἄρα ἀπὸ τῶν ΕΖ, ΖΑ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΕΗ, ΗΓ: ὧν τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΕΗ ἐστιν ἴσον: ἴση γὰρ ἡ ΕΖ τῇ ΕΗ: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΑΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΗ: ἴση ἄρα ἡ ΑΖ τῇ ΓΗ: καί ἐστι τῆς μὲν ΑΖ διπλῆ ἡ ΑΒ, τῆς δὲ ΓΗ διπλῆ ἡ ΓΔ: ἴση ἄρα ἡ ΑΒ τῇ ΓΔ. ἐν κύκλῳ ἄρα αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου, καὶ αἱ ἴσον ἀπέχουσαι ἀπὸ τοῦ κέντρου ἴσαι ἀλλήλαις εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",428]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'3.15'"],["Rule","'Text'","'Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the centre is always greater than the more remote.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐν κύκλῳ μεγίστη μὲν ἡ διάμετρος τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",20]],["List",["Rule","'Book'",1],["Rule","'Theorem'",24]],["List",["Rule","'Book'",3],["Rule","'Definition'",5]],["List",["Rule","'Book'",3],["Rule","'Theorem'",14]]]],["Rule","'Proof'","'Let ABCD be a circle, let AD be its diameter and E the centre; and let BC be nearer to the diameter AD, and FG more remote; I say that AD is greatest and BC greater than FG. For from the centre E let EH, EK  be drawn perpendicular to BC, FG. Then, since BC is nearer to the centre and FG more remote, EK is greater than EH. [III. Def. 5] Let EL be made equal to EH, through L let LM be drawn at right angles to EK and carried through to N, and let ME, EN, FE, EG be joined. Then, since EH is equal to EL, BC is also equal to MN. [III. 14]  Again, since AE is equal to EM, and ED to EN, AD is equal to ME, EN. But ME, EN are greater than MN, [I. 20] and MN is equal to BC; therefore AD is greater than BC. And, since the two sides ME, EN are equal to the two sides FE, EG, and the angle MEN greater than the angle FEG, therefore the base MN is greater than the base FG. [I. 24] But MN was proved equal to BC. Therefore the diameter AD is greatest and BC greater than FG.'"],["Rule","'ProofWordCount'",210],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΑΔ, κέντρον δὲ τὸ Ε, καὶ ἔγγιον μὲν τῆς ΑΔ διαμέτρου ἔστω ἡ ΒΓ, ἀπώτερον δὲ ἡ ΖΗ: λέγω, ὅτι μεγίστη μέν ἐστιν ἡ ΑΔ, μείζων δὲ ἡ ΒΓ τῆς ΖΗ. ἤχθωσαν γὰρ ἀπὸ τοῦ Ε κέντρου ἐπὶ τὰς ΒΓ, ΖΗ κάθετοι αἱ ΕΘ, ΕΚ. καὶ ἐπεὶ ἔγγιον μὲν τοῦ κέντρου ἐστὶν ἡ ΒΓ, ἀπώτερον δὲ ἡ ΖΗ, μείζων ἄρα ἡ ΕΚ τῆς ΕΘ. κείσθω τῇ ΕΘ ἴση ἡ ΕΛ, καὶ διὰ τοῦ Λ τῇ ΕΚ πρὸς ὀρθὰς ἀχθεῖσα ἡ ΛΜ διήχθω ἐπὶ τὸ Ν, καὶ ἐπεζεύχθωσαν αἱ ΜΕ, ΕΝ, ΖΕ, ΕΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΕΘ τῇ ΕΛ, ἴση ἐστὶ καὶ ἡ ΒΓ τῇ ΜΝ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΜ, ἡ δὲ ΕΔ τῇ ΕΝ, ἡ ἄρα ΑΔ ταῖς ΜΕ, ΕΝ ἴση ἐστίν. ἀλλ᾽ αἱ μὲν ΜΕ, ΕΝ τῆς ΜΝ μείζονές εἰσιν [καὶ ἡ ΑΔ τῆς ΜΝ μείζων ἐστίν, ἴση δὲ ἡ ΜΝ τῇ ΒΓ: ἡ ΑΔ ἄρα τῆς ΒΓ μείζων ἐστίν. καὶ ἐπεὶ δύο αἱ ΜΕ, ΕΝ δύο ταῖς ΖΕ, ΕΗ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ ΜΕΝ γωνίας τῆς ὑπὸ ΖΕΗ μείζων ἐστίν, βάσις ἄρα ἡ ΜΝ βάσεως τῆς ΖΗ μείζων ἐστίν. ἀλλὰ ἡ ΜΝ τῇ ΒΓ ἐδείχθη ἴση καὶ ἡ ΒΓ τῆς ΖΗ μείζων ἐστίν. μεγίστη μὲν ἄρα ἡ ΑΔ διάμετρος, μείζων δὲ ἡ ΒΓ τῆς ΖΗ. ἐν κύκλῳ ἄρα μεγίστη μέν ἐστιν ἡ διάμετρος, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",249]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'3.16'"],["Rule","'Text'","'The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed; further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilineal angle.'"],["Rule","'TextWordCount'",56],["Rule","'GreekText'","'ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐκτὸς πεσεῖται τοῦ κύκλου, καὶ εἰς τὸν μεταξὺ τόπον τῆς τε εὐθείας καὶ τῆς περιφερείας ἑτέρα εὐθεῖα οὐ παρεμπεσεῖται, καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἁπάσης γωνίας ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ἐλάττων.'"],["Rule","'GreekTextWordCount'",46],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",17]],["List",["Rule","'Book'",1],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'Let ABC be a circle about D as centre and AB as diameter; I say that the straight line drawn from A at right angles to AB from its extremity will fall outside the circle. For suppose it does not, but, if possible, let it fall within as CA, and let DC be joined. Since DA is equal to DC, the angle DAC is also equal to the angle ACD. [I. 5]  But the angle DAC is right; therefore the angle ACD is also right: thus, in the triangle ACD, the two angles DAC, ACD are equal to two right angles: which is impossible. [I. 17] Therefore the straight line drawn from the point A at right angles to BA will not fall within the circle. Similarly we can prove that neither will it fall on the circumference; therefore it will fall outside. Let it fall as AE; I say next that into the space between the straight line AE and the circumference CHA another straight line cannot be interposed. For, if possible, let another straight line be so interposed, as FA, and let DG be drawn from the point D perpendicular to FA. Then, since the angle AGD is right, and the angle DAG is less than a right angle, AD is greater than DG. [I. 19]  But DA is equal to DH; therefore DH is greater than DG, the less than the greater: which is impossible. Therefore another straight line cannot be interposed into the space between the straight line and the circumference. I say further that the angle of the semicircle contained by the straight line BA and the circumference CHA is greater than any acute rectilineal angle, and the remaining angle contained by the circumference CHA and the straight line AE is less than any acute rectilineal angle. For, if there is any rectilineal angle greater than the angle contained by the straight line BA and the circumference CHA, and any rectilineal angle less than the angle contained by the circumference CHA and the straight line AE, then into the space between the circumference and the straight line AE a straight line will be interposed such as will make an angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, and another angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE. But such a straight line cannot be interposed; therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, nor yet any acute angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE.'"],["Rule","'ProofWordCount'",469],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓ περὶ κέντρον τὸ Δ καὶ διάμετρον τὴν ΑΒ: λέγω, ὅτι ἡ ἀπὸ τοῦ Α τῇ ΑΒ πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐκτὸς πεσεῖται τοῦ κύκλου. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, πιπτέτω ἐντὸς ὡς ἡ ΓΑ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΔΓ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΑΓΔ. ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΓΔ: τριγώνου δὴ τοῦ ΑΓΔ αἱ δύο γωνίαι αἱ ὑπὸ ΔΑΓ, ΑΓΔ δύο ὀρθαῖς ἴσαι εἰσίν: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Α σημείου τῇ ΒΑ πρὸς ὀρθὰς ἀγομένη ἐντὸς πεσεῖται τοῦ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἐπὶ τῆς περιφερείας: ἐκτὸς ἄρα. πιπτέτω ὡς ἡ ΑΕ: λέγω δή, ὅτι εἰς τὸν μεταξὺ τόπον τῆς τε ΑΕ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ἑτέρα εὐθεῖα οὐ παρεμπεσεῖται. εἰ γὰρ δυνατόν, παρεμπιπτέτω ὡς ἡ ΖΑ, καὶ ἤχθω ἀπὸ τοῦ Δ σημείου ἐπὶ τὴν ΖΑ κάθετος ἡ ΔΗ. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΑΗΔ, ἐλάττων δὲ ὀρθῆς ἡ ὑπὸ ΔΑΗ, μείζων ἄρα ἡ ΑΔ τῆς ΔΗ. ἴση δὲ ἡ ΔΑ τῇ ΔΘ: μείζων ἄρα ἡ ΔΘ τῆς ΔΗ, ἡ ἐλάττων τῆς μείζονος: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα εἰς τὸν μεταξὺ τόπον τῆς τε εὐθείας καὶ τῆς περιφερείας ἑτέρα εὐθεῖα παρεμπεσεῖται. λέγω, ὅτι καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἡ περιεχομένη ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ἁπάσης γωνίας ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ἡ περιεχομένη ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας ἁπάσης γωνίας ὀξείας εὐθυγράμμου ἐλάττων ἐστίν. εἰ γὰρ ἐστί τις γωνία εὐθύγραμμος μείζων μὲν τῆς περιεχομένης ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας, ἐλάττων δὲ τῆς περιεχομένης ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας, εἰς τὸν μεταξὺ τόπον τῆς τε ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας εὐθεῖα περεμπεσεῖται, ἥτις ποιήσει μείζονα μὲν τῆς περιεχομένης ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ὑπὸ εὐθειῶν περιεχομένην, ἐλάττονα δὲ τῆς περιεχομένης ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας. οὐ παρεμπίπτει δέ: οὐκ ἄρα τῆς περιεχομένης γωνίας ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ἔσται μείζων ὀξεῖα ὑπὸ εὐθειῶν περιεχομένη, οὐδὲ μὴν ἐλάττων τῆς περιεχομένης ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου καὶ ὅτι εὐθεῖα κύκλου καθ᾽ ἓν μόνον ἐφάπτεται σημεῖον, ἐπειδήπερ καὶ ἡ κατὰ δύο αὐτῷ συμβάλλουσα ἐντὸς αὐτοῦ πίπτουσα ἐδείχθη. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",417]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'3.17'"],["Rule","'Text'","'From a given point to draw a straight line touching a given circle.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἀπὸ τοῦ δοθέντος σημείου τοῦ δοθέντος κύκλου ἐφαπτομένην εὐθεῖαν γραμμὴν ἀγαγεῖν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",3],["Rule","'Theorem'",1]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'Let A be the given point, and BCD the given circle; thus it is required to draw from the point A a straight line touching the circle BCD. For let the centre E of the circle be taken; [III. 1] let AE be joined, and with centre E and distance EA let the circle AFG be described; from D let DF be drawn at right angles to EA, and let EF, AB be joined; I say that AB has been drawn from the point A touching the circle BCD. For, since E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two sides FE, ED: and they contain a common angle, the angle at E; therefore the base DF is equal to the base AB, and the triangle DEF is equal to the triangle BEA, and the remaining angles to the remaining angles; [I. 4]therefore the angle EDF is equal to the angle EBA. But the angle EDF is right; therefore the angle EBA is also right. Now EB is a radius; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16] therefore AB touches the circle BCD. Therefore from the given point A the straight line AB has been drawn touching the circle BCD.'"],["Rule","'ProofWordCount'",236],["Rule","'GreekProof'","'ἔστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ὁ δὲ δοθεὶς κύκλος ὁ ΒΓΔ: δεῖ δὴ ἀπὸ τοῦ Α σημείου τοῦ ΒΓΔ κύκλου ἐφαπτομένην εὐθεῖαν γραμμὴν ἀγαγεῖν. εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ε, καὶ ἐπεζεύχθω ἡ ΑΕ, καὶ κέντρῳ μὲν τῷ Ε διαστήματι δὲ τῷ ΕΑ κύκλος γεγράφθω ὁ ΑΖΗ, καὶ ἀπὸ τοῦ Δ τῇ ΕΑ πρὸς ὀρθὰς ἤχθω ἡ ΔΖ, καὶ ἐπεζεύχθωσαν αἱ ΕΖ, ΑΒ: λέγω, ὅτι ἀπὸ τοῦ Α σημείου τοῦ ΒΓΔ κύκλου ἐφαπτομένη ἦκται ἡ ΑΒ. ἐπεὶ γὰρ τὸ Ε κέντρον ἐστὶ τῶν ΒΓΔ, ΑΖΗ κύκλων, ἴση ἄρα ἐστὶν ἡ μὲν ΕΑ τῇ ΕΖ, ἡ δὲ ΕΔ τῇ ΕΒ: δύο δὴ αἱ ΑΕ, ΕΒ δύο ταῖς ΖΕ, ΕΔ ἴσαι εἰσίν: καὶ γωνίαν κοινὴν περιέχουσι τὴν πρὸς τῷ Ε: βάσις ἄρα ἡ ΔΖ βάσει τῇ ΑΒ ἴση ἐστίν, καὶ τὸ ΔΕΖ τρίγωνον τῷ ΕΒΑ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις: ἴση ἄρα ἡ ὑπὸ ΕΔΖ τῇ ὑπὸ ΕΒΑ. ὀρθὴ δὲ ἡ ὑπὸ ΕΔΖ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΕΒΑ. καί ἐστιν ἡ ΕΒ ἐκ τοῦ κέντρου: ἡ δὲ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου: ἡ ΑΒ ἄρα ἐφάπτεται τοῦ ΒΓΔ κύκλου. ἀπὸ τοῦ ἄρα δοθέντος σημείου τοῦ Α τοῦ δοθέντος κύκλου τοῦ ΒΓΔ ἐφαπτομένη εὐθεῖα γραμμὴ ἦκται ἡ ΑΒ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",216]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'3.18'"],["Rule","'Text'","'If a straight line touch a circle, and a straight line be joined from the centre to the point of contact, the straight line so joined will be perpendicular to the tangent.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τοῦ κέντρου ἐπὶ τὴν ἁφὴν ἐπιζευχθῇ τις εὐθεῖα, ἡ ἐπιζευχθεῖσα κάθετος ἔσται ἐπὶ τὴν ἐφαπτομένην.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",17]],["List",["Rule","'Book'",1],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'For let a straight line DE touch the circle ABC at the point C, let the centre F of the circle ABC be taken, and let FC be joined from F to C; I say that FC is perpendicular to DE. For, if not, let FG be drawn from F perpendicular to DE. Then, since the angle FGC is right, the angle FCG is acute; [I. 17]and the greater angle is subtended by the greater side; [I. 19] therefore FC is greater than FG. But FC is equal to FB; therefore FB is also greater than FG, the less than the greater: which is impossible. Therefore FG is not perpendicular to DE. Similarly we can prove that neither is any other straight line except FC; therefore FC is perpendicular to DE.'"],["Rule","'ProofWordCount'",132],["Rule","'GreekProof'","'κύκλου γὰρ τοῦ ΑΒΓ ἐφαπτέσθω τις εὐθεῖα ἡ ΔΕ κατὰ τὸ Γ σημεῖον, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου τὸ Ζ, καὶ ἀπὸ τοῦ Ζ ἐπὶ τὸ Γ ἐπεζεύχθω ἡ ΖΓ: λέγω, ὅτι ἡ ΖΓ κάθετός ἐστιν ἐπὶ τὴν ΔΕ. εἰ γὰρ μή, ἤχθω ἀπὸ τοῦ Ζ ἐπὶ τὴν ΔΕ κάθετος ἡ ΖΗ. ἐπεὶ οὖν ἡ ὑπὸ ΖΗΓ γωνία ὀρθή ἐστιν, ὀξεῖα ἄρα ἐστὶν ἡ ὑπὸ ΖΓΗ: ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει: μείζων ἄρα ἡ ΖΓ τῆς ΖΗ: ἴση δὲ ἡ ΖΓ τῇ ΖΒ: μείζων ἄρα καὶ ἡ ΖΒ τῆς ΖΗ ἡ ἐλάττων τῆς μείζονος: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ΖΗ κάθετός ἐστιν ἐπὶ τὴν ΔΕ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλη τις πλὴν τῆς ΖΓ: ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΔΕ. ἐὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τοῦ κέντρου ἐπὶ τὴν ἁφὴν ἐπιζευχθῇ τις εὐθεῖα, ἡ ἐπιζευχθεῖσα κάθετος ἔσται ἐπὶ τὴν ἐφαπτομένην: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",157]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'3.19'"],["Rule","'Text'","'If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the tangent, the centre of the circle will be on the straight line so drawn.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς τῇ ἐφαπτομένῃ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἀχθῇ, ἐπὶ τῆς ἀχθείσης ἔσται τὸ κέντρον τοῦ κύκλου.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'For let a straight line DE touch the circle ABC at the point C, and from C let CA be drawn at right angles to DE; I say that the centre of the circle is on AC. For suppose it is not, but, if possible, let F be the centre, and let CF be joined. Since a straight line DE touches the circle ABC, and FC has been joined from the centre to the point of contact, FC is perpendicular to DE; [III. 18]therefore the angle FCE is right. But the angle ACE is also right; therefore the angle FCE is equal to the angle ACE, the less to the greater: which is impossible. Therefore F is not the centre of the circle ABC. Similarly we can prove that neither is any other point except a point on AC.'"],["Rule","'ProofWordCount'",140],["Rule","'GreekProof'","'κύκλου γὰρ τοῦ ΑΒΓ ἐφαπτέσθω τις εὐθεῖα ἡ ΔΕ κατὰ τὸ Γ σημεῖον, καὶ ἀπὸ τοῦ Γ τῇ ΔΕ πρὸς ὀρθὰς ἤχθω ἡ ΓΑ: λέγω, ὅτι ἐπὶ τῆς ΑΓ ἐστι τὸ κέντρον τοῦ κύκλου. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΓΖ. ἐπεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΔΕ, ἀπὸ δὲ τοῦ κέντρου ἐπὶ τὴν ἁφὴν ἐπέζευκται ἡ ΖΓ, ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΔΕ: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΓΕ. ἐστὶ δὲ καὶ ἡ ὑπὸ ΑΓΕ ὀρθή: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΖΓΕ τῇ ὑπὸ ΑΓΕ ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Ζ κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλο τι πλὴν ἐπὶ τῆς ΑΓ. ἐὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς τῇ ἐφαπτομένῃ πρὸς ὀρθὰς εὐθεῖα γραμμὴ ἀχθῇ, ἐπὶ τῆς ἀχθείσης ἔσται τὸ κέντρον τοῦ κύκλου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",153]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'3.20'"],["Rule","'Text'","'In a circle the angle at the centre is double of the angle at the circumference, when the angles have the same circumference as base.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ἐν κύκλῳ ἡ πρὸς τῷ κέντρῳ γωνία διπλασίων ἐστὶ τῆς πρὸς τῇ περιφερείᾳ, ὅταν τὴν αὐτὴν περιφέρειαν βάσιν ἔχωσιν αἱ γωνίαι.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]]]],["Rule","'Proof'","'Let ABC be a circle, let the angle BEC be an angleat its centre, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base; I say that the angle BEC is double ofthe angle BAC. For let AE be joined and drawn through to F. Then, since EA is equal to EB, the angle EAB is also equal to the angle EBA; [I. 5]therefore the angles EAB, EBA are double of the angle EAB. But the angle BEF is equal to the angles EAB, EBA; [I. 32] therefore the angle BEF is also double of the angleEAB. For the same reason the angle FEC is also double of the angle EAC. Therefore the whole angle BEC is double of the whole angle BAC. Again let another straight line be inflected, and let there be another angle BDC; let DE be joined and produced to G. Similarly then we can prove that the angle GEC is double of the angle EDC,of which the angle GEB is double of the angle EDB; therefore the angle BEC which remains is double of the angle BDC.'"],["Rule","'ProofWordCount'",193],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓ, καὶ πρὸς μὲν τῷ κέντρῳ αὐτοῦ γωνία ἔστω ἡ ὑπὸ ΒΕΓ, πρὸς δὲ τῇ περιφερείᾳ ἡ ὑπὸ ΒΑΓ, ἐχέτωσαν δὲ τὴν αὐτὴν περιφέρειαν βάσιν τὴν ΒΓ: λέγω, ὅτι διπλασίων ἐστὶν ἡ ὑπὸ ΒΕΓ γωνία τῆς ὑπὸ ΒΑΓ. ἐπιζευχθεῖσα γὰρ ἡ ΑΕ διήχθω ἐπὶ τὸ Ζ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΕΑ τῇ ΕΒ, ἴση καὶ γωνία ἡ ὑπὸ ΕΑΒ τῇ ὑπὸ ΕΒΑ: αἱ ἄρα ὑπὸ ΕΑΒ, ΕΒΑ γωνίαι τῆς ὑπὸ ΕΑΒ διπλασίους εἰσίν. ἴση δὲ ἡ ὑπὸ ΒΕΖ ταῖς ὑπὸ ΕΑΒ, ΕΒΑ: καὶ ἡ ὑπὸ ΒΕΖ ἄρα τῆς ὑπὸ ΕΑΒ ἐστι διπλῆ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΖΕΓ τῆς ὑπὸ ΕΑΓ ἐστι διπλῆ. ὅλη ἄρα ἡ ὑπὸ ΒΕΓ ὅλης τῆς ὑπὸ ΒΑΓ ἐστι διπλῆ. Κεκλάσθω δὴ πάλιν, καὶ ἔστω ἑτέρα γωνία ἡ ὑπὸ ΒΔΓ, καὶ ἐπιζευχθεῖσα ἡ ΔΕ ἐκβεβλήσθω ἐπὶ τὸ Η. ὁμοίως δὴ δείξομεν, ὅτι διπλῆ ἐστιν ἡ ὑπὸ ΗΕΓ γωνία τῆς ὑπὸ ΕΔΓ, ὧν ἡ ὑπὸ ΗΕΒ διπλῆ ἐστι τῆς ὑπὸ ΕΔΒ: λοιπὴ ἄρα ἡ ὑπὸ ΒΕΓ διπλῆ ἐστι τῆς ὑπὸ ΒΔΓ. ἐν κύκλῳ ἄρα ἡ πρὸς τῷ κέντρῳ γωνία διπλασίων ἐστὶ τῆς πρὸς τῇ περιφερείᾳ, ὅταν τὴν αὐτὴν περιφέρειαν βάσιν ἔχωσιν αἱ γωνίαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",195]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'3.21'"],["Rule","'Text'","'In a circle the angles in the same segment are equal to one another.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'ἐν κύκλῳ αἱ ἐν τῷ αὐτῷ τμήματι γωνίαι ἴσαι ἀλλήλαις εἰσίν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let ABCD be a circle, and let the angles BAD, BED be angles in the same segment BAED; I say that the angles BAD, BED are equal to one another. For let the centre of the circle ABCD be taken, and let it be F; let BF, FD be joined. Now, since the angle BFD is at the centre, and the angle BAD at the circumference, and they have the same circumference BCD as base, therefore the angle BFD is double of the angle BAD. [III. 20]  For the same reason the angle BFD is also double of the angle BED; therefore the angle BAD is equal to the angle BED.'"],["Rule","'ProofWordCount'",111],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν τῷ αὐτῷ τμήματι τῷ ΒΑΕΔ γωνίαι ἔστωσαν αἱ ὑπὸ ΒΑΔ, ΒΕΔ: λέγω, ὅτι αἱ ὑπὸ ΒΑΔ, ΒΕΔ γωνίαι ἴσαι ἀλλήλαις εἰσίν. εἰλήφθω γὰρ τοῦ ΑΒΓΔ κύκλου τὸ κέντρον, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΒΖ, ΖΔ. καὶ ἐπεὶ ἡ μὲν ὑπὸ ΒΖΔ γωνία πρὸς τῷ κέντρῳ ἐστίν, ἡ δὲ ὑπὸ ΒΑΔ πρὸς τῇ περιφερείᾳ, καὶ ἔχουσι τὴν αὐτὴν περιφέρειαν βάσιν τὴν ΒΓΔ, ἡ ἄρα ὑπὸ ΒΖΔ γωνία διπλασίων ἐστὶ τῆς ὑπὸ ΒΑΔ. διὰ τὰ αὐτὰ δὴ ἡ ὑπὸ ΒΖΔ καὶ τῆς ὑπὸ ΒΕΔ ἐστι διπλασίων: ἴση ἄρα ἡ ὑπὸ ΒΑΔ τῇ ὑπὸ ΒΕΔ. ἐν κύκλῳ ἄρα αἱ ἐν τῷ αὐτῷ τμήματι γωνίαι ἴσαι ἀλλήλαις εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",115]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'3.22'"],["Rule","'Text'","'The opposite angles of quadrilaterals in circles are equal to two right angles.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'τῶν ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",3],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'Let ABCD be a circle, and let ABCD be a quadrilateral in it; I say that the opposite angles are equal to two right angles. Let AC, BD be joined. Then, since in any triangle the three angles are equal to two right angles, [I. 32] the three angles CAB, ABC, BCA of the triangle ABC are equal to two right angles. But the angle CAB is equal to the angle BDC, for they are in the same segment BADC; [III. 21] and the angle ACB is equal to the angle ADB, for they are in the same segment ADCB; therefore the whole angle ADC is equal to the angles BAC, ACB. Let the angle ABC be added to each; therefore the angles ABC, BAC, ACB are equal to the angles ABC, ADC. But the angles ABC, BAC, ACB are equal to two right angles; therefore the angles ABC, ADC are also equal to two right angles. Similarly we can prove that the angles BAD, DCB are also equal to two right angles.'"],["Rule","'ProofWordCount'",173],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ τετράπλευρον ἔστω τὸ ΑΒΓΔ: λέγω, ὅτι αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἐπεζεύχθωσαν αἱ ΑΓ, ΒΔ. ἐπεὶ οὖν παντὸς τριγώνου αἱ τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν, τοῦ ΑΒΓ ἄρα τριγώνου αἱ τρεῖς γωνίαι αἱ ὑπὸ ΓΑΒ, ΑΒΓ, ΒΓΑ δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἴση δὲ ἡ μὲν ὑπὸ ΓΑΒ τῇ ὑπὸ ΒΔΓ: ἐν γὰρ τῷ αὐτῷ τμήματί εἰσι τῷ ΒΑΔΓ: ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΑΔΒ: ἐν γὰρ τῷ αὐτῷ τμήματί εἰσι τῷ ΑΔΓΒ: ὅλη ἄρα ἡ ὑπὸ ΑΔΓ ταῖς ὑπὸ ΒΑΓ, ΑΓΒ ἴση ἐστίν. κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ: αἱ ἄρα ὑπὸ ΑΒΓ, ΒΑΓ, ΑΓΒ ταῖς ὑπὸ ΑΒΓ, ΑΔΓ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΑΒΓ, ΒΑΓ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. καὶ αἱ ὑπὸ ΑΒΓ, ΑΔΓ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ ΒΑΔ, ΔΓΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. τῶν ἄρα ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",162]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'3.23'"],["Rule","'Text'","'On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ἐπὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων ὅμοια καὶ ἄνισα οὐ συσταθήσεται ἐπὶ τὰ αὐτὰ μέρη.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",16]],["List",["Rule","'Book'",3],["Rule","'Definition'",11]]]],["Rule","'Proof'","'For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB, ADB be constructed on the same side; let ACD be drawn through, and let CB, DB be joined. Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, [III. Def. 11] the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible. [I. 16]'"],["Rule","'ProofWordCount'",80],["Rule","'GreekProof'","'εἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ δύο τμήματα κύκλων ὅμοια καὶ ἄνισα συνεστάτω ἐπὶ τὰ αὐτὰ μέρη τὰ ΑΓΒ, ΑΔΒ, καὶ διήχθω ἡ ΑΓΔ, καὶ ἐπεζεύχθωσαν αἱ ΓΒ, ΔΒ. ἐπεὶ οὖν ὅμοιόν ἐστι τὸ ΑΓΒ τμῆμα τῷ ΑΔΒ τμήματι, ὅμοια δὲ τμήματα κύκλων ἐστὶ τὰ δεχόμενα γωνίας ἴσας, ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΑΔΒ ἡ ἐκτὸς τῇ ἐντός: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐπὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων ὅμοια καὶ ἄνισα συσταθήσεται ἐπὶ τὰ αὐτὰ μέρη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",88]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'3.24'"],["Rule","'Text'","'Similar segments of circles on equal straight lines are equal to one another.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'τὰ ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Theorem'",10]]]],["Rule","'Proof'","'For let AEB, CFD be similar segments of circles on equal straight lines AB, CD; I say that the segment AEB is equal to the segment CFD. For, if the segment AEB be applied to CFD, and if the point A be placed on C and the straight line AB on CD, the point B will also coincide with the point D, because AB is equal to CD; and, AB coinciding with CD, the segment AEB will also coincide with CFD. For, if the straight line AB coincide with CD but the segment AEB do not coincide with CFD, it will either fall with it, or outside it;or it will fall awry, as CGD, and a circle cuts a circle at more points than two: which is impossible. [III. 10] Therefore, if the straight line AB be applied to CD, the segment AEB will not fail to coincide with CFD also; therefore it will coincide with it and will be equal to it.'"],["Rule","'ProofWordCount'",164],["Rule","'GreekProof'","'ἔστωσαν γὰρ ἐπὶ ἴσων εὐθειῶν τῶν ΑΒ, ΓΔ ὅμοια τμήματα κύκλων τὰ ΑΕΒ, ΓΖΔ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΕΒ τμῆμα τῷ ΓΖΔ τμήματι. Ἐφαρμοζομένου γὰρ τοῦ ΑΕΒ τμήματος ἐπὶ τὸ ΓΖΔ καὶ τιθεμένου τοῦ μὲν Α σημείου ἐπὶ τὸ Γ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΓΔ, ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Δ σημεῖον διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΓΔ: τῆς δὲ ΑΒ ἐπὶ τὴν ΓΔ ἐφαρμοσάσης ἐφαρμόσει καὶ τὸ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ. εἰ γὰρ ἡ ΑΒ εὐθεῖα ἐπὶ τὴν ΓΔ ἐφαρμόσει, τὸ δὲ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ μὴ ἐφαρμόσει, ἤτοι ἐντὸς αὐτοῦ πεσεῖται á¼¢ ἐκτὸς á¼¢ παραλλάξει ὡς τὸ ΓΗΔ, καὶ κύκλος κύκλον τέμνει κατὰ πλείονα σημεῖα á¼¢ δύο: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐφαρμοζομένης τῆς ΑΒ εὐθείας ἐπὶ τὴν ΓΔ οὐκ ἐφαρμόσει καὶ τὸ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ: ἐφαρμόσει ἄρα, καὶ ἴσον αὐτῷ ἔσται. τὰ ἄρα ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",159]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'3.25'"],["Rule","'Text'","'Given a segment of a circle, to describe the complete circle of which it is a segment.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'κύκλου τμήματος δοθέντος προσαναγράψαι τὸν κύκλον, οὗπέρ ἐστι τμῆμα.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",3],["Rule","'Theorem'",9]]]],["Rule","'Proof'","'Let ABC be the given segment of a circle; thus it is required to describe the complete circle belonging to the segment ABC, that is, of which it is a segment. For let AC be bisected at D, let DB be drawn from the point D at right angles to AC, and let AB. be joined; the angle ABD is then greater than, equal to, or less than the angle BAD. First let it be greater; and on the straight line BA, and at the point A on it, let the angle BAE be constructed equal to the angle ABD; let DB be drawn through to E, and let EC be joined. Then, since the angle ABE is equal to the angle BAE, the straight line EB is also equal to EA. [I. 6]  And, since AD is equal to DC, and DE is common, the two sides AD, DE are equal to the two sides CD, DE respectively; and the angle ADE is equal to the angle CDE, for each is right; therefore the base AE is equal to the base CE. But AE was proved equal to BE; therefore BE is also equal to CE; therefore the three straight lines AE, EB, EC are equal to one another. Therefore the circle drawn with centre E and distance one of the straight lines AE, EB, EC will also pass through the remaining points and will have been completed. [III. 9] Therefore, given a segment of a circle, the complete circle has been described. And it is manifest that the segment ABC is less than a semicircle, because the centre E happens to be outside it. Similarly, even if the angle ABD be equal to the angle BAD, AD being equal to each of the two BD, DC, the three straight lines DA, DB, DC will be equal to one another, D will be the centre of the completed circle, and ABC will clearly be a semicircle. But, if the angle ABD be less than the angle BAD, and if we construct, on the straight line BA and at the point A on it, an angle equal to the angle ABD, the centre will fall on DB within the segment ABC, and the segment ABC will clearly be greater than a semicircle. Therefore, given a segment of a circle, the complete circle has been described.'"],["Rule","'ProofWordCount'",395],["Rule","'GreekProof'","'ἔστω τὸ δοθὲν τμῆμα κύκλου τὸ ΑΒΓ: δεῖ δὴ τοῦ ΑΒΓ τμήματος προσαναγράψαι τὸν κύκλον, οὗπέρ ἐστι τμῆμα. τετμήσθω γὰρ ἡ ΑΓ δίχα κατὰ τὸ Δ, καὶ ἤχθω ἀπὸ τοῦ Δ σημείου τῇ ΑΓ πρὸς ὀρθὰς ἡ ΔΒ, καὶ ἐπεζεύχθω ἡ ΑΒ: ἡ ὑπὸ ΑΒΔ γωνία ἄρα τῆς ὑπὸ ΒΑΔ ἤτοι μείζων ἐστὶν á¼¢ ἴση á¼¢ ἐλάττων. ἔστω πρότερον μείζων, καὶ συνεστάτω πρὸς τῇ ΒΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΒΔ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΕ, καὶ διήχθω ἡ ΔΒ ἐπὶ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΕΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία τῇ ὑπὸ ΒΑΕ, ἴση ἄρα ἐστὶ καὶ ἡ ΕΒ εὐθεῖα τῇ ΕΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΓ, κοινὴ δὲ ἡ ΔΕ, δύο δὴ αἱ ΑΔ, ΔΕ δύο ταῖς ΓΔ, ΔΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΔΕ γωνίᾳ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση: ὀρθὴ γὰρ ἑκατέρα: βάσις ἄρα ἡ ΑΕ βάσει τῇ ΓΕ ἐστιν ἴση. ἀλλὰ ἡ ΑΕ τῇ ΒΕ ἐδείχθη ἴση: καὶ ἡ ΒΕ ἄρα τῇ ΓΕ ἐστιν ἴση: αἱ τρεῖς ἄρα αἱ ΑΕ, ΕΒ, ΕΓ ἴσαι ἀλλήλαις εἰσίν: ὁ ἄρα κέντρῳ τῷ Ε διαστήματι δὲ ἑνὶ τῶν ΑΕ, ΕΒ, ΕΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται προσαναγεγραμμένος. κύκλου ἄρα τμήματος δοθέντος προσαναγέγραπται ὁ κύκλος. καὶ δῆλον, ὡς τὸ ΑΒΓ τμῆμα ἔλαττόν ἐστιν ἡμικυκλίου διὰ τὸ τὸ Ε κέντρον ἐκτὸς αὐτοῦ τυγχάνειν. ὁμοίως δὲ κἂν ᾖ ἡ ὑπὸ ΑΒΔ γωνία ἴση τῇ ὑπὸ ΒΑΔ, τῆς ΑΔ ἴσης γενομένης ἑκατέρᾳ τῶν ΒΔ, ΔΓ αἱ τρεῖς αἱ ΔΑ, ΔΒ, ΔΓ ἴσαι ἀλλήλαις ἔσονται, καὶ ἔσται τὸ Δ κέντρον τοῦ προσαναπεπληρωμένου κύκλου, καὶ δηλαδὴ ἔσται τὸ ΑΒΓ ἡμικύκλιον. ἐὰν δὲ ἡ ὑπὸ ΑΒΔ ἐλάττων ᾖ τῆς ὑπὸ ΒΑΔ, καὶ συστησώμεθα πρὸς τῇ ΒΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΒΔ γωνίᾳ ἴσην, ἐντὸς τοῦ ΑΒΓ τμήματος πεσεῖται τὸ κέντρον ἐπὶ τῆς ΔΒ, καὶ ἔσται δηλαδὴ τὸ ΑΒΓ τμῆμα μεῖζον ἡμικυκλίου. κύκλου ἄρα τμήματος δοθέντος προσαναγέγραπται ὁ κύκλος: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",331]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",26]],["Association",["Rule","'VertexLabel'","'3.26'"],["Rule","'Text'","'In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἐν τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",3],["Rule","'Definition'",11]],["List",["Rule","'Book'",3],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'Let ABC, DEF be equal circles, and in them let there be equal angles, namely at the centres the angles BGC, EHF, and at the circumferences the angles BAC, EDF; I say that the circumference BKC is equal to the circumference ELF. For let BC, EF be joined. Now, since the circles ABC, DEF are equal, the radii are equal. Thus the two straight lines BG, GC are equal to the two straight lines EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equal to the base EF. [I. 4]  And, since the angle at A is equal to the angle at D, the segment BAC is similar to the segment EDF; [III. Def. 11]and they are upon equal straight lines. But similar segments of circles on equal straight lines are equal to one another; [III. 24] therefore the segment BAC is equal to EDF. But the whole circle ABC is also equal to the whole circle DEF; therefore the circumference BKC which remains is equal to the circumference ELF.'"],["Rule","'ProofWordCount'",181],["Rule","'GreekProof'","'ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ καὶ ἐν αὐτοῖς ἴσαι γωνίαι ἔστωσαν πρὸς μὲν τοῖς κέντροις αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ: λέγω, ὅτι ἴση ἐστὶν ἡ ΒΚΓ περιφέρεια τῇ ΕΛΖ περιφερείᾳ. ἐπεζεύχθωσαν γὰρ αἱ ΒΓ, ΕΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΑΒΓ, ΔΕΖ κύκλοι, ἴσαι εἰσὶν αἱ ἐκ τῶν κέντρων: δύο δὴ αἱ ΒΗ, ΗΓ δύο ταῖς ΕΘ, ΘΖ ἴσαι: καὶ γωνία ἡ πρὸς τῷ η γωνίᾳ τῇ πρὸς τῷ Θ ἴση: βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ πρὸς τῷ Α γωνία τῇ πρὸς τῷ Δ, ὅμοιον ἄρα ἐστὶ τὸ ΒΑΓ τμῆμα τῷ ΕΔΖ τμήματι: καί εἰσιν ἐπὶ ἴσων εὐθειῶν τῶν ΒΓ, ΕΖ: τὰ δὲ ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν: ἴσον ἄρα τὸ ΒΑΓ τμῆμα τῷ ΕΔΖ. ἔστι δὲ καὶ ὅλος ὁ ΑΒΓ κύκλος ὅλῳ τῷ ΔΕΖ κύκλῳ ἴσος: λοιπὴ ἄρα ἡ ΒΚΓ περιφέρεια τῇ ΕΛΖ περιφερείᾳ ἐστὶν ἴση. ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",185]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",27]],["Association",["Rule","'VertexLabel'","'3.27'"],["Rule","'Text'","'In equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'ἐν τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βεβηκυῖαι γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",3],["Rule","'Theorem'",20]],["List",["Rule","'Book'",3],["Rule","'Theorem'",26]]]],["Rule","'Proof'","'For in equal circles ABC, DEF, on equal circumferences BC, EF, let the angles BGC, EHF stand at the centres G, H, and the angles BAC, EDF at the circumferences; I say that the angle BGC is equal to the angle EHF, and the angle BAC is equal to the angle EDF. For, if the angle BGC is unequal to the angle EHF, one of them is greater. Let the angle BGC be greater: and on the straight line BG, and at the point G on it, let the angle BGK be constructed equal to the angle EHF. [I. 23] Now equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference BK is equal to the circumference EF. But EF is equal to BC; therefore BK is also equal to BC, the less to the greater: which is impossible. Therefore the angle BGC is not unequal to the angle EHF; therefore it is equal to it. And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF; [III. 20] therefore the angle at A is also equal to the angle at D.'"],["Rule","'ProofWordCount'",199],["Rule","'GreekProof'","'ἐν γὰρ ἴσοις κύκλοις τοῖς ΑΒΓ, ΔΕΖ ἐπὶ ἴσων περιφερειῶν τῶν ΒΓ, ΕΖ πρὸς μὲν τοῖς Η, Θ κέντροις γωνίαι βεβηκέτωσαν αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ: λέγω, ὅτι ἡ μὲν ὑπὸ ΒΗΓ γωνία τῇ ὑπὸ ΕΘΖ ἐστιν ἴση, ἡ δὲ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ τῇ ὑπὸ ΕΘΖ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΒΗΓ, καὶ συνεστάτω πρὸς τῇ ΒΗ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Η τῇ ὑπὸ ΕΘΖ γωνίᾳ ἴση ἡ ὑπὸ ΒΗΚ: αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ὅταν πρὸς τοῖς κέντροις ὦσιν: ἴση ἄρα ἡ ΒΚ περιφέρεια τῇ ΕΖ περιφερείᾳ. ἀλλὰ ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση: καὶ ἡ ΒΚ ἄρα τῇ ΒΓ ἐστιν ἴση ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ γωνία τῇ ὑπὸ ΕΘΖ: ἴση ἄρα. καί ἐστι τῆς μὲν ὑπὸ ΒΗΓ ἡμίσεια ἡ πρὸς τῷ Α, τῆς δὲ ὑπὸ ΕΘΖ ἡμίσεια ἡ πρὸς τῷ Δ: ἴση ἄρα καὶ ἡ πρὸς τῷ Α γωνία τῇ πρὸς τῷ Δ. ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βεβηκυῖαι γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",211]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",28]],["Association",["Rule","'VertexLabel'","'3.28'"],["Rule","'Text'","'In equal circles equal straight lines cut off equal circumferences, the greater equal to the greater and the less to the less.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐν τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",3],["Rule","'Theorem'",26]]]],["Rule","'Proof'","'Let ABC, DEF be equal circles, and in the circles let AB, DE be equal straight lines cutting off ACB, DFE as greater circumferences and AGB, DHE as lesser; I say that the greater circumference ACB is equal to the greater circumference DFE, and the less circumference AGB to DHE. For let the centres K, L of the circles be taken, and let AK, KB, DL, LE be joined. Now, since the circles are equal, the radii are also equal; therefore the two sides AK, KB are equal to the two sides DL, LE; and the base AB is equal to the base DE; therefore the angle AKB is equal to the angle DLE. [I. 8]  But equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference AGB is equal to DHE. And the whole circle ABC is also equal to the whole circle DEF; therefore the circumference ACB which remains is also equal to the circumference DFE which remains.'"],["Rule","'ProofWordCount'",167],["Rule","'GreekProof'","'ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ ἐν τοῖς κύκλοις ἴσαι εὐθεῖαι ἔστωσαν αἱ ΑΒ, ΔΕ τὰς μὲν ΑΓΒ, ΔΖΕ περιφερείας μείζονας ἀφαιροῦσαι τὰς δὲ ΑΗΒ, ΔΘΕ ἐλάττονας: λέγω, ὅτι ἡ μὲν ΑΓΒ μείζων περιφέρεια ἴση ἐστὶ τῇ ΔΖΕ μείζονι περιφερείᾳ, ἡ δὲ ΑΗΒ ἐλάττων περιφέρεια τῇ ΔΘΕ. εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων τὰ Κ, Λ, καὶ ἐπεζεύχθωσαν αἱ ΑΚ, ΚΒ, ΔΛ, ΛΕ. καὶ ἐπεὶ ἴσοι κύκλοι εἰσίν, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν κέντρων: δύο δὴ αἱ ΑΚ, ΚΒ δυσὶ ταῖς ΔΛ, ΛΕ ἴσαι εἰσίν: καὶ βάσις ἡ ΑΒ βάσει τῇ ΔΕ ἴση: γωνία ἄρα ἡ ὑπὸ ΑΚΒ γωνίᾳ τῇ ὑπὸ ΔΛΕ ἴση ἐστίν. αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ὅταν πρὸς τοῖς κέντροις ὦσιν: ἴση ἄρα ἡ ΑΗΒ περιφέρεια τῇ ΔΘΕ. ἐστὶ δὲ καὶ ὅλος ὁ ΑΒΓ κύκλος ὅλῳ τῷ ΔΕΖ κύκλῳ ἴσος: καὶ λοιπὴ ἄρα ἡ ΑΓΒ περιφέρεια λοιπῇ τῇ ΔΖΕ περιφερείᾳ ἴση ἐστίν. ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",174]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",29]],["Association",["Rule","'VertexLabel'","'3.29'"],["Rule","'Text'","'In equal circles equal circumferences are subtended by equal straight lines.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἐν τοῖς ἴσοις κύκλοις τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",3],["Rule","'Theorem'",27]]]],["Rule","'Proof'","'Let ABC, DEF be equal circles, and in them let equal circumferences BGC, EHF be cut off; and let the straight lines BC, EF be joined; I say that BC is equal to EF. For let the centres of the circles be taken, and let them be K, L; let BK, KC, EL, LF be joined. Now, since the circumference BGC is equal to the circumference EHF, the angle BKC is also equal to the angle ELF. [III. 27]  And, since the circles ABC, DEF are equal, the radii are also equal; therefore the two sides BK, KC are equal to the two sides EL, LF; and they contain equal angles; therefore the base BC is equal to the base EF. [I. 4]'"],["Rule","'ProofWordCount'",123],["Rule","'GreekProof'","'ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ ἐν αὐτοῖς ἴσαι περιφέρειαι ἀπειλήφθωσαν αἱ ΒΗΓ, ΕΘΖ, καὶ ἐπεζεύχθωσαν αἱ ΒΓ, ΕΖ εὐθεῖαι: λέγω, ὅτι ἴση ἐστὶν ἡ ΒΓ τῇ ΕΖ. εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων, καὶ ἔστω τὰ Κ, Λ, καὶ ἐπεζεύχθωσαν αἱ ΒΚ, ΚΓ, ΕΛ, ΛΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΗΓ περιφέρεια τῇ ΕΘΖ περιφερείᾳ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΚΓ τῇ ὑπὸ ΕΛΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΑΒΓ, ΔΕΖ κύκλοι, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν κέντρων: δύο δὴ αἱ ΒΚ, ΚΓ δυσὶ ταῖς ΕΛ, ΛΖ ἴσαι εἰσίν: καὶ γωνίας ἴσας περιέχουσιν: βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν. ἐν ἄρα τοῖς ἴσοις κύκλοις τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",120]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",30]],["Association",["Rule","'VertexLabel'","'3.30'"],["Rule","'Text'","'To bisect a given circumference.'"],["Rule","'TextWordCount'",5],["Rule","'GreekText'","'τὴν δοθεῖσαν περιφέρειαν δίχα τεμεῖν.'"],["Rule","'GreekTextWordCount'",5],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",3],["Rule","'Theorem'",28]]]],["Rule","'Proof'","'Let ADB be the given circumference; thus it is required to bisect the circumference ADB. Let AB be joined and bisected at C; from the point C let CD be drawn at right angles to the straight line AB, and let AD, DB be joined. Then, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD; and the angle ACD is equal to the angle BCD, for each is right; therefore the base AD is equal to the base DB. [I. 4]  But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less; [III. 28] and each of the circumferences AD, DB is less than a semicircle; therefore the circumference AD is equal to the circumference DB. Therefore the given circumference has been bisected at the point D.'"],["Rule","'ProofWordCount'",150],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα περιφέρεια ἡ ΑΔΒ: δεῖ δὴ τὴν ΑΔΒ περιφέρειαν δίχα τεμεῖν. ἐπεζεύχθω ἡ ΑΒ, καὶ τετμήσθω δίχα κατὰ τὸ Γ, καὶ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ ΓΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, κοινὴ δὲ ἡ ΓΔ, δύο δὴ αἱ ΑΓ, ΓΔ δυσὶ ταῖς ΒΓ, ΓΔ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση: ὀρθὴ γὰρ ἑκατέρα: βάσις ἄρα ἡ ΑΔ βάσει τῇ ΔΒ ἴση ἐστίν. αἱ δὲ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι: καί ἐστιν ἑκατέρα τῶν ΑΔ, ΔΒ περιφερειῶν ἐλάττων ἡμικυκλίου: ἴση ἄρα ἡ ΑΔ περιφέρεια τῇ ΔΒ περιφερείᾳ. ἡ ἄρα δοθεῖσα περιφέρεια δίχα τέτμηται κατὰ τὸ Δ σημεῖον: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",132]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",31]],["Association",["Rule","'VertexLabel'","'3.31'"],["Rule","'Text'","'In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle, and the angle of the less segment less than a right angle.'"],["Rule","'TextWordCount'",57],["Rule","'GreekText'","'ἐν κύκλῳ ἡ μὲν ἐν τῷ ἡμικυκλίῳ γωνία ὀρθή ἐστιν, ἡ δὲ ἐν τῷ μείζονι τμήματι ἐλάττων ὀρθῆς, ἡ δὲ ἐν τῷ ἐλάττονι τμήματι μείζων ὀρθῆς: καὶ ἔτι ἡ μὲν τοῦ μείζονος τμήματος γωνία μείζων ἐστὶν ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος τμήματος γωνία ἐλάττων ὀρθῆς.'"],["Rule","'GreekTextWordCount'",45],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",17]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",3],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'Let ABCD be a circle, let BC be its diameter, and E its centre, and let BA, AC, AD, DC be joined; I say that the angle BAC in the semicircle BAC is right, the angle ABC in the segment ABC greater than the semicircle is less than a right angle, and the angle ADC in the segment ADC less than the semicircle is greater than a right angle. Let AE be joined, and let BA be carried through to F. Then, since BE is equal to EA, the angle ABE is also equal to the angle BAE. [I. 5]  Again, since CE is equal to EA, the angle ACE is also equal to the angle CAE. [I. 5]  Therefore the whole angle BAC is equal to the two angles ABC, ACB. But the angle FAC exterior to the triangle ABC is also equal to the two angles ABC, ACB; [I. 32] therefore the angle BAC is also equal to the angle FAC; therefore each is right; [I. Def. 10]therefore the angle BAC in the semicircle BAC is right. Next, since in the triangle ABC the two angles ABC, BAC are less than two right angles, [I. 17] and the angle BAC is a right angle, the angle ABC is less than a right angle; and it is the angle in the segment ABC greater than the semicircle. Next, since ABCD is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [III. 22] while the angle ABC is less than a right angle, therefore the angle ADC which remains is greater than a right angle; and it is the angle in the segment ADC less than the semicircle. I say further that the angle of the greater segment, namely that contained by the circumference ABC and the straight line AC, is greater than a right angle; and the angle of the less segment, namely that contained by the circumference ADC and the straight line AC, is less than a right angle. This is at once manifest. For, since the angle contained by the straight lines BA, AC is right, the angle contained by the circumference ABC and the straight line AC is greater than a right angle. Again, since the angle contained by the straight lines AC, AF is right, the angle contained by the straight line CA and the circumference ADC is less than a right angle.'"],["Rule","'ProofWordCount'",409],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΒΓ, κέντρον δὲ τὸ Ε, καὶ ἐπεζεύχθωσαν αἱ ΒΑ, ΑΓ, ΑΔ, ΔΓ: λέγω, ὅτι ἡ μὲν ἐν τῷ ΒΑΓ ἡμικυκλίῳ γωνία ἡ ὑπὸ ΒΑΓ ὀρθή ἐστιν, ἡ δὲ ἐν τῷ ΑΒΓ μείζονι τοῦ ἡμικυκλίου τμήματι γωνία ἡ ὑπὸ ΑΒΓ ἐλάττων ἐστὶν ὀρθῆς, ἡ δὲ ἐν τῷ ΑΔΓ ἐλάττονι τοῦ ἡμικυκλίου τμήματι γωνία ἡ ὑπὸ ΑΔΓ μείζων ἐστὶν ὀρθῆς. ἐπεζεύχθω ἡ ΑΕ, καὶ διήχθω ἡ ΒΑ ἐπὶ τὸ Ζ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΑ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΒΕ τῇ ὑπὸ ΒΑΕ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΓΕ τῇ ΕΑ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΓΑΕ: ὅλη ἄρα ἡ ὑπὸ ΒΑΓ δυσὶ ταῖς ὑπὸ ΑΒΓ, ΑΓΒ ἴση ἐστίν. ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΑΓ ἐκτὸς τοῦ ΑΒΓ τριγώνου δυσὶ ταῖς ὑπὸ ΑΒΓ, ΑΓΒ γωνίαις ἴση: ἴση ἄρα καὶ ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΖΑΓ: ὀρθὴ ἄρα ἑκατέρα: ἡ ἄρα ἐν τῷ ΒΑΓ ἡμικυκλίῳ γωνία ἡ ὑπὸ ΒΑΓ ὀρθή ἐστιν. καὶ ἐπεὶ τοῦ ΑΒΓ τριγώνου δύο γωνίαι αἱ ὑπὸ ΑΒΓ, ΒΑΓ δύο ὀρθῶν ἐλάττονές εἰσιν, ὀρθὴ δὲ ἡ ὑπὸ ΒΑΓ, ἐλάττων ἄρα ὀρθῆς ἐστιν ἡ ὑπὸ ΑΒΓ γωνία: καί ἐστιν ἐν τῷ ΑΒΓ μείζονι τοῦ ἡμικυκλίου τμήματι. καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΑΒΓΔ, τῶν δὲ ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν αἱ ἄρα ὑπὸ ΑΒΓ, ΑΔΓ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν, καί ἐστιν ἡ ὑπὸ ΑΒΓ ἐλάττων ὀρθῆς: λοιπὴ ἄρα ἡ ὑπὸ ΑΔΓ γωνία μείζων ὀρθῆς ἐστιν: καί ἐστιν ἐν τῷ ΑΔΓ ἐλάττονι τοῦ ἡμικυκλίου τμήματι. λέγω, ὅτι καὶ ἡ μὲν τοῦ μείζονος τμήματος γωνία ἡ περιεχομένη ὑπό τε τῆς ΑΒΓ περιφερείας καὶ τῆς ΑΓ εὐθείας μείζων ἐστὶν ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος τμήματος γωνία ἡ περιεχομένη ὑπό τε τῆς ΑΔΓ περιφερείας καὶ τῆς ΑΓ εὐθείας ἐλάττων ἐστὶν ὀρθῆς. καί ἐστιν αὐτόθεν φανερόν. ἐπεὶ γὰρ ἡ ὑπὸ τῶν ΒΑ, ΑΓ εὐθειῶν ὀρθή ἐστιν, ἡ ἄρα ὑπὸ τῆς ΑΒΓ περιφερείας καὶ τῆς ΑΓ εὐθείας περιεχομένη μείζων ἐστὶν ὀρθῆς. πάλιν, ἐπεὶ ἡ ὑπὸ τῶν ΑΓ, ΑΖ εὐθειῶν ὀρθή ἐστιν, ἡ ἄρα ὑπὸ τῆς ΓΑ εὐθείας καὶ τῆς ΑΔΓ περιφερείας περιεχομένη ἐλάττων ἐστὶν ὀρθῆς. ἐν κύκλῳ ἄρα ἡ μὲν ἐν τῷ ἡμικυκλίῳ γωνία ὀρθή ἐστιν, ἡ δὲ ἐν τῷ μείζονι τμήματι ἐλάττων ὀρθῆς, ἡ δὲ ἐν τῷ ἐλάττονι τμήματι μείζων ὀρθῆς, καὶ ἔτι ἡ μὲν τοῦ μείζονος τμήματος γωνία μείζων ἐστὶν ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος τμήματος γωνία ἐλάττων ὀρθῆς: ὅπερ ἔδει δεῖξαι. πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἡ μία γωνία τριγώνου ταῖς δυσὶν ἴση ᾖ, ὀρθή ἐστιν ἡ γωνία διὰ τὸ καὶ τὴν ἐκείνης ἐκτὸς ταῖς αὐταῖς ἴσην εἶναι: ἐὰν δὲ αἱ ἐφεξῆς ἴσαι ὦσιν, ὀρθαί εἰσιν.'"],["Rule","'GreekProofWordCount'",440]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",32]],["Association",["Rule","'VertexLabel'","'3.32'"],["Rule","'Text'","'If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle.'"],["Rule","'TextWordCount'",47],["Rule","'GreekText'","'ἐὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς εἰς τὸν κύκλον διαχθῇ τις εὐθεῖα τέμνουσα τὸν κύκλον, ἃς ποιεῖ γωνίας πρὸς τῇ ἐφαπτομένῃ, ἴσαι ἔσονται ταῖς ἐν τοῖς ἐναλλὰξ τοῦ κύκλου τμήμασι γωνίαις.'"],["Rule","'GreekTextWordCount'",34],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",3],["Rule","'Theorem'",19]],["List",["Rule","'Book'",3],["Rule","'Theorem'",22]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]]]],["Rule","'Proof'","'For let a straight line EF touch the circle ABCD at the point B, and from the point B let there be drawn across, in the circle ABCD, a straight line BD cutting it; I say that the angles which BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle, that is, that the angle FBD is equal to the angle constructed in the segment BAD, and the angle EBD is equal to the angle constructed in the segment DCB. For let BA be drawn from B at right angles to EF, let a point C be taken at random on the circumference BD, and let AD, DC, CB be joined. Then, since a straight line EF touches the circle ABCD at B, and BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle ABCD is on BA. [III. 19] Therefore BA is a diameter of the circle ABCD; therefore the angle ADB, being an angle in a semicircle, is right. [III. 31]  Therefore the remaining angles BAD, ABD are equal to one right angle. [I. 32] But the angle ABF is also right; therefore the angle ABF is equal to the angles BAD, ABD. Let the angle ABD be subtracted from each; therefore the angle DBF which remains is equal to the angle BAD in the alternate segment of the circle. Next, since ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [III. 22] But the angles DBF, DBE are also equal to two right angles; therefore the angles DBF, DBE are equal to the angles BAD, BCD, of which the angle BAD was proved equal to the angle DBF; therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle.'"],["Rule","'ProofWordCount'",317],["Rule","'GreekProof'","'κύκλου γὰρ τοῦ ΑΒΓΔ ἐφαπτέσθω τις εὐθεῖα ἡ ΕΖ κατὰ τὸ Β σημεῖον, καὶ ἀπὸ τοῦ Β σημείου διήχθω τις εὐθεῖα εἰς τὸν ΑΒΓΔ κύκλον τέμνουσα αὐτὸν ἡ ΒΔ. λέγω, ὅτι ἃς ποιεῖ γωνίας ἡ ΒΔ μετὰ τῆς ΕΖ ἐφαπτομένης, ἴσαι ἔσονται ταῖς ἐν τοῖς ἐναλλὰξ τμήμασι τοῦ κύκλου γωνίαις, τουτέστιν, ὅτι ἡ μὲν ὑπὸ ΖΒΔ γωνία ἴση ἐστὶ τῇ ἐν τῷ ΒΑΔ τμήματι συνισταμένῃ γωνίᾳ, ἡ δὲ ὑπὸ ΕΒΔ γωνία ἴση ἐστὶ τῇ ἐν τῷ ΔΓΒ τμήματι συνισταμένῃ γωνίᾳ. ἤχθω γὰρ ἀπὸ τοῦ Β τῇ ΕΖ πρὸς ὀρθὰς ἡ ΒΑ, καὶ εἰλήφθω ἐπὶ τῆς ΒΔ περιφερείας τυχὸν σημεῖον τὸ Γ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ, ΓΒ. καὶ ἐπεὶ κύκλου τοῦ ΑΒΓΔ ἐφάπτεταί τις εὐθεῖα ἡ ΕΖ κατὰ τὸ Β, καὶ ἀπὸ τῆς ἁφῆς ἦκται τῇ ἐφαπτομένῃ πρὸς ὀρθὰς ἡ ΒΑ, ἐπὶ τῆς ΒΑ ἄρα τὸ κέντρον ἐστὶ τοῦ ΑΒΓΔ κύκλου. ἡ ΒΑ ἄρα διάμετρός ἐστι τοῦ ΑΒΓΔ κύκλου: ἡ ἄρα ὑπὸ ΑΔΒ γωνία ἐν ἡμικυκλίῳ οὖσα ὀρθή ἐστιν. λοιπαὶ ἄρα αἱ ὑπὸ ΒΑΔ, ΑΒΔ μιᾷ ὀρθῇ ἴσαι εἰσίν. ἐστὶ δὲ καὶ ἡ ὑπὸ ΑΒΖ ὀρθή: ἡ ἄρα ὑπὸ ΑΒΖ ἴση ἐστὶ ταῖς ὑπὸ ΒΑΔ, ΑΒΔ. κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΑΒΔ: λοιπὴ ἄρα ἡ ὑπὸ ΔΒΖ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τμήματι τοῦ κύκλου γωνίᾳ τῇ ὑπὸ ΒΑΔ. καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΑΒΓΔ, αἱ ἀπεναντίον αὐτοῦ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. εἰσὶ δὲ καὶ αἱ ὑπὸ ΔΒΖ, ΔΒΕ δυσὶν ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΔΒΖ, ΔΒΕ ταῖς ὑπὸ ΒΑΔ, ΒΓΔ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΒΑΔ τῇ ὑπὸ ΔΒΖ ἐδείχθη ἴση: λοιπὴ ἄρα ἡ ὑπὸ ΔΒΕ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι τῷ ΔΓΒ τῇ ὑπὸ ΔΓΒ γωνίᾳ ἐστὶν ἴση. ἐὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς εἰς τὸν κύκλον διαχθῇ τις εὐθεῖα τέμνουσα τὸν κύκλον, ἃς ποιεῖ γωνίας πρὸς τῇ ἐφαπτομένῃ, ἴσαι ἔσονται ταῖς ἐν τοῖς ἐναλλὰξ τοῦ κύκλου τμήμασι γωνίαις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",314]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",33]],["Association",["Rule","'VertexLabel'","'3.33'"],["Rule","'Text'","'On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'ἐπὶ τῆς δοθείσης εὐθείας γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",3],["Rule","'Theorem'",32]]]],["Rule","'Proof'","'Let AB be the given straight line, and the angle at C the given rectilineal angle; thus it is required to describe on the given straight line AB a segment of a circle admitting an angle equal to the angle at C. The angle at C is then acute, or right, or obtuse. First let it be acute, and, as in the first figure, on the straight line AB, and at the point A, let the angle BAD be constructed equal to the angle at C; therefore the angle BAD is also acute. Let AE be drawn at right angles to DA, let AB be bisected at F, let FG be drawn from the point F at right angles to AB, and let GB be joined. Then, since AF is equal to FB, and FG is common, the two sides AF, FG are equal to the two sides BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base BG. [I. 4]  Therefore the circle described with centre G and distance GA will pass through B also. Let it be drawn, and let it be ABE; let EB be joined. Now, since AD is drawn from A, the extremity of the diameter AE, at right angles to AE, therefore AD touches the circle ABE. [III. 16]  Since then a straight line AD touches the circle ABE, and from the point of contact at A a straight line AB is drawn across in the circle ABE, the angle DAB is equal to the angle AEB in the alternate segment of the circle. [III. 32]  But the angle DAB is equal to the angle at C; therefore the angle at C is also equal to the angle AEB. Therefore on the given straight line AB the segment AEB of a circle has been described admitting the angle AEB equal to the given angle, the angle at C. Next let the angle at C be right; and let it be again required to describe on AB a segment of a circle admitting an angle equal to the right angle at C. Let the angle BAD be constructed equal to the right angle at C, as is the case in the second figure; let AB be bisected at F, and with centre F and distance either FA or FB let the circle AEB be described. Therefore the straight line AD touches the circle ABE, because the angle at A is right. [III. 16] And the angle BAD is equal to the angle in the segment AEB, for the latter too is itself a right angle, being an angle in a semicircle. [III. 31] But the angle BAD is also equal to the angle at C. Therefore the angle AEB is also equal to the angle at C. Therefore again the segment AEB of a circle has been described on AB admitting an angle equal to the angle at C. Next, let the angle at C be obtuse; and on the straight line AB, and at the point A, let the angle BAD be constructed equal to it, as is the case in the third figure; let AE be drawn at right angles to AD, let AB be again bisected at F, let FG be drawn at right angles to AB, and let GB be joined. Then, since AF is again equal to FB, and FG is common, the two sides AF, FG are equal to the two sides BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base BG. [I. 4]  Therefore the circle described with centre G and distance GA will pass through B also; let it so pass, as AEB. Now, since AD is drawn at right angles to the diameter AE from its extremity, AD touches the circle AEB. [III. 16]  And AB has been drawn across from the point of contact at A; therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [III. 32]  But the angle BAD is equal to the angle at C. Therefore the angle in the segment AHB is also equal to the angle at C. Therefore on the given straight line AB the segment AHB of a circle has been described admitting an angle equal to the angle at C.'"],["Rule","'ProofWordCount'",735],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ πρὸς τῷ Γ: δεῖ δὴ ἐπὶ τῆς δοθείσης εὐθείας τῆς ΑΒ γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ. ἡ δὴ πρὸς τῷ Γ γωνία ἤτοι ὀξεῖά ἐστιν á¼¢ ὀρθὴ á¼¢ ἀμβλεῖα: ἔστω πρότερον ὀξεῖα, καὶ ὡς ἐπὶ τῆς πρώτης καταγραφῆς συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ α σημείῳ τῇ πρὸς τῷ Γ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΔ: ὀξεῖα ἄρα ἐστὶ καὶ ἡ ὑπὸ ΒΑΔ. ἤχθω τῇ ΔΑ πρὸς ὀρθὰς ἡ ΑΕ, καὶ τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ ἤχθω ἀπὸ τοῦ Ζ σημείου τῇ ΑΒ πρὸς ὀρθὰς ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΗΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΗ, δύο δὴ αἱ ΑΖ, ΖΗ δύο ταῖς ΒΖ, ΖΗ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΖΗ γωνίᾳ τῇ ὑπὸ ΒΖΗ ἴση: βάσις ἄρα ἡ ΑΗ βάσει τῇ ΒΗ ἴση ἐστίν. ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ τῷ ΗΑ κύκλος γραφόμενος ἥξει καὶ διὰ τοῦ Β. γεγράφθω καὶ ἔστω ὁ ΑΒΕ, καὶ ἐπεζεύχθω ἡ ΕΒ. ἐπεὶ οὖν ἀπ᾽ ἄκρας τῆς ΑΕ διαμέτρου ἀπὸ τοῦ Α τῇ ΑΕ πρὸς ὀρθάς ἐστιν ἡ ΑΔ, ἡ ΑΔ ἄρα ἐφάπτεται τοῦ ΑΒΕ κύκλου: ἐπεὶ οὖν κύκλου τοῦ ΑΒΕ ἐφάπτεταί τις εὐθεῖα ἡ ΑΔ, καὶ ἀπὸ τῆς κατὰ τὸ Α ἁφῆς εἰς τὸν ΑΒΕ κύκλον διῆκταί τις εὐθεῖα ἡ ΑΒ, ἡ ἄρα ὑπὸ ΔΑΒ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ ΑΕΒ. ἀλλ᾽ ἡ ὑπὸ ΔΑΒ τῇ πρὸς τῷ Γ ἐστιν ἴση: καὶ ἡ πρὸς τῷ Γ ἄρα γωνία ἴση ἐστὶ τῇ ὑπὸ ΑΕΒ. ἐπὶ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τμῆμα κύκλου γέγραπται τὸ ΑΕΒ δεχόμενον γωνίαν τὴν ὑπὸ ΑΕΒ ἴσην τῇ δοθείσῃ τῇ πρὸς τῷ Γ. ἀλλὰ δὴ ὀρθὴ ἔστω ἡ πρὸς τῷ Γ: καὶ δέον πάλιν ἔστω ἐπὶ τῆς ΑΒ γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ ὀρθῇ γωνίᾳ. συνεστάτω πάλιν τῇ πρὸς τῷ Γ ὀρθῇ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΔ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ κέντρῳ τῷ Ζ, διαστήματι δὲ ὁποτέρῳ τῶν ΖΑ, ΖΒ, κύκλος γεγράφθω ὁ ΑΕΒ. ἐφάπτεται ἄρα ἡ ΑΔ εὐθεῖα τοῦ ΑΒΕ κύκλου διὰ τὸ ὀρθὴν εἶναι τὴν πρὸς τῷ Α γωνίαν. καὶ ἴση ἐστὶν ἡ ὑπὸ ΒΑΔ γωνία τῇ ἐν τῷ ΑΕΒ τμήματι: ὀρθὴ γὰρ καὶ αὐτὴ ἐν ἡμικυκλίῳ οὖσα. ἀλλὰ καὶ ἡ ὑπὸ ΒΑΔ τῇ πρὸς τῷ Γ ἴση ἐστίν. καὶ ἡ ἐν τῷ ΑΕΒ ἄρα ἴση ἐστὶ τῇ πρὸς τῷ Γ. γέγραπται ἄρα πάλιν ἐπὶ τῆς ΑΒ τμῆμα κύκλου τὸ ΑΕΒ δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ. ἀλλὰ δὴ ἡ πρὸς τῷ Γ ἀμβλεῖα ἔστω: καὶ συνεστάτω αὐτῇ ἴση πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ Α σημείῳ ἡ ὑπὸ ΒΑΔ, ὡς ἔχει ἐπὶ τῆς τρίτης καταγραφῆς, καὶ τῇ ΑΔ πρὸς ὀρθὰς ἤχθω ἡ ΑΕ, καὶ τετμήσθω πάλιν ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΗΒ. καὶ ἐπεὶ πάλιν ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, καὶ κοινὴ ἡ ΖΗ, δύο δὴ αἱ ΑΖ, ΖΗ δύο ταῖς ΒΖ, ΖΗ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΖΗ γωνίᾳ τῇ ὑπὸ ΒΖΗ ἴση: βάσις ἄρα ἡ ΑΗ βάσει τῇ ΒΗ ἴση ἐστίν: ὁ ἄρα κέντρῳ μὲν τῷ η διαστήματι δὲ τῷ ΗΑ κύκλος γραφόμενος ἥξει καὶ διὰ τοῦ Β. ἐρχέσθω ὡς ὁ ΑΕΒ. καὶ ἐπεὶ τῇ ΑΕ διαμέτρῳ ἀπ᾽ ἄκρας πρὸς ὀρθάς ἐστιν ἡ ΑΔ, ἡ ΑΔ ἄρα ἐφάπτεται τοῦ ΑΕΒ κύκλου. καὶ ἀπὸ τῆς κατὰ τὸ Α ἐπαφῆς διῆκται ἡ ΑΒ: ἡ ἄρα ὑπὸ ΒΑΔ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι τῷ ΑΘΒ συνισταμένῃ γωνίᾳ. ἀλλ᾽ ἡ ὑπὸ ΒΑΔ γωνία τῇ πρὸς τῷ Γ ἴση ἐστίν. καὶ ἡ ἐν τῷ ΑΘΒ ἄρα τμήματι γωνία ἴση ἐστὶ τῇ πρὸς τῷ Γ. ἐπὶ τῆς ἄρα δοθείσης εὐθείας τῆς ΑΒ γέγραπται τμῆμα κύκλου τὸ ΑΘΒ δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",656]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",34]],["Association",["Rule","'VertexLabel'","'3.34'"],["Rule","'Text'","'From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'ἀπὸ τοῦ δοθέντος κύκλου τμῆμα ἀφελεῖν δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",3],["Rule","'Theorem'",32]]]],["Rule","'Proof'","'Let ABC be the given circle, and the angle at D the given rectilineal angle; thus it is required to cut off from the circle ABC a segment admitting an angle equal to the given rectilineal angle, the angle at D. Let EF be drawn touching ABC at the point B, and on the straight line FB, and at the point B on it, let the angle FBC be constructed equal to the angle at D. [I. 23] Then, since a straight line EF touches the circle ABC, and BC has been drawn across from the point of contact at B, the angle FBC is equal to the angle constructed in the alternate segment BAC. [III. 32] But the angle FBC is equal to the angle at D; therefore the angle in the segment BAC is equal to the angle at D.  Therefore from the given circle ABC the segment BAC. has been cut off admitting an angle equal to the given rectilineal angle, the angle at D.'"],["Rule","'ProofWordCount'",168],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ πρὸς τῷ Δ: δεῖ δὴ ἀπὸ τοῦ ΑΒΓ κύκλου τμῆμα ἀφελεῖν δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ πρὸς τῷ Δ. ἤχθω τοῦ ΑΒΓ ἐφαπτομένη ἡ ΕΖ κατὰ τὸ Β σημεῖον, καὶ συνεστάτω πρὸς τῇ ΖΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β τῇ πρὸς τῷ Δ γωνίᾳ ἴση ἡ ὑπὸ ΖΒΓ. ἐπεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΕΖ, καὶ ἀπὸ τῆς κατὰ τὸ Β ἐπαφῆς διῆκται ἡ ΒΓ, ἡ ὑπὸ ΖΒΓ ἄρα γωνία ἴση ἐστὶ τῇ ἐν τῷ ΒΑΓ ἐναλλὰξ τμήματι συνισταμένῃ γωνίᾳ. ἀλλ᾽ ἡ ὑπὸ ΖΒΓ τῇ πρὸς τῷ Δ ἐστιν ἴση: καὶ ἡ ἐν τῷ ΒΑΓ ἄρα τμήματι ἴση ἐστὶ τῇ πρὸς τῷ Δ γωνίᾳ. ἀπὸ τοῦ δοθέντος ἄρα κύκλου τοῦ ΑΒΓ τμῆμα ἀφῄρηται τὸ ΒΑΓ δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ πρὸς τῷ Δ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",151]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",35]],["Association",["Rule","'VertexLabel'","'3.35'"],["Rule","'Text'","'If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὸ ὑπὸ τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθογωνίῳ.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",2],["Rule","'Theorem'",5]],["List",["Rule","'Book'",3],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'For in the circle ABCD let the two straight lines AC, BD cut one another at the point E; I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB. If now AC, BD are through the centre, so that E is the centre of the circle ABCD, it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB. Next let AC, DB not be through the centre; let the centre of ABCD be taken, and let it be F; from F let FG, FH be drawn perpendicular to the straight lines AC, DB, and let FB, FC, FE be joined. Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3]therefore AG is equal to GC. Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E, the rectangle contained by AE, EC together with the square on EG is equal to the square on GC; [II. 5] Let the square on GF be added; therefore the rectangle AE, EC together with the squares on GE, GF is equal to the squares on CG, GF. But the square on FE is equal to the squares on EG, GF, and the square on FC is equal to the squares on CG, GF; [I. 47] therefore the rectangle AE, EC together with the square on FE is equal to the square on FC.  And FC is equal to FB; therefore the rectangle AE, EC together with the square on EF is equal to the square on FB. For the same reason, also, the rectangle DE, EB together with the square on FE is equal to the square on FB. But the rectangle AE, EC together with the square on FE was also proved equal to the square on FB; therefore the rectangle AE, EC together with the square on FE is equal to the rectangle DE, EB together with the square on FE. Let the square on FE be subtracted from each; therefore the rectangle contained by AE, EC which remains is equal to the rectangle contained by DE, EB.'"],["Rule","'ProofWordCount'",387],["Rule","'GreekProof'","'ἐν γὰρ κύκλῳ τῷ ΑΒΓΔ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. εἰ μὲν οὖν αἱ ΑΓ, ΒΔ διὰ τοῦ κέντρου εἰσὶν ὥστε τὸ Ε κέντρον εἶναι τοῦ ΑΒΓΔ κύκλου, φανερόν, ὅτι ἴσων οὐσῶν τῶν ΑΕ, ΕΓ, ΔΕ, ΕΒ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. μὴ ἔστωσαν δὴ αἱ ΑΓ, ΔΒ διὰ τοῦ κέντρου, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ, καὶ ἔστω τὸ Ζ, καὶ ἀπὸ τοῦ Ζ ἐπὶ τὰς ΑΓ, ΔΒ εὐθείας κάθετοι ἤχθωσαν αἱ ΖΗ, ΖΘ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΓ, ΖΕ. καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΗΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει: ἴση ἄρα ἡ ΑΗ τῇ ΗΓ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΓ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς δὲ ἄνισα κατὰ τὸ Ε, τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΓ: κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΗΖ: τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τῶν ἀπὸ τῶν ΗΕ, ΗΖ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΓΗ, ΗΖ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΕΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΕ, τοῖς δὲ ἀπὸ τῶν ΓΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΓ: τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΓ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ: τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΕΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΒ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΒ. ἐδείχθη δὲ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον τῷ ἀπὸ τῆς ΖΒ: τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ ἀπὸ τῆς ΖΕ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΖΕ: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. ἐὰν ἄρα ἐν κύκλῳ εὐθεῖαι δύο τέμνωσιν ἀλλήλας, τὸ ὑπὸ τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθογωνίῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",390]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",36]],["Association",["Rule","'VertexLabel'","'3.36'"],["Rule","'Text'","'If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent.'"],["Rule","'TextWordCount'",70],["Rule","'GreekText'","'ἐὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπ᾽ αὐτοῦ πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφαπτομένης τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",50],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",2],["Rule","'Theorem'",6]],["List",["Rule","'Book'",3],["Rule","'Theorem'",3]],["List",["Rule","'Book'",3],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'For let a point D be taken outside the circle ABC, and from D let the two straight lines DCA, DB fall on the circle ABC; let DCA cut the circle ABC and let BD touch it; I say that the rectangle contained by AD, DC is equal to the square on DB. Then DCA is either through the centre or not through the centre. First let it be through the centre, and let F be the centre of the circle ABC; let FB be joined; therefore the angle FBD is right. [III. 18]  And, since AC has been bisected at F, and CD is added to it, the rectangle AD, DC together with the square on FC is equal to the square on FD. [II. 6] But FC is equal to FB; therefore the rectangle AD, DC together with the square on FB is equal to the square on FD. And the squares on FB, BD are equal to the square on FD; [I. 47] therefore the rectangle AD, DC together with the square on FB is equal to the squares on FB, BD. Let the square on FB be subtracted from each; therefore the rectangle AD, DC which remains is equal to the square on the tangent DB. Again, let DCA not be through the centre of the circle ABC; let the centre E be taken, and from E let EF be drawn perpendicular to AC; let EB, EC, ED be joined. Then the angle EBD is right. [III. 18] And, since a straight line EF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3]therefore AF is equal to FC. Now, since the straight line AC has been bisected at the point F, and CD is added to it, the rectangle contained by AD, DC together with the square on FC is equal to the square on FD. [II. 6] Let the square on FE be added to each; therefore the rectangle AD, DC together with the squares on CF, FE is equal to the squares on FD, FE. But the square on EC is equal to the squares on CF, FE, for the angle EFC is right; [I. 47] and the square on ED is equal to the squares on DF, FE; therefore the rectangle AD, DC together with the square on EC is equal to the square on ED. And EC is equal to EB; therefore the rectangle AD, DC together with the square on EB is equal to the square on ED. But the squares on EB, BD are equal to the square on ED, for the angle EBD is right; [I. 47] therefore the rectangle AD, DC together with the square on EB is equal to the squares on EB, BD. Let the square on EB be subtracted from each; therefore the rectangle AD, DC which remains is equal to the square on DB.'"],["Rule","'ProofWordCount'",493],["Rule","'GreekProof'","'κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο εὐθεῖαι αἱ ΔΓΑ, ΔΒ: καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν ΑΒΓ κύκλον, ἡ δὲ ΒΔ ἐφαπτέσθω: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ τετραγώνῳ. ἡ ἄρα ΔΓΑ ἤτοι διὰ τοῦ κέντρου ἐστὶν á¼¢ οὔ. ἔστω πρότερον διὰ τοῦ κέντρου, καὶ ἔστω τὸ Ζ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἐπεζεύχθω ἡ ΖΒ: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΒΔ. καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ δίχα τέτμηται κατὰ τὸ Ζ, πρόσκειται δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. τῷ δὲ ἀπὸ τῆς ΖΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΖΒ, ΒΔ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΒ, ΒΔ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΖΒ: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ ἐφαπτομένης. ἀλλὰ δὴ ἡ ΔΓΑ μὴ ἔστω διὰ τοῦ κέντρου τοῦ ΑΒΓ κύκλου, καὶ εἰλήφθω τὸ κέντρον τὸ Ε, καὶ ἀπὸ τοῦ Ε ἐπὶ τὴν ΑΓ κάθετος ἤχθω ἡ ΕΖ, καὶ ἐπεζεύχθωσαν αἱ ΕΒ, ΕΓ, ΕΔ: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΕΒΔ. καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΕΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει: ἡ ΑΖ ἄρα τῇ ΖΓ ἐστιν ἴση. καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ζ σημεῖον, πρόσκειται δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΖΕ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τῶν ἀπὸ τῶν ΓΖ, ΖΕ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΔ, ΖΕ. τοῖς δὲ ἀπὸ τῶν ΓΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΓ: ὀρθὴ γὰρ ἐστιν ἡ ὑπὸ ΕΖΓ γωνία: τοῖς δὲ ἀπὸ τῶν ΔΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΔ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΔ. ἴση δὲ ἡ ΕΓ τῇ ΕΒ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΔ. τῷ δὲ ἀπὸ τῆς ΕΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΕΒ, ΒΔ: ὀρθὴ γὰρ ἡ ὑπὸ ΕΒΔ γωνία: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΕΒ, ΒΔ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΕΒ: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ. ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπ᾽ αὐτοῦ πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφαπτομένης τετραγώνῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",500]]],["Rule",["Association",["Rule","'Book'",3],["Rule","'Theorem'",37]],["Association",["Rule","'VertexLabel'","'3.37'"],["Rule","'Text'","'If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle.'"],["Rule","'TextWordCount'",90],["Rule","'GreekText'","'ἐὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ προσπίπτῃ, ᾖ δὲ τὸ ὑπὸ τῆς ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς προσπιπτούσης, ἡ προσπίπτουσα ἐφάψεται τοῦ κύκλου.'"],["Rule","'GreekTextWordCount'",56],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]],["List",["Rule","'Book'",3],["Rule","'Theorem'",18]],["List",["Rule","'Book'",3],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'For let a point D be taken outside the circle ABC; from D let the two straight lines DCA, DB fall on the circle ACB; let DCA cut the circle and DB fall on it; and let the rectangle AD, DC be equal to the square on DB. I say that DB touches the circle ABC. For let DE be drawn touching ABC; let the centre of the circle ABC be taken, and let it be F; let FE, FB, FD be joined. Thus the angle FED is right. [III. 18] Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE. [III. 36] But the rectangle AD, DC was also equal to the square on DB; therefore the square on DE is equal to the square on DB; therefore DE is equal to DB. And FE is equal to FB; therefore the two sides DE, EF are equal to the two sides DB, BF; and FD is the common base of the triangles; therefore the angle DEF is equal to the angle DBF. [I. 8] But the angle DEF is right; therefore the angle DBF is also right. And FB produced is a diameter; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16] therefore DB touches the circle. Similarly this can be proved to be the case even if the centre be on AC.'"],["Rule","'ProofWordCount'",249],["Rule","'GreekProof'","'κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο εὐθεῖαι αἱ ΔΓΑ, ΔΒ, καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν κύκλον, ἡ δὲ ΔΒ προσπιπτέτω, ἔστω δὲ τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΔΒ. λέγω, ὅτι ἡ ΔΒ ἐφάπτεται τοῦ ΑΒΓ κύκλου. ἤχθω γὰρ τοῦ ΑΒΓ ἐφαπτομένη ἡ ΔΕ, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΕ, ΖΒ, ΖΔ. ἡ ἄρα ὑπὸ ΖΕΔ ὀρθή ἐστιν. καὶ ἐπεὶ ἡ ΔΕ ἐφάπτεται τοῦ ΑΒΓ κύκλου, τέμνει δὲ ἡ ΔΓΑ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΕ. ἦν δὲ καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΔΒ: τὸ ἄρα ἀπὸ τῆς ΔΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ: ἴση ἄρα ἡ ΔΕ τῇ ΔΒ. ἐστὶ δὲ καὶ ἡ ΖΕ τῇ ΖΒ ἴση: δύο δὴ αἱ ΔΕ, ΕΖ δύο ταῖς ΔΒ, ΒΖ ἴσαι εἰσίν: καὶ βάσις αὐτῶν κοινὴ ἡ ΖΔ: γωνία ἄρα ἡ ὑπὸ ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΒΖ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΔΕΖ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΔΒΖ. καί ἐστιν ἡ ΖΒ ἐκβαλλομένη διάμετρος: ἡ δὲ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου: ἡ ΔΒ ἄρα ἐφάπτεται τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δειχθήσεται, κἂν τὸ κέντρον ἐπὶ τῆς ΑΓ τυγχάνῃ. ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ προσπίπτῃ, ᾖ δὲ τὸ ὑπὸ ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς προσπιπτούσης, ἡ προσπίπτουσα ἐφάψεται τοῦ κύκλου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",283]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'4.1'"],["Rule","'Text'","'Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the circle.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'εἰς τὸν δοθέντα κύκλον τῇ δοθείσῃ εὐθείᾳ μὴ μείζονι οὔσῃ τῆς τοῦ κύκλου διαμέτρου ἴσην εὐθεῖαν ἐναρμόσαι.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List"]],["Rule","'Proof'","'Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle; thus it is required to fit into the circle ABC a straight line equal to the straight line D. Let a diameter BC of the circle ABC be drawn. Then, if BC is equal to D, that which was enjoined will have been done; for BC has been fitted into the circle ABC equal to the straight line D. But, if BC is greater than D, let CE be made equal to D, and with centre C and distance CE let the circle EAF be described; let CA be joined. Then, since the point C is the centre of the circle EAF, CA is equal to CE. But CE is equal to D; therefore D is also equal to CA. Therefore into the given circle ABC there has been fitted CA equal to the given straight line D.'"],["Rule","'ProofWordCount'",158],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα εὐθεῖα μὴ μείζων τῆς τοῦ κύκλου διαμέτρου ἡ Δ. δεῖ δὴ εἰς τὸν ΑΒΓ κύκλον τῇ Δ εὐθείᾳ ἴσην εὐθεῖαν ἐναρμόσαι. ἤχθω τοῦ ΑΒΓ κύκλου διάμετρος ἡ ΒΓ. εἰ μὲν οὖν ἴση ἐστὶν ἡ ΒΓ τῇ Δ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν: ἐνήρμοσται γὰρ εἰς τὸν ΑΒΓ κύκλον τῇ Δ εὐθείᾳ ἴση ἡ ΒΓ. εἰ δὲ μείζων ἐστὶν ἡ ΒΓ τῆς Δ, κείσθω τῇ Δ ἴση ἡ ΓΕ, καὶ κέντρῳ τῷ Γ διαστήματι δὲ τῷ ΓΕ κύκλος γεγράφθω ὁ ΕΑΖ, καὶ ἐπεζεύχθω ἡ ΓΑ. ἐπεὶ οὖν τὸ Γ σημεῖον κέντρον ἐστὶ τοῦ ΕΑΖ κύκλου, ἴση ἐστὶν ἡ ΓΑ τῇ ΓΕ. ἀλλὰ τῇ Δ ἡ ΓΕ ἐστιν ἴση: καὶ ἡ Δ ἄρα τῇ ΓΑ ἐστιν ἴση. εἰς ἄρα τὸν δοθέντα κύκλον τὸν ΑΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ Δ ἴση ἐνήρμοσται ἡ ΓΑ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",143]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'4.2'"],["Rule","'Text'","'In a given circle to inscribe a triangle equiangular with a given triangle.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'εἰς τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον ἐγγράψαι.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]],["List",["Rule","'Book'",3],["Rule","'Theorem'",32]]]],["Rule","'Proof'","'Let ABC be the given circle, and DEF the given triangle; thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF. Let GH be drawn touching the circle ABC at A [III. 16]; on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23] let BC be joined. Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32] But the angle HAC is equal to the angle DEF; therefore the angle ABC is also equal to the angle DEF. For the same reason the angle ACB is also equal to the angle DFE; therefore the remaining angle BAC is also equal to the remaining angle EDF. [I. 32] Therefore in the given circle there has been inscribed a triangle equiangular with the given triangle.'"],["Rule","'ProofWordCount'",206],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τρίγωνον τὸ ΔΕΖ: δεῖ δὴ εἰς τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ ἰσογώνιον τρίγωνον ἐγγράψαι. ἤχθω τοῦ ΑΒΓ κύκλου ἐφαπτομένη ἡ ΗΘ κατὰ τὸ Α, καὶ συνεστάτω πρὸς τῇ ΑΘ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΔΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΘΑΓ, πρὸς δὲ τῇ ΑΗ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΔΖΕ γωνίᾳ ἴση ἡ ὑπὸ ΗΑΒ, καὶ ἐπεζεύχθω ἡ ΒΓ. ἐπεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΑΘ, καὶ ἀπὸ τῆς κατὰ τὸ Α ἐπαφῆς εἰς τὸν κύκλον διῆκται εὐθεῖα ἡ ΑΓ, ἡ ἄρα ὑπὸ ΘΑΓ ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ ΑΒΓ. ἀλλ᾽ ἡ ὑπὸ ΘΑΓ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση: καὶ ἡ ὑπὸ ΑΒΓ ἄρα γωνία τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση: καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΑΓ λοιπῇ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ, καὶ ἐγγέγραπται εἰς τὸν ΑΒΓ κύκλον. εἰς τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",195]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'4.3'"],["Rule","'Text'","'About a given circle to circumscribe a triangle equiangular with a given triangle.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'περὶ τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον περιγράψαι.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",13]],["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",3],["Rule","'Theorem'",1]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]],["List",["Rule","'Book'",3],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let ABC be the given circle, and DEF the given triangle;thus it is required to circumscribe about the circle ABC a triangle equiangular with the triangle DEF. Let EF be produced in both directions to the points G, H, let the centre K of the circle ABC be taken [III. 1], and letthe straight line KB be drawn across at random; on the straight line KB, and at the point K on it, let the angle BKA be constructed equal to the angle DEG, and the angle BKC equal to the angle DFH; [I. 23] and through the points A, B, C let LAM, MBN, NCL bedrawn touching the circle ABC. [III. 16] Now, since LM, MN, NL touch the circle ABC at the points A, B, C, and KA, KB, KC have been joined from the centre K to the points A, B, C,therefore the angles at the points A, B, C are right. [III. 18] And, since the four angles of the quadrilateral AMBK are equal to four right angles, inasmuch as AMBK is in fact divisible into two triangles, and the angles KAM, KBM are right,  therefore the remaining angles AKB, AMB are equal to two right angles. But the angles DEG, DEF are also equal to two right angles; [I. 13] therefore the angles AKB, AMB are equal to the anglesDEG, DEF, of which the angle AKB is equal to the angle DEG; therefore the angle AMB which remains is equal to the angle DEF which remains. Similarly it can be proved that the angle LNB is alsoequal to the angle DFE; therefore the remaining angle MLN is equal to the angle EDF. [I. 32]  Therefore the triangle LMN is equiangular with the triangle DEF; and it has been circumscribed about thecircle ABC. Therefore about a given circle there has been circumscribed a triangle equiangular with the given triangle.'"],["Rule","'ProofWordCount'",315],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τρίγωνον τὸ ΔΕΖ: δεῖ δὴ περὶ τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ ἰσογώνιον τρίγωνον περιγράψαι. Ἐκβεβλήσθω ἡ ΕΖ ἐφ᾽ ἑκάτερα τὰ μέρη κατὰ τὰ Η, Θ σημεῖα, καὶ εἰλήφθω τοῦ ΑΒΓ κύκλου κέντρον τὸ Κ, καὶ διήχθω, ὡς ἔτυχεν, εὐθεῖα ἡ ΚΒ, καὶ συνεστάτω πρὸς τῇ ΚΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Κ τῇ μὲν ὑπὸ ΔΕΗ γωνίᾳ ἴση ἡ ὑπὸ ΒΚΑ, τῇ δὲ ὑπὸ ΔΖΘ ἴση ἡ ὑπὸ ΒΚΓ, καὶ διὰ τῶν Α, Β, Γ σημείων ἤχθωσαν ἐφαπτόμεναι τοῦ ΑΒΓ κύκλου αἱ ΛΑΜ, ΜΒΝ, ΝΓΛ. καὶ ἐπεὶ ἐφάπτονται τοῦ ΑΒΓ κύκλου αἱ ΛΜ, ΜΝ, ΝΛ κατὰ τὰ Α, Β, Γ σημεῖα, ἀπὸ δὲ τοῦ Κ κέντρου ἐπὶ τὰ Α, Β, Γ σημεῖα ἐπεζευγμέναι εἰσὶν αἱ ΚΑ, ΚΒ, ΚΓ, ὀρθαὶ ἄρα εἰσὶν αἱ πρὸς τοῖς Α, Β, Γ σημείοις γωνίαι. καὶ ἐπεὶ τοῦ ΑΜΒΚ τετραπλεύρου αἱ τέσσαρες γωνίαι τέτρασιν ὀρθαῖς ἴσαι εἰσίν, ἐπειδήπερ καὶ εἰς δύο τρίγωνα διαιρεῖται τὸ ΑΜΒΚ, καί εἰσιν ὀρθαὶ αἱ ὑπὸ ΚΑΜ, ΚΒΜ γωνίαι, λοιπαὶ ἄρα αἱ ὑπὸ ΑΚΒ, ΑΜΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. εἰσὶ δὲ καὶ αἱ ὑπὸ ΔΕΗ, ΔΕΖ δυσὶν ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΑΚΒ, ΑΜΒ ταῖς ὑπὸ ΔΕΗ, ΔΕΖ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΑΚΒ τῇ ὑπὸ ΔΕΗ ἐστιν ἴση: λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἡ ὑπὸ ΛΝΒ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση: καὶ λοιπὴ ἄρα ἡ ὑπὸ ΜΛΝ λοιπῇ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΛΜΝ τρίγωνον τῷ ΔΕΖ τριγώνῳ: καὶ περιγέγραπται περὶ τὸν ΑΒΓ κύκλον. περὶ τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον περιγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",275]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'4.4'"],["Rule","'Text'","'In a given triangle to inscribe a circle.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'εἰς τὸ δοθὲν τρίγωνον κύκλον ἐγγράψαι.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",9]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]],["List",["Rule","'Book'",4],["Rule","'Definition'",5]]]],["Rule","'Proof'","'Let ABC be the given triangle; thus it is required to inscribe a circle in the triangle ABC. Let the angles ABC, ACB  be bisected by the straight lines BD, CD [I. 9], and let these meet one another at the point D; from D let DE, DF, DG be drawn perpendicular to the straightlines AB, BC, CA. Now, since the angle ABD is equal to the angle CBD, and the right angle BED is also equal to the right angle BFD,EBD, FBD are two triangles having two angles equal to two angles and one side equal to one side, namely that subtending one of the equal angles, which is BD common to the triangles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26]  therefore DE is equal to DF. For the same reason DG is also equal to DF. Therefore the three straight lines DE, DF, DG are equalto one another; therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will pass also through the remaining points, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right. For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle: which was proved absurd; [III. 16] therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA;  therefore it will touch them, and will be the circle inscribed in the triangle ABC. [IV. Def. 5]  Let it be inscribed, as FGE. Therefore in the given triangle ABC the circle EFG has been inscribed.'"],["Rule","'ProofWordCount'",300],["Rule","'GreekProof'","'ἔστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ: δεῖ δὴ εἰς τὸ ΑΒΓ τρίγωνον κύκλον ἐγγράψαι. τετμήσθωσαν αἱ ὑπὸ ΑΒΓ, ΑΓΒ γωνίαι δίχα ταῖς ΒΔ, ΓΔ εὐθείαις, καὶ συμβαλλέτωσαν ἀλλήλαις κατὰ τὸ Δ σημεῖον, καὶ ἤχθωσαν ἀπὸ τοῦ Δ ἐπὶ τὰς ΑΒ, ΒΓ, ΓΑ εὐθείας κάθετοι αἱ ΔΕ, ΔΖ, ΔΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΑΒΔ γωνία τῇ ὑπὸ ΓΒΔ, ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΒΕΔ ὀρθῇ τῇ ὑπὸ ΒΖΔ ἴση, δύο δὴ τρίγωνά ἐστι τὰ ΕΒΔ, ΖΒΔ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν κοινὴν αὐτῶν τὴν ΒΔ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν: ἴση ἄρα ἡ ΔΕ τῇ ΔΖ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΔΗ τῇ ΔΖ ἐστιν ἴση. αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΔΕ, ΔΖ, ΔΗ ἴσαι ἀλλήλαις εἰσίν: ὁ ἄρα κέντρῳ τῷ Δ καὶ διαστήματι ἑνὶ τῶν Ε, Ζ, Η κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Ε, Ζ, Η σημείοις γωνίας. εἰ γὰρ τεμεῖ αὐτάς, ἔσται ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐντὸς πίπτουσα τοῦ κύκλου: ὅπερ ἄτοπον ἐδείχθη: οὐκ ἄρα ὁ κέντρῳ τῷ Δ διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Η γραφόμενος κύκλος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΑ εὐθείας: ἐφάψεται ἄρα αὐτῶν, καὶ ἔσται ὁ κύκλος ἐγγεγραμμένος εἰς τὸ ΑΒΓ τρίγωνον. ἐγγεγράφθω ὡς ὁ ΖΗΕ. εἰς ἄρα τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλος ἐγγέγραπται ὁ ΕΖΗ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",254]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'4.5'"],["Rule","'Text'","'About a given triangle to circumscribe a circle.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'περὶ τὸ δοθὲν τρίγωνον κύκλον περιγράψαι.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",10]]]],["Rule","'Proof'","'Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC. Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC. First let them meet within at F, and let FB, FC, FA be joined. Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4] Similarly we can prove that CF is also equal to AF; so that FB is also equal to FC; therefore the three straight lines FA, FB, FC are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle ABC. Let it be circumscribed, as ABC. Next, let DF, EF meet on the straight line BC at F, as is the case in the second figure; and let AF be joined. Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC. Again, let DF, EF meet outside the triangle ABC at F, as is the case in the third figure, and let AF, BF, CF be joined. Then again, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I. 4] Similarly we can prove that CF is also equal to AF; so that BF is also equal to FC; therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and will have been circumscribed about the triangle ABC. Therefore about the given triangle a circle has been circumscribed.'"],["Rule","'ProofWordCount'",343],["Rule","'GreekProof'","'ἔστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ: δεῖ δὴ περὶ τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλον περιγράψαι. τετμήσθωσαν αἱ ΑΒ, ΑΓ εὐθεῖαι δίχα κατὰ τὰ Δ, Ε σημεῖα, καὶ ἀπὸ τῶν Δ, Ε σημείων ταῖς ΑΒ, ΑΓ πρὸς ὁρθὰς ἤχθωσαν αἱ ΔΖ, ΕΖ: συμπεσοῦνται δὴ ἤτοι ἐντὸς τοῦ ΑΒΓ τριγώνου á¼¢ ἐπὶ τῆς ΒΓ εὐθείας á¼¢ ἐκτὸς τῆς ΒΓ. συμπιπτέτωσαν πρότερον ἐντὸς κατὰ τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΓ, ΖΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις ἄρα ἡ ΑΖ βάσει τῇ ΖΒ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση: ὥστε καὶ ἡ ΖΒ τῇ ΖΓ ἐστιν ἴση: αἱ τρεῖς ἄρα αἱ ΖΑ, ΖΒ, ΖΓ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν Α, Β, Γ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων, καὶ ἔσται περιγεγραμμένος ὁ κύκλος περὶ τὸ ΑΒΓ τρίγωνον. περιγεγράφθω ὡς ὁ ΑΒΓ. ἀλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐπὶ τῆς ΒΓ εὐθείας κατὰ τὸ Ζ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ ἐπεζεύχθω ἡ ΑΖ. ὁμοίως δὴ δείξομεν, ὅτι τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ περὶ τὸ ΑΒΓ τρίγωνον περιγραφομένου κύκλου. ἀλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐκτὸς τοῦ ΑΒΓ τριγώνου κατὰ τὸ Ζ πάλιν, ὡς ἔχει ἐπὶ τῆς τρίτης καταγραφῆς, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΒΖ, ΓΖ. καὶ ἐπεὶ πάλιν ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις ἄρα ἡ ΑΖ βάσει τῇ ΒΖ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση: ὥστε καὶ ἡ ΒΖ τῇ ΖΓ ἐστιν ἴση: ὁ ἄρα πάλιν κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν ΖΑ, ΖΒ, ΖΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων, καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓ τρίγωνον. περὶ τὸ δοθὲν ἄρα τρίγωνον κύκλος περιγέγραπται: ὅπερ ἔδει ποιῆσαι. Πόρισμα καὶ φανερόν, ὅτι, ὅτε μὲν ἐντὸς τοῦ τριγώνου πίπτει τὸ κέντρον τοῦ κύκλου, ἡ ὑπὸ ΒΑΓ γωνία ἐν μείζονι τμήματι τοῦ ἡμικυκλίου τυγχάνουσα ἐλάττων ἐστὶν ὀρθῆς: ὅτε δὲ ἐπὶ τῆς ΒΓ εὐθείας τὸ κέντρον πίπτει, ἡ ὑπὸ ΒΑΓ γωνία ἐν ἡμικυκλίῳ τυγχάνουσα ὀρθή ἐστιν: ὅτε δὲ τὸ κέντρον τοῦ κύκλου ἐκτὸς τοῦ τριγώνου πίπτει, ἡ ὑπὸ ΒΑΓ ἐν ἐλάττονι τμήματι τοῦ ἡμικυκλίου τυγχάνουσα μείζων ἐστὶν ὀρθῆς. ὥστε καὶ ὅταν ἐλάττων ὀρθῆς τυγχάνῃ ἡ διδομένη γωνία, ἐντὸς τοῦ τριγώνου πεσοῦνται αἱ ΔΖ, ΕΖ, ὅταν δὲ ὀρθή, ἐπὶ τῆς ΒΓ, ὅταν δὲ μείζων ὀρθῆς, ἐκτὸς τῆς ΒΓ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",402]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'4.6'"],["Rule","'Text'","'In a given circle to inscribe a square.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'εἰς τὸν δοθέντα κύκλον τετράγωνον ἐγγράψαι.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",22]],["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]]]],["Rule","'Proof'","'Let ABCD be the given circle; thus it is required to inscribe a square in the circle ABCD. Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and let AB, BC, CD, DA be joined. Then, since BE is equal to ED, for E is the centre, and EA is common and at right angles, therefore the base AB is equal to the base AD. [I. 4] For the same reason each of the straight lines BC, CD is also equal to each of the straight lines AB, AD; therefore the quadrilateral ABCD is equilateral. I say next that it is also right-angled. For, since the straight line BD is a diameter of the circle ABCD, therefore BAD is a semicircle; therefore the angle BAD is right. [III. 31]  For the same reason each of the angles ABC, BCD, CDA is also right; therefore the quadrilateral ABCD is right-angled. But it was also proved equilateral; therefore it is a square; [I. Def. 22] and it has been inscribed in the circle ABCD. Therefore in the given circle the square ABCD has been inscribed.'"],["Rule","'ProofWordCount'",191],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ: δεῖ δὴ εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον ἐγγράψαι. ἤχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΔ: κέντρον γὰρ τὸ Ε: κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΕΑ, βάσις ἄρα ἡ ΑΒ βάσει τῇ ΑΔ ἴση ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ΒΓ, ΓΔ ἑκατέρᾳ τῶν ΑΒ, ΑΔ ἴση ἐστίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔ τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ ἡ ΒΔ εὐθεῖα διάμετρός ἐστι τοῦ ΑΒΓΔ κύκλου, ἡμικύκλιον ἄρα ἐστὶ τὸ ΒΑΔ: ὀρθὴ ἄρα ἡ ὑπὸ ΒΑΔ γωνία. διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΑ ὀρθή ἐστιν: ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔ τετράπλευρον. ἐδείχθη δὲ καὶ ἰσόπλευρον: τετράγωνον ἄρα ἐστίν. καὶ ἐγγέγραπται εἰς τὸν ΑΒΓΔ κύκλον. εἰς ἄρα τὸν δοθέντα κύκλον τετράγωνον ἐγγέγραπται τὸ ΑΒΓΔ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",152]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'4.7'"],["Rule","'Text'","'About a given circle to circumscribe a square.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'περὶ τὸν δοθέντα κύκλον τετράγωνον περιγράψαι.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",28]],["List",["Rule","'Book'",1],["Rule","'Theorem'",30]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]],["List",["Rule","'Book'",3],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let ABCD be the given circle; thus it is required to circumscribe a square about the circle ABCD. Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and through the points A, B, C, D let FG, GH, HK, KF be drawn touching the circle ABCD. [III. 16] Then, since FG touches the circle ABCD, and EA has been joined from the centre E to the point of contact at A, therefore the angles at A are right. [III. 18]  For the same reason the angles at the points B, C, D are also right. Now, since the angle AEB is right, and the angle EBG is also right, therefore GH is parailel to AC. [I. 28]  For the same reason AC is also parallel to FK, so that GH is also parallel to FK. [I. 30]  Similarly we can prove that each of the straight lines GF, HK is parallel to BED. Therefore GK, GC, AK, FB, BK are parallelograms; therefore GF is equal to HK, and GH to FK. [I. 34] And, since AC is equal to BD, and AC is also equal to each of the straight lines GH, FK, while BD is equal to each of the straight lines GF, HK, [I. 34]therefore the quadrilateral FGHK is equilateral. I say next that it is also right-angled. For, since GBEA is a parallelogram, and the angle AEB is right, therefore the angle AGB is also right. [I. 34] Similarly we can prove that the angles at H, K, F are also right. Therefore FGHK is right-angled. But it was also proved equilateral; therefore it is a square; and it has been circumscribed about the circle ABCD. Therefore about the given circle a square has been circumscribed.'"],["Rule","'ProofWordCount'",298],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ: δεῖ δὴ περὶ τὸν ΑΒΓΔ κύκλον τετράγωνον περιγράψαι. ἤχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ διὰ τῶν Α, Β, Γ, Δ σημείων ἤχθωσαν ἐφαπτόμεναι τοῦ ΑΒΓΔ κύκλου αἱ ΖΗ, ΗΘ, ΘΚ, ΚΖ. ἐπεὶ οὖν ἐφάπτεται ἡ ΖΗ τοῦ ΑΒΓΔ κύκλου, ἀπὸ δὲ τοῦ Ε κέντρου ἐπὶ τὴν κατὰ τὸ Α ἐπαφὴν ἐπέζευκται ἡ ΕΑ, αἱ ἄρα πρὸς τῷ Α γωνίαι ὀρθαί εἰσιν. διὰ τὰ αὐτὰ δὴ καὶ αἱ πρὸς τοῖς Β, Γ, Δ σημείοις γωνίαι ὀρθαί εἰσιν. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΑΕΒ γωνία, ἐστὶ δὲ ὀρθὴ καὶ ἡ ὑπὸ ΕΒΗ, παράλληλος ἄρα ἐστὶν ἡ ΗΘ τῇ ΑΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΑΓ τῇ ΖΚ ἐστι παράλληλος. ὥστε καὶ ἡ ΗΘ τῇ ΖΚ ἐστι παράλληλος. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΗΖ, ΘΚ τῇ ΒΕΔ ἐστι παράλληλος. παραλληλόγραμμα ἄρα ἐστὶ τὰ ΗΚ, ΗΓ, ΑΚ, ΖΒ, ΒΚ: ἴση ἄρα ἐστὶν ἡ μὲν ΗΖ τῇ ΘΚ, ἡ δὲ ΗΘ τῇ ΖΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΒΔ, ἀλλὰ καὶ ἡ μὲν ΑΓ ἑκατέρᾳ τῶν ΗΘ, ΖΚ, ἡ δὲ ΒΔ ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση καὶ ἑκατέρα ἄρα τῶν ΗΘ, ΖΚ ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση, ἰσόπλευρον ἄρα ἐστὶ τὸ ΖΗΘΚ τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΗΒΕΑ, καί ἐστιν ὀρθὴ ἡ ὑπὸ ΑΕΒ, ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΗΒ. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ πρὸς τοῖς Θ, Κ, Ζ γωνίαι ὀρθαί εἰσιν. ὀρθογώνιον ἄρα ἐστὶ τὸ ΖΗΘΚ. ἐδείχθη δὲ καὶ ἰσόπλευρον: τετράγωνον ἄρα ἐστίν. καὶ περιγέγραπται περὶ τὸν ΑΒΓΔ κύκλον. περὶ τὸν δοθέντα ἄρα κύκλον τετράγωνον περιγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",274]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'4.8'"],["Rule","'Text'","'In a given square to inscribe a circle.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'εἰς τὸ δοθὲν τετράγωνον κύκλον ἐγγράψαι.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",10]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'Let ABCD be the given square; thus it is required to inscribe a circle in the given square ABCD. Let the straight lines AD, AB be bisected at the points E, F respectively [I. 10], through E let EH be drawn parallel to either AB or CD, and through F let FK be drawn parallel to either AD or BC; [I. 31] therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are evidently equal. [I. 34] Now, since AD is equal to AB, and AE is half of AD, and AF half of AB, therefore AE is equal to AF, so that the opposite sides are also equal;  therefore FG is equal to GE. Similarly we can prove that each of the straight lines GH, GK is equal to each of the straight lines FG, GE; therefore the four straight lines GE, GF, GH, GK are equal to one another. Therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will pass also through the remaining points. And it will touch the straight lines AB, BC, CD, DA, because the angles at E, F, H, K are right. For, if the circle cuts AB, BC, CD, DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle: which was proved absurd; [III. 16] therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will not cut the straight lines AB, BC, CD, DA. Therefore it will touch them, and will have been inscribed in the square ABCD. Therefore in the given square a circle has been inscribed.'"],["Rule","'ProofWordCount'",295],["Rule","'GreekProof'","'ἔστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ: δεῖ δὴ εἰς τὸ ΑΒΓΔ τετράγωνον κύκλον ἐγγράψαι. τετμήσθω ἑκατέρα τῶν ΑΔ, ΑΒ δίχα κατὰ τὰ Ε, Ζ σημεῖα, καὶ διὰ μὲν τοῦ Ε ὁποτέρᾳ τῶν ΑΒ, ΓΔ παράλληλος ἤχθω ἡ ΕΘ, διὰ δὲ τοῦ Ζ ὁποτέρᾳ τῶν ΑΔ, ΒΓ παράλληλος ἤχθω ἡ ΖΚ: παραλληλόγραμμον ἄρα ἐστὶν ἕκαστον τῶν ΑΚ, ΚΒ, ΑΘ, ΘΔ, ΑΗ, ΗΓ, ΒΗ, ΗΔ, καὶ αἱ ἀπεναντίον αὐτῶν πλευραὶ δηλονότι ἴσαι εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΒ, καί ἐστι τῆς μὲν ΑΔ ἡμίσεια ἡ ΑΕ, τῆς δὲ ΑΒ ἡμίσεια ἡ ΑΖ, ἴση ἄρα καὶ ἡ ΑΕ τῇ ΑΖ: ὥστε καὶ αἱ ἀπεναντίον: ἴση ἄρα καὶ ἡ ΖΗ τῇ ΗΕ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΗΘ, ΗΚ ἑκατέρᾳ τῶν ΖΗ, ΗΕ ἐστιν ἴση: αἱ τέσσαρες ἄρα αἱ ΗΕ, ΗΖ, ΗΘ, ΗΚ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Θ, Κ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων: καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Ε, Ζ, Θ, Κ γωνίας: εἰ γὰρ τεμεῖ ὁ κύκλος τὰς ΑΒ, ΒΓ, ΓΔ, ΔΑ, ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐντὸς πεσεῖται τοῦ κύκλου: ὅπερ ἄτοπον ἐδείχθη. οὐκ ἄρα ὁ κέντρῳ τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Θ, Κ κύκλος γραφόμενος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΑ εὐθείας. ἐφάψεται ἄρα αὐτῶν καὶ ἔσται ἐγγεγραμμένος εἰς τὸ ΑΒΓΔ τετράγωνον. εἰς ἄρα τὸ δοθὲν τετράγωνον κύκλος ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",250]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'4.9'"],["Rule","'Text'","'About a given square to circumscribe a circle.'"],["Rule","'TextWordCount'",8],["Rule","'GreekText'","'περὶ τὸ δοθὲν τετράγωνον κύκλον περιγράψαι.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let ABCD be the given square; thus it is required to circumscribe a circle about the square ABCD. For let AC, BD be joined, and let them cut one another at E. Then, since DA is equal to AB, and AC is common, therefore the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8]  Therefore the angle DAB is bisected by AC. Similarly we can prove that each of the angles ABC, BCD, CDA is bisected by the straight lines AC, DB. Now, since the angle DAB is equal to the angle ABC, and the angle EAB is half the angle DAB, and the angle EBA half the angle ABC, therefore the angle EAB is also equal to the angle EBA; so that the side EA is also equal to EB. [I. 6] Similarly we can prove that each of the straight lines EA, EB is equal to each of the straight lines EC, ED. Therefore the four straight lines EA, EB, EC, ED are equal to one another. Therefore the circle described with centre E and distance one of the straight lines EA, EB, EC, ED will pass also through the remaining points; and it will have been circumscribed about the square ABCD. Let it be circumscribed, as ABCD. Therefore about the given square a circle has been circumscribed.'"],["Rule","'ProofWordCount'",245],["Rule","'GreekProof'","'ἔστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ: δεῖ δὴ περὶ τὸ ΑΒΓΔ τετράγωνον κύκλον περιγράψαι. ἐπιζευχθεῖσαι γὰρ αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο δὴ αἱ ΔΑ, ΑΓ δυσὶ ταῖς ΒΑ, ΑΓ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΓ βάσει τῇ ΒΓ ἴση: γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ ἴση ἐστίν: ἡ ἄρα ὑπὸ ΔΑΒ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΓ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΑ δίχα τέτμηται ὑπὸ τῶν ΑΓ, ΔΒ εὐθειῶν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΑΒ γωνία τῇ ὑπὸ ΑΒΓ, καί ἐστι τῆς μὲν ὑπὸ ΔΑΒ ἡμίσεια ἡ ὑπὸ ΕΑΒ, τῆς δὲ ὑπὸ ΑΒΓ ἡμίσεια ἡ ὑπὸ ΕΒΑ, καὶ ἡ ὑπὸ ΕΑΒ ἄρα τῇ ὑπὸ ΕΒΑ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΕΑ τῇ ΕΒ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΕΑ, ΕΒ εὐθειῶν ἑκατέρᾳ τῶν ΕΓ, ΕΔ ἴση ἐστίν. αἱ τέσσαρες ἄρα αἱ ΕΑ, ΕΒ, ΕΓ, ΕΔ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ε καὶ διαστήματι ἑνὶ τῶν Α, Β, Γ, Δ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓΔ τετράγωνον. περιγεγράφθω ὡς ὁ ΑΒΓΔ. περὶ τὸ δοθὲν ἄρα τετράγωνον κύκλος περιγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",211]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'4.10'"],["Rule","'Text'","'To construct an isosceles triangle having each of the angles at the base double of the remaining one.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'ἰσοσκελὲς τρίγωνον συστήσασθαι ἔχον ἑκατέραν τῶν πρὸς τῇ βάσει γωνιῶν διπλασίονα τῆς λοιπῆς.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Times",["Rule","'Book'",2],["Rule","'Book'",3]],["Times",["Rule","'Theorem'",11],["Rule","'Theorem'",32]]],["List",["Rule","'Book'",3],["Rule","'Theorem'",37]],["List",["Rule","'Book'",4],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'Let any straight line AB be set out, and let it be cut at the point C so that the rectangle contained by AB, BC is equal to the square on CA; [II. 11] with centre A and distance AB let the circle BDE be described, and let there be fitted in the circle BDE the straight line BD equal to the straight line AC which is not greater than the diameter of the circle BDE. [IV. 1] Let AD, DC be joined, and let the circle ACD be circumscribed about the triangle ACD. [IV. 5] Then, since the rectangle AB, BC is equal to the square on AC, and AC is equal to BD, therefore the rectangle AB, BC is equal to the square on BD. And, since a point B has been taken outside the circle ACD, and from B the two straight lines BA, BD have fallen on the circle ACD, and one of them cuts it, while the other falls on it, and the rectangle AB, BC is equal to the square on BD, therefore BD touches the circle ACD. [III. 37]  Since, then, BD touches it, and DC is drawn across from the point of contact at D, therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III. 32] Since, then, the angle BDC is equal to the angle DAC, let the angle CDA be added to each; therefore the whole angle BDA is equal to the two angles CDA, DAC. But the exterior angle BCD is equal to the angles CDA, DAC; [I. 32] therefore the angle BDA is also equal to the angle BCD. But the angle BDA is equal to the angle CBD, since the side AD is also equal to AB; [I. 5] so that the angle DBA is also equal to the angle BCD. Therefore the three angles BDA, DBA, BCD are equal to one another. And, since the angle DBC is equal to the angle BCD, the side BD is also equal to the side DC. [I. 6]  But BD is by hypothesis equal to CA; therefore CA is also equal to CD, so that the angle CDA is also equal to the angle DAC; [I. 5] therefore the angles CDA, DAC are double of the angle DAC. But the angle BCD is equal to the angles CDA, DAC; therefore the angle BCD is also double of the angle CAD. But the angle BCD is equal to each of the angles BDA, DBA; therefore each of the angles BDA, DBA is also double of the angle DAB. Therefore the isosceles triangle ABD has been constructed having each of the angles at the base DB double of the remaining one.'"],["Rule","'ProofWordCount'",458],["Rule","'GreekProof'","'Ἐκκείσθω τις εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω κατὰ τὸ Γ σημεῖον, ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ: καὶ κέντρῳ τῷ Α καὶ διαστήματι τῷ ΑΒ κύκλος γεγράφθω ὁ ΒΔΕ, καὶ ἐνηρμόσθω εἰς τὸν ΒΔΕ κύκλον τῇ ΑΓ εὐθείᾳ μὴ μείζονι οὔσῃ τῆς τοῦ ΒΔΕ κύκλου διαμέτρου ἴση εὐθεῖα ἡ ΒΔ: καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ, καὶ περιγεγράφθω περὶ τὸ ΑΓΔ τρίγωνον κύκλος ὁ ΑΓΔ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ, ἴση δὲ ἡ ΑΓ τῇ ΒΔ, τὸ ἄρα ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΔ. καὶ ἐπεὶ κύκλου τοῦ ΑΓΔ εἴληπταί τι σημεῖον ἐκτὸς τὸ Β, καὶ ἀπὸ τοῦ Β πρὸς τὸν ΑΓΔ κύκλον προσπεπτώκασι δύο εὐθεῖαι αἱ ΒΑ, ΒΔ, καὶ ἡ μὲν αὐτῶν τέμνει, ἡ δὲ προσπίπτει, καί ἐστι τὸ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον τῷ ἀπὸ τῆς ΒΔ, ἡ ΒΔ ἄρα ἐφάπτεται τοῦ ΑΓΔ κύκλου. ἐπεὶ οὖν ἐφάπτεται μὲν ἡ ΒΔ, ἀπὸ δὲ τῆς κατὰ τὸ Δ ἐπαφῆς διῆκται ἡ ΔΓ, ἡ ἄρα ὑπὸ ΒΔΓ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ ΔΑΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΒΔΓ τῇ ὑπὸ ΔΑΓ, κοινὴ προσκείσθω ἡ ὑπὸ ΓΔΑ: ὅλη ἄρα ἡ ὑπὸ ΒΔΑ ἴση ἐστὶ δυσὶ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ. ἀλλὰ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ ἴση ἐστὶν ἡ ἐκτὸς ἡ ὑπὸ ΒΓΔ: καὶ ἡ ὑπὸ ΒΔΑ ἄρα ἴση ἐστὶ τῇ ὑπὸ ΒΓΔ. ἀλλὰ ἡ ὑπὸ ΒΔΑ τῇ ὑπὸ ΓΒΔ ἐστιν ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΑΔ τῇ ΑΒ ἐστιν ἴση: ὥστε καὶ ἡ ὑπὸ ΔΒΑ τῇ ὑπὸ ΒΓΔ ἐστιν ἴση. αἱ τρεῖς ἄρα αἱ ὑπὸ ΒΔΑ, ΔΒΑ, ΒΓΔ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΒΓΔ, ἴση ἐστὶ καὶ πλευρὰ ἡ ΒΔ πλευρᾷ τῇ ΔΓ. ἀλλὰ ἡ ΒΔ τῇ ΓΑ ὑπόκειται ἴση: καὶ ἡ ΓΑ ἄρα τῇ ΓΔ ἐστιν ἴση: ὥστε καὶ γωνία ἡ ὑπὸ ΓΔΑ γωνίᾳ τῇ ὑπὸ ΔΑΓ ἐστιν ἴση: αἱ ἄρα ὑπὸ ΓΔΑ, ΔΑΓ τῆς ὑπὸ ΔΑΓ εἰσι διπλασίους. ἴση δὲ ἡ ὑπὸ ΒΓΔ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ: καὶ ἡ ὑπὸ ΒΓΔ ἄρα τῆς ὑπὸ ΓΑΔ ἐστι διπλῆ. ἴση δὲ ἡ ὑπὸ ΒΓΔ ἑκατέρᾳ τῶν ὑπὸ ΒΔΑ, ΔΒΑ: καὶ ἑκατέρα ἄρα τῶν ὑπὸ ΒΔΑ, ΔΒΑ τῆς ὑπὸ ΔΑΒ ἐστι διπλῆ. ἰσοσκελὲς ἄρα τρίγωνον συνέσταται τὸ ΑΒΔ ἔχον ἑκατέραν τῶν πρὸς τῇ ΔΒ βάσει γωνιῶν διπλασίονα τῆς λοιπῆς: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",396]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'4.11'"],["Rule","'Text'","'In a given circle to inscribe an equilateral and equiangular pentagon.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'εἰς τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",9]],["List",["Rule","'Book'",3],["Rule","'Theorem'",26]],["List",["Rule","'Book'",3],["Rule","'Theorem'",27]],["List",["Rule","'Book'",3],["Rule","'Theorem'",29]],["List",["Rule","'Book'",4],["Rule","'Theorem'",2]],["List",["Rule","'Book'",4],["Rule","'Theorem'",10]]]],["Rule","'Proof'","'Let ABCDE be the given circle; thus it is required to inscribe in the circle ABCDE an equilateral and equiangular pentagon. Let the isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F; [IV. 10] let there be inscribed in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angle CAD is equal to the angle at F and the angles at G, H respectively equal to the angles ACD, CDA; [IV. 2] therefore each of the angles ACD, CDA is also double of the angle CAD. Now let the angles ACD, CDA be bisected respectively by the straight lines CE, DB [I. 9], and let AB, BC, DE, EA be joined. Then, since each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another. But equal angles stand on equal circumferences; [III. 26] therefore the five circumferences AB, BC, CD, DE, EA are equal to one another. But equal circumferences are subtended by equal straight lines; [III. 29] therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; therefore the pentagon ABCDE is equilateral. I say next that it is also equiangular. For, since the circumference AB is equal to the circumference DE, let BCD be added to each; therefore the whole circumference ABCD is equal to the whole circumference EDCB. And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is also equal to the angle AED. [III. 27] For the same reason each of the angles ABC, BCD, CDE is also equal to each of the angles BAE, AED; therefore the pentagon ABCDE is equiangular. But it was also proved equilateral; therefore in the given circle an equilateral and equiangular pentagon has been inscribed.'"],["Rule","'ProofWordCount'",334],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ: δεῖ δὴ εἰς τὸν ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι. Ἐκκείσθω τρίγωνον ἰσοσκελὲς τὸ ΖΗΘ διπλασίονα ἔχον ἑκατέραν τῶν πρὸς τοῖς Η, Θ γωνιῶν τῆς πρὸς τῷ Ζ, καὶ ἐγγεγράφθω εἰς τὸν ΑΒΓΔΕ κύκλον τῷ ΖΗΘ τριγώνῳ ἰσογώνιον τρίγωνον τὸ ΑΓΔ, ὥστε τῇ μὲν πρὸς τῷ Ζ γωνίᾳ ἴσην εἶναι τὴν ὑπὸ ΓΑΔ, ἑκατέραν δὲ τῶν πρὸς τοῖς Η, Θ ἴσην ἑκατέρᾳ τῶν ὑπὸ ΑΓΔ, ΓΔΑ: καὶ ἑκατέρα ἄρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ τῆς ὑπὸ ΓΑΔ ἐστι διπλῆ. τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ δίχα ὑπὸ ἑκατέρας τῶν ΓΕ, ΔΒ εὐθειῶν, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ. ἐπεὶ οὖν ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ γωνιῶν διπλασίων ἐστὶ τῆς ὑπὸ ΓΑΔ, καὶ τετμημέναι εἰσὶ δίχα ὑπὸ τῶν ΓΕ, ΔΒ εὐθειῶν, αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΔΑΓ, ΑΓΕ, ΕΓΔ, ΓΔΒ, ΒΔΑ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν: αἱ πέντε ἄρα περιφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν. ὑπὸ δὲ τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν: αἱ πέντε ἄρα εὐθεῖαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἡ ΑΒ περιφέρεια τῇ ΔΕ περιφερείᾳ ἐστὶν ἴση, κοινὴ προσκείσθω ἡ ΒΓΔ: ὅλη ἄρα ἡ ΑΒΓΔ περιφέρεια ὅλῃ τῇ ΕΔΓΒ περιφερείᾳ ἐστὶν ἴση. καὶ βέβηκεν ἐπὶ μὲν τῆς ΑΒΓΔ περιφερείας γωνία ἡ ὑπὸ ΑΕΔ, ἐπὶ δὲ τῆς ΕΔΓΒ περιφερείας γωνία ἡ ὑπὸ ΒΑΕ: καὶ ἡ ὑπὸ ΒΑΕ ἄρα γωνία τῇ ὑπὸ ΑΕΔ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΕ γωνιῶν ἑκατέρᾳ τῶν ὑπὸ ΒΑΕ, ΑΕΔ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον. εἰς ἄρα τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",296]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'4.12'"],["Rule","'Text'","'About a given circle to circumscribe an equilateral and equiangular pentagon.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'περὶ τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγράψαι.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Theorem'",1]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]],["List",["Rule","'Book'",3],["Rule","'Theorem'",18]],["List",["Rule","'Book'",3],["Rule","'Theorem'",27]],["List",["Rule","'Book'",4],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'Let ABCDE be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE. Let A, B, C, D, E be conceived to be the angular points of the inscribed pentagon, so that the circumferences AB, BC, CD, DE, EA are equal; [IV. 11] through A, B, C, D, E let GH, HK, KL, LM, MG be drawn touching the circle; [III. 16] let the centre F of the circle ABCDE be taken [III. 1], and let FB, FK, FC, FL, FD be joined. Then, since the straight line KL touches the circle ABCDE at C, and FC has been joined from the centre F to the point of contact at C, therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right. For the same reason the angles at the points B, D are also right. And, since the angle FCK is right, therefore the square on FK is equal to the squares on FC, CK. For the same reason [I. 47] the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB; therefore the square on CK which remains is equal to the square on BK. Therefore BK is equal to CK. And, since FB is equal to FC, and FK common, the two sides BF, FK are equal to the two sides CF, FK; and the base BK equal to the base CK; therefore the angle BFK is equal to the angle KFC, [I. 8]and the angle BKF to the angle FKC. Therefore the angle BFC is double of the angle KFC, and the angle BKC of the angle FKC. For the same reason the angle CFD is also double of the angle CFL, and the angle DLC of the angle FLC. Now, since the circumference BC is equal to CD, the angle BFC is also equal to the angle CFD. [III. 27] And the angle BFC is double of the angle KFC, and the angle DFC of the angle LFC; therefore the angle KFC is also equal to the angle LFC. But the angle FCK is also equal to the angle FCL; therefore FKC, FLC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [I. 26] therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC. And, since KC is equal to CL, therefore KL is double of KC. For the same reason it can be proved that HK is also double of BK.  And BK is equal to KC; therefore HK is also equal to KL. Similarly each of the straight lines HG, GM, ML can also be proved equal to each of the straight lines HK, KL; therefore the pentagon GHKLM is equilateral. I say next that it is also equiangular. For, since the angle FKC is equal to the angle FLC, and the angle HKL was proved double of the angle FKC, and the angle KLM double of the angle FLC, therefore the angle HKL is also equal to the angle KLM. Similarly each of the angles KHG, HGM, GML can also be proved equal to each of the angles HKL, KLM; therefore the five angles GHK, HKL, KLM, LMG, MGH are equal to one another. Therefore the pentagon GHKLM is equiangular. And it was also proved equilateral; and it has been circumscribed about the circle ABCDE.'"],["Rule","'ProofWordCount'",631],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ: δεῖ δὴ περὶ τὸν ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγράψαι. νενοήσθω τοῦ ἐγγεγραμμένου πενταγώνου τῶν γωνιῶν σημεῖα τὰ Α, Β, Γ, Δ, Ε, ὥστε ἴσας εἶναι τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ περιφερείας: καὶ διὰ τῶν Α, Β, Γ, Δ, Ε ἤχθωσαν τοῦ κύκλου ἐφαπτόμεναι αἱ ΗΘ, ΘΚ, ΚΛ, ΛΜ, ΜΗ, καὶ εἰλήφθω τοῦ ΑΒΓΔΕ κύκλου κέντρον τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΚ, ΖΓ, ΖΛ, ΖΔ. καὶ ἐπεὶ ἡ μὲν ΚΛ εὐθεῖα ἐφάπτεται τοῦ ΑΒΓΔΕ κατὰ τὸ Γ, ἀπὸ δὲ τοῦ Ζ κέντρου ἐπὶ τὴν κατὰ τὸ Γ ἐπαφὴν ἐπέζευκται ἡ ΖΓ, ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΚΛ: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν πρὸς τῷ Γ γωνιῶν. διὰ τὰ αὐτὰ δὴ καὶ αἱ πρὸς τοῖς Β, Δ σημείοις γωνίαι ὀρθαί εἰσιν. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΖΓΚ γωνία, τὸ ἄρα ἀπὸ τῆς ΖΚ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΓ, ΓΚ. διὰ τὰ αὐτὰ δὴ καὶ τοῖς ἀπὸ τῶν ΖΒ, ΒΚ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΚ: ὥστε τὰ ἀπὸ τῶν ΖΓ, ΓΚ τοῖς ἀπὸ τῶν ΖΒ, ΒΚ ἐστιν ἴσα, ὧν τὸ ἀπὸ τῆς ΖΓ τῷ ἀπὸ τῆς ΖΒ ἐστιν ἴσον: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΓΚ τῷ ἀπὸ τῆς ΒΚ ἐστιν ἴσον. ἴση ἄρα ἡ ΒΚ τῇ ΓΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΖΒ τῇ ΖΓ, καὶ κοινὴ ἡ ΖΚ, δύο δὴ αἱ ΒΖ, ΖΚ δυσὶ ταῖς ΓΖ, ΖΚ ἴσαι εἰσίν: καὶ βάσις ἡ ΒΚ βάσει τῇ ΓΚ ἐστιν ἴση: γωνία ἄρα ἡ μὲν ὑπὸ ΒΖΚ γωνίᾳ τῇ ὑπὸ ΚΖΓ ἐστιν ἴση: ἡ δὲ ὑπὸ ΒΚΖ τῇ ὑπὸ ΖΚΓ: διπλῆ ἄρα ἡ μὲν ὑπὸ ΒΖΓ τῆς ὑπὸ ΚΖΓ, ἡ δὲ ὑπὸ ΒΚΓ τῆς ὑπὸ ΖΚΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ μὲν ὑπὸ ΓΖΔ τῆς ὑπὸ ΓΖΛ ἐστι διπλῆ, ἡ δὲ ὑπὸ ΔΛΓ τῆς ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ περιφέρεια τῇ ΓΔ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΖΓ τῇ ὑπὸ ΓΖΔ. καί ἐστιν ἡ μὲν ὑπὸ ΒΖΓ τῆς ὑπὸ ΚΖΓ διπλῆ, ἡ δὲ ὑπὸ ΔΖΓ τῆς ὑπὸ ΛΖΓ: ἴση ἄρα καὶ ἡ ὑπὸ ΚΖΓ τῇ ὑπὸ ΛΖΓ: ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΓΚ γωνία τῇ ὑπὸ ΖΓΛ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΖΚΓ, ΖΛΓ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ: ἴση ἄρα ἡ μὲν ΚΓ εὐθεῖα τῇ ΓΛ, ἡ δὲ ὑπὸ ΖΚΓ γωνία τῇ ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΚΓ τῇ ΓΛ, διπλῆ ἄρα ἡ ΚΛ τῆς ΚΓ. διὰ τὰ αὐτὰ δὴ δειχθήσεται καὶ ἡ ΘΚ τῆς ΒΚ διπλῆ. καί ἐστιν ἡ ΒΚ τῇ ΚΓ ἴση: καὶ ἡ ΘΚ ἄρα τῇ ΚΛ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται καὶ ἑκάστη τῶν ΘΗ, ΗΜ, ΜΛ ἑκατέρᾳ τῶν ΘΚ, ΚΛ ἴση: ἰσόπλευρον ἄρα ἐστὶ τὸ ΗΘΚΛΜ πεντάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΖΚΓ γωνία τῇ ὑπὸ ΖΛΓ, καὶ ἐδείχθη τῆς μὲν ὑπὸ ΖΚΓ διπλῆ ἡ ὑπὸ ΘΚΛ, τῆς δὲ ὑπὸ ΖΛΓ διπλῆ ἡ ὑπὸ ΚΛΜ, καὶ ἡ ὑπὸ ΘΚΛ ἄρα τῇ ὑπὸ ΚΛΜ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται καὶ ἑκάστη τῶν ὑπὸ ΚΘΗ, ΘΗΜ, ΗΜΛ ἑκατέρᾳ τῶν ὑπὸ ΘΚΛ, ΚΛΜ ἴση: αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΗΘΚ, ΘΚΛ, ΚΛΜ, ΛΜΗ, ΜΗΘ ἴσαι ἀλλήλαις εἰσίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΗΘΚΛΜ πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον, καὶ περιγέγραπται περὶ τὸν ΑΒΓΔΕ κύκλον. Περὶ τὸν δοθέντα ἄρα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",576]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'4.13'"],["Rule","'Text'","'In a given pentagon, which is equilateral and equiangular, to inscribe a circle.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'εἰς τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλον ἐγγράψαι.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'Let ABCDE be the given equilateral and equiangular pentagon; thus it is required to inscribe a circle in the pentagon ABCDE. For let the angles BCD, CDE be bisected by the straight lines CF, DF respectively; and from the point F, at which the straight lines CF, DF meet one another, let the straight lines FB, FA, FE be joined. Then, since BC is equal to CD, and CF common, the two sides BC, CF are equal to the two sides DC, CF; and the angle BCF is equal to the angle DCF; therefore the base BF is equal to the base DF, and the triangle BCF is equal to the triangle DCF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle CBF is equal to the angle CDF. And, since the angle CDE is double of the angle CDF, and the angle CDE is equal to the angle ABC, while the angle CDF is equal to the angle CBF; therefore the angle CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle FBC; therefore the angle ABC has been bisected by the straight line BF. Similarly it can be proved that the angles BAE, AED have also been bisected by the straight lines FA, FE respectively. Now let FG, FH, FK, FL, FM be drawn from the point F perpendicular to the straight lines AB, BC, CD, DE, EA. Then, since the angle HCF is equal to the angle KCF, and the right angle FHC is also equal to the angle FKC, FHC, FKC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore the perpendicular FH is equal to the perpendicular FK. Similarly it can be proved that each of the straight lines FL, FM, FG is also equal to each of the straight lines FH, FK; therefore the five straight lines FG, FH, FK, FL, FM are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will pass also through the remaining points; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right. For, if it does not touch them. but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle: which was proved absurd. [III. 16] Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will not cut the straight lines AB, BC, CD, DE, EA; therefore it will touch them. Let it be described, as GHKLM. Therefore in the given pentagon, which is equilateral and equiangular, a circle has been inscribed.'"],["Rule","'ProofWordCount'",519],["Rule","'GreekProof'","'ἔστω τὸ δοθὲν πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον τὸ ΑΒΓΔΕ: δεῖ δὴ εἰς τὸ ΑΒΓΔΕ πεντάγωνον κύκλον ἐγγράψαι. τετμήσθω γὰρ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ εὐθειῶν: καὶ ἀπὸ τοῦ Ζ σημείου, καθ᾽ ὃ συμβάλλουσιν ἀλλήλαις αἱ ΓΖ, ΔΖ εὐθεῖαι, ἐπεζεύχθωσαν αἱ ΖΒ, ΖΑ, ΖΕ εὐθεῖαι. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΓΔ, κοινὴ δὲ ἡ ΓΖ, δύο δὴ αἱ ΒΓ, ΓΖ δυσὶ ταῖς ΔΓ, ΓΖ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΒΓΖ γωνίᾳ τῇ ὑπὸ ΔΓΖ ἐστιν ἴση: βάσις ἄρα ἡ ΒΖ βάσει τῇ ΔΖ ἐστιν ἴση, καὶ τὸ ΒΓΖ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἐστιν ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἡ ὑπὸ ΓΒΖ γωνία τῇ ὑπὸ ΓΔΖ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ὑπὸ ΓΔΕ τῆς ὑπὸ ΓΔΖ, ἴση δὲ ἡ μὲν ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΓ, ἡ δὲ ὑπὸ ΓΔΖ τῇ ὑπὸ ΓΒΖ, καὶ ἡ ὑπὸ ΓΒΑ ἄρα τῆς ὑπὸ ΓΒΖ ἐστι διπλῆ: ἴση ἄρα ἡ ὑπὸ ΑΒΖ γωνία τῇ ὑπὸ ΖΒΓ: ἡ ἄρα ὑπὸ ΑΒΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΒΖ εὐθείας. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκατέρα τῶν ὑπὸ ΒΑΕ, ΑΕΔ δίχα τέτμηται ὑπὸ ἑκατέρας τῶν ΖΑ, ΖΕ εὐθειῶν. ἤχθωσαν δὴ ἀπὸ τοῦ Ζ σημείου ἐπὶ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας κάθετοι αἱ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΘΓΖ γωνία τῇ ὑπὸ ΚΓΖ, ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΖΘΓ ὀρθῇ τῇ ὑπὸ ΖΚΓ ἴση, δύο δὴ τρίγωνά ἐστι τὰ ΖΘΓ, ΖΚΓ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει: ἴση ἄρα ἡ ΖΘ κάθετος τῇ ΖΚ καθέτῳ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΛ, ΖΜ, ΖΗ ἑκατέρᾳ τῶν ΖΘ, ΖΚ ἴση ἐστίν: αἱ πέντε ἄρα εὐθεῖαι αἱ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν Η, Θ, Κ, Λ, Μ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Η, Θ, Κ, Λ, Μ σημείοις γωνίας. εἰ γὰρ οὐκ ἐφάψεται αὐτῶν, ἀλλὰ τεμεῖ αὐτάς, συμβήσεται τὴν τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένην ἐντὸς πίπτειν τοῦ κύκλου: ὅπερ ἄτοπον ἐδείχθη. οὐκ ἄρα ὁ κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν η, Θ, Κ, Λ, Μ σημείων γραφόμενος κύκλος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας: ἐφάψεται ἄρα αὐτῶν. γεγράφθω ὡς ὁ ΗΘΚΛΜ. εἰς ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλος ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",446]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'4.14'"],["Rule","'Text'","'About a given pentagon, which is equilateral and equiangular, to circumscribe a circle.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'περὶ τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλον περιγράψαι.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'Let ABCDE be the given pentagon, which is equilateral and equiangular; thus it is required to circumscribe a circle about the pentagon ABCDE. Let the angles BCD, CDE be bisected by the straight lines CF, DF respectively, and from the point F, at which the straight lines meet, let the straight lines FB, FA, FE be joined to the points B, A, E. Then in manner similar to the preceding it can be proved that the angles CBA, BAE, AED have also been bisected by the straight lines FB, FA, FE respectively. Now, since the angle BCD is equal to the angle CDE, and the angle FCD is half of the angle BCD, and the angle CDF half of the angle CDE, therefore the angle FCD is also equal to the angle CDF, so that the side FC is also equal to the side FD. [I. 6]  Similarly it can be proved that each of the straight lines FB, FA, FE is also equal to each of the straight lines FC, FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC, FD, FE will pass also through the remaining points, and will have been circumscribed. Let it be circumscribed, and let it be ABCDE. Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed.'"],["Rule","'ProofWordCount'",242],["Rule","'GreekProof'","'ἔστω τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, τὸ ΑΒΓΔΕ: δεῖ δὴ περὶ τὸ ΑΒΓΔΕ πεντάγωνον κύκλον περιγράψαι. τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ, καὶ ἀπὸ τοῦ Ζ σημείου, καθ᾽ ὃ συμβάλλουσιν αἱ εὐθεῖαι, ἐπὶ τὰ Β, Α, Ε σημεῖα ἐπεζεύχθωσαν εὐθεῖαι αἱ ΖΒ, ΖΑ, ΖΕ. ὁμοίως δὴ τῷ πρὸ τούτου δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΓΒΑ, ΒΑΕ, ΑΕΔ γωνιῶν δίχα τέτμηται ὑπὸ ἑκάστης τῶν ΖΒ, ΖΑ, ΖΕ εὐθειῶν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΓΔ γωνία τῇ ὑπὸ ΓΔΕ, καί ἐστι τῆς μὲν ὑπὸ ΒΓΔ ἡμίσεια ἡ ὑπὸ ΖΓΔ, τῆς δὲ ὑπὸ ΓΔΕ ἡμίσεια ἡ ὑπὸ ΓΔΖ, καὶ ἡ ὑπὸ ΖΓΔ ἄρα τῇ ὑπὸ ΖΔΓ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΖΓ πλευρᾷ τῇ ΖΔ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΒ, ΖΑ, ΖΕ ἑκατέρᾳ τῶν ΖΓ, ΖΔ ἐστιν ἴση: αἱ πέντε ἄρα εὐθεῖαι αἱ ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ καὶ διαστήματι ἑνὶ τῶν ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται περιγεγραμμένος. περιγεγράφθω καὶ ἔστω ὁ ΑΒΓΔΕ. περὶ ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλος περιγέγραπται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",205]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'4.15'"],["Rule","'Text'","'In a given circle to inscribe an equilateral and equiangular hexagon.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'εἰς τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",15]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",3],["Rule","'Theorem'",26]],["List",["Rule","'Book'",3],["Rule","'Theorem'",27]],["List",["Rule","'Book'",3],["Rule","'Theorem'",29]]]],["Rule","'Proof'","'Let ABCDEF be the given circle; thus it is required to inscribe an equilateral and equiangular hexagon in the circle ABCDEF. Let the diameter AD of the circle ABCDEF be drawn; let the centre G of the circle be taken, and with centre D and distance DG let the circle EGCH be described; let EG, CG be joined and carried through to the points B, F, and let AB, BC, CD, DE, EF, FA be joined. I say that the hexagon ABCDEF is equilateral and equiangular. For, since the point G is the centre of the circle ABCDEF, GE is equal to GD. Again, since the point D is the centre of the circle GCH, DE is equal to DG. But GE was proved equal to GD; therefore GE is also equal to ED; therefore the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, inasmuch as, in isosceles triangles, the angles at the base are equal to one another. [I. 5] And the three angles of the triangle are equal to two right angles; [I. 32] therefore the angle EGD is one-third of two right angles. Similarly, the angle DGC can also be proved to be onethird of two right angles. And, since the straight line CG standing on EB makes the adjacent angles EGC, CGB equal to two right angles, therefore the remaining angle CGB is also one-third of two right angles. Therefore the angles EGD, DGC, CGB are equal to one another; so that the angles vertical to them, the angles BGA, AGF, FGE are equal. [I. 15] Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another. But equal angles stand on equal circumferences; [III. 26] therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another. And equal circumferences are subtended by equal straight lines; [III. 29] therefore the six straight lines are equal to one another; therefore the hexagon ABCDEF is equilateral. I say next that it is also equiangular. For, since the circumference FA is equal to the circumference ED, let the circumference ABCD be added to each; therefore the whole FABCD is equal to the whole EDCBA; and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA; therefore the angle AFE is equal to the angle DEF. [III. 27] Similarly it can be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE, FED; therefore the hexagon ABCDEF is equiangular. But it was also proved equilateral; and it has been inscribed in the circle ABCDEF. Therefore in the given circle an equilateral and equiangular hexagon has been inscribed.'"],["Rule","'ProofWordCount'",459],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕΖ: δεῖ δὴ εἰς τὸν ΑΒΓΔΕΖ κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι. ἤχθω τοῦ ΑΒΓΔΕΖ κύκλου διάμετρος ἡ ΑΔ, καὶ εἰλήφθω τὸ κέντρον τοῦ κύκλου τὸ Η, καὶ κέντρῳ μὲν τῷ Δ διαστήματι δὲ τῷ ΔΗ κύκλος γεγράφθω ὁ ΕΗΓΘ, καὶ ἐπιζευχθεῖσαι αἱ ΕΗ, ΓΗ διήχθωσαν ἐπὶ τὰ Β, Ζ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ: λέγω, ὅτι τὸ ΑΒΓΔΕΖ ἑξάγωνον ἰσόπλευρόν τέ ἐστι καὶ ἰσογώνιον. ἐπεὶ γὰρ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓΔΕΖ κύκλου, ἴση ἐστὶν ἡ ΗΕ τῇ ΗΔ. πάλιν, ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ ΗΓΘ κύκλου, ἴση ἐστὶν ἡ ΔΕ τῇ ΔΗ. ἀλλ᾽ ἡ ΗΕ τῇ ΗΔ ἐδείχθη ἴση: καὶ ἡ ΗΕ ἄρα τῇ ΕΔ ἴση ἐστίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΕΗΔ τρίγωνον: καὶ αἱ τρεῖς ἄρα αὐτοῦ γωνίαι αἱ ὑπὸ ΕΗΔ, ΗΔΕ, ΔΕΗ ἴσαι ἀλλήλαις εἰσίν, ἐπειδήπερ τῶν ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει γωνίαι ἴσαι ἀλλήλαις εἰσίν: καί εἰσιν αἱ τρεῖς τοῦ τριγώνου γωνίαι δυσὶν ὀρθαῖς ἴσαι: ἡ ἄρα ὑπὸ ΕΗΔ γωνία τρίτον ἐστὶ δύο ὀρθῶν. ὁμοίως δὴ δειχθήσεται καὶ ἡ ὑπὸ ΔΗΓ τρίτον δύο ὀρθῶν. καὶ ἐπεὶ ἡ ΓΗ εὐθεῖα ἐπὶ τὴν ΕΒ σταθεῖσα τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΕΗΓ, ΓΗΒ δυσὶν ὀρθαῖς ἴσας ποιεῖ, καὶ λοιπὴ ἄρα ἡ ὑπὸ ΓΗΒ τρίτον ἐστὶ δύο ὀρθῶν: αἱ ἄρα ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ γωνίαι ἴσαι ἀλλήλαις εἰσίν: ὥστε καὶ αἱ κατὰ κορυφὴν αὐταῖς αἱ ὑπὸ ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι εἰσίν ταῖς ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ. αἱ ἓξ ἄρα γωνίαι αἱ ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ, ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν: αἱ ἓξ ἄρα περιφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ ἴσαι ἀλλήλαις εἰσίν. ὑπὸ δὲ τὰς ἴσας περιφερείας αἱ ἴσαι εὐθεῖαι ὑποτείνουσιν: αἱ ἓξ ἄρα εὐθεῖαι ἴσαι ἀλλήλαις εἰσίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ ἑξάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΖΑ περιφέρεια τῇ ΕΔ περιφερείᾳ, κοινὴ προσκείσθω ἡ ΑΒΓΔ περιφέρεια: ὅλη ἄρα ἡ ΖΑΒΓΔ ὅλῃ τῇ ΕΔΓΒΑ ἐστιν ἴση: καὶ βέβηκεν ἐπὶ μὲν τῆς ΖΑΒΓΔ περιφερείας ἡ ὑπὸ ΖΕΔ γωνία, ἐπὶ δὲ τῆς ΕΔΓΒΑ περιφερείας ἡ ὑπὸ ΑΖΕ γωνία: ἴση ἄρα ἡ ὑπὸ ΑΖΕ γωνία τῇ ὑπὸ ΔΕΖ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ λοιπαὶ γωνίαι τοῦ ΑΒΓΔΕΖ ἑξαγώνου κατὰ μίαν ἴσαι εἰσὶν ἑκατέρᾳ τῶν ὑπὸ ΑΖΕ, ΖΕΔ γωνιῶν: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ ἑξάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον: καὶ ἐγγέγραπται εἰς τὸν ΑΒΓΔΕΖ κύκλον. εἰς ἄρα τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἡ τοῦ ἑξαγώνου πλευρὰ ἴση ἐστὶ τῇ ἐκ τοῦ κέντρου τοῦ κύκλου. ὁμοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἀκολούθως τοῖς ἐπὶ τοῦ πενταγώνου εἰρημένοις. καὶ ἔτι διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου εἰρημένοις εἰς τὸ δοθὲν ἑξάγωνον κύκλον ἐγγράψομέν τε καὶ περιγράψομεν: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",483]]],["Rule",["Association",["Rule","'Book'",4],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'4.16'"],["Rule","'Text'","'In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'εἰς τὸν δοθέντα κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Theorem'",30]]]],["Rule","'Proof'","'Let ABCD be the given circle; thus it is required to inscribe in the circle ABCD a fifteenangled figure which shall be both equilateral and equiangular. In the circle ABCD let there be inscribed a side AC of the equilateral triangle inscribed in it, and a side AB of an equilateral pentagon; therefore, of the equal segments of which there are fifteen in the circle ABCD, there will be five in the circumference ABC which is one-third of the circle, and there will be three in the circumference AB which is one-fifth of the circle; therefore in the remainder BC there will be two of the equal segments. Let BC be bisected at E; [III. 30] therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD. If therefore we join BE, EC and fit into the circle ABCD straight lines equal to them and in contiguity, a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it.'"],["Rule","'ProofWordCount'",165],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ: δεῖ δὴ εἰς τὸν ΑΒΓΔ κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι. ἐγγεγράφθω εἰς τὸν ΑΒΓΔ κύκλον τριγώνου μὲν ἰσοπλεύρου τοῦ εἰς αὐτὸν ἐγγραφομένου πλευρὰ ἡ ΑΓ, πενταγώνου δὲ ἰσοπλεύρου ἡ ΑΒ: οἵων ἄρα ἐστὶν ὁ ΑΒΓΔ κύκλος ἴσων τμημάτων δεκαπέντε, τοιούτων ἡ μὲν ΑΒΓ περιφέρεια τρίτον οὖσα τοῦ κύκλου ἔσται πέντε, ἡ δὲ ΑΒ περιφέρεια πέμπτον οὖσα τοῦ κύκλου ἔσται τριῶν: λοιπὴ ἄρα ἡ ΒΓ τῶν ἴσων δύο. τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε: ἑκατέρα ἄρα τῶν ΒΕ, ΕΓ περιφερειῶν πεντεκαιδέκατόν ἐστι τοῦ ΑΒΓΔ κύκλου. ἐὰν ἄρα ἐπιζεύξαντες τὰς ΒΕ, ΕΓ ἴσας αὐταῖς κατὰ τὸ συνεχὲς εὐθείας ἐναρμόσωμεν εἰς τὸν ΑΒΓΔΕ κύκλον, ἔσται εἰς αὐτὸν ἐγγεγραμμένον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον: ὅπερ ἔδει ποιῆσαι. ὁμοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον. ἔτι δὲ διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου δείξεων καὶ εἰς τὸ δοθὲν πεντεκαιδεκάγωνον κύκλον ἐγγράψομέν τε καὶ περιγράψομεν: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",171]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'5.1'"],["Rule","'Text'","'If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted, the remainder will also be the same multiple of the remainder that the whole is of the whole.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα τῶν πάντων.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List"]],["Rule","'Proof'","'Let any number of magnitudes whatever AB, CD be respectively equimultiples of any magnitudes E, F equal in multitude; I say that, whatever multiple AB is of E, that multiple will AB, CD also be of E, F. For, since AB is the same multiple of E that CD is of F, as many magnitudes as there are in AB equal to E, so many also are there in CD equal to F. Let AB be divided into the magnitudes AG, GB equal to E, and CD into CH, HD equal to F; then the multitude of the magnitudes AG, GB will be equal to the multitude of the magnitudes CH, HD. Now, since AG is equal to E, and CH to F, therefore AG is equal to E, and AG, CH to E, F. For the same reason GB is equal to E, and GB, HD to E, F; therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, CD equal to E, F; therefore, whatever multiple AB is of E, that multiple will AB, CD also be of E, F.'"],["Rule","'ProofWordCount'",191],["Rule","'GreekProof'","'ἔστω ὁποσαοῦν μεγέθη τὰ ΑΒ, ΓΔ ὁποσωνοῦν μεγεθῶν τῶν Ε, Ζ ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον: λέγω, ὅτι ὁσαπλάσιόν ἐστι τὸ ΑΒ τοῦ Ε, τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ. ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΓΔ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ Ε, τοσαῦτα καὶ ἐν τῷ ΓΔ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ εἰς τὰ τῷ Ε μεγέθη ἴσα τὰ ΑΗ, ΗΒ, τὸ δὲ ΓΔ εἰς τὰ τῷ Ζ ἴσα τὰ ΓΘ, ΘΔ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει τῶν ΓΘ, ΘΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΗ τῷ Ε, τὸ δὲ ΓΘ τῷ Ζ, ἴσον ἄρα τὸ ΑΗ τῷ Ε, καὶ τὰ ΑΗ, ΓΘ τοῖς Ε, Ζ. διὰ τὰ αὐτὰ δὴ ἴσον ἐστὶ τὸ ΗΒ τῷ ε, καὶ τὰ ΗΒ, ΘΔ τοῖς Ε, Ζ: ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Ε, τοσαῦτα καὶ ἐν τοῖς ΑΒ, ΓΔ ἴσα τοῖς Ε, Ζ: ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΒ τοῦ Ε, τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ. ἐὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα τῶν πάντων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",213]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'5.2'"],["Rule","'Text'","'If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth.'"],["Rule","'TextWordCount'",64],["Rule","'GreekText'","'ἐὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολλαπλάσιον καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τετάρτου.'"],["Rule","'GreekTextWordCount'",33],["Rule","'References'",["List"]],["Rule","'Proof'","'Let a first magnitude, AB, be the same multiple of a second, C, that a third, DE, is of a fourth, F, and let a fifth, BG, also be the same multiple of the second, C, that a sixth, EH, is of the fourth F; I say that the sum of the first and fifth, AG, will be the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F. For, since AB is the same multiple of C that DE is of F, therefore, as many magnitudes as there are in AB equal to C, so many also are there in DE equal to F. For the same reason also, as many as there are in BG equal to C, so many are there also in EH equal to F; therefore, as many as there are in the whole AG equal to C, so many also are there in the whole DH equal to F. Therefore, whatever multiple AG is of C, that multiple also is DH of F. Therefore the sum of the first and fifth, AG, is the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F.'"],["Rule","'ProofWordCount'",210],["Rule","'GreekProof'","'πρῶτον γὰρ τὸ ΑΒ δευτέρου τοῦ Γ ἰσάκις ἔστω πολλαπλάσιον καὶ τρίτον τὸ ΔΕ τετάρτου τοῦ Ζ, ἔστω δὲ καὶ πέμπτον τὸ ΒΗ δευτέρου τοῦ Γ ἰσάκις πολλαπλάσιον καὶ ἕκτον τὸ ΕΘ τετάρτου τοῦ Ζ: λέγω, ὅτι καὶ συντεθὲν πρῶτον καὶ πέμπτον τὸ ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ. ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ ἴσα τῷ Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὅσα ἐστὶν ἐν τῷ ΒΗ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΕΘ ἴσα τῷ Ζ: ὅσα ἄρα ἐστὶν ἐν ὅλῳ τῷ ΑΗ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν ὅλῳ τῷ ΔΘ ἴσα τῷ Ζ: ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΗ τοῦ Γ, τοσαυταπλάσιον ἔσται καὶ τὸ ΔΘ τοῦ Ζ. καὶ συντεθὲν ἄρα πρῶτον καὶ πέμπτον τὸ ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ. ἐὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολλαπλάσιον καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τετάρτου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",204]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'5.3'"],["Rule","'Text'","'If a first magnitude be the same multiple of a second that a third is of a fourth, and if equimultiples be taken of the first and third, then also ex aequali the magnitudes taken will be equimultiples respectively, the one of the second and the other of the fourth.'"],["Rule","'TextWordCount'",50],["Rule","'GreekText'","'ἐὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ληφθῇ δὲ ἰσάκις πολλαπλάσια τοῦ τε πρώτου καὶ τρίτου, καὶ δι᾽ ἴσου τῶν ληφθέντων ἑκάτερον ἑκατέρου ἰσάκις ἔσται πολλαπλάσιον τὸ μὲν τοῦ δευτέρου τὸ δὲ τοῦ τετάρτου.'"],["Rule","'GreekTextWordCount'",37],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let a first magnitude A be the same multiple of a second B that a third C is of a fourth D, and let equimultiples EF, GH be taken of A, C; I say that EF is the same multiple of B that GH is of D. For, since EF is the same multiple of A that GH is of C, therefore, as many magnitudes as there are in EF equal to A, so many also are there in GH equal to C. Let EF be divided into the magnitudes EK, KF equal to A, and GH into the magnitudes GL, LH equal to C; then the multitude of the magnitudes EK, KF will be equal to the multitude of the magnitudes GL, LH. And, since A is the same multiple of B that C is of D, while EK is equal to A, and GL to C, therefore EK is the same multiple of B that GL is of D. For the same reason  KF is the same multiple of B that LH is of D. Since, then, a first magnitude EK is the same multiple of a second B that a third GL is of a fourth D, and a fifth KF is also the same multiple of the second B that a sixth LH is of the fourth D, therefore the sum of the first and fifth, EF, is also the same multiple of the second B that the sum of the third and sixth, GH, is of the fourth D. [V. 2]'"],["Rule","'ProofWordCount'",256],["Rule","'GreekProof'","'πρῶτον γὰρ τὸ Α δευτέρου τοῦ Β ἰσάκις ἔστω πολλαπλάσιον καὶ τρίτον τὸ Γ τετάρτου τοῦ Δ, καὶ εἰλήφθω τῶν Α, Γ ἰσάκις πολλαπλάσια τὰ ΕΖ, ΗΘ: λέγω, ὅτι ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΕΖ τοῦ Β καὶ τὸ ΗΘ τοῦ Δ. ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΕΖ τοῦ Α καὶ τὸ ΗΘ τοῦ Γ, ὅσα ἄρα ἐστὶν ἐν τῷ ΕΖ ἴσα τῷ Α, τοσαῦτα καὶ ἐν τῷ ΗΘ ἴσα τῷ Γ. διῃρήσθω τὸ μὲν ΕΖ εἰς τὰ τῷ Α μεγέθη ἴσα τὰ ΕΚ, ΚΖ, τὸ δὲ ΗΘ εἰς τὰ τῷ Γ ἴσα τὰ ΗΛ, ΛΘ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΕΚ, ΚΖ τῷ πλήθει τῶν ΗΛ, ΛΘ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Α τοῦ Β καὶ τὸ Γ τοῦ Δ, ἴσον δὲ τὸ μὲν ΕΚ τῷ Α, τὸ δὲ ΗΛ τῷ Γ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΚ τοῦ Β καὶ τὸ ΗΛ τοῦ Δ. διὰ τὰ αὐτὰ δὴ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΚΖ τοῦ Β καὶ τὸ ΛΘ τοῦ Δ. ἐπεὶ οὖν πρῶτον τὸ ΕΚ δευτέρου τοῦ Β ἰσάκις ἐστὶ πολλαπλάσιον καὶ τρίτον τὸ ΗΛ τετάρτου τοῦ Δ, ἔστι δὲ καὶ πέμπτον τὸ ΚΖ δευτέρου τοῦ Β ἰσάκις πολλαπλάσιον καὶ ἕκτον τὸ ΛΘ τετάρτου τοῦ Δ, καὶ συντεθὲν ἄρα πρῶτον καὶ πέμπτον τὸ ΕΖ δευτέρου τοῦ Β ἰσάκις ἐστὶ πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΗΘ τετάρτου τοῦ Δ. ἐὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ληφθῇ δὲ τοῦ πρώτου καὶ τρίτου ἰσάκις πολλαπλάσια, καὶ δι᾽ ἴσου τῶν ληφθέντων ἑκάτερον ἑκατέρου ἰσάκις ἔσται πολλαπλάσιον τὸ μὲν τοῦ δευτέρου τὸ δὲ τοῦ τετάρτου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",263]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'5.4'"],["Rule","'Text'","'If a first magnitude have to a second the same ratio as a third to a fourth, any equimultiples whatever of the first and third will also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order.'"],["Rule","'TextWordCount'",45],["Rule","'GreekText'","'ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, καὶ τὰ ἰσάκις πολλαπλάσια τοῦ τε πρώτου καὶ τρίτου πρὸς τὰ ἰσάκις πολλαπλάσια τοῦ δευτέρου καὶ τετάρτου καθ᾽ ὁποιονοῦν πολλαπλασιασμὸν τὸν αὐτὸν ἕξει λόγον ληφθέντα κατάλληλα.'"],["Rule","'GreekTextWordCount'",39],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",5]],["List",["Rule","'Book'",5],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'For let a first magnitude A have to a second B the same ratio as a third C to a fourth D; and let equimultiples E, F be taken of A, C, and G, H other, chance, equimultiples of B, D; I say that, as E is to G, so is F to H. For let equimultiples K, L be taken of E, F, and other, chance, equimultiples M, N of G, H. Since E is the same multiple of A that F is of C, and equimultiples K, L of E, F have been taken, therefore K is the same multiple of A that L is of C. [V. 3] For the same reason M is the same multiple of B that N is of D.  And, since, as A is to B, so is C to D, and of A, C equimultiples K, L have been taken, and of B, D other, chance, equimultiples M, N, therefore, if K is in excess of M, L also is in excess of N, if it is equal, equal, and if less, less. [V. Def. 5] And K, L are equimultiples of E, F, and M, N other, chance, equimultiples of G, H; therefore, as E is to G, so is F to H. [V. Def. 5]'"],["Rule","'ProofWordCount'",216],["Rule","'GreekProof'","'πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, καὶ εἰλήφθω τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Ε, Ζ, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Η, Θ: λέγω, ὅτι ἐστὶν ὡς τὸ Ε πρὸς τὸ Η, οὕτως τὸ Ζ πρὸς τὸ Θ. εἰλήφθω γὰρ τῶν μὲν Ε, Ζ ἰσάκις πολλαπλάσια τὰ Κ, Λ, τῶν δὲ Η, Θ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν. Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ μὲν Ε τοῦ Α, τὸ δὲ Ζ τοῦ Γ, καὶ εἴληπται τῶν Ε, Ζ ἰσάκις πολλαπλάσια τὰ Κ, Λ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ Κ τοῦ α καὶ τὸ Λ τοῦ Γ. διὰ τὰ αὐτὰ δὴ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Μ τοῦ Β καὶ τὸ Ν τοῦ Λ. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Κ, Λ, τῶν δὲ Β, Δ ἄλλα ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Κ τοῦ Μ, ὑπερέχει καὶ τὸ Λ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Κ, Λ τῶν Ε, Ζ ἰσάκις πολλαπλάσια, τὰ δὲ Μ, Ν τῶν Η, Θ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Ε πρὸς τὸ Η, οὕτως τὸ Ζ πρὸς τὸ Θ. ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, καὶ τὰ ἰσάκις πολλαπλάσια τοῦ τε πρώτου καὶ τρίτου πρὸς τὰ ἰσάκις πολλαπλάσια τοῦ δευτέρου καὶ τετάρτου τὸν αὐτὸν ἕξει λόγον καθ᾽ ὁποιονοῦν πολλαπλασιασμὸν ληφθέντα κατάλληλα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",274]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'5.5'"],["Rule","'Text'","'If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted, the remainder will also be the same multiple of the remainder that the whole is of the whole.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, ὅπερ ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι τὸ ὅλον τοῦ ὅλου.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'For let the magnitude AB be the same multiple of the magnitude CD that the part AE subtracted is of the part CF subtracted; I say that the remainder EB is also the same multiple of the remainder FD that the whole AB is of the whole CD. For, whatever multiple AE is of CF, let EB be made that multiple of CG. Then, since AE is the same multiple of CF that EB is of GC, therefore AE is the same multiple of CF that AB is of GF. [V. 1] But, by the assumption, AE is the same multiple of CF that AB is of CD. Therefore AB is the same multiple of each of the magnitudes GF, CD; therefore GF is equal to CD. Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD. And, since AE is the same multiple of CF that EB is of GC, and GC is equal to DF,therefore AE is the same multiple of CF that EB is of FD. But, by hypothesis,  AE is the same multiple of CF that AB is of CD; therefore EB is the same multiple of FD that AB is of CD. That is, the remainder EB will be the same multiple ofthe remainder FD that the whole AB is of the whole CD.'"],["Rule","'ProofWordCount'",226],["Rule","'GreekProof'","'μέγεθος γὰρ τὸ ΑΒ μεγέθους τοῦ ΓΔ ἰσάκις ἔστω πολλαπλάσιον, ὅπερ ἀφαιρεθὲν τὸ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ: λέγω, ὅτι καὶ λοιπὸν τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. ὁσαπλάσιον γάρ ἐστι τὸ ΑΕ τοῦ ΓΖ, τοσαυταπλάσιον γεγονέτω καὶ τὸ ΕΒ τοῦ ΓΗ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΗΓ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΗΖ. κεῖται δὲ ἰσάκις πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ. ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΒ ἑκατέρου τῶν ΗΖ, ΓΔ: ἴσον ἄρα τὸ ΗΖ τῷ ΓΔ. κοινὸν ἀφῃρήσθω τὸ ΓΖ: λοιπὸν ἄρα τὸ ΗΓ λοιπῷ τῷ ΖΔ ἴσον ἐστίν. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΗΓ, ἴσον δὲ τὸ ΗΓ τῷ ΔΖ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΖΔ. ἰσάκις δὲ ὑπόκειται πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΒ τοῦ ΖΔ καὶ τὸ ΑΒ τοῦ ΓΔ. καὶ λοιπὸν ἄρα τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. ἐὰν ἄρα μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, ὅπερ ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι καὶ τὸ ὅλον τοῦ ὅλου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",226]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'5.6'"],["Rule","'Text'","'If two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted from them be equimultiples of the same, the remainders also are either equal to the same or equimultiples of them.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολλαπλάσια, καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολλαπλάσια, καὶ τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν á¼¢ ἰσάκις αὐτῶν πολλαπλάσια.'"],["Rule","'GreekTextWordCount'",28],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'For let two magnitudes AB, CD be equimultiples of two magnitudes E, F, and let AG, CH subtracted from them be equimultiples of the same two E, F; I say that the remainders also, GB, HD, are either equal to E, F or equimultiples of them. For, first, let GB be equal to E; I say that HD is also equal to F. For let CK be made equal to F. Since AG is the same multiple of E that CH is of F, while GB is equal to E and KC to F, therefore AB is the same multiple of E that KH is of F. [V. 2] But, by hypothesis, AB is the same multiple of E that CD is of F; therefore KH is the same multiple of F that CD is of F. Since then each of the magnitudes KH, CD is the same multiple of F, therefore KH is equal to CD. Let CH be subtracted from each; therefore the remainder KC is equal to the remainder HD. But F is equal to KC; therefore HD is also equal to F. Hence, if GB is equal to E, HD is also equal to F. Similarly we can prove that, even if GB be a multiple of E, HD is also the same multiple of F.'"],["Rule","'ProofWordCount'",220],["Rule","'GreekProof'","'δύο γὰρ μεγέθη τὰ ΑΒ, ΓΔ δύο μεγεθῶν τῶν Ε, Ζ ἰσάκις ἔστω πολλαπλάσια, καὶ ἀφαιρεθέντα τὰ ΑΗ, ΓΘ τῶν αὐτῶν τῶν Ε, Ζ ἰσάκις ἔστω πολλαπλάσια: λέγω, ὅτι καὶ λοιπὰ τὰ ΗΒ, ΘΔ τοῖς Ε, Ζ ἤτοι ἴσα ἐστὶν á¼¢ ἰσάκις αὐτῶν πολλαπλάσια. ἔστω γὰρ πρότερον τὸ ΗΒ τῷ Ε ἴσον. λέγω, ὅτι καὶ τὸ ΘΔ τῷ Ζ ἴσον ἐστίν. κείσθω γὰρ τῷ Ζ ἴσον τὸ ΓΚ. ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΗ τοῦ Ε καὶ τὸ ΓΘ τοῦ Ζ, ἴσον δὲ τὸ μὲν ΗΒ τῷ Ε, τὸ δὲ ΚΓ τῷ Ζ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΚΘ τοῦ Ζ. ἰσάκις δὲ ὑπόκειται πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΓΔ τοῦ Ζ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΚΘ τοῦ Ζ καὶ τὸ ΓΔ τοῦ Ζ. ἐπεὶ οὖν ἑκάτερον τῶν ΚΘ, ΓΔ τοῦ Ζ ἰσάκις ἐστὶ πολλαπλάσιον, ἴσον ἄρα ἐστὶ τὸ ΚΘ τῷ ΓΔ. κοινὸν ἀφῃρήσθω τὸ ΓΘ: λοιπὸν ἄρα τὸ ΚΓ λοιπῷ τῷ ΘΔ ἴσον ἐστίν. ἀλλὰ τὸ Ζ τῷ ΚΓ ἐστιν ἴσον: καὶ τὸ ΘΔ ἄρα τῷ Ζ ἴσον ἐστίν. ὥστε εἰ τὸ ΗΒ τῷ Ε ἴσον ἐστίν, καὶ τὸ ΘΔ ἴσον ἔσται τῷ Ζ. ὁμοίως δὴ δείξομεν, ὅτι, κἂν πολλαπλάσιον ᾖ τὸ ΗΒ τοῦ Ε, τοσαυταπλάσιον ἔσται καὶ τὸ ΘΔ τοῦ Ζ. ἐὰν ἄρα δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολλαπλάσια, καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολλαπλάσια, καὶ τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν á¼¢ ἰσάκις αὐτῶν πολλαπλάσια: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",243]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'5.7'"],["Rule","'Text'","'Equal magnitudes have to the same the same ratio, as also has the same to equal magnitudes.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'τὰ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ αὐτὸ πρὸς τὰ ἴσα.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",5]]]],["Rule","'Proof'","'Let A, B be equal magnitudes and C any other, chance, magnitude; I say that each of the magnitudes A, B has the same ratio to C, and C has the same ratio to each of the magnitudes A, B. For let equimultiples D, E of A, B be taken, and of C another, chance, multiple F. Then, since D is the same multiple of A that E is of B, while A is equal to B, therefore D is equal to E.  But F is another, chance, magnitude. If therefore D is in excess of F, E is also in excess of F, if equal to it, equal; and, if less, less. And D, E are equimultiples of A, B, while F is another, chance, multiple of C; therefore, as A is to C, so is B to C. [V. Def. 5]  I say next that C also has the same ratio to each of the magnitudes A, B. For, with the same construction, we can prove similarly that D is equal to E; and F is some other magnitude. If therefore F is in excess of D, it is also in excess of E, if equal, equal; and, if less, less. And F is a multiple of C, while D, E are other, chance, equimultiples of A, B; therefore, as C is to A, so is C to B. [V. Def. 5]'"],["Rule","'ProofWordCount'",234],["Rule","'GreekProof'","'ἔστω ἴσα μεγέθη τὰ Α, Β, ἄλλο δέ τι, ὃ ἔτυχεν, μέγεθος τὸ Γ: λέγω, ὅτι ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν ἔχει λόγον, καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β. εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Δ, Ε, τοῦ δὲ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον τὸ Ζ. ἐπεὶ οὖν ἰσάκις ἐστὶ πολλαπλάσιον τὸ Δ τοῦ Α καὶ τὸ Ε τοῦ Β, ἴσον δὲ τὸ Α τῷ Β, ἴσον ἄρα καὶ τὸ Δ τῷ Ε. ἄλλο δέ, ὃ ἔτυχεν, τὸ Ζ. εἰ ἄρα ὑπερέχει τὸ Δ τοῦ Ζ, ὑπερέχει καὶ τὸ Ε τοῦ Ζ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Δ, Ε τῶν Α, Β ἰσάκις πολλαπλάσια, τὸ δὲ Ζ τοῦ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον: ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Γ. λέγω δή, ὅτι καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν ἔχει λόγον. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἴσον ἐστὶ τὸ Δ τῷ Ε: ἄλλο δέ τι τὸ Ζ: εἰ ἄρα ὑπερέχει τὸ Ζ τοῦ Δ, ὑπερέχει καὶ τοῦ Ε, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὸ μὲν Ζ τοῦ Γ πολλαπλάσιον, τὰ δὲ Δ, Ε τῶν Α, Β ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως τὸ Γ πρὸς τὸ Β. τὰ ἴσα ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ αὐτὸ πρὸς τὰ ἴσα. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν μεγέθη τινὰ ἀνάλογον ᾖ, καὶ ἀνάπαλιν ἀνάλογον ἔσται. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",263]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'5.8'"],["Rule","'Text'","'Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'τῶν ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",4]],["List",["Rule","'Book'",5],["Rule","'Definition'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let AB, C be unequal magnitudes, and let AB be greater; let D be another, chance, magnitude; I say that AB has to D a greater ratio than C has to D, and D has to C a greater ratio than it has to AB. For, since AB is greater than C, let BE be made equal to C; then the less of the magnitudes AE, EB, if multiplied, will sometime be greater than D. [V. Def. 4] [Case I.] First, let AE be less than EB; let AE be multiplied, and let FG be a multiple of it which is greater than D; then, whatever multiple FG is of AE, let GH be made the same multiple of EB and K of C; and let L be taken double of D, M triple of it, and successive multiples increasing by one, until what is taken is a multiple of D and the first that is greater than K. Let it be taken, and let it be N which is quadruple of D and the first multiple of it that is greather than K. Then, since K is less than N first, therefore K is not less than M. And, since FG is the same multiple of AE that GH is of EB, therefore FG is the same multiple of AE that FH is of AB. [V. 1] But FG is the same multiple of AE that K is of C; therefore FH is the same multiple of AB that K is of C; therefore FH, K are equimultiples of AB, C.  Again, since GH is the same multiple of EB that K is of C, and EB is equal to C, therefore GH is equal to K.  But K is not less than M; therefore neither is GH less than M.  And FG is greater than D; therefore the whole FH is greater than D, M together. But D, M together are equal to N, inasmuch as M is triple of D, and M, D together are quadruple of D, while N is also quadruple of D; whence M, D together are equal to N. But FH is greater than M, D; therefore FH is in excess of N, while K is not in excess of N. And FH, K are equimultiples of AB, C, while N is another, chance, multiple of D; therefore AB has to D a greater ratio than C has to D. [V. Def. 7]  I say next, that D also has to C a greater ratio than D has to AB. For, with the same construction, we can prove similarly that N is in excess of K, while N is not in excess of FH. And N is a multiple of D, while FH, K are other, chance, equimultiples of AB, C; therefore D has to C a greater ratio than D has to AB. [V. Def. 7]  [Case 2.] Again, let AE be greater than EB. Then the less, EB, if multiplied, will sometime be greater than D. [V. Def. 4] Let it be multiplied, and let GH be a multiple of EB and greater than D; and, whatever multiple GH is of EB, let FG be made the same multiple of AE, and K  of C. Then we can prove similarly that FH, K are equimultiples of AB, C; and, similarly, let N be taken a multiple of D but the first that is greater than FG, so that FG is again not less than M. But GH is greater than D; therefore the whole FH is in excess of D, M, that is, of N. Now K is not in excess of N, inasmuch as FG also, which is greater than GH, that is, than K, is not in excess of N. And in the same manner, by following the above argument, we complete the demonstration.'"],["Rule","'ProofWordCount'",645],["Rule","'GreekProof'","'καὶ τὸ αὐτὸ πρὸς τὸ ἔλαττον μείζονα λόγον ἔχει ἤπερ πρὸς τὸ μεῖζον. ἔστω ἄνισα μεγέθη τὰ ΑΒ, Γ, καὶ ἔστω μεῖζον τὸ ΑΒ, ἄλλο δέ, ὃ ἔτυχεν, τὸ Δ: λέγω, ὅτι τὸ ΑΒ πρὸς τὸ Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Δ, καὶ τὸ Δ πρὸς τὸ Γ μείζονα λόγον ἔχει ἤπερ πρὸς τὸ ΑΒ. ἐπεὶ γὰρ μεῖζόν ἐστι τὸ ΑΒ τοῦ Γ, κείσθω τῷ Γ ἴσον τὸ ΒΕ: τὸ δὴ ἔλασσον τῶν ΑΕ, ΕΒ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ Δ μεῖζον. ἔστω πρότερον τὸ ΑΕ ἔλαττον τοῦ ΕΒ, καὶ πεπολλαπλασιάσθω τὸ ΑΕ, καὶ ἔστω αὐτοῦ πολλαπλάσιον τὸ ΖΗ μεῖζον ὂν τοῦ δ, καὶ ὁσαπλάσιόν ἐστι τὸ ΖΗ τοῦ ΑΕ, τοσαυταπλάσιον γεγονέτω καὶ τὸ μὲν ΗΘ τοῦ ΕΒ τὸ δὲ Κ τοῦ Γ: καὶ εἰλήφθω τοῦ Δ διπλάσιον μὲν τὸ Λ, τριπλάσιον δὲ τὸ Μ, καὶ ἑξῆς ἑνὶ πλεῖον, ἕως ἂν τὸ λαμβανόμενον πολλαπλάσιον μὲν γένηται τοῦ Δ, πρώτως δὲ μεῖζον τοῦ Κ. εἰλήφθω, καὶ ἔστω τὸ Ν τετραπλάσιον μὲν τοῦ Δ, πρώτως δὲ μεῖζον τοῦ Κ. ἐπεὶ οὖν τὸ Κ τοῦ Ν πρώτως ἐστὶν ἔλαττον, τὸ Κ ἄρα τοῦ Μ οὔκ ἐστιν ἔλαττον. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ ΗΘ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ ΖΘ τοῦ ΑΒ. ἰσάκις δέ ἐστι πολλαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ Κ τοῦ Γ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΖΘ τοῦ ΑΒ καὶ τὸ Κ τοῦ Γ. τὰ ΖΘ, Κ ἄρα τῶν ΑΒ, Γ ἰσάκις ἐστὶ πολλαπλάσια. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΕΒ καὶ τὸ Κ τοῦ Γ, ἴσον δὲ τὸ ΕΒ τῷ Γ, ἴσον ἄρα καὶ τὸ ΗΘ τῷ Κ: τὸ δὲ Κ τοῦ Μ οὔκ ἐστιν ἔλαττον: οὐδ᾽ ἄρα τὸ ΗΘ τοῦ μ ἔλαττόν ἐστιν. μεῖζον δὲ τὸ ΖΗ τοῦ Δ: ὅλον ἄρα τὸ ΖΘ συναμφοτέρων τῶν Δ, Μ μεῖζόν ἐστιν. ἀλλὰ συναμφότερα τὰ Δ, Μ τῷ Ν ἐστιν ἴσα, ἐπειδήπερ τὸ Μ τοῦ Δ τριπλάσιόν ἐστιν, συναμφότερα δὲ τὰ Μ, Δ τοῦ Δ ἐστι τετραπλάσια, ἔστι δὲ καὶ τὸ Ν τοῦ Δ τετραπλάσιον: συναμφότερα ἄρα τὰ Μ, Δ τῷ Ν ἴσα ἐστίν. ἀλλὰ τὸ ΖΘ τῶν Μ, Δ μεῖζόν ἐστιν: τὸ ΖΘ ἄρα τοῦ Ν ὑπερέχει: τὸ δὲ Κ τοῦ Ν οὐχ ὑπερέχει. καί ἐστι τὰ μὲν ΖΘ, Κ τῶν ΑΒ, Γ ἰσάκις πολλαπλάσια, τὸ δὲ Ν τοῦ Δ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον: τὸ ΑΒ ἄρα πρὸς τὸ Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Δ. λέγω δή, ὅτι καὶ τὸ Δ πρὸς τὸ Γ μείζονα λόγον ἔχει ἤπερ τὸ Δ πρὸς τὸ ΑΒ. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι τὸ μὲν Ν τοῦ Κ ὑπερέχει, τὸ δὲ Ν τοῦ ΖΘ οὐχ ὑπερέχει. καί ἐστι τὸ μὲν Ν τοῦ Δ πολλαπλάσιον, τὰ δὲ ΖΘ, Κ τῶν ΑΒ, Γ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: τὸ Δ ἄρα πρὸς τὸ Γ μείζονα λόγον ἔχει ἤπερ τὸ Δ πρὸς τὸ ΑΒ. ἀλλὰ δὴ τὸ ΑΕ τοῦ ΕΒ μεῖζον ἔστω. τὸ δὴ ἔλαττον τὸ ΕΒ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ Δ μεῖζον. πεπολλαπλασιάσθω, καὶ ἔστω τὸ ΗΘ πολλαπλάσιον μὲν τοῦ ΕΒ, μεῖζον δὲ τοῦ Δ: καὶ ὁσαπλάσιόν ἐστι τὸ ΗΘ τοῦ ΕΒ, τοσαυταπλάσιον γεγονέτω καὶ τὸ μὲν ΖΗ τοῦ ΑΕ, τὸ δὲ Κ τοῦ Γ. ὁμοίως δὴ δείξομεν, ὅτι τὰ ΖΘ, Κ τῶν ΑΒ, Γ ἰσάκις ἐστὶ πολλαπλάσια: καὶ εἰλήφθω ὁμοίως τὸ Ν πολλαπλάσιον μὲν τοῦ Δ, πρώτως δὲ μεῖζον τοῦ ΖΗ: ὥστε πάλιν τὸ ΖΗ τοῦ Μ οὔκ ἐστιν ἔλασσον. μεῖζον δὲ τὸ ΗΘ τοῦ Δ: ὅλον ἄρα τὸ ΖΘ τῶν Δ, Μ, τουτέστι τοῦ Ν, ὑπερέχει. τὸ δὲ Κ τοῦ Ν οὐχ ὑπερέχει, ἐπειδήπερ καὶ τὸ ΖΗ μεῖζον ὂν τοῦ ΗΘ, τουτέστι τοῦ Κ, τοῦ Ν οὐχ ὑπερέχει. καὶ ὡσαύτως κατακολουθοῦντες τοῖς ἐπάνω περαίνομεν τὴν ἀπόδειξιν. τῶν ἄρα ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον: καὶ τὸ αὐτὸ πρὸς τὸ ἔλαττον μείζονα λόγον ἔχει ἤπερ πρὸς τὸ μεῖζον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",647]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'5.9'"],["Rule","'Text'","'Magnitudes which have the same ratio to the same are equal to one another; and magnitudes to which the same has the same ratio are equal.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'τὰ πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λόγον ἴσα ἀλλήλοις ἐστίν : καὶ πρὸς ἃ τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, ἐκεῖνα ἴσα ἐστίν.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'For let each of the magnitudes A, B have the same ratio to C; I say that A is equal to B. For, otherwise, each of the magnitudes A, B would not have had the same ratio to C; [V. 8] but it has; therefore A is equal to B. Again, let C have the same ratio to each of the magnitudes A, B; I say that A is equal to B. For, otherwise, C would not have had the same ratio to each of the magnitudes A, B; [V. 8] but it has; therefore A is equal to B.'"],["Rule","'ProofWordCount'",100],["Rule","'GreekProof'","'ἐχέτω γὰρ ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν λόγον: λέγω, ὅτι ἴσον ἐστὶ τὸ Α τῷ Β. εἰ γὰρ μή, οὐκ ἂν ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν εἶχε λόγον: ἔχει δέ: ἴσον ἄρα ἐστὶ τὸ Α τῷ Β. ἐχέτω δὴ πάλιν τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν λόγον: λέγω, ὅτι ἴσον ἐστὶ τὸ Α τῷ Β. εἰ γὰρ μή, οὐκ ἂν τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν εἶχε λόγον: ἔχει δέ: ἴσον ἄρα ἐστὶ τὸ Α τῷ Β. τὰ ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λόγον ἴσα ἀλλήλοις ἐστίν: καὶ πρὸς ἃ τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, ἐκεῖνα ἴσα ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",118]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'5.10'"],["Rule","'Text'","'Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'τῶν πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον ἐκεῖνο μεῖζόν ἐστιν: πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλαττόν ἐστιν.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'For let A have to C a greater ratio than B has to C; I say that A is greater than B. For, if not, A is either equal to B or less. Now A is not equal to B; for in that case each of the magnitudes A, B would have had the same ratio to C; [V. 7] but they have not; therefore A is not equal to B. Nor again is A less than B; for in that case A would have had to C a less ratio than B has to C; [V. 8] but it has not; therefore A is not less than B.  But it was proved not to be equal either; therefore A is greater than B.  Again, let C have to B a greater ratio than C has to A; I say that B is less than A. For, if not, it is either equal or greater. Now B is not equal to A; for in that case C would have had the same ratio to each of the magnitudes A, B; [V. 7] but it has not; therefore A is not equal to B.  Nor again is B greater than A; for in that case C would have had to B a less ratio than it has to A; [V. 8] but it has not; therefore B is not greater than A.  But it was proved that it is not equal either; therefore B is less than A.'"],["Rule","'ProofWordCount'",247],["Rule","'GreekProof'","'ἐχέτω γὰρ τὸ Α πρὸς τὸ Γ μείζονα λόγον ἤπερ τὸ Β πρὸς τὸ Γ: λέγω, ὅτι μεῖζόν ἐστι τὸ Α τοῦ Β. εἰ γὰρ μή, ἤτοι ἴσον ἐστὶ τὸ Α τῷ Β á¼¢ ἔλασσον. ἴσον μὲν οὖν οὔκ ἐστι τὸ Α τῷ Β: ἑκάτερον γὰρ ἂν τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ: οὐκ ἄρα ἴσον ἐστὶ τὸ Α τῷ Β. οὐδὲ μὴν ἔλασσόν ἐστι τὸ Α τοῦ Β: τὸ Α γὰρ ἂν πρὸς τὸ Γ ἐλάσσονα λόγον εἶχεν ἤπερ τὸ Β πρὸς τὸ Γ. οὐκ ἔχει δέ: οὐκ ἄρα ἔλασσόν ἐστι τὸ Α τοῦ Β. ἐδείχθη δὲ οὐδὲ ἴσον: μεῖζον ἄρα ἐστὶ τὸ Α τοῦ Β. ἐχέτω δὴ πάλιν τὸ Γ πρὸς τὸ Β μείζονα λόγον ἤπερ τὸ Γ πρὸς τὸ Α: λέγω, ὅτι ἔλασσόν ἐστι τὸ Β τοῦ Α. εἰ γὰρ μή, ἤτοι ἴσον ἐστὶν á¼¢ μεῖζον. ἴσον μὲν οὖν οὔκ ἐστι τὸ Β τῷ Α: τὸ Γ γὰρ ἂν πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ: οὐκ ἄρα ἴσον ἐστὶ τὸ Α τῷ Β. οὐδὲ μὴν μεῖζόν ἐστι τὸ Β τοῦ Α: τὸ Γ γὰρ ἂν πρὸς τὸ Β ἐλάσσονα λόγον εἶχεν ἤπερ πρὸς τὸ Α. οὐκ ἔχει δέ: οὐκ ἄρα μεῖζόν ἐστι τὸ Β τοῦ Α. ἐδείχθη δέ, ὅτι οὐδὲ ἴσον: ἔλαττον ἄρα ἐστὶ τὸ Β τοῦ Α. τῶν ἄρα πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον μεῖζόν ἐστιν: καὶ πρὸς ὃ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλαττόν ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",251]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'5.11'"],["Rule","'Text'","'Ratios which are the same with the same ratio are also the same with one another.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'οἱ τῷ αὐτῷ λόγῳ οἱ αὐτοὶ καὶ ἀλλήλοις εἰσὶν οἱ αὐτοί.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List"]],["Rule","'Proof'","'For, as A is to B, so let C be to D, and, as C is to D, so let E be to F; I say that, as A is to B, so is E to F. For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N. Then since, as A is to B, so is C to D, and of A, C equimultiples G, H have been taken, and of B, D other, chance, equimultiples L, M, therefore, if G is in excess of L, H is also in excess of M, if equal, equal, and if less, less. Again, since, as C is to D, so is E to F, and of C, E equimultiples H, K have been taken, and of D, F other, chance, equimultiples M, N, therefore, if H is in excess of M, K is also in excess of N, if equal, equal, and if less, less. But we saw that, if H was in excess of M, G was also in excess of L; if equal, equal; and if less, less; so that, in addition, if G is in excess of L, K is also in excess of N, if equal, equal, and if less, less. And G, K are equimultiples of A, E, while L, N are other, chance, equimultiples of B, F; therefore, as A is to B, so is E to F.'"],["Rule","'ProofWordCount'",245],["Rule","'GreekProof'","'ἔστωσαν γὰρ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, ὡς δὲ τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ: λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. εἰλήφθω γὰρ τῶν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, Κ, τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Η, Θ, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, εἰ ἄρα ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Θ τοῦ Μ, καὶ εἰ ἴσον ἐστίν, ἴσον, καὶ εἰ ἐλλείπει, ἐλλείπει. πάλιν, ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ, καὶ εἴληπται τῶν Γ, Ε ἰσάκις πολλαπλάσια τὰ Θ, Κ, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Θ τοῦ Μ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ἀλλὰ εἰ ὑπερεῖχε τὸ Θ τοῦ Μ, ὑπερεῖχε καὶ τὸ Η τοῦ Λ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον: ὥστε καὶ εἰ ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Η, Κ τῶν Α, Ε ἰσάκις πολλαπλάσια, τὰ δὲ Λ, Ν τῶν Β, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. οἱ ἄρα τῷ αὐτῷ λόγῳ οἱ αὐτοὶ καὶ ἀλλήλοις εἰσὶν οἱ αὐτοί: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",282]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'5.12'"],["Rule","'Text'","'If any number of magnitudes be proportional, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐὰν ᾖ ὁποσαοῦν μεγέθη ἀνάλογον, ἔσται ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",5]],["List",["Rule","'Book'",5],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let any number of magnitudes A, B, C, D, E, F be proportional, so that, as A is to B, so is C to D and E to F; I say that, as A is to B, so are A, C, E to B, D, F. For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N. Then since, as A is to B, so is C to D, and E to F, and of A, C, E equimultiples G, H, K have been taken, and of B, D, F other, chance, equimultiples L, M, N, therefore, if G is in excess of L, H is also in excess of M, and K of N, if equal, equal, and if less, less; so that, in addition, if G is in excess of L, then G, H, K are in excess of L, M, N, if equal, equal, and if less, less. Now G and G, H, K are equimultiples of A and A, C, E, since, if any number of magnitudes whatever are respectively equimultiples of any magnitudes equal in multitude, whatever multiple one of the magnitudes is of one, that multiple also will all be of all. [V. 1] For the same reason L and L, M, N are also equimultiples of B and B, D, F; therefore, as A is to B, so are A, C, E to B, D, F. [V. Def. 5]'"],["Rule","'ProofWordCount'",248],["Rule","'GreekProof'","'ἔστωσαν ὁποσαοῦν μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ε, Ζ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ τὸ Ε πρὸς τὸ Ζ: λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὰ Α, Γ, Ε πρὸς τὰ Β, Δ, Ζ. εἰλήφθω γὰρ τῶν μὲν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, Κ, τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ τὸ Ε πρὸς τὸ Ζ, καὶ εἴληπται τῶν μὲν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, Κ τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Θ τοῦ Μ, καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ὥστε καὶ εἰ ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὰ Η, Θ, Κ τῶν Λ, Μ, Ν, καὶ εἰ ἴσον, ἴσα, καὶ εἰ ἔλαττον, ἐλάττονα. καί ἐστι τὸ μὲν Η καὶ τὰ Η, Θ, Κ τοῦ Α καὶ τῶν Α, Γ, Ε ἰσάκις πολλαπλάσια, ἐπειδήπερ ἐὰν ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα τῶν πάντων. διὰ τὰ αὐτὰ δὴ καὶ τὸ Λ καὶ τὰ Λ, Μ, Ν τοῦ Β καὶ τῶν β, Δ, Ζ ἰσάκις ἐστὶ πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὰ Α, Γ, Ε πρὸς τὰ Β, Δ, Ζ. ἐὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ἀνάλογον, ἔσται ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",287]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'5.13'"],["Rule","'Text'","'If a first magnitude have to a second the same ratio as a third to a fourth, and the third have to the fourth a greater ratio than a fifth has to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth.'"],["Rule","'TextWordCount'",51],["Rule","'GreekText'","'ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα λόγον ἔχῃ á¼¢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον μείζονα λόγον ἕξει á¼¢ πέμπτον πρὸς ἕκτον.'"],["Rule","'GreekTextWordCount'",34],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",5]],["List",["Rule","'Book'",5],["Rule","'Definition'",7]]]],["Rule","'Proof'","'For let a first magnitude A have to a second B the same ratio as a third C has to a fourth D, and let the third C have to the fourth D a greater ratio than a fifth E has to a sixth F; I say that the first A will also have to the second B a greater ratio than the fifth E to the sixth F. For, since there are some equimultiples of C, E, and of D, F other, chance, equimultiples, such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, [V. Def. 7] let them be taken, and let G, H be equimultiples of C, E, and K, L other, chance, equimultiples of D, F, so that G is in excess of K, but H is not in excess of L; and, whatever multiple G is of C, let M be also that multiple of A, and, whatever multiple K is of D, let N be also that multiple of B. Now, since, as A is to B, so is C to D, and of A, C equimultiples M, G have been taken, and of B, D other, chance, equimultiples N, K, therefore, if M is in excess of N, G is also in excess of K, if equal, equal, and if less, less. [V. Def. 5] But G is in excess of K; therefore M is also in excess of N. But H is not in excess of L; and M, H are equimultiples of A, E, and N, L other, chance, equimultiples of B, F; therefore A has to B a greater ratio than E has to F. [V. Def. 7]'"],["Rule","'ProofWordCount'",296],["Rule","'GreekProof'","'πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, τρίτον δὲ τὸ Γ πρὸς τέταρτον τὸ Δ μείζονα λόγον ἐχέτω á¼¢ πέμπτον τὸ Ε πρὸς ἕκτον τὸ Ζ. λέγω, ὅτι καὶ πρῶτον τὸ Α πρὸς δεύτερον τὸ Β μείζονα λόγον ἕξει ἤπερ πέμπτον τὸ Ε πρὸς ἕκτον τὸ Ζ. ἐπεὶ γὰρ ἔστι τινὰ τῶν μὲν Γ, Ε ἰσάκις πολλαπλάσια, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια, καὶ τὸ μὲν τοῦ Γ πολλαπλάσιον τοῦ τοῦ Δ πολλαπλασίου ὑπερέχει, τὸ δὲ τοῦ Ε πολλαπλάσιον τοῦ τοῦ Ζ πολλαπλασίου οὐχ ὑπερέχει, εἰλήφθω, καὶ ἔστω τῶν μὲν Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, Λ, ὥστε τὸ μὲν Η τοῦ Κ ὑπερέχειν, τὸ δὲ Θ τοῦ Λ μὴ ὑπερέχειν: καὶ ὁσαπλάσιον μέν ἐστι τὸ Η τοῦ Γ, τοσαυταπλάσιον ἔστω καὶ τὸ Μ τοῦ Α, ὁσαπλάσιον δὲ τὸ Κ τοῦ Δ, τοσαυταπλάσιον ἔστω καὶ τὸ Ν τοῦ Β. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Μ, Η, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Ν, Κ, εἰ ἄρα ὑπερέχει τὸ Μ τοῦ Ν, ὑπερέχει καὶ τὸ Η τοῦ Κ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ὑπερέχει δὲ τὸ Η τοῦ Κ: ὑπερέχει ἄρα καὶ τὸ Μ τοῦ Ν. τὸ δὲ Θ τοῦ Λ οὐχ ὑπερέχει: καί ἐστι τὰ μὲν Μ, Θ τῶν Α, Ε ἰσάκις πολλαπλάσια, τὰ δὲ Ν, Λ τῶν Β, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: τὸ ἄρα Α πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Ε πρὸς τὸ Ζ. ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα λόγον ἔχῃ á¼¢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον μείζονα λόγον ἕξει á¼¢ πέμπτον πρὸς ἕκτον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",322]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'5.14'"],["Rule","'Text'","'If a first magnitude have to a second the same ratio as a third has to a fourth, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less.'"],["Rule","'TextWordCount'",42],["Rule","'GreekText'","'ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",8]],["List",["Rule","'Book'",5],["Rule","'Theorem'",10]],["List",["Rule","'Book'",5],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'For let a first magnitude A have the same ratio to a second B as a third C has to a fourth D; and let A be greater than C; I say that B is also greater than D. For, since A is greater than C, and B is another, chance, magnitude, therefore A has to B a greater ratio than C has to B. [V. 8] But, as A is to B, so is C to D; therefore C has also to D a greater ratio than C has to B. [V. 13]  But that to which the same has a greater ratio is less; [V. 10] therefore D is less than B; so that B is greater than D.  Similarly we can prove that, if A be equal to C, B will also be equal to D; and, if A be less than C, B will also be less than D.'"],["Rule","'ProofWordCount'",153],["Rule","'GreekProof'","'πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, μεῖζον δὲ ἔστω τὸ Α τοῦ Γ: λέγω, ὅτι καὶ τὸ Β τοῦ Δ μεῖζόν ἐστιν. ἐπεὶ γὰρ τὸ Α τοῦ Γ μεῖζόν ἐστιν, ἄλλο δέ, ὃ ἔτυχεν, μέγεθος τὸ Β, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: καὶ τὸ Γ ἄρα πρὸς τὸ Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλασσόν ἐστιν: ἔλασσον ἄρα τὸ Δ τοῦ Β: ὥστε μεῖζόν ἐστι τὸ Β τοῦ Δ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Β τῷ Δ, κἂν ἔλασσον ᾖ τὸ Α τοῦ Γ, ἔλασσον ἔσται καὶ τὸ Β τοῦ Δ. ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",187]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'5.15'"],["Rule","'Text'","'Parts have the same ratio as the same multiples of them taken in corresponding order.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'τὰ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον ληφθέντα κατάλληλα.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'For let AB be the same multiple of C that DE is of F; I say that, as C is to F, so is AB to DE. For, since AB is the same multiple of C that DE is of F, as many magnitudes as there are in AB equal to C, so many are there also in DE equal to F. Let AB be divided into the magnitudes AG, GH, HB equal to C, and DE into the magnitudes DK, KL, LE equal to F; then the multitude of the magnitudes AG, GH, HB will be equal to the multitude of the magnitudes DK, KL, LE. And, since AG. GH, HB are equal to one another, and DK, KL, LE are also equal to one another, therefore, as AG is to DK, so is GH to KL, and HB to LE. [V. 7] Therefore, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [V. 12] therefore, as AG is to DK, so is AB to DE. But AG is equal to C and DK to F; therefore, as C is to F, so is AB to DE.'"],["Rule","'ProofWordCount'",200],["Rule","'GreekProof'","'ἔστω γὰρ ἰσάκις πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ ΔΕ τοῦ Ζ: λέγω, ὅτι ἐστὶν ὡς τὸ Γ πρὸς τὸ Ζ, οὕτως τὸ ΑΒ πρὸς τὸ ΔΕ. ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ εἰς τὰ τῷ Γ ἴσα τὰ ΑΗ, ΗΘ, ΘΒ, τὸ δὲ ΔΕ εἰς τὰ τῷ Ζ ἴσα τὰ ΔΚ, ΚΛ, ΛΕ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΘ, ΘΒ τῷ πλήθει τῶν ΔΚ, ΚΛ, ΛΕ. καὶ ἐπεὶ ἴσα ἐστὶ τὰ ΑΗ, ΗΘ, ΘΒ ἀλλήλοις, ἔστι δὲ καὶ τὰ ΔΚ, ΚΛ, ΛΕ ἴσα ἀλλήλοις, ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ ΔΚ, οὕτως τὸ ΗΘ πρὸς τὸ ΚΛ, καὶ τὸ ΘΒ πρὸς τὸ ΛΕ. ἔσται ἄρα καὶ ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ ΔΚ, οὕτως τὸ ΑΒ πρὸς τὸ ΔΕ. ἴσον δὲ τὸ μὲν ΑΗ τῷ Γ, τὸ δὲ ΔΚ τῷ Ζ: ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ οὕτως τὸ ΑΒ πρὸς τὸ ΔΕ. τὰ ἄρα μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον ληφθέντα κατάλληλα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",212]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'5.16'"],["Rule","'Text'","'If four magnitudes be proportional, they will also be proportional alternately.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἐὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ ἀνάλογον ἔσται.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",5]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",14]],["List",["Rule","'Book'",5],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let A, B, C, D be four proportional magnitudes, so that, as A is to B, so is C to D; I say that they will also be so alternately, that is, as A is to C, so is B to D. For of A, B let equimultiples E, F be taken, and of C, D other, chance, equimultiples G, H. Then, since E is the same multiple of A that F is of B, and parts have the same ratio as the same multiples of them, [V. 15] therefore, as A is to B, so is E to F. But as A is to B, so is C to D; therefore also, as C is to D, so is E to F. [V. 11] Again, since G, H are equimultiples of C, D, therefore, as C is to D, so is G to H. [V. 15] But, as C is to D, so is E to F; therefore also, as E is to F, so is G to H. [V. 11] But, if four magnitudes be proportional, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less. [V. 14] Therefore, if E is in excess of G, F is also in excess of H, if equal, equal, and if less, less. Now E, F are equimultiples of A, B, and G, H other, chance, equimultiples of C, D; therefore, as A is to C, so is B to D. [V. Def. 5]'"],["Rule","'ProofWordCount'",257],["Rule","'GreekProof'","'ἔστω τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: λέγω, ὅτι καὶ ἐναλλὰξ ἀνάλογον ἔσται, ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Δ. εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Ε, Ζ, τῶν δὲ Γ, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Η, Θ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Ε τοῦ Α καὶ τὸ Ζ τοῦ Β, τὰ δὲ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: καὶ ὡς ἄρα τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ. πάλιν, ἐπεὶ τὰ Η, Θ τῶν Γ, Δ ἰσάκις ἐστὶ πολλαπλάσια, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Η πρὸς τὸ Θ. ὡς δὲ τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ: καὶ ὡς ἄρα τὸ Ε πρὸς τὸ Ζ, οὕτως τὸ Η πρὸς τὸ Θ. ἐὰν δὲ τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. εἰ ἄρα ὑπερέχει τὸ Ε τοῦ Η, ὑπερέχει καὶ τὸ Ζ τοῦ Θ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Ε, Ζ τῶν Α, Β ἰσάκις πολλαπλάσια, τὰ δὲ Η, Θ τῶν Γ, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Δ. ἐὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",276]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'5.17'"],["Rule","'Text'","'If magnitudes be proportional componendo, they will also be proportional separando.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἐὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",1]],["List",["Rule","'Book'",5],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let AB, BE, CD, DF be magnitudes proportional componendo, so that, as AB is to BE, so is CD to DF; I say that they will also be proportional separando, that is, as AE is to EB, so is CF to DF. For of AE, EB, CF, FD let equimultiples GH, HK, LM, MN be taken, and of EB, FD other, chance, equimultiples, KO, NP. Then, since GH is the same multiple of AE that HK is of EB, therefore GH is the same multiple of AE that GK is of AB. [V. 1] But GH is the same multiple of AE that LM is of CF; therefore GK is the same multiple of AB that LM is of CF. Again, since LM is the same multiple of CF that MN is of FD, therefore LM is the same multiple of CF that LN is of CD. [V. 1] But LM was the same multiple of CF that GK is of AB; therefore GK is the same multiple of AB that LN is of CD. Therefore GK, LN are equimultiples of AB, CD. Again, since HK is the same multiple of EB that MN is of FD, and KO is also the same multiple of EB that NP is of FD, therefore the sum HO is also the same multiple of EB that MP is of FD. [V. 2]  And, since, as AB is to BE, so is CD to DF, and of AB, CD equimultiples GK, LN have been taken, and of EB, FD equimultiples HO, MP, therefore, if GK is in excess of HO, LN is also in excess of MP, if equal, equal, and if less, less. Let GK be in excess of HO; then, if HK be subtracted from each, GH is also in excess of KO. But we saw that, if GK was in excess of HO, LN was also in excess of MP; therefore LN is also in excess of MP,  and, if MN be subtracted from each, LM is also in excess of NP; so that, if GH is in excess of KO, LM is also in excess of NP. Similarly we can prove that, if GH be equal to KO, LM will also be equal to NP, and if less, less. And GH, LM are equimultiples of AE, CF, while KO, NP are other, chance, equimultiples of EB, FD; therefore, as AE is to EB, so is CF to FD.'"],["Rule","'ProofWordCount'",408],["Rule","'GreekProof'","'ἔστω συγκείμενα μεγέθη ἀνάλογον τὰ ΑΒ, ΒΕ, ΓΔ, ΔΖ, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ: λέγω, ὅτι καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΔΖ. εἰλήφθω γὰρ τῶν μὲν ΑΕ, ΕΒ, ΓΖ, ΖΔ ἰσάκις πολλαπλάσια τὰ ΗΘ, ΘΚ, ΛΜ, ΜΝ, τῶν δὲ ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ ΚΞ, ΝΠ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΘΚ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΗΚ τοῦ ΑΒ. ἰσάκις δέ ἐστι πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΛΜ τοῦ ΓΖ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΜ τοῦ ΓΖ. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΜΝ τοῦ ΖΔ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΛΝ τοῦ ΓΔ. ἰσάκις δὲ ἦν πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΗΚ τοῦ ΑΒ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΝ τοῦ ΓΔ. τὰ ΗΚ, ΛΝ ἄρα τῶν ΑΒ, ΓΔ ἰσάκις ἐστὶ πολλαπλάσια. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΘΚ τοῦ ΕΒ καὶ τὸ ΜΝ τοῦ ΖΔ, ἔστι δὲ καὶ τὸ ΚΞ τοῦ ΕΒ ἰσάκις πολλαπλάσιον καὶ τὸ ΝΠ τοῦ ΖΔ, καὶ συντεθὲν τὸ ΘΞ τοῦ ΕΒ ἰσάκις ἐστὶ πολλαπλάσιον καὶ τὸ ΜΠ τοῦ ΖΔ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, καὶ εἴληπται τῶν μὲν ΑΒ, ΓΔ ἰσάκις πολλαπλάσια τὰ ΗΚ, ΛΝ, τῶν δὲ ΕΒ, ΖΔ ἰσάκις πολλαπλάσια τὰ ΘΞ, ΜΠ, εἰ ἄρα ὑπερέχει τὸ ΗΚ τοῦ ΘΞ, ὑπερέχει καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ὑπερεχέτω δὴ τὸ ΗΚ τοῦ ΘΞ, καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΘΚ ὑπερέχει ἄρα καὶ τὸ ΗΘ τοῦ ΚΞ. ἀλλὰ εἰ ὑπερεῖχε τὸ ΗΚ τοῦ ΘΞ, ὑπερεῖχε καὶ τὸ ΛΝ τοῦ ΜΠ: ὑπερέχει ἄρα καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΜΝ ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ: ὥστε εἰ ὑπερέχει τὸ ΗΘ τοῦ ΚΞ, ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ ΗΘ τῷ ΚΞ, ἴσον ἔσται καὶ τὸ ΛΜ τῷ ΝΠ, κἂν ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν ΗΘ, ΛΜ τῶν ΑΕ, ΓΖ ἰσάκις πολλαπλάσια, τὰ δὲ ΚΞ, ΝΠ τῶν ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. ἐὰν ἄρα συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",413]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'5.18'"],["Rule","'Text'","'If magnitudes be proportional separando, they will also be proportional componendo.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἐὰν διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα ἀνάλογον ἔσται.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",14]],["List",["Rule","'Book'",5],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'Let AE, EB, CF, FD be magnitudes proportional separando, so that, as AE is to EB, so is CF to FD; I say that they will also be proportional componendo, that is, as AB is to BE, so is CD to FD. For, if CD be not to DF as AB to BE, then, as AB is to BE, so will CD be either to some magnitude less than DF or to a greater. First, let it be in that ratio to a less magnitude DG. Then, since, as AB is to BE, so is CD to DG, they are magnitudes proportional componendo; so that they will also be proportional separando. [V. 17]  Therefore, as AE is to EB, so is CG to GD. But also, by hypothesis, as AE is to EB, so is CF to FD. Therefore also, as CG is to GD, so is CF to FD. [V. 11] But the first CG is greater than the third CF; therefore the second GD is also greater than the fourth FD. [V. 14]  But it is also less: which is impossible. Therefore, as AB is to BE, so is not CD to a less magnitude than FD. Similarly we can prove that neither is it in that ratio to a greater; it is therefore in that ratio to FD itself.'"],["Rule","'ProofWordCount'",222],["Rule","'GreekProof'","'ἔστω διῃρημένα μεγέθη ἀνάλογον τὰ ΑΕ, ΕΒ, ΓΖ, ΖΔ, ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ: λέγω, ὅτι καὶ συντεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΖΔ. εἰ γὰρ μή ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, ἔσται ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ ἤτοι πρὸς ἔλασσόν τι τοῦ ΔΖ á¼¢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ ΔΗ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΗ, συγκείμενα μεγέθη ἀνάλογόν ἐστιν: ὥστε καὶ διαιρεθέντα ἀνάλογον ἔσται. ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΗ πρὸς τὸ ΗΔ. ὑπόκειται δὲ καὶ ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. καὶ ὡς ἄρα τὸ ΓΗ πρὸς τὸ ΗΔ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. μεῖζον δὲ τὸ πρῶτον τὸ ΓΗ τοῦ τρίτου τοῦ ΓΖ: μεῖζον ἄρα καὶ τὸ δεύτερον τὸ ΗΔ τοῦ τετάρτου τοῦ ΖΔ. ἀλλὰ καὶ ἔλαττον: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς ἔλασσον τοῦ ΖΔ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ πρὸς μεῖζον: πρὸς αὐτὸ ἄρα. ἐὰν ἄρα διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",213]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'5.19'"],["Rule","'Text'","'If, as a whole is to a whole, so is a part subtracted to a part subtracted, the remainder will also be to the remainder as whole to whole.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'ἐὰν ᾖ ὡς ὅλον πρὸς ὅλον, οὕτως ἀφαιρεθὲν πρὸς ἀφαιρεθέν, καὶ τὸ λοιπὸν πρὸς τὸ λοιπὸν ἔσται ὡς ὅλον πρὸς ὅλον.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'For, as the whole AB is to the whole CD, so let the part AE subtracted be to the part CF subtracted; I say that the remainder EB will also be to the remainder FD as the whole AB to the whole CD. For since, as AB is to CD, so is AE to CF, alternately also, as BA is to AE, so is DC to CF. [V. 16] And, since the magnitudes are proportional componendo, they will also be proportional separando, [V. 17] that is, as BE is to EA, so is DF to CF, and, alternately, as BE is to DF, so is EA to FC. [V. 16] But, as AE is to CF, so by hypothesis is the whole AB to the whole CD. Therefore also the remainder EB will be to the remainder FD as the whole AB is to the whole CD. [V. 11]'"],["Rule","'ProofWordCount'",149],["Rule","'GreekProof'","'ἔστω γὰρ ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ, οὕτως ἀφαιρεθὲν τὸ ΑΕ πρὸς ἀφαιρεθὲν τὸ ΓΖ: λέγω, ὅτι καὶ λοιπὸν τὸ ΕΒ πρὸς λοιπὸν τὸ ΖΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. ἐπεὶ γάρ ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΑΕ πρὸς τὸ ΓΖ, καὶ ἐναλλὰξ ὡς τὸ ΒΑ πρὸς τὸ ΑΕ, οὕτως τὸ ΔΓ πρὸς τὸ ΓΖ. καὶ ἐπεὶ συγκείμενα μεγέθη ἀνάλογόν ἐστιν, καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΒΕ πρὸς τὸ ΕΑ, οὕτως τὸ ΔΖ πρὸς τὸ ΓΖ: καὶ ἐναλλάξ, ὡς τὸ ΒΕ πρὸς τὸ ΔΖ, οὕτως τὸ ΕΑ πρὸς τὸ ΖΓ. ὡς δὲ τὸ ΑΕ πρὸς τὸ ΓΖ, οὕτως ὑπόκειται ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. καὶ λοιπὸν ἄρα τὸ ΕΒ πρὸς λοιπὸν τὸ ΖΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. ἐὰν ἄρα ᾖ ὡς ὅλον πρὸς ὅλον, οὕτως ἀφαιρεθὲν πρὸς ἀφαιρεθέν, καὶ τὸ λοιπὸν πρὸς τὸ λοιπὸν ἔσται ὡς ὅλον πρὸς ὅλον ὅπερ ἔδει δεῖξαι. Καὶ ἐπεὶ ἐδείχθη ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΕΒ πρὸς τὸ ΖΔ, καὶ ἐναλλὰξ ὡς τὸ ΑΒ πρὸς τὸ ΒΕ οὕτως τὸ ΓΔ πρὸς τὸ ΖΔ, συγκείμενα ἄρα μεγέθη ἀνάλογόν ἐστιν: ἐδείχθη δὲ ὡς τὸ ΒΑ πρὸς τὸ ΑΕ, οὕτως τὸ ΔΓ πρὸς τὸ ΓΖ: καί ἐστιν ἀναστρέψαντι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ ἀναστρέψαντι ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",230]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'5.20'"],["Rule","'Text'","'If there be three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if ex aequali the first be greater than the third, the fourth will also be greater than the sixth; if equal, equal; and, if less, less.'"],["Rule","'TextWordCount'",49],["Rule","'GreekText'","'ἐὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, δι᾽ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον.'"],["Rule","'GreekTextWordCount'",40],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",8]],["List",["Rule","'Book'",5],["Rule","'Theorem'",10]],["List",["Rule","'Book'",5],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and as B is to C, so is E to F; and let A be greater than C ex aequali; I say that D will also be greater than F; if A is equal to C, equal; and, if less, less. For, since A is greater than C, and B is some other magnitude, and the greater has to the same a greater ratio than the less has, [V. 8] therefore A has to B a greater ratio than C has to B. But, as A is to B, so is D to E, and, as C is to B, inversely, so is F to E; therefore D has also to E a greater ratio than F has to E. [V. 13] But, of magnitudes which have a ratio to the same, that which has a greater ratio is greater; [V. 10] therefore D is greater than F.  Similarly we can prove that, if A be equal to C, D will also be equal to F; and if less, less.'"],["Rule","'ProofWordCount'",210],["Rule","'GreekProof'","'ἔστω τρία μεγέθη τὰ Α, Β, Γ, καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Ε πρὸς τὸ Ζ, δι᾽ ἴσου δὲ μεῖζον ἔστω τὸ Α τοῦ Γ: λέγω, ὅτι καὶ τὸ Δ τοῦ Ζ μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. ἐπεὶ γὰρ μεῖζόν ἐστι τὸ Α τοῦ Γ, ἄλλο δέ τι τὸ Β, τὸ δὲ μεῖζον πρὸς τὸ αὐτὸ μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. ἀλλ᾽ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς δὲ τὸ Γ πρὸς τὸ Β, ἀνάπαλιν οὕτως τὸ Ζ πρὸς τὸ Ε: καὶ τὸ Δ ἄρα πρὸς τὸ Ε μείζονα λόγον ἔχει ἤπερ τὸ Ζ πρὸς τὸ Ε. τῶν δὲ πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον μεῖζόν ἐστιν. μεῖζον ἄρα τὸ Δ τοῦ Ζ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Δ τῷ Ζ, κἂν ἔλαττον, ἔλαττον. ἐὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, δι᾽ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",243]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'5.21'"],["Rule","'Text'","'If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, then, if ex aequali the first magnitude is greater than the third, the fourth will also be greater than the sixth; if equal, equal; and if less, less.'"],["Rule","'TextWordCount'",58],["Rule","'GreekText'","'ἐὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, δι᾽ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον.'"],["Rule","'GreekTextWordCount'",46],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",8]],["List",["Rule","'Book'",5],["Rule","'Theorem'",10]],["List",["Rule","'Book'",5],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, and let the proportion of them be perturbed, so that, as A is to B, so is E to F, and, as B is to C, so is D to E, and let A be greater than C   ex aequali; I say that D will also be greater than F; if A is equal to C, equal; and if less, less. For, since A is greater than C, and B is some other magnitude, therefore A has to B a greater ratio than C has to B. [V. 8] But, as A is to B, so is E to F, and, as C is to B, inversely, so is E to D. Therefore also E has to F a greater ratio than E has to D. [V. 13] But that to which the same has a greater ratio is less; [V. 10] therefore F is less than D; therefore D is greater than F.  Similarly we can prove that, if A be equal to C, D will also be equal to F; and if less, less.'"],["Rule","'ProofWordCount'",204],["Rule","'GreekProof'","'ἔστω τρία μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ἔστω δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ε, δι᾽ ἴσου δὲ τὸ Α τοῦ Γ μεῖζον ἔστω: λέγω, ὅτι καὶ τὸ Δ τοῦ Ζ μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. ἐπεὶ γὰρ μεῖζόν ἐστι τὸ Α τοῦ Γ, ἄλλο δέ τι τὸ Β, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. ἀλλ᾽ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Γ πρὸς τὸ Β, ἀνάπαλιν οὕτως τὸ Ε πρὸς τὸ Δ. καὶ τὸ Ε ἄρα πρὸς τὸ Ζ μείζονα λόγον ἔχει ἤπερ τὸ Ε πρὸς τὸ Δ. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλασσόν ἐστιν: ἔλασσον ἄρα ἐστὶ τὸ Ζ τοῦ Δ: μεῖζον ἄρα ἐστὶ τὸ Δ τοῦ Ζ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Δ τῷ Ζ, κἂν ἔλαττον, ἔλαττον. ἐὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, δι᾽ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",250]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'5.22'"],["Rule","'Text'","'If there be any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, they will also be in the same ratio ex aequali.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν ᾖ ὁποσαοῦν μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",5]],["List",["Rule","'Book'",5],["Rule","'Theorem'",4]],["List",["Rule","'Book'",5],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let there be any number of magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two together are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F; I say that they will also be in the same ratio ex aequali, <that is, as A is to C, so is D to F>. For of A, D let equimultiples G, H be taken, and of B, E other, chance, equimultiples K, L; and, further, of C, F other, chance, equimultiples M, N. Then, since, as A is to B, so is D to E, and of A, D equimultiples G, H have been taken, and of B, E other, chance, equimultiples K, L, therefore, as G is to K, so is H to L. [V. 4]  For the same reason also, as K is to M, so is L to N.  Since, then, there are three magnitudes G, K, M, and others H, L, N equal to them in multitude, which taken two and two together are in the same ratio, therefore, ex aequali, if G is in excess of M, H is also in excess of N; if equal, equal; and if less, less. [V. 20] And G, H are equimultiples of A, D, and M, N other, chance, equimultiples of C, F.  Therefore, as A is to C, so is D to F. [V. Def. 5]'"],["Rule","'ProofWordCount'",254],["Rule","'GreekProof'","'ἔστω ὁποσαοῦν μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Ε πρὸς τὸ Ζ: λέγω, ὅτι καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται. εἰλήφθω γὰρ τῶν μὲν Α, Δ ἰσάκις πολλαπλάσια τὰ η, Θ, τῶν δὲ Β, Ε ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, Λ, καὶ ἔτι τῶν Γ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, καὶ εἴληπται τῶν μὲν Α, Δ ἰσάκις πολλαπλάσια τὰ η, Θ, τῶν δὲ Β, Ε ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, Λ, ἔστιν ἄρα ὡς τὸ Η πρὸς τὸ Κ, οὕτως τὸ Θ πρὸς τὸ Λ. διὰ τὰ αὐτὰ δὴ καὶ ὡς τὸ Κ πρὸς τὸ Μ, οὕτως τὸ Λ πρὸς τὸ Ν. ἐπεὶ οὖν τρία μεγέθη ἐστὶ τὰ Η, Κ, Μ, καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Θ, Λ, Ν, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, δι᾽ ἴσου ἄρα, εἰ ὑπερέχει τὸ Η τοῦ Μ, ὑπερέχει καὶ τὸ Θ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Η, Θ τῶν Α, Δ ἰσάκις πολλαπλάσια, τὰ δὲ Μ, Ν τῶν Γ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια. ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ζ. ἐὰν ἄρα ᾖ ὁποσαοῦν μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",281]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'5.23'"],["Rule","'Text'","'If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, they will also be in the same ratio ex aequali.'"],["Rule","'TextWordCount'",40],["Rule","'GreekText'","'ἐὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",15]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'Let there be three magnitudes A, B, C, and others equal to them in multitude, which, taken two and two together, are in the same proportion, namely D, E, F; and let the proportion of them be perturbed, so that, as A is to B, so is E to F, and, as B is to C, so is D to E; I say that, as A is to C, so is D to F. Of A, B, D let equimultiples G, H, K be taken, and of C, E, F other, chance, equimultiples L, M, N. Then, since G, H are equimultiples of A, B, and parts have the same ratio as the same multiples of them, [V. 15] therefore, as A is to B, so is G to H.  For the same reason also, as E is to F, so is M to N. And, as A is to B, so is E to F; therefore also, as G is to H, so is M to N. [V. 11]  Next, since, as B is to C, so is D to E, alternately, also, as B is to D, so is C to E. [V. 16] And, since H, K are equimultiples of B, D, and parts have the same ratio as their equimultiples, therefore, as B is to D, so is H to K. [V. 15]  But, as B is to D, so is C to E; therefore also, as H is to K, so is C to E. [V. 11]  Again, since L, M are equimultiples of C, E, therefore, as C is to E, so is L to M. [V. 15]  But, as C is to E, so is H to K; therefore also, as H is to K, so is L to M, [V. 11]and, alternately, as H is to L, so is K to M. [V. 16] But it was also proved that, as G is to H, so is M to N.  Since, then, there are three magnitudes G, H, L, and others equal to them in multitude K, M, N, which taken two and two together are in the same ratio, and the proportion of them is perturbed, therefore, ex aequali, if G is in excess of L, K is also in excess of N; if equal, equal; and if less, less. [V. 21] And G, K are equimultiples of A, D, and L, N of C, F. Therefore, as A is to C, so is D to F.'"],["Rule","'ProofWordCount'",417],["Rule","'GreekProof'","'ἔστω τρία μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ τὰ Δ, Ε, Ζ, ἔστω δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ε: λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ζ. εἰλήφθω τῶν μὲν Α, Β, Δ ἰσάκις πολλαπλάσια τὰ Η, Θ, Κ, τῶν δὲ Γ, Ε, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσια τὰ Η, Θ τῶν Α, Β, τὰ δὲ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Η πρὸς τὸ Θ. διὰ τὰ αὐτὰ δὴ καὶ ὡς τὸ Ε πρὸς τὸ Ζ, οὕτως τὸ Μ πρὸς τὸ Ν: καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ: καὶ ὡς ἄρα τὸ Η πρὸς τὸ Θ, οὕτως τὸ Μ πρὸς τὸ Ν. καὶ ἐπεί ἐστιν ὡς τὸ Β πρὸς τὸ Γ, οὕτως τὸ δ πρὸς τὸ Ε, καὶ ἐναλλὰξ ὡς τὸ Β πρὸς τὸ Δ, οὕτως τὸ Γ πρὸς τὸ Ε. καὶ ἐπεὶ τὰ Θ, Κ τῶν Β, Δ ἰσάκις ἐστὶ πολλαπλάσια, τὰ δὲ μέρη τοῖς ἰσάκις πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ Β πρὸς τὸ Δ, οὕτως τὸ Θ πρὸς τὸ Κ. ἀλλ᾽ ὡς τὸ Β πρὸς τὸ Δ, οὕτως τὸ Γ πρὸς τὸ Ε: καὶ ὡς ἄρα τὸ Θ πρὸς τὸ Κ, οὕτως τὸ Γ πρὸς τὸ Ε. πάλιν, ἐπεὶ τὰ Λ, Μ τῶν Γ, Ε ἰσάκις ἐστι πολλαπλάσια, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ε, οὕτως τὸ Λ πρὸς τὸ Μ. ἀλλ᾽ ὡς τὸ Γ πρὸς τὸ Ε, οὕτως τὸ Θ πρὸς τὸ Κ: καὶ ὡς ἄρα τὸ Θ πρὸς τὸ Κ, οὕτως τὸ Λ πρὸς τὸ Μ, καὶ ἐναλλὰξ ὡς τὸ Θ πρὸς τὸ Λ, τὸ Κ πρὸς τὸ Μ. ἐδείχθη δὲ καὶ ὡς τὸ Η πρὸς τὸ Θ, οὕτως τὸ Μ πρὸς τὸ Ν. ἐπεὶ οὖν τρία μεγέθη ἐστὶ τὰ Η, Θ, Λ, καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Κ, Μ, Ν σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, καί ἐστιν αὐτῶν τεταραγμένη ἡ ἀναλογία, δι᾽ ἴσου ἄρα, εἰ ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Η, Κ τῶν Α, Δ ἰσάκις πολλαπλάσια, τὰ δὲ Λ, Ν τῶν Γ, Ζ. ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ζ. ἐὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",471]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'5.24'"],["Rule","'Text'","'If a first magnitude have to a second the same ratio as a third has to a fourth, and also a fifth have to the second the same ratio as a sixth to the fourth, the first and fifth added together will have to the second the same ratio as the third and sixth have to the fourth.'"],["Rule","'TextWordCount'",58],["Rule","'GreekText'","'ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, ἔχῃ δὲ καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν λόγον καὶ ἕκτον πρὸς τέταρτον, καὶ συντεθὲν πρῶτον καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν ἕξει λόγον καὶ τρίτον καὶ ἕκτον πρὸς τέταρτον.'"],["Rule","'GreekTextWordCount'",42],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",18]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'Let a first magnitude AB have to a second C the same ratio as a third DE has to a fourth F; and let also a fifth BG have to the second C the same ratio as a sixth EH has to the fourth F; I say that the first and fifth added together, AG, will have to the second C the same ratio as the third and sixth, DH, has to the fourth F. For since, as BG is to C, so is EH to F, inversely, as C is to BG, so is F to EH. Since, then, as AB is to C, so is DE to F, and, as C is to BG, so is F to EH, therefore, ex aequali, as AB is to BG, so is DE to EH. [V. 22] And, since the magnitudes are proportional separando, they will also be proportional componendo; [V. 18] therefore, as AG is to GB, so is DH to HE. But also, as BG is to C, so is EH to F; therefore, ex aequali, as AG is to C, so is DH to F. [V. 22]'"],["Rule","'ProofWordCount'",189],["Rule","'GreekProof'","'πρῶτον γὰρ τὸ ΑΒ πρὸς δεύτερον τὸ Γ τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ ΔΕ πρὸς τέταρτον τὸ Ζ, ἐχέτω δὲ καὶ πέμπτον τὸ ΒΗ πρὸς δεύτερον τὸ Γ τὸν αὐτὸν λόγον καὶ ἕκτον τὸ ΕΘ πρὸς τέταρτον τὸ Ζ: λέγω, ὅτι καὶ συντεθὲν πρῶτον καὶ πέμπτον τὸ ΑΗ πρὸς δεύτερον τὸ Γ τὸν αὐτὸν ἕξει λόγον, καὶ τρίτον καὶ ἕκτον τὸ ΔΘ πρὸς τέταρτον τὸ Ζ. ἐπεὶ γάρ ἐστιν ὡς τὸ ΒΗ πρὸς τὸ Γ, οὕτως τὸ ΕΘ πρὸς τὸ Ζ, ἀνάπαλιν ἄρα ὡς τὸ Γ πρὸς τὸ ΒΗ, οὕτως τὸ Ζ πρὸς τὸ ΕΘ. ἐπεὶ οὖν ἐστιν ὡς τὸ ΑΒ πρὸς τὸ Γ, οὕτως τὸ ΔΕ πρὸς τὸ Ζ, ὡς δὲ τὸ Γ πρὸς τὸ ΒΗ, οὕτως τὸ Ζ πρὸς τὸ ΕΘ, δι᾽ ἴσου ἄρα ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΗ, οὕτως τὸ ΔΕ πρὸς τὸ ΕΘ. καὶ ἐπεὶ διῃρημένα μεγέθη ἀνάλογόν ἐστιν, καὶ συντεθέντα ἀνάλογον ἔσται: ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ ΗΒ, οὕτως τὸ ΔΘ πρὸς τὸ ΘΕ. ἔστι δὲ καὶ ὡς τὸ ΒΗ πρὸς τὸ Γ, οὕτως τὸ ΕΘ πρὸς τὸ Ζ: δι᾽ ἴσου ἄρα ἐστὶν ὡς τὸ ΑΗ πρὸς τὸ Γ, οὕτως τὸ ΔΘ πρὸς τὸ Ζ. ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, ἔχῃ δὲ καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν λόγον καὶ ἕκτον πρὸς τέταρτον, καὶ συντεθὲν πρῶτον καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν ἕξει λόγον καὶ τρίτον καὶ ἕκτον πρὸς τέταρτον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",244]]],["Rule",["Association",["Rule","'Book'",5],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'5.25'"],["Rule","'Text'","'If four magnitudes be proportional, the greatest and the least are greater than the remaining two.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'ἐὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ μέγιστον αὐτῶν καὶ τὸ ἐλάχιστον δύο τῶν λοιπῶν μείζονά ἐστιν.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'Let the four magnitudes AB, CD, E, F be proportional so that, as AB is to CD, so is E to F, and let AB be the greatest of them and F the least; I say that AB, F are greater than CD, E. For let AG be made equal to E, and CH equal to F. Since, as AB is to CD, so is E to F, and E is equal to AG, and F to CH, therefore, as AB is to CD, so is AG to CH. And since, as the whole AB is to the whole CD, so is the part AG subtracted to the part CH subtracted, the remainder GB will also be to the remainder HD as the whole AB is to the whole CD. [V. 19]  But AB is greater than CD; therefore GB is also greater than HD. And, since AG is equal to E, and CH to F, therefore AG, F are equal to CH, E. And if, GB, HD being unequal, and GB greater, AG, F be added to GB and CH, E be added to HD, it follows that AB, F are greater than CD, E.'"],["Rule","'ProofWordCount'",196],["Rule","'GreekProof'","'ἔστω τέσσαρα μεγέθη ἀνάλογον τὰ ΑΒ, ΓΔ, Ε, Ζ, ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ Ε πρὸς τὸ Ζ, ἔστω δὲ μέγιστον μὲν αὐτῶν τὸ ΑΒ, ἐλάχιστον δὲ τὸ Ζ: λέγω, ὅτι τὰ ΑΒ, Ζ τῶν ΓΔ, Ε μείζονά ἐστιν. κείσθω γὰρ τῷ μὲν Ε ἴσον τὸ ΑΗ, τῷ δὲ Ζ ἴσον τὸ ΓΘ. ἐπεὶ οὖν ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ Ε πρὸς τὸ Ζ, ἴσον δὲ τὸ μὲν Ε τῷ ΑΗ, τὸ δὲ Ζ τῷ ΓΘ, ἔστιν ἄρα ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΑΗ πρὸς τὸ ΓΘ. καὶ ἐπεί ἐστιν ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ, οὕτως ἀφαιρεθὲν τὸ ΑΗ πρὸς ἀφαιρεθὲν τὸ ΓΘ, καὶ λοιπὸν ἄρα τὸ ΗΒ πρὸς λοιπὸν τὸ ΘΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. μεῖζον δὲ τὸ ΑΒ τοῦ ΓΔ: μεῖζον ἄρα καὶ τὸ ΗΒ τοῦ ΘΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΗ τῷ Ε, τὸ δὲ ΓΘ τῷ Ζ, τὰ ἄρα ΑΗ, Ζ ἴσα ἐστὶ τοῖς ΓΘ, ε. καὶ ἐπεὶ ἐὰν ἀνίσοις ἴσα προστεθῇ, τὰ ὅλα ἄνισά ἐστιν, ἐὰν ἄρα τῶν ΗΒ, ΘΔ ἀνίσων ὄντων καὶ μείζονος τοῦ ΗΒ τῷ μὲν ΗΒ προστεθῇ τὰ ΑΗ, Ζ, τῷ δὲ ΘΔ προστεθῇ τὰ ΓΘ, Ε, συνάγεται τὰ ΑΒ, Ζ μείζονα τῶν ΓΔ, Ε. ἐὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ μέγιστον αὐτῶν καὶ τὸ ἐλάχιστον δύο τῶν λοιπῶν μείζονά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",233]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'6.1'"],["Rule","'Text'","'Triangles and parallelograms which are under the same height are to one another as their bases.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'τὰ τρίγωνα καὶ τὰ παραλληλόγραμμα, τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",38]],["List",["Rule","'Book'",1],["Rule","'Theorem'",41]],["List",["Rule","'Book'",5],["Rule","'Definition'",5]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let ABC, ACD be triangles and EC, CF parallelograms under the same height;I say that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF. For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH bemade equal to the base BC, and any number of straight lines DK, KL equal to the base CD; let AG, AH, AK, AL be joined. Then, since CB, BG, GH are equal to one another, the triangles ABC, AGB, AHG are also equal to one another. [I. 38] Therefore, whatever multiple the base HC is of the base BC, that multiple also is the triangle AHC of the triangle ABC. For the same reason,whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC  is also in excess of the triangle ACL, and, if less, less. Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and thetriangle ABC, namely the base HC and the triangle AHC, and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC; and it has been proved that, if the base HC is in excess of the base CL, the triangle AHC  is also in excess of the triangle ALC; if equal, equal; and, if less, less. Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. [V. Def. 5] Next, since the parallelogram EC is double of the triangleABC, [I. 41] and the parallelogram FC is double of the triangle ACD, while parts have the same ratio as the same multiples of them, [V. 15] therefore, as the triangle ABC is to the triangle ACD, so isthe parallelogram EC to the parallelogram FC. Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD, and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF,therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11]'"],["Rule","'ProofWordCount'",436],["Rule","'GreekProof'","'ἔστω τρίγωνα μὲν τὰ ΑΒΓ, ΑΓΔ, παραλληλόγραμμα δὲ τὰ ΕΓ, ΓΖ ὑπὸ τὸ αὐτὸ ὕψος τὸ ΑΓ: λέγω, ὅτι ἐστὶν ὡς ἡ ΒΓ βάσις πρὸς τὴν ΓΔ βάσιν, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, καὶ τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον. Ἐκβεβλήσθω γὰρ ἡ ΒΔ ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ τὰ Θ, Λ σημεῖα, καὶ κείσθωσαν τῇ μὲν ΒΓ βάσει ἴσαι ὁσαιδηποτοῦν αἱ ΒΗ, ΗΘ, τῇ δὲ ΓΔ βάσει ἴσαι ὁσαιδηποτοῦν αἱ ΔΚ, ΚΛ, καὶ ἐπεζεύχθωσαν αἱ ΑΗ, ΑΘ, ΑΚ, ΑΛ. καὶ ἐπεὶ ἴσαι εἰσὶν αἱ ΓΒ, ΒΗ, ΗΘ ἀλλήλαις, ἴσα ἐστὶ καὶ τὰ ΑΘΗ, ΑΗΒ, ΑΒΓ τρίγωνα ἀλλήλοις. ὁσαπλασίων ἄρα ἐστὶν ἡ ΘΓ βάσις τῆς ΒΓ βάσεως, τοσαυταπλάσιόν ἐστι καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΒΓ τριγώνου. διὰ τὰ αὐτὰ δὴ ὁσαπλασίων ἐστὶν ἡ ΛΓ βάσις τῆς ΓΔ βάσεως, τοσαυταπλάσιόν ἐστι καὶ τὸ ΑΛΓ τρίγωνον τοῦ ΑΓΔ τριγώνου: καὶ εἰ ἴση ἐστὶν ἡ ΘΓ βάσις τῇ ΓΛ βάσει, ἴσον ἐστὶ καὶ τὸ ΑΘΓ τρίγωνον τῷ ΑΓΛ τριγώνῳ, καὶ εἰ ὑπερέχει ἡ ΘΓ βάσις τῆς ΓΛ βάσεως, ὑπερέχει καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΓΛ τριγώνου, καὶ εἰ ἐλάσσων, ἔλασσον. τεσσάρων δὴ ὄντων μεγεθῶν δύο μὲν βάσεων τῶν ΒΓ, ΓΔ, δύο δὲ τριγώνων τῶν ΑΒΓ, ΑΓΔ εἴληπται ἰσάκις πολλαπλάσια τῆς μὲν ΒΓ βάσεως καὶ τοῦ ΑΒΓ τριγώνου á¼¥ τε ΘΓ βάσις καὶ τὸ ΑΘΓ τρίγωνον, τῆς δὲ ΓΔ βάσεως καὶ τοῦ ΑΔΓ τριγώνου ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια á¼¥ τε ΛΓ βάσις καὶ τὸ ΑΛΓ τρίγωνον: καὶ δέδεικται, ὅτι, εἰ ὑπερέχει ἡ ΘΓ βάσις τῆς ΓΛ βάσεως, ὑπερέχει καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΛΓ τριγώνου, καὶ εἰ ἴση, ἴσον, καὶ εἰ ἐλάσσων, ἔλασσον: ἔστιν ἄρα ὡς ἡ ΒΓ βάσις πρὸς τὴν ΓΔ βάσιν, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον. καὶ ἐπεὶ τοῦ μὲν ΑΒΓ τριγώνου διπλάσιόν ἐστι τὸ ΕΓ παραλληλόγραμμον, τοῦ δὲ ΑΓΔ τριγώνου διπλάσιόν ἐστι τὸ ΖΓ παραλληλόγραμμον, τὰ δὲ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, οὕτως τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΖΓ παραλληλόγραμμον. ἐπεὶ οὖν ἐδείχθη, ὡς μὲν ἡ ΒΓ βάσις πρὸς τὴν ΓΔ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, ὡς δὲ τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, οὕτως τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον, καὶ ὡς ἄρα ἡ ΒΓ βάσις πρὸς τὴν ΓΔ βάσιν, οὕτως τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΖΓ παραλληλόγραμμον. τὰ ἄρα τρίγωνα καὶ τὰ παραλληλόγραμμα τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",406]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'6.2'"],["Rule","'Text'","'If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle.'"],["Rule","'TextWordCount'",51],["Rule","'GreekText'","'ἐὰν τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις εὐθεῖα, ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς: καὶ ἐὰν αἱ τοῦ τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς τομὰς ἐπιζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ τριγώνου πλευράν.'"],["Rule","'GreekTextWordCount'",36],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",38]],["List",["Rule","'Book'",1],["Rule","'Theorem'",39]],["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'For let DE be drawn parallel to BC, one of the sides of the triangle ABC; I say that, as BD is to DA, so is CE to EA. For let BE, CD be joined. Therefore the triangle BDE is equal to the triangle CDE; for they are on the same base DE and in the same parallels DE, BC. [I. 38] And the triangle ADE is another area. But equals have the same ratio to the same; [V. 7] therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. But, as the triangle BDE is to ADE, so is BD to DA; for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. [VI. 1] For the same reason also, as the triangle CDE is to ADE, so is CE to EA. Therefore also, as BD is to DA, so is CE to EA. [V. 11] Again, let the sides AB, AC of the triangle ABC be cut proportionally, so that, as BD is to DA, so is CE to EA; and let DE be joined. I say that DE is parallel to BC. For, with the same construction, since, as BD is to DA, so is CE to EA, but, as BD is to DA, so is the triangle BDE to the triangle ADE, and, as CE is to EA, so is the triangle CDE to the triangle ADE, [VI. 1] therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11]  Therefore each of the triangles BDE, CDE has the same ratio to ADE. Therefore the triangle BDE is equal to the triangle CDE; [V. 9] and they are on the same base DE. But equal triangles which are on the same base are also in the same parallels. [I. 39] Therefore DE is parallel to BC.'"],["Rule","'ProofWordCount'",330],["Rule","'GreekProof'","'τριγώνου γὰρ τοῦ ΑΒΓ παράλληλος μιᾷ τῶν πλευρῶν τῇ ΒΓ ἤχθω ἡ ΔΕ: λέγω, ὅτι ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΔ. ἴσον ἄρα ἐστὶ ΒΔΕ τρίγωνον τῷ ΓΔΕ τριγώνῳ: ἐπὶ γὰρ τῆς αὐτῆς βάσεώς ἐστι τῆς ΔΕ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΔΕ, ΒΓ: ἄλλο δέ τι τὸ ΑΔΕ τρίγωνον. τὰ δὲ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον: ἔστιν ἄρα ὡς τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον. ἀλλ᾽ ὡς μὲν τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως ἡ ΒΔ πρὸς τὴν ΔΑ: ὑπὸ γὰρ τὸ αὐτὸ ὕψος ὄντα τὴν ἀπὸ τοῦ Ε ἐπὶ τὴν ΑΒ κάθετον ἀγομένην πρὸς ἄλληλά εἰσιν ὡς αἱ βάσεις. διὰ τὰ αὐτὰ δὴ ὡς τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ: καὶ ὡς ἄρα ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ. ἀλλὰ δὴ αἱ τοῦ ΑΒΓ τριγώνου πλευραὶ αἱ ΑΒ, ΑΓ ἀνάλογον τετμήσθωσαν, ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ, καὶ ἐπεζεύχθω ἡ ΔΕ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΔΕ τῇ ΒΓ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ, ἀλλ᾽ ὡς μὲν ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, ὡς δὲ ἡ ΓΕ πρὸς τὴν ΕΑ, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, καὶ ὡς ἄρα τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον. ἑκάτερον ἄρα τῶν ΒΔΕ, ΓΔΕ τριγώνων πρὸς τὸ ΑΔΕ τὸν αὐτὸν ἔχει λόγον. ἴσον ἄρα ἐστὶ τὸ ΒΔΕ τρίγωνον τῷ ΓΔΕ τριγώνῳ: καί εἰσιν ἐπὶ τῆς αὐτῆς βάσεως τῆς ΔΕ. τὰ δὲ ἴσα τρίγωνα καὶ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. παράλληλος ἄρα ἐστὶν ἡ ΔΕ τῇ ΒΓ. ἐὰν ἄρα τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις εὐθεῖα, ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς: καὶ ἐὰν αἱ τοῦ τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς τομὰς ἐπιζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ τριγώνου πλευράν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",356]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'6.3'"],["Rule","'Text'","'If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.'"],["Rule","'TextWordCount'",73],["Rule","'GreekText'","'ἐὰν τριγώνου ἡ γωνία δίχα τμηθῇ, ἡ δὲ τέμνουσα τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς: καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τεμεῖ τὴν τοῦ τριγώνου γωνίαν.'"],["Rule","'GreekTextWordCount'",59],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that, as BD is to CD, so is BA to AC. For let CE be drawn through C parallel to DA, and let BA  be carried through and meet it at E. Then, since the straight line AC falls upon the parallels AD, EC, the angle ACE is equal to the angle CAD. [I. 29]  But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE. Again, since the straight line BAE falls upon the parallels AD, EC, the exterior angle BAD is equal to the interior angle AEC. [I. 29]  But the angle ACE was also proved equal to the angle BAD; therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC. [I. 6]  And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE. But AE is equal to AC; [VI. 2] therefore, as BD is to DC, so is BA to AC. Again, let BA be to AC as BD to DC, and let AD be joined; I say that the angle BAC has been bisected by the straight line A.D. For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE: for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE. [V. 11] Therefore AC is equal to AE, [V. 9] so that the angle AEC is also equal to the angle ACE. [I. 5] But the angle AEC is equal to the exterior angle BAD, [I. 29] and the angle ACE is equal to the alternate angle CAD; [id.]therefore the angle BAD is also equal to the angle CAD. Therefore the angle BAC has been bisected by the straight line AD.'"],["Rule","'ProofWordCount'",366],["Rule","'GreekProof'","'ἔστω τρίγωνον τὸ ΑΒΓ, καὶ τετμήσθω ἡ ὑπὸ ΒΑΓ γωνία δίχα ὑπὸ τῆς ΑΔ εὐθείας: λέγω, ὅτι ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΓΔ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ. ἤχθω γὰρ διὰ τοῦ Γ τῇ ΔΑ παράλληλος ἡ ΓΕ καὶ διαχθεῖσα ἡ ΒΑ συμπιπτέτω αὐτῇ κατὰ τὸ Ε. καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΔ, ΕΓ εὐθεῖα ἐνέπεσεν ἡ ΑΓ, ἡ ἄρα ὑπὸ ΑΓΕ γωνία ἴση ἐστὶ τῇ ὑπὸ ΓΑΔ. ἀλλ᾽ ἡ ὑπὸ ΓΑΔ τῇ ὑπὸ ΒΑΔ ὑπόκειται ἴση: καὶ ἡ ὑπὸ ΒΑΔ ἄρα τῇ ὑπὸ ΑΓΕ ἐστιν ἴση. πάλιν, ἐπεὶ εἰς παραλλήλους τὰς ΑΔ, ΕΓ εὐθεῖα ἐνέπεσεν ἡ ΒΑΕ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΑΔ ἴση ἐστὶ τῇ ἐντὸς τῇ ὑπὸ ΑΕΓ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΔ ἴση: καὶ ἡ ὑπὸ ΑΓΕ ἄρα γωνία τῇ ὑπὸ ΑΕΓ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΑΕ πλευρᾷ τῇ ΑΓ ἐστιν ἴση. καὶ ἐπεὶ τριγώνου τοῦ ΒΓΕ παρὰ μίαν τῶν πλευρῶν τὴν ΕΓ ἦκται ἡ ΑΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΕ. ἴση δὲ ἡ ΑΕ τῇ ΑΓ: ὡς ἄρα ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ. ἀλλὰ δὴ ἔστω ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ, καὶ ἐπεζεύχθω ἡ ΑΔ: λέγω, ὅτι δίχα τέτμηται ἡ ὑπὸ ΒΑΓ γωνία ὑπὸ τῆς ΑΔ εὐθείας. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ, ἀλλὰ καὶ ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἐστὶν ἡ ΒΑ πρὸς τὴν ΑΕ: τριγώνου γὰρ τοῦ ΒΓΕ παρὰ μίαν τὴν ΕΓ ἦκται ἡ ΑΔ: καὶ ὡς ἄρα ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΕ. ἴση ἄρα ἡ ΑΓ τῇ ΑΕ: ὥστε καὶ γωνία ἡ ὑπὸ ΑΕΓ τῇ ὑπὸ ΑΓΕ ἐστιν ἴση. ἀλλ᾽ ἡ μὲν ὑπὸ ΑΕΓ τῇ ἐκτὸς τῇ ὑπὸ ΒΑΔ ἐστιν ἴση, ἡ δὲ ὑπὸ ΑΓΕ τῇ ἐναλλὰξ τῇ ὑπὸ ΓΑΔ ἐστιν ἴση: καὶ ἡ ὑπὸ ΒΑΔ ἄρα τῇ ὑπὸ ΓΑΔ ἐστιν ἴση. ἡ ἄρα ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΔ εὐθείας. ἐὰν ἄρα τριγώνου ἡ γωνία δίχα τμηθῇ, ἡ δὲ τέμνουσα τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς: καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τέμνει τὴν τοῦ τριγώνου γωνίαν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",407]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'6.4'"],["Rule","'Text'","'In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'τῶν ἰσογωνίων τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Postulate'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",17]],["List",["Rule","'Book'",1],["Rule","'Theorem'",28]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED; I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles. For let BC be placed in a straight line with CE. Then, since the angles ABC, ACB are less than two right angles, [I. 17] and the angle ACB is equal to the angle DEC, therefore the angles ABC, DEC are less than two right angles; therefore BA, ED, when produced, will meet. [I. Post. 5] Let them be produced and meet at F. Now, since the angle DCE is equal to the angle ABC, BF is parallel to CD. [I. 28]  Again, since the angle ACB is equal to the angle DEC, AC is parallel to FE. [I. 28]  Therefore FACD is a parallelogram; therefore FA is equal to DC, and AC to FD. [I. 34] And, since AC has been drawn parallel to FE, one side of the triangle FBE, therefore, as BA is to AF, so is BC to CE. [VI. 2] But AF is equal to CD; therefore, as BA is to CD, so is BC to CE, and alternately, as AB is to BC, so is DC to CE. [V. 16] Again, since CD is parallel to BF, therefore, as BC is to CE, so is FD to DE. [VI. 2] But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16] Since then it was proved that, as AB is to BC, so is DC to CE, and, as BC is to CA, so is CE to ED; therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22]'"],["Rule","'ProofWordCount'",326],["Rule","'GreekProof'","'ἔστω ἰσογώνια τρίγωνα τὰ ΑΒΓ, ΔΓΕ ἴσην ἔχοντα τὴν μὲν ὑπὸ ΑΒΓ γωνίαν τῇ ὑπὸ ΔΓΕ, τὴν δὲ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΕ καὶ ἔτι τὴν ὑπὸ ΑΓΒ τῇ ὑπὸ ΓΕΔ: λέγω, ὅτι τῶν ΑΒΓ, ΔΓΕ τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι. κείσθω γὰρ ἐπ᾽ εὐθείας ἡ ΒΓ τῇ ΓΕ. καὶ ἐπεὶ αἱ ὑπὸ ΑΒΓ, ΑΓΒ γωνίαι δύο ὀρθῶν ἐλάττονές εἰσιν, ἴση δὲ ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΕΓ, αἱ ἄρα ὑπὸ ΑΒΓ, ΔΕΓ δύο ὀρθῶν ἐλάττονές εἰσιν: αἱ ΒΑ, ΕΔ ἄρα ἐκβαλλόμεναι συμπεσοῦνται. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Ζ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΓΕ γωνία τῇ ὑπὸ ΑΒΓ, παράλληλός ἐστιν ἡ ΒΖ τῇ ΓΔ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΕΓ, παράλληλός ἐστιν ἡ ΑΓ τῇ ΖΕ. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΑΓΔ: ἴση ἄρα ἡ μὲν ΖΑ τῇ ΔΓ, ἡ δὲ ΑΓ τῇ ΖΔ. καὶ ἐπεὶ τριγώνου τοῦ ΖΒΕ παρὰ μίαν τὴν ΖΕ ἦκται ἡ ΑΓ, ἔστιν ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΖ, οὕτως ἡ ΒΓ πρὸς τὴν ΓΕ. ἴση δὲ ἡ ΑΖ τῇ ΓΔ: ὡς ἄρα ἡ ΒΑ πρὸς τὴν ΓΔ, οὕτως ἡ ΒΓ πρὸς τὴν ΓΕ, καὶ ἐναλλὰξ ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΓ πρὸς τὴν ΓΕ. πάλιν, ἐπεὶ παράλληλός ἐστιν ἡ ΓΔ τῇ ΒΖ, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΕ, οὕτως ἡ ΖΔ πρὸς τὴν ΔΕ. ἴση δὲ ἡ ΖΔ τῇ ΑΓ: ὡς ἄρα ἡ ΒΓ πρὸς τὴν ΓΕ, οὕτως ἡ ΑΓ πρὸς τὴν ΔΕ, καὶ ἐναλλὰξ ὡς ἡ ΒΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΔ. ἐπεὶ οὖν ἐδείχθη ὡς μὲν ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΓ πρὸς τὴν ΓΕ, ὡς δὲ ἡ ΒΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΔ, δι᾽ ἴσου ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΓΔ πρὸς τὴν ΔΕ. τῶν ἄρα ἰσογωνίων τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",336]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'6.5'"],["Rule","'Text'","'If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα τὰς πλευρὰς ἀνάλογον ἔχῃ, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let ABC, DEF be two triangles having their sides proportional, so that, as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF, the angle BCA to the angle EFD, and further the angle BAC to the angle EDF. For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23] therefore the remaining angle at A is equal to the remaining angle at G. [I. 32]  Therefore the triangle ABC is equiangular with the triangle GEF. Therefore in the triangles ABC, GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [VI. 4] therefore, as AB is to BC, so is GE to EF. But, as AB is to BC, so by hypothesis is DE to EF; therefore, as DE is to EF, so is GE to EF. [V. 11]  Therefore each of the straight lines DE, GE has the same ratio to EF; therefore DE is equal to GE. [V. 9]  For the same reason DF is also equal to GF. Since then DE is equal to EG, and EF is common, the two sides DE, EF are equal to the two sides GE, EF; and the base DF is equal to the base FG; therefore the angle DEF is equal to the angle GEF, [I. 8] and the triangle DEF is equal to the triangle GEF, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle DFE is also equal to the angle GFE, and the angle EDF to the angle EGF. And, since the angle FED is equal to the angle GEF, while the angle GEF is equal to the angle ABC, therefore the angle ABC is also equal to the angle DEF. For the same reason the angle ACB is also equal to the angle DFE, and further, the angle at A to the angle at D; therefore the triangle ABC is equiangular with the triangle DEF.'"],["Rule","'ProofWordCount'",411],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς πλευρὰς ἀνάλογον ἔχοντα, ὡς μὲν τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, ὡς δὲ τὴν ΒΓ πρὸς τὴν ΓΑ, οὕτως τὴν ΕΖ πρὸς τὴν ΖΔ, καὶ ἔτι ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, οὕτως τὴν ΕΔ πρὸς τὴν ΔΖ. λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ καὶ ἴσας ἕξουσι τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν, τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ καὶ ἔτι τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ. συνεστάτω γὰρ πρὸς τῇ ΕΖ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Ε, Ζ τῇ μὲν ὑπὸ ΑΒΓ γωνίᾳ ἴση ἡ ὑπὸ ΖΕΗ, τῇ δὲ ὑπὸ ΑΓΒ ἴση ἡ ὑπὸ ΕΖΗ: λοιπὴ ἄρα ἡ πρὸς τῷ Α λοιπῇ τῇ πρὸς τῷ Η ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΕΗΖ τριγώνῳ. τῶν ἄρα ΑΒΓ, ΕΗΖ τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι: ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΖ. ἀλλ᾽ ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ὑπόκειται ἡ ΔΕ πρὸς τὴν ΕΖ: ὡς ἄρα ἡ ΔΕ πρὸς τὴν ΕΖ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΖ. ἑκατέρα ἄρα τῶν ΔΕ, ΗΕ πρὸς τὴν ΕΖ τὸν αὐτὸν ἔχει λόγον: ἴση ἄρα ἐστὶν ἡ ΔΕ τῇ ΗΕ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΔΖ τῇ ΗΖ ἐστιν ἴση. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΕ τῇ ΕΗ, κοινὴ δὲ ἡ ΕΖ, δύο δὴ αἱ ΔΕ, ΕΖ δυσὶ ταῖς ΗΕ, ΕΖ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΗ ἐστιν ἴση: γωνία ἄρα ἡ ὑπὸ ΔΕΖ γωνίᾳ τῇ ὑπὸ ΗΕΖ ἐστιν ἴση, καὶ τὸ ΔΕΖ τρίγωνον τῷ ΗΕΖ τριγώνῳ ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν. ἴση ἄρα ἐστὶ καὶ ἡ μὲν ὑπὸ ΔΖΕ γωνία τῇ ὑπὸ ΗΖΕ, ἡ δὲ ὑπὸ ΕΔΖ τῇ ὑπὸ ΕΗΖ. καὶ ἐπεὶ ἡ μὲν ὑπὸ ΖΕΔ τῇ ὑπὸ ΗΕΖ ἐστιν ἴση, ἀλλ᾽ ἡ ὑπὸ ΗΕΖ τῇ ὑπὸ ΑΒΓ, καὶ ἡ ὑπὸ ΑΒΓ ἄρα γωνία τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση, καὶ ἔτι ἡ πρὸς τῷ Α τῇ πρὸς τῷ Δ: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἐὰν ἄρα δύο τρίγωνα τὰς πλευρὰς ἀνάλογον ἔχῃ, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",415]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'6.6'"],["Rule","'Text'","'If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν.'"],["Rule","'GreekTextWordCount'",33],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE. For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23] therefore the remaining angle at B is equal to the remaining angle at G. [I. 32]  Therefore the triangle ABC is equiangular with the triangle DGF. Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4] But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF. [V. 11] Therefore ED is equal to DG; [V. 9] and DF is common; therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle DFG is equal to the angle DFE, and the angle DGF to the angle DEF. But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE. And, by hypothesis, the angle BAC is also equal to the angle EDF; therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32] therefore the triangle ABC is equiangular with the triangle DEF.'"],["Rule","'ProofWordCount'",329],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ μίαν γωνίαν τὴν ὑπὸ ΒΑΓ μιᾷ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, οὕτως τὴν ΕΔ πρὸς τὴν ΔΖ: λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ καὶ ἴσην ἕξει τὴν ὑπὸ ΑΒΓ γωνίαν τῇ ὑπὸ ΔΕΖ, τὴν δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ. συνεστάτω γὰρ πρὸς τῇ ΔΖ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Δ, Ζ ὁποτέρᾳ μὲν τῶν ὑπὸ ΒΑΓ, ΕΔΖ ἴση ἡ ὑπὸ ΖΔΗ, τῇ δὲ ὑπὸ ΑΓΒ ἴση ἡ ὑπὸ ΔΖΗ: λοιπὴ ἄρα ἡ πρὸς τῷ Β γωνία λοιπῇ τῇ πρὸς τῷ Η ἴση ἐστίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΗΖ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΗΔ πρὸς τὴν ΔΖ. ὑπόκειται δὲ καὶ ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΕΔ πρὸς τὴν ΔΖ: καὶ ὡς ἄρα ἡ ΕΔ πρὸς τὴν ΔΖ, οὕτως ἡ ΗΔ πρὸς τὴν ΔΖ. ἴση ἄρα ἡ ΕΔ τῇ ΔΗ: καὶ κοινὴ ἡ ΔΖ: δύο δὴ αἱ ΕΔ, ΔΖ δυσὶ ταῖς ΗΔ, ΔΖ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΕΔΖ γωνίᾳ τῇ ὑπὸ ΗΔΖ ἐστιν ἴση: βάσις ἄρα ἡ ΕΖ βάσει τῇ ΗΖ ἐστιν ἴση, καὶ τὸ ΔΕΖ τρίγωνον τῷ ΗΔΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν. ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΔΖΗ τῇ ὑπὸ ΔΖΕ, ἡ δὲ ὑπὸ ΔΗΖ τῇ ὑπὸ ΔΕΖ. ἀλλ᾽ ἡ ὑπὸ ΔΖΗ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση: καὶ ἡ ὑπὸ ΑΓΒ ἄρα τῇ ὑπὸ ΔΖΕ ἐστιν ἴση. ὑπόκειται δὲ καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ἴση: καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Β λοιπῇ τῇ πρὸς τῷ Ε ἴση ἐστίν: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἐὰν ἄρα δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",337]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'6.7'"],["Rule","'Text'","'If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional.'"],["Rule","'TextWordCount'",48],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ λοιπῶν ἑκατέραν ἅμα ἤτοι ἐλάσσονα á¼¢ μὴ ἐλάσσονα ὀρθῆς, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, περὶ ἃς ἀνάλογόν εἰσιν αἱ πλευραί.'"],["Rule","'GreekTextWordCount'",42],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",13]],["List",["Rule","'Book'",1],["Rule","'Theorem'",17]],["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle; I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F. For, if the angle ABC is unequal to the angle DEF, one of them is greater. Let the angle ABC be greater; and on the straight line AB, and at the point B on it, let the angle ABG be constructed equal to the angle DEF. [I. 23] Then, since the angle A is equal to D, and the angle ABG to the angle DEF, therefore the remaining angle AGB is equal to the remaining angle DFE. [I. 32] Therefore the triangle ABG is equiangular with the triangle DEF. Therefore, as AB is to BG, so is DE to EF [VI. 4] But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11] therefore BC is equal to BG, [V. 9]so that the angle at C is also equal to the angle BGC. [I. 5] But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [I. 13] And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle. But it is by hypothesis less than a right angle: which is absurd. Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal to it. But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32] Therefore the triangle ABC is equiangular with the triangle DEF. But, again, let each of the angles at C, F be supposed not less than a right angle; I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF. For, with the same construction, we can prove similarly that BC is equal to BG; so that the angle at C is also equal to the angle BGC. [I. 5] But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle. Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17] Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it. But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32] Therefore the triangle ABC is equiangular with the triangle DEF.'"],["Rule","'ProofWordCount'",546],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ, περὶ δὲ ἄλλας γωνίας τὰς ὑπὸ ΑΒΓ, ΔΕΖ τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, τῶν δὲ λοιπῶν τῶν πρὸς τοῖς Γ, Ζ πρότερον ἑκατέραν ἅμα ἐλάσσονα ὀρθῆς: λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ, καὶ ἴση ἔσται ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ, καὶ λοιπὴ δηλονότι ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση. εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΑΒΓ. καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β τῇ ὑπὸ ΔΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΑΒΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν Α γωνία τῇ Δ, ἡ δὲ ὑπὸ ΑΒΗ τῇ ὑπὸ ΔΕΖ, λοιπὴ ἄρα ἡ ὑπὸ ΑΗΒ λοιπῇ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΒΗ, οὕτως ἡ ΔΕ πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΔΕ πρὸς τὴν ΕΖ, οὕτως ὑπόκειται ἡ ΑΒ πρὸς τὴν ΒΓ: ἡ ΑΒ ἄρα πρὸς ἑκατέραν τῶν ΒΓ, ΒΗ τὸν αὐτὸν ἔχει λόγον: ἴση ἄρα ἡ ΒΓ τῇ ΒΗ. ὥστε καὶ γωνία ἡ πρὸς τῷ Γ γωνίᾳ τῇ ὑπὸ ΒΗΓ ἐστιν ἴση. ἐλάττων δὲ ὀρθῆς ὑπόκειται ἡ πρὸς τῷ Γ: ἐλάττων ἄρα ἐστὶν ὀρθῆς καὶ ἡ ὑπὸ ΒΗΓ: ὥστε ἡ ἐφεξῆς αὐτῇ γωνία ἡ ὑπὸ ΑΗΒ μείζων ἐστὶν ὀρθῆς. καὶ ἐδείχθη ἴση οὖσα τῇ πρὸς τῷ Ζ: καὶ ἡ πρὸς τῷ Ζ ἄρα μείζων ἐστὶν ὀρθῆς. ὑπόκειται δὲ ἐλάσσων ὀρθῆς: ὅπερ ἐστὶν ἄτοπον. οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ: ἴση ἄρα. ἔστι δὲ καὶ ἡ πρὸς τῷ Α ἴση τῇ πρὸς τῷ Δ: καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση ἐστίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἀλλὰ δὴ πάλιν ὑποκείσθω ἑκατέρα τῶν πρὸς τοῖς Γ, Ζ μὴ ἐλάσσων ὀρθῆς: λέγω πάλιν, ὅτι καὶ οὕτως ἐστὶν ἰσογώνιον τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ: ὥστε καὶ γωνία ἡ πρὸς τῷ Γ τῇ ὑπὸ ΒΗΓ ἴση ἐστίν. οὐκ ἐλάττων δὲ ὀρθῆς ἡ πρὸς τῷ Γ: οὐκ ἐλάττων ἄρα ὀρθῆς οὐδὲ ἡ ὑπὸ ΒΗΓ. τριγώνου δὴ τοῦ ΒΗΓ αἱ δύο γωνίαι δύο ὀρθῶν οὔκ εἰσιν ἐλάττονες: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα πάλιν ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ: ἴση ἄρα. ἔστι δὲ καὶ ἡ πρὸς τῷ Α τῇ πρὸς τῷ Δ ἴση: λοιπὴ ἄρα ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση ἐστίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἐὰν ἄρα δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ λοιπῶν ἑκατέραν ἅμα ἐλάττονα á¼¢ μὴ ἐλάττονα ὀρθῆς, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, περὶ ἃς ἀνάλογόν εἰσιν αἱ πλευραί: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",498]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'6.8'"],["Rule","'Text'","'If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, τὰ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another. For, since the angle BAC is equal to the angle ADB, for each is right, and the angle at B is common to the two triangles ABC and ABD, therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32] therefore the triangle ABC is equiangular with the triangle ABD. Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD, so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD, and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4] Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional. Therefore the triangle ABC is similar to the triangle ABD. [VI. Def. 1] Similarly we can prove that the triangle ABC is also similar to the triangle ADC; therefore each of the triangles ABD, ADC is similar to the whole ABC. I say next that the triangles ABD, ADC are also similar to one another. For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32] therefore the triangle ABD is equiangular with the triangle ADC. Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1]'"],["Rule","'ProofWordCount'",384],["Rule","'GreekProof'","'ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν, καὶ ἤχθω ἀπὸ τοῦ Α ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ: λέγω, ὅτι ὅμοιόν ἐστιν ἑκάτερον τῶν ΑΒΔ, ΑΔΓ τριγώνων ὅλῳ τῷ ΑΒΓ καὶ ἔτι ἀλλήλοις. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΑΔΒ: ὀρθὴ γὰρ ἑκατέρα: καὶ κοινὴ τῶν δύο τριγώνων τοῦ τε ΑΒΓ καὶ τοῦ ΑΒΔ ἡ πρὸς τῷ Β, λοιπὴ ἄρα ἡ ὑπὸ ΑΓΒ λοιπῇ τῇ ὑπὸ ΒΑΔ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΒΓ ὑποτείνουσα τὴν ὀρθὴν τοῦ ΑΒΓ τριγώνου πρὸς τὴν ΒΑ ὑποτείνουσαν τὴν ὀρθὴν τοῦ ΑΒΔ τριγώνου, οὕτως αὐτὴ ἡ ΑΒ ὑποτείνουσα τὴν πρὸς τῷ Γ γωνίαν τοῦ ΑΒΓ τριγώνου πρὸς τὴν ΒΔ ὑποτείνουσαν τὴν ἴσην τὴν ὑπὸ ΒΑΔ τοῦ ΑΒΔ τριγώνου, καὶ ἔτι ἡ ΑΓ πρὸς τὴν ΑΔ ὑποτείνουσαν τὴν πρὸς τῷ Β γωνίαν κοινὴν τῶν δύο τριγώνων. τὸ ΑΒΓ ἄρα τρίγωνον τῷ ΑΒΔ τριγώνῳ ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. ὅμοιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τῷ ΑΔΓ τριγώνῳ ὅμοιόν ἐστι τὸ ΑΒΓ τρίγωνον: ἑκάτερον ἄρα τῶν ΑΒΔ, ΑΔΓ τριγώνων ὅμοιόν ἐστιν ὅλῳ τῷ ΑΒΓ. λέγω δή, ὅτι καὶ ἀλλήλοις ἐστὶν ὅμοια τὰ ΑΒΔ, ΑΔΓ τρίγωνα. ἐπεὶ γὰρ ὀρθὴ ἡ ὑπὸ ΒΔΑ ὀρθῇ τῇ ὑπὸ ΑΔΓ ἐστιν ἴση, ἀλλὰ μὴν καὶ ἡ ὑπὸ ΒΑΔ τῇ πρὸς τῷ Γ ἐδείχθη ἴση, καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Β λοιπῇ τῇ ὑπὸ ΔΑΓ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΑΔΓ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΒΔ τοῦ ΑΒΔ τριγώνου ὑποτείνουσα τὴν ὑπὸ ΒΑΔ πρὸς τὴν ΔΑ τοῦ ΑΔΓ τριγώνου ὑποτείνουσαν τὴν πρὸς τῷ Γ ἴσην τῇ ὑπὸ ΒΑΔ, οὕτως αὐτὴ ἡ ΑΔ τοῦ ΑΒΔ τριγώνου ὑποτείνουσα τὴν πρὸς τῷ Β γωνίαν πρὸς τὴν ΔΓ ὑποτείνουσαν τὴν ὑπὸ ΔΑΓ τοῦ ΑΔΓ τριγώνου ἴσην τῇ πρὸς τῷ Β, καὶ ἔτι ἡ ΒΑ πρὸς τὴν ΑΓ ὑποτείνουσαι τὰς ὀρθάς: ὅμοιον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΑΔΓ τριγώνῳ. ἐὰν ἄρα ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, τὰ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, ἡ ἀχθεῖσα τῶν τῆς βάσεως τμημάτων μέση ἀνάλογόν ἐστιν: ὅπερ ἔδει δεῖξαι καὶ ἔτι τῆς βάσεως καὶ ἑνὸς ὁποιουοῦν τῶν τμημάτων ἡ πρὸς τῷ τμήματι πλευρὰ μέση ἀνάλογόν ἐστιν.'"],["Rule","'GreekProofWordCount'",406]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'6.9'"],["Rule","'Text'","'From a given straight line to cut off a prescribed part.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'τῆς δοθείσης εὐθείας τὸ προσταχθὲν μέρος ἀφελεῖν.'"],["Rule","'GreekTextWordCount'",7],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let AB be the given straight line; thus it is required to cut off from AB a prescribed part. Let the third part be that prescribed. Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3] Let BC be joined, and through D let DF be drawn parallel to it. [I. 31] Then, since FD has been drawn parallel to BC, one of the sides of the triangle ABC, therefore, proportionally, as CD is to DA, so is BF to FA. [VI. 2] But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF. Therefore from the given straight line AB the prescribed third part AF has been cut off.'"],["Rule","'ProofWordCount'",144],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ: δεῖ δὴ τῆς ΑΒ τὸ προσταχθὲν μέρος ἀφελεῖν. Ἐπιτετάχθω δὴ τὸ τρίτον. καὶ διήχθω τις ἀπὸ τοῦ α εὐθεῖα ἡ ΑΓ γωνίαν περιέχουσα μετὰ τῆς ΑΒ τυχοῦσαν: καὶ εἰλήφθω τυχὸν σημεῖον ἐπὶ τῆς ΑΓ τὸ Δ, καὶ κείσθωσαν τῇ ΑΔ ἴσαι αἱ ΔΕ, ΕΓ. καὶ ἐπεζεύχθω ἡ ΒΓ, καὶ διὰ τοῦ Δ παράλληλος αὐτῇ ἤχθω ἡ ΔΖ. ἐπεὶ οὖν τριγώνου τοῦ ΑΒΓ παρὰ μίαν τῶν πλευρῶν τὴν ΒΓ ἦκται ἡ ΖΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΒΖ πρὸς τὴν ΖΑ. διπλῆ δὲ ἡ ΓΔ τῆς ΔΑ: διπλῆ ἄρα καὶ ἡ ΒΖ τῆς ΖΑ: τριπλῆ ἄρα ἡ ΒΑ τῆς ΑΖ. τῆς ἄρα δοθείσης εὐθείας τῆς ΑΒ τὸ ἐπιταχθὲν τρίτον μέρος ἀφῄρηται τὸ ΑΖ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",127]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'6.10'"],["Rule","'Text'","'To cut a given uncut straight line similarly to a given cut straight line.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'τὴν δοθεῖσαν εὐθεῖαν ἄτμητον τῇ δοθείσῃ τετμημένῃ ὁμοίως τεμεῖν.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31] Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34] Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2] But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF. Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2] But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA. Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC.'"],["Rule","'ProofWordCount'",222],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἄτμητος ἡ ΑΒ, ἡ δὲ τετμημένη ἡ ΑΓ κατὰ τὰ Δ, Ε σημεῖα, καὶ κείσθωσαν ὥστε γωνίαν τυχοῦσαν περιέχειν, καὶ ἐπεζεύχθω ἡ ΓΒ, καὶ διὰ τῶν Δ, Ε τῇ ΒΓ παράλληλοι ἤχθωσαν αἱ ΔΖ, ΕΗ, διὰ δὲ τοῦ Δ τῇ ΑΒ παράλληλος ἤχθω ἡ ΔΘΚ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΖΘ, ΘΒ: ἴση ἄρα ἡ μὲν ΔΘ τῇ ΖΗ, ἡ δὲ ΘΚ τῇ ΗΒ. καὶ ἐπεὶ τριγώνου τοῦ ΔΚΓ παρὰ μίαν τῶν πλευρῶν τὴν ΚΓ εὐθεῖα ἦκται ἡ ΘΕ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΚΘ πρὸς τὴν ΘΔ. ἴση δὲ ἡ μὲν ΚΘ τῇ ΒΗ, ἡ δὲ ΘΔ τῇ ΗΖ. ἔστιν ἄρα ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ. πάλιν, ἐπεὶ τριγώνου τοῦ ΑΗΕ παρὰ μίαν τῶν πλευρῶν τὴν ΗΕ ἦκται ἡ ΖΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΕΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ. ἐδείχθη δὲ καὶ ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ: ἔστιν ἄρα ὡς μὲν ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ, ὡς δὲ ἡ ΕΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ. ἡ ἄρα δοθεῖσα εὐθεῖα ἄτμητος ἡ ΑΒ τῇ δοθείσῃ εὐθείᾳ τετμημένῃ τῇ ΑΓ ὁμοίως τέτμηται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",215]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'6.11'"],["Rule","'Text'","'To two given straight lines to find a third proportional.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'δύο δοθεισῶν εὐθειῶν τρίτην ἀνάλογον προσευρεῖν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",3]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let BA, AC be the two given straight lines, and let them be placed so as to contain any angle; thus it is required to find a third proportional to BA, AC. For let them be produced to the points D, E, and let BD be made equal to AC; [I. 3] let BC be joined, and through D let DE be drawn parallel to it. [I. 31] Since, then, BC has been drawn parallel to DE, one of the sides of the triangle ADE, proportionally, as AB is to BD, so is AC to CE. [VI. 2] But BD is equal to AC; therefore, as AB is to AC, so is AC to CE. Therefore to two given straight lines AB, AC a third proportional to them, CE, has been found.'"],["Rule","'ProofWordCount'",132],["Rule","'GreekProof'","'ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΒΑ, ΑΓ καὶ κείσθωσαν γωνίαν περιέχουσαι τυχοῦσαν. δεῖ δὴ τῶν ΒΑ, ΑΓ τρίτην ἀνάλογον προσευρεῖν. ἐκβεβλήσθωσαν γὰρ ἐπὶ τὰ Δ, Ε σημεῖα, καὶ κείσθω τῇ ΑΓ ἴση ἡ ΒΔ, καὶ ἐπεζεύχθω ἡ ΒΓ, καὶ διὰ τοῦ Δ παράλληλος αὐτῇ ἤχθω ἡ ΔΕ. ἐπεὶ οὖν τριγώνου τοῦ ΑΔΕ παρὰ μίαν τῶν πλευρῶν τὴν ΔΕ ἦκται ἡ ΒΓ, ἀνάλογόν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΔ, οὕτως ἡ ΑΓ πρὸς τὴν ΓΕ. ἴση δὲ ἡ ΒΔ τῇ ΑΓ. ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΑΓ, οὕτως ἡ ΑΓ πρὸς τὴν ΓΕ. δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΑΓ τρίτη ἀνάλογον αὐταῖς προσεύρηται ἡ ΓΕ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",112]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'6.12'"],["Rule","'Text'","'To three given straight lines to find a fourth proportional.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'τριῶν δοθεισῶν εὐθειῶν τετάρτην ἀνάλογον προσευρεῖν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C. Let two straight lines DE, DF be set out containing any angle EDF; let DG be made equal to A, GE equal to B, and further DH equal to C; let GH be joined, and let EF be drawn through E parallel to it. [I. 31] Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF, therefore, as DG is to GE, so is DH to HF. [VI. 2] But DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF. Therefore to the three given straight lines A, B, C a fourth proportional HF has been found.'"],["Rule","'ProofWordCount'",140],["Rule","'GreekProof'","'ἔστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ: δεῖ δὴ τῶν Α, Β, Γ τετάρτην ἀνάλογον προσευρεῖν. Ἐκκείσθωσαν δύο εὐθεῖαι αἱ ΔΕ, ΔΖ γωνίαν περιέχουσαι τυχοῦσαν τὴν ὑπὸ ΕΔΖ: καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΗ, τῇ δὲ Β ἴση ἡ ΗΕ, καὶ ἔτι τῇ Γ ἴση ἡ ΔΘ: καὶ ἐπιζευχθείσης τῆς ΗΘ παράλληλος αὐτῇ ἤχθω διὰ τοῦ Ε ἡ ΕΖ. ἐπεὶ οὖν τριγώνου τοῦ ΔΕΖ παρὰ μίαν τὴν ΕΖ ἦκται ἡ ΗΘ, ἔστιν ἄρα ὡς ἡ ΔΗ πρὸς τὴν ΗΕ, οὕτως ἡ ΔΘ πρὸς τὴν ΘΖ. ἴση δὲ ἡ μὲν ΔΗ τῇ Α, ἡ δὲ ΗΕ τῇ Β, ἡ δὲ ΔΘ τῇ Γ: ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν ΘΖ. τριῶν ἄρα δοθεισῶν εὐθειῶν τῶν Α, Β, Γ τετάρτη ἀνάλογον προσεύρηται ἡ ΘΖ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",136]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'6.13'"],["Rule","'Text'","'To two given straight lines to find a mean proportional.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'δύο δοθεισῶν εὐθειῶν μέσην ἀνάλογον προσευρεῖν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",6],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let AB, BC be the two given straight lines; thus it is required to find a mean proportional to AB, BC. Let them be placed in a straight line, and let the semicircle ADC be described on AC; let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined. Since the angle ADC is an angle in a semicircle, it is right. [III. 31] And, since, in the right-angled triangle ADC, DB has been drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between the segments of the base, AB, BC. [VI. 8] Therefore to the two given straight lines AB, BC a mean proportional DB has been found.'"],["Rule","'ProofWordCount'",126],["Rule","'GreekProof'","'ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΒ, ΒΓ: δεῖ δὴ τῶν ΑΒ, ΒΓ μέσην ἀνάλογον προσευρεῖν. κείσθωσαν ἐπ᾽ εὐθείας, καὶ γεγράφθω ἐπὶ τῆς ΑΓ ἡμικύκλιον τὸ ΑΔΓ, καὶ ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΑΓ εὐθείᾳ πρὸς ὀρθὰς ἡ ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ. ἐπεὶ ἐν ἡμικυκλίῳ γωνία ἐστὶν ἡ ὑπὸ ΑΔΓ, ὀρθή ἐστιν. καὶ ἐπεὶ ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΔΓ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἦκται ἡ ΔΒ, ἡ ΔΒ ἄρα τῶν τῆς βάσεως τμημάτων τῶν ΑΒ, ΒΓ μέση ἀνάλογόν ἐστιν. δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΒΓ μέση ἀνάλογον προσεύρηται ἡ ΔΒ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",102]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'6.14'"],["Rule","'Text'","'In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'τῶν ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: καὶ ὧν ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα.'"],["Rule","'GreekTextWordCount'",29],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",14]],["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore FB, BG are also in a straight line. [I. 14] I say that, in AB, BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE, so is GB to BF. For let the parallelogram FE be completed. Since, then, the parallelogram AB is equal to the parallelogram BC, and FE is another area, therefore, as AB is to FE, so is BC to FE. [V. 7] But, as AB is to FE, so is DB to BE, [VI. 1] and, as BC is to FE, so is GB to BF. [id.]therefore also, as DB is to BE, so is GB to BF. [V. 11] Therefore in the parallelograms AB, BC the sides about the equal angles are reciprocally proportional. Next, let GB be to BF as DB to BE; I say that the parallelogram AB is equal to the parallelogram BC. For since, as DB is to BE, so is GB to BF, while, as DB is to BE, so is the parallelogram AB to the parallelogram FE, [VI. 1] and, as GB is to BF, so is the parallelogram BC to the parallelogram FE, [VI. 1] therefore also, as AB is to FE, so is BC to FE; [V. 11] therefore the parallelogram AB is equal to the parallelogram BC. [V. 9]'"],["Rule","'ProofWordCount'",249],["Rule","'GreekProof'","'ἔστω ἴσα τε καὶ ἰσογώνια παραλληλόγραμμα τὰ ΑΒ, ΒΓ ἴσας ἔχοντα τὰς πρὸς τῷ Β γωνίας, καὶ κείσθωσαν ἐπ᾽ εὐθείας αἱ ΔΒ, ΒΕ: ἐπ᾽ εὐθείας ἄρα εἰσὶ καὶ αἱ ΖΒ, ΒΗ. λέγω, ὅτι τῶν ΑΒ, ΒΓ ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, τουτέστιν, ὅτι ἐστὶν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ. συμπεπληρώσθω γὰρ τὸ ΖΕ παραλληλόγραμμον. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ, ἄλλο δέ τι τὸ ΖΕ, ἔστιν ἄρα ὡς τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως τὸ ΒΓ πρὸς τὸ ΖΕ. ἀλλ᾽ ὡς μὲν τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως ἡ ΔΒ πρὸς τὴν ΒΕ, ὡς δὲ τὸ ΒΓ πρὸς τὸ ΖΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ: καὶ ὡς ἄρα ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ. τῶν ἄρα ΑΒ, ΒΓ παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ἀλλὰ δὴ ἔστω ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ. ἐπεὶ γάρ ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ, ἀλλ᾽ ὡς μὲν ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως τὸ ΑΒ παραλληλόγραμμον πρὸς τὸ ΖΕ παραλληλόγραμμον, ὡς δὲ ἡ ΗΒ πρὸς τὴν ΒΖ, οὕτως τὸ ΒΓ παραλληλόγραμμον πρὸς τὸ ΖΕ παραλληλόγραμμον, καὶ ὡς ἄρα τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως τὸ ΒΓ πρὸς τὸ ΖΕ: ἴσον ἄρα ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ. τῶν ἄρα ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: καὶ ὧν ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",278]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'6.15'"],["Rule","'Text'","'In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal.'"],["Rule","'TextWordCount'",45],["Rule","'GreekText'","'τῶν ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: καὶ ὧν μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα.'"],["Rule","'GreekTextWordCount'",36],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",14]],["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let ABC, ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE; I say that in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD, so is EA to AB. For let them be placed so that CA is in a straight line with AD; therefore EA is also in a straight line with AB. [I. 14] Let BD be joined. Since then the triangle ABC is equal to the triangle ADE, and BAD is another area, therefore, as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 7] But, as CAB is to BAD, so is CA to AD, [VI. 1] and, as EAD is to BAD, so is EA to AB. [id.] Therefore also, as CA is to AD, so is EA to AB. [V. 11] Therefore in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional. Next, let the sides of the triangles ABC, ADE be reciprocally proportional, that is to say, let EA be to AB as CA to AD; I say that the triangle ABC is equal to the triangle ADE. For, if BD be again joined, since, as CA is to AD, so is EA to AB, while, as CA is to AD, so is the triangle ABC to the triangle BAD, and, as EA is to AB, so is the triangle EAD to the triangle BAD, [VI. 1] therefore, as the triangle ABC is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 11] Therefore each of the triangles ABC, EAD has the same ratio to BAD. Therefore the triangle ABC is equal to the triangle EAD. [V. 9]'"],["Rule","'ProofWordCount'",307],["Rule","'GreekProof'","'ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΑΔΕ μίαν μιᾷ ἴσην ἔχοντα γωνίαν τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΔΑΕ: λέγω, ὅτι τῶν ΑΒΓ, ΑΔΕ τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, τουτέστιν, ὅτι ἐστὶν ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. κείσθω γὰρ ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΓΑ τῇ ΑΔ: ἐπ᾽ εὐθείας ἄρα ἐστὶ καὶ ἡ ΕΑ τῇ ΑΒ. καὶ ἐπεζεύχθω ἡ ΒΔ. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΔΕ τριγώνῳ, ἄλλο δέ τι τὸ ΒΑΔ, ἔστιν ἄρα ὡς τὸ ΓΑΒ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον. ἀλλ᾽ ὡς μὲν τὸ ΓΑΒ πρὸς τὸ ΒΑΔ, οὕτως ἡ ΓΑ πρὸς τὴν ΑΔ, ὡς δὲ τὸ ΕΑΔ πρὸς τὸ ΒΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. καὶ ὡς ἄρα ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. τῶν ΑΒΓ, ΑΔΕ ἄρα τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ἀλλὰ δὴ ἀντιπεπονθέτωσαν αἱ πλευραὶ τῶν ΑΒΓ, ΑΔΕ τριγώνων, καὶ ἔστω ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΔΕ τριγώνῳ. Ἐπιζευχθείσης γὰρ πάλιν τῆς ΒΔ, ἐπεί ἐστιν ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ, ἀλλ᾽ ὡς μὲν ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, ὡς δὲ ἡ ΕΑ πρὸς τὴν ΑΒ, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον. ἑκάτερον ἄρα τῶν ΑΒΓ, ΕΑΔ πρὸς τὸ ΒΑΔ τὸν αὐτὸν ἔχει λόγον. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΕΑΔ τριγώνῳ. τῶν ἄρα ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: καὶ ὧν μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἐκεῖνα ἴσα ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",321]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'6.16'"],["Rule","'Text'","'If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines will be proportional.'"],["Rule","'TextWordCount'",45],["Rule","'GreekText'","'ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ: κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται.'"],["Rule","'GreekTextWordCount'",39],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",14]]]],["Rule","'Proof'","'Let the four straight lines AB, CD, E, F be proportional, so that, as AB is to CD, so is E to F; I say that the rectangle contained by AB, F is equal to the rectangle contained by CD, E. Let AG, CH be drawn from the points A, C at right angles to the straight lines AB, CD, and let AG be made equal to F, and CH equal to E. Let the parallelograms BG, DH be completed. Then since, as AB is to CD, so is E to F, while E is equal to CH, and F to AG, therefore, as AB is to CD, so is CH to AG. Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional. But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [VI. 14] therefore the parallelogram BG is equal to the parallelogram DH. And BG is the rectangle AB, F, for AG is equal to F; and DH is the rectangle CD, E, for E is equal to CH; therefore the rectangle contained by AB, F is equal to the rectangle contained by CD, E. Next, let the rectangle contained by AB, F be equal to the rectangle contained by CD, E; I say that the four straight lines will be proportional, so that, as AB is to CD, so is E to F. For, with the same construction, since the rectangle AB, F is equal to the rectangle CD, E, and the rectangle AB, F is BG, for AG is equal to F, and the rectangle CD, E is DH, for CH is equal to E, therefore BG is equal to DH. And they are equiangular But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [VI. 14] Therefore, as AB is to CD, so is CH to AG. But CH is equal to E, and AG to F; therefore, as AB is to CD, so is E to F.'"],["Rule","'ProofWordCount'",340],["Rule","'GreekProof'","'Ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, Ε, Ζ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. Ἤχθωσαν [γὰρ] ἀπὸ τῶν Α, Γ σημείων ταῖς ΑΒ, ΓΔ εὐθείαις πρὸς ὀρθὰς αἱ ΑΗ, ΓΘ, καὶ κείσθω τῇ μὲν Ζ ἴση ἡ ΑΓ, τῇ δὲ Ε ἴση ἡ ΓΘ. καὶ συμπεπληρώσθω τὰ ΒΗ, ΔΘ παραλληλόγραμμα.Καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ, ἴση δὲ ἡ μὲν Ε τῇ ΓΘ, ἡ δὲ Ζ τῇ ΑΗ, ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΘ πρὸς τὴν ΑΗ.  τῶν ΒΗ, ΔΘ ἄρα παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας.  ὧν δὲ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα:  ἴσον ἄρα ἐστὶ τὸ ΒΗ παραλληλόγραμμον τῷ ΔΘ παραλληλογράμμῳ. καί ἐστι τὸ μὲν ΒΗ τὸ ὑπὸ τῶν ΑΒ, Ζ:  ἴση γὰρ ἡ ΑΗ τῇ Ζ: τὸ δὲ ΔΘ τὸ ὑπὸ τῶν ΓΔ, Ε: ἴση γὰρ ἡ Ε τῇ ΓΘ: τὸ ἄρα ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. Ἀλλὰ δὴ τὸ ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἔστω τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ: λέγω, ὅτι αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ. Τῶν γὰρ αὐτῶν κατασκευασθέντων,  ἐπεὶ τὸ ὑπὸ τῶν ΑΒ, Ζ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε, καί ἐστι τὸ μὲν ὑπὸ τῶν ΑΒ, Ζ τὸ ΒΗ: ἴση γάρ ἐστιν ἡ ΑΗ τῇ Ζ:  τὸ δὲ ὑπὸ τῶν ΓΔ, Ε τὸ ΔΘ: ἴση γὰρ ἡ ΓΘ τῇ Ε: τὸ ἄρα ΒΗ ἴσον ἐστὶ τῷ ΔΘ.  καί ἐστιν ἰσογώνια.  τῶν δὲ ἴσων καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας.  ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΘ πρὸς τὴν ΑΗ. ἴση δὲ ἡ μὲν ΓΘ τῇ Ε, ἡ δὲ ΑΗ τῇ Ζ:  ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ. Ἐὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ: κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται:  ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",391]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'6.17'"],["Rule","'Text'","'If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines will be proportional. '"],["Rule","'TextWordCount'",43],["Rule","'GreekText'","'ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης τετραγώνῳ: κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι ἀνάλογον ἔσονται.'"],["Rule","'GreekTextWordCount'",37],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'Let the three straight lines A, B, C be proportional, so that, as A is to B, so is B to C; I say that the rectangle contained by A, C is equal to the square on B. Let D be made equal to B. Then, since, as A is to B, so is B to C, and B is equal to D, therefore, as A is to B, so is D to C. But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means.[VI. 16] Therefore the rectangle A, C is equal to the rectangle B, D. But the rectangle B, D is the square on B, for B is equal to D; therefore the rectangle contained by A, C is equal to the square on B. Next, let the rectangle A, C be equal to the square on B; I say that, as A is to B, so is B to C. For, with the same construction, since the rectangle A, C is equal to the square on B, while the square on B is the rectangle B, D, for B is equal to D, therefore the rectangle A, C is equal to the rectangle B, D. But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [VI. 16] Therefore, as A is to B, so is D to C. But B is equal to D; therefore, as A is to B, so is B to C.'"],["Rule","'ProofWordCount'",263],["Rule","'GreekProof'","'ἔστωσαν τρεῖς εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ: λέγω, ὅτι τὸ ὑπὸ τῶν Α, Γ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς β τετραγώνῳ. κείσθω τῇ Β ἴση ἡ Δ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ, ἴση δὲ ἡ Β τῇ Δ, ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, ἡ Δ πρὸς τὴν Γ. ἐὰν δὲ τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ. τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Β, Δ. ἀλλὰ τὸ ὑπὸ τῶν Β, Δ τὸ ἀπὸ τῆς Β ἐστιν: ἴση γὰρ ἡ Β τῇ Δ: τὸ ἄρα ὑπὸ τῶν Α, Γ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς Β τετραγώνῳ. ἀλλὰ δὴ τὸ ὑπὸ τῶν Α, Γ ἴσον ἔστω τῷ ἀπὸ τῆς Β: λέγω, ὅτι ἐστὶν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ τὸ ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ἀπὸ τῆς Β, ἀλλὰ τὸ ἀπὸ τῆς Β τὸ ὑπὸ τῶν Β, Δ ἐστιν: ἴση γὰρ ἡ Β τῇ Δ: τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Β, Δ. ἐὰν δὲ τὸ ὑπὸ τῶν ἄκρων ἴσον ᾖ τῷ ὑπὸ τῶν μέσων, αἱ τέσσαρες εὐθεῖαι ἀνάλογόν εἰσιν. ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Δ πρὸς τὴν Γ. ἴση δὲ ἡ Β τῇ Δ: ὡς ἄρα ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ. ἐὰν ἄρα τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης τετραγώνῳ: κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι ἀνάλογον ἔσονται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",306]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'6.18'"],["Rule","'Text'","'On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἀπὸ τῆς δοθείσης εὐθείας τῷ δοθέντι εὐθυγράμμῳ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγράψαι.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let AB be the given straight line and CE the given rectilineal figure; thus it is required to describe on the straight line AB a rectilineal figure similar and similarly situated to the rectilineal figure CE. Let DF be joined, and on the straight line AB, and at the points A, B on it, let the angle GAB be constructed equal to the angle at C, and the angle ABG equal to the angle CDF. [I. 23] Therefore the remaining angle CFD is equal to the angle AGB; [I. 32] therefore the triangle FCD is equiangular with the triangle GAB. Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB. Again, on the straight line BG, and at the points B, G on it, let the angle BGH be constructed equal to the angle DFE, and the angle GBH equal to the angle FDE. [I. 23] Therefore the remaining angle at E is equal to the remaining angle at H; [I. 32] therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4]  But it was also proved that, as FD is to GB, so is FC to GA, and CD to AB; therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB. And, since the angle CFD is equal to the angle AGB, and the angle DFE to the angle BGH, therefore the whole angle CFE is equal to the whole angle AGH. For the same reason the angle CDE is also equal to the angle ABH. And the angle at C is also equal to the angle at A, and the angle at E to the angle at H.  Therefore AH is equiangular with CE; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE. [VI. Def. 1] Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE.'"],["Rule","'ProofWordCount'",360],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν εὐθύγραμμον τὸ ΓΕ: δεῖ δὴ ἀπὸ τῆς ΑΒ εὐθείας τῷ ΓΕ εὐθυγράμμῳ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγράψαι. ἐπεζεύχθω ἡ ΔΖ, καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Α, Β τῇ μὲν πρὸς τῷ Γ γωνίᾳ ἴση ἡ ὑπὸ ΗΑΒ, τῇ δὲ ὑπὸ ΓΔΖ ἴση ἡ ὑπὸ ΑΒΗ. λοιπὴ ἄρα ἡ ὑπὸ ΓΖΔ τῇ ὑπὸ ΑΗΒ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΖΓΔ τρίγωνον τῷ ΗΑΒ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΓ πρὸς τὴν ΗΑ, καὶ ἡ ΓΔ πρὸς τὴν ΑΒ. πάλιν συνεστάτω πρὸς τῇ ΒΗ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Β, Η τῇ μὲν ὑπὸ ΔΖΕ γωνίᾳ ἴση ἡ ὑπὸ ΒΗΘ, τῇ δὲ ὑπὸ ΖΔΕ ἴση ἡ ὑπὸ ΗΒΘ. λοιπὴ ἄρα ἡ πρὸς τῷ Ε λοιπῇ τῇ πρὸς τῷ Θ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΖΔΕ τρίγωνον τῷ ΗΘΒ τριγώνῳ: ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΕ πρὸς τὴν ΗΘ καὶ ἡ ΕΔ πρὸς τὴν ΘΒ. ἐδείχθη δὲ καὶ ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΓ πρὸς τὴν ΗΑ καὶ ἡ ΓΔ πρὸς τὴν ΑΒ: καὶ ὡς ἄρα ἡ ΖΓ πρὸς τὴν ΑΗ, οὕτως á¼¥ τε ΓΔ πρὸς τὴν ΑΒ καὶ ἡ ΖΕ πρὸς τὴν ΗΘ καὶ ἔτι ἡ ΕΔ πρὸς τὴν ΘΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΓΖΔ γωνία τῇ ὑπὸ ΑΗΒ, ἡ δὲ ὑπὸ ΔΖΕ τῇ ὑπὸ ΒΗΘ, ὅλη ἄρα ἡ ὑπὸ ΓΖΕ ὅλῃ τῇ ὑπὸ ΑΗΘ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΘ ἐστιν ἴση. ἔστι δὲ καὶ ἡ μὲν πρὸς τῷ Γ τῇ πρὸς τῷ Α ἴση, ἡ δὲ πρὸς τῷ Ε τῇ πρὸς τῷ Θ. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΘ τῷ ΓΕ: καὶ τὰς περὶ τὰς ἴσας γωνίας αὐτῶν πλευρὰς ἀνάλογον ἔχει: ὅμοιον ἄρα ἐστὶ τὸ ΑΘ εὐθύγραμμον τῷ ΓΕ εὐθυγράμμῳ. ἀπὸ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ ΓΕ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγέγραπται τὸ ΑΘ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",342]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'6.19'"],["Rule","'Text'","'Similar triangles are to one another in the duplicate ratio of the corresponding sides.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'τὰ ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",9]],["List",["Rule","'Book'",5],["Rule","'Definition'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, sois DE to EF, so that BC corresponds to EF; [V. Def. 11] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF. For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [VI. 11]and let AG be joined. Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [V. 16] But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [V. 11] Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional. But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [VI. 15]therefore the triangle ABG is equal to the triangle DEF. Now since, as BC is to EF, so is EF to BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [V. Def. 9] therefore BC has to BG a ratio duplicate of that which CB  has to EF. But, as CB is to BG, so is the triangle ABC to the triangle ABG; [VI. 1] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF. But the triangle ABG is equal to the triangle DEF; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF.'"],["Rule","'ProofWordCount'",313],["Rule","'GreekProof'","'ἔστω ὅμοια τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἴσην ἔχοντα τὴν πρὸς τῷ Β γωνίαν τῇ πρὸς τῷ Ε, ὡς δὲ τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, ὥστε ὁμόλογον εἶναι τὴν ΒΓ τῇ ΕΖ: λέγω, ὅτι τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. εἰλήφθω γὰρ τῶν ΒΓ, ΕΖ τρίτη ἀνάλογον ἡ ΒΗ, ὥστε εἶναι ὡς τὴν ΒΓ πρὸς τὴν ΕΖ, οὕτως τὴν ΕΖ πρὸς τὴν ΒΗ: καὶ ἐπεζεύχθω ἡ ΑΗ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΕ πρὸς τὴν ΕΖ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΔΕ, οὕτως ἡ ΒΓ πρὸς τὴν ΕΖ. ἀλλ᾽ ὡς ἡ ΒΓ πρὸς ΕΖ, οὕτως ἐστὶν ἡ ΕΖ πρὸς ΒΗ. καὶ ὡς ἄρα ἡ ΑΒ πρὸς ΔΕ, οὕτως ἡ ΕΖ πρὸς ΒΗ: τῶν ΑΒΗ, ΔΕΖ ἄρα τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ὧν δὲ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα. ἴσον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΕΖ, οὕτως ἡ ΕΖ πρὸς τὴν ΒΗ, ἐὰν δὲ τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἡ πρώτη πρὸς τὴν τρίτην διπλασίονα λόγον ἔχει ἤπερ πρὸς τὴν δευτέραν, ἡ ΒΓ ἄρα πρὸς τὴν ΒΗ διπλασίονα λόγον ἔχει ἤπερ ἡ ΓΒ πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΓΒ πρὸς τὴν ΒΗ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον: καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΑΒΗ διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἴσον δὲ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ: καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. τὰ ἄρα ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον ἐπείπερ ἐδείχθη, ὡς ἡ ΓΒ πρὸς ΒΗ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον, τουτέστι τὸ ΔΕΖ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",355]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'6.20'"],["Rule","'Text'","'Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding  side.'"],["Rule","'TextWordCount'",42],["Rule","'GreekText'","'τὰ ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν.'"],["Rule","'GreekTextWordCount'",34],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",5],["Rule","'Theorem'",12]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",6]],["List",["Rule","'Book'",6],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'Let ABCDE, FGHKL be similar polygons, and let AB correspond to FG; I say that the polygons ABCDE, FGHKL are divided into similar triangles, and into triangles equal in multitude and inthe same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG. Let BE, EC, GL, LH be joined. Now, since the polygon ABCDE is similar to the polygonFGHKL, the angle BAE is equal to the angle GFL; and, as BA is to AE, so is GF to FL. [VI. Def. 1]  Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal anglesproportional, therefore the triangle ABE is equiangular with the triangle FGL; [VI. 6] so that it is also similar; [VI. 4 and Def. 1]therefore the angle ABE is equal to the angle FGL. But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons; therefore the remaining angle EBC is equal to the angle LGH. And, since, because of the similarity of the triangles ABE,FGL, as EB is to BA, so is LG to GF, and moreover also, because of the similarity of the polygons, as AB is to BC, so is FG to GH, therefore, ex aequali, as EB is to BC, so is LG to GH; [V. 22]that is, the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is equiangular with the triangle LGH, [VI. 6] so that the triangle EBC is also similar to the triangle LGH. [VI. 4 and Def. 1]  For the same reason the triangle ECD is also similar to the triangle LHK. Therefore the similar polygons ABCDE, FGHKL have been divided into similar triangles, and into triangles equal inmultitude. I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, ECD are antecedents, while FGL, LGH, LHK are their consequents, and that the polygon ABCDE  has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG. For let AC, FH be joined. Then since, because of the similarity of the polygons,the angle ABC is equal to the angle FGH, and, as AB is to BC, so is FG to GH, the triangle ABC is equiangular with the triangle FGH; [VI. 6]therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF. And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangleFGN. Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH. Therefore, proportionally, as AM is to MB, so is FN to NG,and, as BM is to MC, so is GN to NH; so that, in addition, ex aequali, as AM is to MC, so is FN to NH. But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC; for they are to one another as theirbases. [VI. 1] Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle AMB is to BMC, so is ABE to CBE. But, as AMB is to BMC, so is AM to MC; therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC. For the same reason also, as FN is to NH, so is the triangle FGL to the triangleGLH. And, as AM is to MC, so is FN to NH; therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH; and, alternately, as the triangle ABE is to the triangle FGL,so is the triangle BEC to the triangle GLH. Similarly we can prove, if BD, GK be joined, that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK. And since, as the triangle ABE is to the triangle FGL,so is EBC to LGH, and further ECD to LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL. But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG; for similar triangles are in the duplicate ratio of the corresponding sides. [VI. 19] Therefore the polygon ABCDE also has to the polygonFGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG.'"],["Rule","'ProofWordCount'",855],["Rule","'GreekProof'","'ἔστω ὅμοια πολύγωνα τὰ ΑΒΓΔΕ, ΖΗΘΚΛ, ὁμόλογος δὲ ἔστω ἡ ΑΒ τῇ ΖΗ: λέγω, ὅτι τὰ ΑΒΓΔΕ, ΖΗΘΚΛ πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΖΗ. ἐπεζεύχθωσαν αἱ ΒΕ, ΕΓ, ΗΛ, ΛΘ. καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓΔΕ πολύγωνον τῷ ΖΗΘΚΛ πολυγώνῳ, ἴση ἐστὶν ἡ ὑπὸ ΒΑΕ γωνία τῇ ὑπὸ ΗΖΛ. καί ἐστιν ὡς ἡ ΒΑ πρὸς ΑΕ, οὕτως ἡ ΗΖ πρὸς ΖΛ. ἐπεὶ οὖν δύο τρίγωνά ἐστι τὰ ΑΒΕ, ΖΗΛ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΗΛ τριγώνῳ: ὥστε καὶ ὅμοιον: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία τῇ ὑπὸ ΖΗΛ. ἔστι δὲ καὶ ὅλη ἡ ὑπὸ ΑΒΓ ὅλῃ τῇ ὑπὸ ΖΗΘ ἴση διὰ τὴν ὁμοιότητα τῶν πολυγώνων: λοιπὴ ἄρα ἡ ὑπὸ ΕΒΓ γωνία τῇ ὑπὸ ΛΗΘ ἐστιν ἴση. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν ΑΒΕ, ΖΗΛ τριγώνων ἐστὶν ὡς ἡ ΕΒ πρὸς ΒΑ, οὕτως ἡ ΛΗ πρὸς ΗΖ, ἀλλὰ μὴν καὶ διὰ τὴν ὁμοιότητα τῶν πολυγώνων ἐστὶν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς ΗΘ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ ΕΒ πρὸς ΒΓ, οὕτως ἡ ΛΗ πρὸς ΗΘ, καὶ περὶ τὰς ἴσας γωνίας τὰς ὑπὸ ΕΒΓ, ΛΗΘ αἱ πλευραὶ ἀνάλογόν εἰσιν: ἰσογώνιον ἄρα ἐστὶ τὸ ΕΒΓ τρίγωνον τῷ ΛΗΘ τριγώνῳ: ὥστε καὶ ὅμοιόν ἐστι τὸ ΕΒΓ τρίγωνον τῷ ΛΗΘ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΕΓΔ τρίγωνον ὅμοιόν ἐστι τῷ ΛΘΚ τριγώνῳ. τὰ ἄρα ὅμοια πολύγωνα τὰ ΑΒΓΔΕ, ΖΗΘΚΛ εἴς τε ὅμοια τρίγωνα διῄρηται καὶ εἰς ἴσα τὸ πλῆθος. λέγω, ὅτι καὶ ὁμόλογα τοῖς ὅλοις, τουτέστιν ὥστε ἀνάλογον εἶναι τὰ τρίγωνα, καὶ ἡγούμενα μὲν εἶναι τὰ ΑΒΕ, ΕΒΓ, ΕΓΔ, ἑπόμενα δὲ αὐτῶν τὰ ΖΗΛ, ΛΗΘ, ΛΘΚ, καὶ ὅτι τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἡ ΑΒ πρὸς τὴν ΖΗ. ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΖΘ. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν πολυγώνων ἴση ἐστὶν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΖΗΘ, καί ἐστιν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς ΗΘ, ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΖΗΘ τριγώνῳ: ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΗΖΘ, ἡ δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΗΘΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΑΜ γωνία τῇ ὑπὸ ΗΖΝ, ἔστι δὲ καὶ ἡ ὑπὸ ΑΒΜ τῇ ὑπὸ ΖΗΝ ἴση, καὶ λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΖΝΗ ἴση ἐστίν: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΜ τρίγωνον τῷ ΖΗΝ τριγώνῳ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τὸ ΒΜΓ τρίγωνον ἰσογώνιόν ἐστι τῷ ΗΝΘ τριγώνῳ. ἀνάλογον ἄρα ἐστίν, ὡς μὲν ἡ ΑΜ πρὸς ΜΒ, οὕτως ἡ ΖΝ πρὸς ΝΗ, ὡς δὲ ἡ ΒΜ πρὸς ΜΓ, οὕτως ἡ ΗΝ πρὸς ΝΘ: ὥστε καὶ δι᾽ ἴσου, ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ. ἀλλ᾽ ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΜ τρίγωνον πρὸς τὸ ΜΒΓ, καὶ τὸ ΑΜΕ πρὸς τὸ ΕΜΓ: πρὸς ἄλληλα γάρ εἰσιν ὡς αἱ βάσεις. καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ὡς ἄρα τὸ ΑΜΒ τρίγωνον πρὸς τὸ ΒΜΓ, οὕτως τὸ ΑΒΕ πρὸς τὸ ΓΒΕ. ἀλλ᾽ ὡς τὸ ΑΜΒ πρὸς τὸ ΒΜΓ, οὕτως ἡ ΑΜ πρὸς ΜΓ: καὶ ὡς ἄρα ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΕΒΓ τρίγωνον. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΖΝ πρὸς ΝΘ, οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον. καί ἐστιν ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ: καὶ ὡς ἄρα τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΒΕΓ τρίγωνον, οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον, καὶ ἐναλλὰξ, ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον. ὁμοίως δὴ δείξομεν ἐπιζευχθεισῶν τῶν ΒΔ, ΗΚ, ὅτι καὶ ὡς τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΛΗΘ τρίγωνον, οὕτως τὸ ΕΓΔ τρίγωνον πρὸς τὸ ΛΘΚ τρίγωνον. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΕΒΓ πρὸς τὸ ΛΗΘ, καὶ ἔτι τὸ ΕΓΔ πρὸς τὸ ΛΘΚ, καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ἔστιν ἄρα ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. ἀλλὰ τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν: τὰ γὰρ ὅμοια τρίγωνα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν. καὶ τὸ ΑΒΓΔΕ ἄρα πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν. τὰ ἄρα ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν: ὅπερ ἔδει δεῖξαι. Πόρισμα ὡσαύτως δὲ καὶ ἐπὶ τῶν ὁμοίων τετραπλεύρων δειχθήσεται, ὅτι ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ἐδείχθη δὲ καὶ ἐπὶ τῶν τριγώνων: ὥστε καὶ καθόλου τὰ ὅμοια εὐθύγραμμα σχήματα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ὅπερ ἔδει δεῖξαι. πόρισμα β # καὶ ἐὰν τῶν ΑΒ, ΖΗ τρίτην ἀνάλογον λάβωμεν τὴν Ξ, ἡ ΒΑ πρὸς τὴν Ξ διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΖΗ. ἔχει δὲ καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον á¼¢ τὸ τετράπλευρον πρὸς τὸ τετράπλευρον διπλασίονα λόγον ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἡ ΑΒ πρὸς τὴν ΖΗ: ἐδείχθη δὲ τοῦτο καὶ ἐπὶ τῶν τριγώνων: ὥστε καὶ καθόλου φανερόν, ὅτι, ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔσται ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον.'"],["Rule","'GreekProofWordCount'",950]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'6.21'"],["Rule","'Text'","'Figures which are similar to the same rectilineal figure are also similar to one another.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'τὰ τῷ αὐτῷ εὐθυγράμμῳ ὅμοια καὶ ἀλλήλοις ἐστὶν ὅμοια.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Definition'",1]]]],["Rule","'Proof'","'For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B. For, since A is similar to C, it is equiangular with it and has the sides about the equal angles proportional. [VI. Def. 1] Again, since B is similar to C, it is equiangular with it and has the sides about the equal angles proportional. Therefore each of the figures A, B is equiangular with C and with C has the sides about the equal angles proportional; therefore A is similar to B.'"],["Rule","'ProofWordCount'",95],["Rule","'GreekProof'","'ἔστω γὰρ ἑκάτερον τῶν Α, Β εὐθυγράμμων τῷ Γ ὅμοιον: λέγω, ὅτι καὶ τὸ Α τῷ Β ἐστιν ὅμοιον. ἐπεὶ γὰρ ὅμοιόν ἐστι τὸ Α τῷ Γ, ἰσογώνιόν τέ ἐστιν αὐτῷ καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. πάλιν, ἐπεὶ ὅμοιόν ἐστι τὸ Β τῷ Γ, ἰσογώνιόν τέ ἐστιν αὐτῷ καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. ἑκάτερον ἄρα τῶν Α, Β τῷ Γ ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει ὥστε καὶ τὸ Α τῷ Β ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. ὅμοιον ἄρα ἐστὶ τὸ Α τῷ Β: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",108]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'6.22'"],["Rule","'Text'","'If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and, if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional.'"],["Rule","'TextWordCount'",40],["Rule","'GreekText'","'ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπ᾽ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ἔσται: κἂν τὰ ἀπ᾽ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ᾖ, καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται.'"],["Rule","'GreekTextWordCount'",37],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",6],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",12]],["List",["Rule","'Book'",6],["Rule","'Theorem'",18]],["List",["Rule","'Book'",6],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH; I say that, as KAB is to LCD, so is MF to NH. For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11] Then since, as AB is to CD, so is EF to GH, and, as CD is to O, so is GH to P, therefore, ex aequali, as AB is to O, so is EF to P. [V. 22] But, as AB is to O, so is KAB to LCD, [VI. 19] and, as EF is to P, so is MF to NH; therefore also, as KAB is to LCD, so is MF to NH. [V. 11] Next, let MF be to NH as KAB is to LCD; I say also that, as AB is to CD, so is EF to GH. For, if EF is not to GH as AB to CD, let EF be to QR as AB to CD, [VI. 12]and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF, NH. [VI. 18] Since then, as AB is to CD, so is EF to QR, and there have been described on AB, CD the similar and similarly situated figures KAB, LCD, and on EF, QR the similar and similarly situated figures MF, SR, therefore, as KAB is to LCD, so is MF to SR. But also, by hypothesis, as KAB is to LCD, so is MF to NH; therefore also, as MF is to SR, so is MF to NH. [V. 11] Therefore MF has the same ratio to each of the figures NH, SR; therefore NH is equal to SR. [V. 9]  But it is also similar and similarly situated to it; therefore GH is equal to QR. And, since, as AB is to CD, so is EF to QR, while QR is equal to GH, therefore, as AB is to CD, so is EF to GH.'"],["Rule","'ProofWordCount'",379],["Rule","'GreekProof'","'ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, ΕΖ, ΗΘ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, καὶ ἀναγεγράφθωσαν ἀπὸ μὲν τῶν ΑΒ, ΓΔ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΚΑΒ, ΛΓΔ, ἀπὸ δὲ τῶν ΕΖ, ΗΘ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΜΖ, ΝΘ: λέγω, ὅτι ἐστὶν ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. εἰλήφθω γὰρ τῶν μὲν ΑΒ, ΓΔ τρίτη ἀνάλογον ἡ Ξ, τῶν δὲ ΕΖ, ΗΘ τρίτη ἀνάλογον ἡ Ο. καὶ ἐπεί ἐστιν ὡς μὲν ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, ὡς δὲ ἡ ΓΔ πρὸς τὴν Ξ, οὕτως ἡ ΗΘ πρὸς τὴν Ο, δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν Ξ, οὕτως ἡ ΕΖ πρὸς τὴν Ο. ἀλλ᾽ ὡς μὲν ἡ ΑΒ πρὸς τὴν Ξ, οὕτως καὶ τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, ὡς δὲ ἡ ΕΖ πρὸς τὴν Ο, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ: καὶ ὡς ἄρα τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. ἀλλὰ δὴ ἔστω ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ: λέγω, ὅτι ἐστὶ καὶ ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. εἰ γὰρ μή ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, ἔστω ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΠΡ ὁποτέρῳ τῶν ΜΖ, ΝΘ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον τὸ ΣΡ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, καὶ ἀναγέγραπται ἀπὸ μὲν τῶν ΑΒ, ΓΔ ὅμοιά τε καὶ ὁμοίως κείμενα τὰ ΚΑΒ, ΛΓΔ, ἀπὸ δὲ τῶν ΕΖ, ΠΡ ὅμοιά τε καὶ ὁμοίως κείμενα τὰ ΜΖ, ΣΡ, ἔστιν ἄρα ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΣΡ. ὑπόκειται δὲ καὶ ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ: καὶ ὡς ἄρα τὸ ΜΖ πρὸς τὸ ΣΡ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. τὸ ΜΖ ἄρα πρὸς ἑκάτερον τῶν ΝΘ, ΣΡ τὸν αὐτὸν ἔχει λόγον: ἴσον ἄρα ἐστὶ τὸ ΝΘ τῷ ΣΡ. ἔστι δὲ αὐτῷ καὶ ὅμοιον καὶ ὁμοίως κείμενον: ἴση ἄρα ἡ ΗΘ τῇ ΠΡ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, ἴση δὲ ἡ ΠΡ τῇ ΗΘ, ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. ἐὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπ᾽ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ἔσται: κἂν τὰ ἀπ᾽ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ᾖ, καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται: ὅπερ ἔδει δεῖξαι. λῆμμα ὅτι δέ, ἐὰν εὐθύγραμμα ἴσα ᾖ καὶ ὅμοια, αἱ ὁμόλογοι αὐτῶν πλευραὶ ἴσαι ἀλλήλαις εἰσίν, δείξομεν οὕτως. ἔστω ἴσα καὶ ὅμοια εὐθύγραμμα τὰ ΝΘ, ΣΡ, καὶ ἔστω ὡς ἡ ΘΗ πρὸς τὴν ΗΝ, οὕτως ἡ ΡΠ πρὸς τὴν ΠΣ: λέγω, ὅτι ἴση ἐστὶν ἡ ΡΠ τῇ ΘΗ. εἰ γὰρ ἄνισοί εἰσιν, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ΡΠ τῆς ΘΗ. καὶ ἐπεί ἐστιν ὡς ἡ ΡΠ πρὸς ΠΣ, οὕτως ἡ ΘΗ πρὸς τὴν ΗΝ, καὶ ἐναλλάξ, ὡς ἡ ΡΠ πρὸς τὴν ΘΗ, οὕτως ἡ ΠΣ πρὸς τὴν ΗΝ, μείζων δὲ ἡ ΠΡ τῆς ΘΗ, μείζων ἄρα καὶ ἡ ΠΣ τῆς ΗΝ: ὥστε καὶ τὸ ΡΣ μεῖζόν ἐστι τοῦ ΘΝ. ἀλλὰ καὶ ἴσον: ὅπερ ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ΠΡ τῇ ΗΘ: ἴση ἄρα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",574]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'6.23'"],["Rule","'Text'","'Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'τὰ ἰσογώνια παραλληλόγραμμα πρὸς ἄλληλα λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG;I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides. For let them be placed so that BC is in a straight line with CG; therefore DC is also in a straight line with CE. Let the parallelogram DG be completed; let a straight line K be set out, and let it be contrived that, as BC is to CG, so is K to L, and, as DC is to CE, so is L to M. [VI. 12] Then the ratios of K to L and of L to M are the sameas the ratios of the sides, namely of BC to CG and of DC to CE. But the ratio of K to M is compounded of the ratio of K to L and of that of L to M; so that K has also to M the ratio compounded of the ratiosof the sides. Now since, as BC is to CG, so is the parallelogram AC to the parallelogram CH, [VI. 1] while, as BC is to CG, so is K to L, therefore also, as K is to L, so is AC to CH. [V. 11] Again, since, as DC is to CE, so is the parallelogram CH to CF, [VI. 1] while, as DC is to CE, so is L to M, therefore also, as L is to M, so is the parallelogram CH to the parallelogram CF. [V. 11] Since then it was proved that, as K is to L, so is the parallelogram AC to the parallelogram CH, and, as L is to M, so is the parallelogram CH to the parallelogram CF, therefore, ex aequali, as K is to M, so is AC to the parallelogram  CF. But K has to M the ratio compounded of the ratios of the sides; therefore AC also has to CF the ratio compounded of the ratios of the sides.'"],["Rule","'ProofWordCount'",338],["Rule","'GreekProof'","'ἔστω ἰσογώνια παραλληλόγραμμα τὰ ΑΓ, ΓΖ ἴσην ἔχοντα τὴν ὑπὸ ΒΓΔ γωνίαν τῇ ὑπὸ ΕΓΗ: λέγω, ὅτι τὸ ΑΓ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν. κείσθω γὰρ ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΒΓ τῇ ΓΗ: ἐπ᾽ εὐθείας ἄρα ἐστὶ καὶ ἡ ΔΓ τῇ ΓΕ. καὶ συμπεπληρώσθω τὸ ΔΗ παραλληλόγραμμον, καὶ ἐκκείσθω τις εὐθεῖα ἡ Κ, καὶ γεγονέτω ὡς μὲν ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως ἡ Κ πρὸς τὴν Λ, ὡς δὲ ἡ ΔΓ πρὸς τὴν ΓΕ, οὕτως ἡ Λ πρὸς τὴν Μ. οἱ ἄρα λόγοι τῆς τε Κ πρὸς τὴν Λ καὶ τῆς Λ πρὸς τὴν Μ οἱ αὐτοί εἰσι τοῖς λόγοις τῶν πλευρῶν, τῆς τε ΒΓ πρὸς τὴν ΓΗ καὶ τῆς ΔΓ πρὸς τὴν ΓΕ. ἀλλ᾽ ὁ τῆς Κ πρὸς Μ λόγος σύγκειται ἔκ τε τοῦ τῆς Κ πρὸς Λ λόγου καὶ τοῦ τῆς Λ πρὸς Μ: ὥστε καὶ ἡ Κ πρὸς τὴν Μ λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως τὸ ΑΓ παραλληλόγραμμον πρὸς τὸ ΓΘ, ἀλλ᾽ ὡς ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως ἡ Κ πρὸς τὴν λ, καὶ ὡς ἄρα ἡ Κ πρὸς τὴν Λ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΘ. πάλιν, ἐπεί ἐστιν ὡς ἡ ΔΓ πρὸς τὴν ΓΕ, οὕτως τὸ ΓΘ παραλληλόγραμμον πρὸς τὸ ΓΖ, ἀλλ᾽ ὡς ἡ ΔΓ πρὸς τὴν ΓΕ, οὕτως ἡ Λ πρὸς τὴν Μ, καὶ ὡς ἄρα ἡ Λ πρὸς τὴν Μ, οὕτως τὸ ΓΘ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον. ἐπεὶ οὖν ἐδείχθη, ὡς μὲν ἡ Κ πρὸς τὴν Λ, οὕτως τὸ ΑΓ παραλληλόγραμμον πρὸς τὸ ΓΘ παραλληλόγραμμον, ὡς δὲ ἡ Λ πρὸς τὴν Μ, οὕτως τὸ ΓΘ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον, δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ Κ πρὸς τὴν Μ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΖ παραλληλόγραμμον. ἡ δὲ Κ πρὸς τὴν Μ λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν: καὶ τὸ ΑΓ ἄρα πρὸς τὸ ΓΖ λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν. τὰ ἄρα ἰσογώνια παραλληλόγραμμα πρὸς ἄλληλα λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",346]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'6.24'"],["Rule","'Text'","'In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'παντὸς παραλληλογράμμου τὰ περὶ τὴν διάμετρον παραλληλόγραμμα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",18]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]],["List",["Rule","'Book'",6],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'Let ABCD be a parallelogram, and AC its diameter, and let EG, HK be parallelograms about AC; I say that each of the parallelograms EG, HK is similar both to the whole ABCD and to the other. For, since EF has been drawn parallel to BC, one of the sides of the triangle ABC, proportionally, as BE is to EA, so is CF to FA. [VI. 2]  Again, since FG has been drawn parallel to CD, one of the sides of the triangle ACD, proportionally, as CF is to FA, so is DG to GA. [VI. 2]  But it was proved that, as CF is to FA, so also is BE to EA; therefore also, as BE is to EA, so is DG to GA, and therefore, componendo, as BA is to AE, so is DA to AG, [V. 18]and, alternately, as BA is to AD, so is EA to AG. [V. 16]  Therefore in the parallelograms ABCD, EG, the sides about the common angle BAD are proportional. And, since GF is parallel to DC, the angle AFG is equal to the angle DCA; and the angle DAC is common to the two triangles ADC, AGF; therefore the triangle ADC is equiangular with the triangle AGF. For the same reason the triangle ACB is also equiangular with the triangle AFE, and the whole parallelogram ABCD is equiangular with the parallelogram EG. Therefore, proportionally, as AD is to DC, so is AG to GF, as DC is to CA, so is GF to FA, as AC is to CB, so is AF to FE, and further, as CB is to BA, so is FE to EA. And, since it was proved that, as DC is to CA, so is GF to FA, and, as AC is to CB, so is AF to FE, therefore, ex aequali, as DC is to CB, so is GF to FE. [V. 22] Therefore in the parallelograms ABCD, EG the sides about the equal angles are proportional; therefore the parallelogram ABCD is similar to the parallelogram EG. [VI. Def. 1] For the same reason the parallelogram ABCD is also similar to the parallelogram KH; therefore each of the parallelograms EG, HK is similar to ABCD. But figures similar to the same rectilineal figure are also similar to one another; [VI. 21] therefore the parallelogram EG is also similar to the parallelogram HK.'"],["Rule","'ProofWordCount'",397],["Rule","'GreekProof'","'ἔστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἡ ΑΓ, περὶ δὲ τὴν ΑΓ παραλληλόγραμμα ἔστω τὰ ΕΗ, ΘΚ: λέγω, ὅτι ἑκάτερον τῶν ΕΗ, ΘΚ παραλληλογράμμων ὅμοιόν ἐστι ὅλῳ τῷ ΑΒΓΔ καὶ ἀλλήλοις. ἐπεὶ γὰρ τριγώνου τοῦ ΑΒΓ παρὰ μίαν τῶν πλευρῶν τὴν ΒΓ ἦκται ἡ ΕΖ, ἀνάλογόν ἐστιν ὡς ἡ ΒΕ πρὸς τὴν ΕΑ, οὕτως, ἡ ΓΖ πρὸς τὴν ΖΑ. πάλιν, ἐπεὶ τριγώνου τοῦ ΑΓΔ παρὰ μίαν τὴν ΓΔ ἦκται ἡ ΖΗ, ἀνάλογόν ἐστιν ὡς ἡ ΓΖ πρὸς τὴν ΖΑ, οὕτως ἡ ΔΗ πρὸς τὴν ΗΑ. ἀλλ᾽ ὡς ἡ ΓΖ πρὸς τὴν ΖΑ, οὕτως ἐδείχθη καὶ ἡ ΒΕ πρὸς τὴν ΕΑ: καὶ ὡς ἄρα ἡ ΒΕ πρὸς τὴν ΕΑ, οὕτως ἡ ΔΗ πρὸς τὴν ΗΑ, καὶ συνθέντι ἄρα ὡς ἡ ΒΑ πρὸς ΑΕ, οὕτως ἡ ΔΑ πρὸς ΑΗ, καὶ ἐναλλὰξ ὡς ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΗ. τῶν ἄρα ΑΒΓΔ, ΕΗ παραλληλογράμμων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὴν κοινὴν γωνίαν τὴν ὑπὸ ΒΑΔ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΗΖ τῇ ΔΓ, ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΖΗ γωνία τῇ ὑπὸ ΔΓΑ: καὶ κοινὴ τῶν δύο τριγώνων τῶν ΑΔΓ, ΑΗΖ ἡ ὑπὸ ΔΑΓ γωνία: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΔΓ τρίγωνον τῷ ΑΗΖ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΓΒ τρίγωνον ἰσογώνιόν ἐστι τῷ ΑΖΕ τριγώνῳ, καὶ ὅλον τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΗ παραλληλογράμμῳ ἰσογώνιόν ἐστιν. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΑΗ πρὸς τὴν ΗΖ, ὡς δὲ ἡ ΔΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ, ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΑΖ πρὸς τὴν ΖΕ, καὶ ἔτι ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΖΕ πρὸς τὴν ΕΑ. καὶ ἐπεὶ ἐδείχθη ὡς μὲν ἡ ΔΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ, ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΑΖ πρὸς τὴν ΖΕ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ ΔΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΕ. τῶν ἄρα ΑΒΓΔ, ΕΗ παραλληλογράμμων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: ὅμοιον ἄρα ἐστὶ τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΗ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ τὸ ΑΒΓΔ παραλληλόγραμμον καὶ τῷ ΚΘ παραλληλογράμμῳ ὅμοιόν ἐστιν: ἑκάτερον ἄρα τῶν ΕΗ, ΘΚ παραλληλογράμμων τῷ ΑΒΓΔ παραλληλογράμμῳ ὅμοιόν ἐστιν. τὰ δὲ τῷ αὐτῷ εὐθυγράμμῳ ὅμοια καὶ ἀλλήλοις ἐστὶν ὅμοια: καὶ τὸ ΕΗ ἄρα παραλληλόγραμμον τῷ ΘΚ παραλληλογράμμῳ ὅμοιόν ἐστιν. παντὸς ἄρα παραλληλογράμμου τὰ περὶ τὴν διάμετρον παραλληλόγραμμα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",408]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'6.25'"],["Rule","'Text'","'To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'τῷ δοθέντι εὐθυγράμμῳ ὅμοιον καὶ ἄλλῳ τῷ δοθέντι ἴσον τὸ αὐτὸ συστήσασθαι.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",44]],["List",["Rule","'Book'",1],["Rule","'Theorem'",45]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",13]],["List",["Rule","'Book'",6],["Rule","'Theorem'",18]],["List",["Rule","'Book'",6],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'Let ABC be the given rectilineal figure to which the figure to be constructed must be similar, and D that to which it must be equal; thus it is required to construct one and the same figure similar to ABC and equal to D. Let there be applied to BC the parallelogram BE equal to the triangle ABC [I. 44], and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL. [I. 45] Therefore BC is in a straight line with CF, and LE with EM. Now let GH be taken a mean proportional to BC, CF [VI. 13], and on GH let KGH be described similar and similarly situated to ABC. [VI. 18] Then, since, as BC is to GH, so is GH to CF, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [VI. 19] therefore, as BC is to CF, so is the triangle ABC to the triangle KGH. But, as BC is to CF, so also is the parallelogram BE to the parallelogram EF. [VI. 1] Therefore also, as the triangle ABC is to the triangle KGH, so is the parallelogram BE to the parallelogram EF; therefore, alternately, as the triangle ABC is to the parallelogram BE, so is the triangle KGH to the parallelogram EF. [V. 16] But the triangle ABC is equal to the parallelogram BE; therefore the triangle KGH is also equal to the parallelogram EF. But the parallelogram EF is equal to D; therefore KGH is also equal to D. And KGH is also similar to ABC. Therefore one and the same figure KGH has been constructed similar to the given rectilineal figure ABC and equal to the other given figure D.'"],["Rule","'ProofWordCount'",312],["Rule","'GreekProof'","'ἔστω τὸ μὲν δοθὲν εὐθύγραμμον, ᾧ δεῖ ὅμοιον συστήσασθαι, τὸ ΑΒΓ, ᾧ δὲ δεῖ ἴσον, τὸ Δ: δεῖ δὴ τῷ μὲν ΑΒΓ ὅμοιον, τῷ δὲ Δ ἴσον τὸ αὐτὸ συστήσασθαι. παραβεβλήσθω γὰρ παρὰ μὲν τὴν ΒΓ τῷ ΑΒΓ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΒΕ, παρὰ δὲ τὴν ΓΕ τῷ Δ ἴσον παραλληλόγραμμον τὸ ΓΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΓΕ, á¼¥ ἐστιν ἴση τῇ ὑπὸ ΓΒΛ. ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ μὲν ΒΓ τῇ ΓΖ, ἡ δὲ ΛΕ τῇ ΕΜ. καὶ εἰλήφθω τῶν ΒΓ, ΓΖ μέση ἀνάλογον ἡ ΗΘ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΗΘ τῷ ΑΒΓ ὅμοιόν τε καὶ ὁμοίως κείμενον τὸ ΚΗΘ. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΗΘ, οὕτως ἡ ΗΘ πρὸς τὴν ΓΖ, ἐὰν δὲ τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΖ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΚΗΘ τρίγωνον. ἀλλὰ καὶ ὡς ἡ ΒΓ πρὸς τὴν ΓΖ, οὕτως τὸ ΒΕ παραλληλόγραμμον πρὸς τὸ ΕΖ παραλληλόγραμμον. καὶ ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΚΗΘ τρίγωνον, οὕτως τὸ ΒΕ παραλληλόγραμμον πρὸς τὸ ΕΖ παραλληλόγραμμον: ἐναλλὰξ ἄρα ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΒΕ παραλληλόγραμμον, οὕτως τὸ ΚΗΘ τρίγωνον πρὸς τὸ ΕΖ παραλληλόγραμμον. ἴσον δὲ τὸ ΑΒΓ τρίγωνον τῷ ΒΕ παραλληλογράμμῳ: ἴσον ἄρα καὶ τὸ ΚΗΘ τρίγωνον τῷ ΕΖ παραλληλογράμμῳ. ἀλλὰ τὸ ΕΖ παραλληλόγραμμον τῷ Δ ἐστιν ἴσον: καὶ τὸ ΚΗΘ ἄρα τῷ Δ ἐστιν ἴσον. ἔστι δὲ τὸ ΚΗΘ καὶ τῷ ΑΒΓ ὅμοιον. τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓ ὅμοιον καὶ ἄλλῳ τῷ δοθέντι τῷ Δ ἴσον τὸ αὐτὸ συνέσταται τὸ ΚΗΘ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",277]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",26]],["Association",["Rule","'VertexLabel'","'6.26'"],["Rule","'Text'","'If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, it is about the same diameter with the whole.'"],["Rule","'TextWordCount'",33],["Rule","'GreekText'","'ἐὰν ἀπὸ παραλληλογράμμου παραλληλόγραμμον ἀφαιρεθῇ ὅμοιόν τε τῷ ὅλῳ καὶ ὁμοίως κείμενον κοινὴν γωνίαν ἔχον αὐτῷ, περὶ τὴν αὐτὴν διάμετρόν ἐστι τῷ ὅλῳ.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'For suppose it is not, but, if possible, let AHC be the diameter < of ABCD >, let GF be produced and carried through to H, and let HK be drawn through H parallel to either of the straight lines AD, BC. [I. 31] Since, then, ABCD is about the same diameter with KG, therefore, as DA is to AB, so is GA to AK. [VI. 24] But also, because of the similarity of ABCD, EG, as DA is to AB, so is GA to AE; therefore also, as GA is to AK, so is GA to AE. [V. 11] Therefore GA has the same ratio to each of the straight lines AK, AE. Therefore AE is equal to AK [V. 9], the less to the greater: which is impossible. Therefore ABCD cannot but be about the same diameter with AF; therefore the parallelogram ABCD is about the same diameter with the parallelogram AF.'"],["Rule","'ProofWordCount'",152],["Rule","'GreekProof'","'ἀπὸ γὰρ παραλληλογράμμου τοῦ ΑΒΓΔ παραλληλόγραμμον ἀφῃρήσθω τὸ ΑΖ ὅμοιον τῷ ΑΒΓΔ καὶ ὁμοίως κείμενον κοινὴν γωνίαν ἔχον αὐτῷ τὴν ὑπὸ ΔΑΒ: λέγω, ὅτι περὶ τὴν αὐτὴν διάμετρόν ἐστι τὸ ΑΒΓΔ τῷ ΑΖ. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἔστω αὐτῶν διάμετρος ἡ ΑΘΓ, καὶ ἐκβληθεῖσα ἡ ΗΖ διήχθω ἐπὶ τὸ Θ, καὶ ἤχθω διὰ τοῦ Θ ὁποτέρᾳ τῶν ΑΔ, ΒΓ παράλληλος ἡ ΘΚ. ἐπεὶ οὖν περὶ τὴν αὐτὴν διάμετρόν ἐστι τὸ ΑΒΓΔ τῷ ΚΗ, ἔστιν ἄρα ὡς ἡ ΔΑ πρὸς τὴν ΑΒ, οὕτως ἡ ΗΑ πρὸς τὴν ΑΚ. ἔστι δὲ καὶ διὰ τὴν ὁμοιότητα τῶν ΑΒΓΔ, ΕΗ καὶ ὡς ἡ ΔΑ πρὸς τὴν ΑΒ, οὕτως ἡ ΗΑ πρὸς τὴν ΑΕ: καὶ ὡς ἄρα ἡ ΗΑ πρὸς τὴν ΑΚ, οὕτως ἡ ΗΑ πρὸς τὴν ΑΕ. ἡ ΗΑ ἄρα πρὸς ἑκατέραν τῶν ΑΚ, ΑΕ τὸν αὐτὸν ἔχει λόγον. ἴση ἄρα ἐστὶν ἡ ΑΕ τῇ ΑΚ ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οὔκ ἐστι περὶ τὴν αὐτὴν διάμετρον τὸ ΑΒΓΔ τῷ ΑΖ: περὶ τὴν αὐτὴν ἄρα ἐστὶ διάμετρον τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΑΖ παραλληλογράμμῳ. ἐὰν ἄρα ἀπὸ παραλληλογράμμου παραλληλόγραμμον ἀφαιρεθῇ ὅμοιόν τε τῷ ὅλῳ καὶ ὁμοίως κείμενον κοινὴν γωνίαν ἔχον αὐτῷ, περὶ τὴν αὐτὴν διάμετρόν ἐστι τῷ ὅλῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",202]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",27]],["Association",["Rule","'VertexLabel'","'6.27'"],["Rule","'Text'","'Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.'"],["Rule","'TextWordCount'",49],["Rule","'GreekText'","'πάντων τῶν παρὰ τὴν αὐτὴν εὐθεῖαν παραβαλλομένων παραλληλογράμμων καὶ ἐλλειπόντων εἴδεσι παραλληλογράμμοις ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ἀπὸ τῆς ἡμισείας ἀναγραφομένῳ μέγιστόν ἐστι τὸ ἀπὸ τῆς ἡμισείας παραβαλλόμενον παραλληλόγραμμον ὅμοιον ὂν τῷ ἐλλείμματι.'"],["Rule","'GreekTextWordCount'",34],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",36]],["List",["Rule","'Book'",1],["Rule","'Theorem'",43]],["List",["Rule","'Book'",6],["Rule","'Theorem'",26]]]],["Rule","'Proof'","'Let AB be a straight line and let it be bisected at C; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB, that is, CB; I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest. For let there be applied to the straight line AB the parallelogram AF deficient by the parallelogrammic figure FB similar and similarly situated to DB; I say that AD is greater than AF. For, since the parallelogram DB is similar to the parallelogram FB, they are about the same diameter. [VI. 26]  Let their diameter DB be drawn, and let the figure be described. Then, since CF is equal to FE, [I. 43] and FB is common, therefore the whole CH is equal to the whole KE. But CH is equal to CG, since AC is also equal to CB. [I. 36] Therefore GC is also equal to EK. Let CF be added to each; therefore the whole AF is equal to the gnomon LMN; so that the parallelogram DB, that is, AD, is greater than the parallelogram AF.'"],["Rule","'ProofWordCount'",203],["Rule","'GreekProof'","'ἔστω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω δίχα κατὰ τὸ Γ, καὶ παραβεβλήσθω παρὰ τὴν ΑΒ εὐθεῖαν τὸ ΑΔ παραλληλόγραμμον ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΔΒ ἀναγραφέντι ἀπὸ τῆς ἡμισείας τῆς ΑΒ, τουτέστι τῆς ΓΒ: λέγω, ὅτι πάντων τῶν παρὰ τὴν ΑΒ παραβαλλομένων παραλληλογράμμων καὶ ἐλλειπόντων εἴδεσι παραλληλογράμμοις ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ΔΒ μέγιστόν ἐστι τὸ ΑΔ. παραβεβλήσθω γὰρ παρὰ τὴν ΑΒ εὐθεῖαν τὸ ΑΖ παραλληλόγραμμον ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΖΒ ὁμοίῳ τε καὶ ὁμοίως κειμένῳ τῷ ΔΒ: λέγω, ὅτι μεῖζόν ἐστι τὸ ΑΔ τοῦ ΑΖ. ἐπεὶ γὰρ ὅμοιόν ἐστι τὸ ΔΒ παραλληλόγραμμον τῷ ΖΒ παραλληλογράμμῳ, περὶ τὴν αὐτήν εἰσι διάμετρον. ἤχθω αὐτῶν διάμετρος ἡ ΔΒ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΓΖ τῷ ΖΕ, κοινὸν δὲ τὸ ΖΒ, ὅλον ἄρα τὸ ΓΘ ὅλῳ τῷ ΚΕ ἐστιν ἴσον. ἀλλὰ τὸ ΓΘ τῷ ΓΗ ἐστιν ἴσον, ἐπεὶ καὶ ἡ ΑΓ τῇ ΓΒ. καὶ τὸ ΗΓ ἄρα τῷ ΕΚ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΓΖ: ὅλον ἄρα τὸ ΑΖ τῷ ΛΜΝ γνώμονί ἐστιν ἴσον: ὥστε τὸ ΔΒ παραλληλόγραμμον, τουτέστι τὸ ΑΔ, τοῦ ΑΖ παραλληλογράμμου μεῖζόν ἐστιν. πάντων ἄρα τῶν παρὰ τὴν αὐτὴν εὐθεῖαν παραβαλλομένων παραλληλογράμμων καὶ ἐλλειπόντων εἴδεσι παραλληλογράμμοις ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ἀπὸ τῆς ἡμισείας ἀναγραφομένῳ μέγιστόν ἐστι τὸ ἀπὸ τῆς ἡμισείας παραβληθέν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",210]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",28]],["Association",["Rule","'VertexLabel'","'6.28'"],["Rule","'Text'","'To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one: thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.'"],["Rule","'TextWordCount'",51],["Rule","'GreekText'","'παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ἐλλεῖπον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ δοθέντι: δεῖ δὲ τὸ διδόμενον εὐθύγραμμον ᾧ δεῖ ἴσον παραβαλεῖν μὴ μεῖζον εἶναι τοῦ ἀπὸ τῆς ἡμισείας ἀναγραφομένου ὁμοίου τῷ ἐλλείμματι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ ᾧ δεῖ ὅμοιον ἐλλείπειν.'"],["Rule","'GreekTextWordCount'",46],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",36]],["List",["Rule","'Book'",6],["Rule","'Theorem'",18]],["List",["Rule","'Book'",6],["Rule","'Theorem'",21]],["List",["Rule","'Book'",6],["Rule","'Theorem'",25]],["List",["Rule","'Book'",6],["Rule","'Theorem'",26]]]],["Rule","'Proof'","'Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D. Let AB be bisected at the point E, and on EB let EBFG be described similar and similarly situated to D; [VI. 18] let the parallelogram AG be completed. If then AG is equal to C, that which was enjoined will have been done; for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D.  But, if not, let HE be greater than C. Now HE is equal to GB; therefore GB is also greater than C.  Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D. [VI. 25] But D is similar to GB; therefore KM is also similar to GB. [VI. 21]  Let, then, KL correspond to GE, and LM to GF. Now, since GB is equal to C, KM, therefore GB is greater than KM; therefore also GE is greater than KL, and GF than LM. Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed; therefore it is equal and similar to KM. Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26] Let GQB be their diameter, and let the figure be described. Then, since BG is equal to C, KM, and in them GQ is equal to KM, therefore the remainder, the gnomon UWV, is equal to the remainder C. And, since PR is equal to OS, let QB be added to each; therefore the whole PB is equal to the whole OB. But OB is equal to TE, since the side AE is also equal to the side EB; [I. 36] therefore TE is also equal to PB. Let OS be added to each; therefore the whole TS is equal to the whole, the gnomon VWU. But the gnomon VWU was proved equal to C; therefore TS is also equal to C.  Therefore to the given straight line AB there has been applied the parallelogram ST equal to the given rectilineal figure C and deficient by a parallelogrammic figure QB which is similar to D.'"],["Rule","'ProofWordCount'",465],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν εὐθύγραμμον, ᾧ δεῖ ἴσον παρὰ τὴν ΑΒ παραβαλεῖν, τὸ Γ μὴ μεῖζον ὂν τοῦ ἀπὸ τῆς ἡμισείας τῆς ΑΒ ἀναγραφομένου ὁμοίου τῷ ἐλλείμματι, ᾧ δὲ δεῖ ὅμοιον ἐλλείπειν, τὸ Δ: δεῖ δὴ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβαλεῖν ἐλλεῖπον εἴδει παραλληλογράμμῳ ὁμοίῳ ὄντι τῷ Δ. τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε σημεῖον, καὶ ἀναγεγράφθω ἀπὸ τῆς ΕΒ τῷ Δ ὅμοιον καὶ ὁμοίως κείμενον τὸ ΕΒΖΗ, καὶ συμπεπληρώσθω τὸ ΑΗ παραλληλόγραμμον. εἰ μὲν οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ Γ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν: παραβέβληται γὰρ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον τὸ ΑΗ ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΗΒ ὁμοίῳ ὄντι τῷ Δ. εἰ δὲ οὔ, μεῖζον ἔστω τὸ ΘΕ τοῦ Γ. ἴσον δὲ τὸ ΘΕ τῷ ΗΒ: μεῖζον ἄρα καὶ τὸ ΗΒ τοῦ Γ. ᾧ δὴ μεῖζόν ἐστι τὸ ΗΒ τοῦ Γ, ταύτῃ τῇ ὑπεροχῇ ἴσον, τῷ δὲ Δ ὅμοιον καὶ ὁμοίως κείμενον τὸ αὐτὸ συνεστάτω τὸ ΚΛΜΝ. ἀλλὰ τὸ Δ τῷ ΗΒ ἐστιν ὅμοιον: καὶ τὸ ΚΜ ἄρα τῷ ΗΒ ἐστιν ὅμοιον. ἔστω οὖν ὁμόλογος ἡ μὲν ΚΛ τῇ ΗΕ, ἡ δὲ ΛΜ τῇ ΗΖ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΗΒ τοῖς Γ, ΚΜ, μεῖζον ἄρα ἐστὶ τὸ ΗΒ τοῦ ΚΜ: μείζων ἄρα ἐστὶ καὶ ἡ μὲν ΗΕ τῆς ΚΛ, ἡ δὲ ΗΖ τῆς ΛΜ. κείσθω τῇ μὲν ΚΛ ἴση ἡ ΗΞ, τῇ δὲ ΛΜ ἴση ἡ ΗΟ, καὶ συμπεπληρώσθω τὸ ΞΗΟΠ παραλληλόγραμμον: ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΗΠ τῷ ΚΜ ἀλλὰ τὸ ΚΜ τῷ ΗΒ ὅμοιόν ἐστιν. καὶ τὸ ΗΠ ἄρα τῷ ΗΒ ὅμοιόν ἐστιν: περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὸ ΗΠ τῷ ΗΒ. ἔστω αὐτῶν διάμετρος ἡ ΗΠΒ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΒΗ τοῖς Γ, ΚΜ, ὧν τὸ ΗΠ τῷ ΚΜ ἐστιν ἴσον, λοιπὸς ἄρα ὁ ΥΧΦ γνώμων λοιπῷ τῷ Γ ἴσος ἐστίν. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΟΡ τῷ ΞΣ, κοινὸν προσκείσθω τὸ ΠΒ: ὅλον ἄρα τὸ ΟΒ ὅλῳ τῷ ΞΒ ἴσον ἐστίν. ἀλλὰ τὸ ΞΒ τῷ ΤΕ ἐστιν ἴσον, ἐπεὶ καὶ πλευρὰ ἡ ΑΕ πλευρᾷ τῇ ΕΒ ἐστιν ἴση: καὶ τὸ ΤΕ ἄρα τῷ ΟΒ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΞΣ: ὅλον ἄρα τὸ ΤΣ ὅλῳ τῷ ΦΧΥ γνώμονί ἐστιν ἴσον. ἀλλ᾽ ὁ ΦΧΥ γνώμων τῷ Γ ἐδείχθη ἴσος: καὶ τὸ ΤΣ ἄρα τῷ Γ ἐστιν ἴσον. παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΣΤ ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΠΒ ὁμοίῳ ὄντι τῷ Δ ἐπειδήπερ τὸ ΠΒ τῷ ΗΠ ὅμοιόν ἐστιν: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",433]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",29]],["Association",["Rule","'VertexLabel'","'6.29'"],["Rule","'Text'","'To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ὑπερβάλλον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ δοθέντι.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",36]],["List",["Rule","'Book'",1],["Rule","'Theorem'",43]],["List",["Rule","'Book'",6],["Rule","'Theorem'",21]],["List",["Rule","'Book'",6],["Rule","'Theorem'",24]],["List",["Rule","'Book'",6],["Rule","'Theorem'",25]],["List",["Rule","'Book'",6],["Rule","'Theorem'",26]]]],["Rule","'Proof'","'Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, and D that to which the excess is required to be similar; thus it is required to apply to the straight line AB a parallelogram equal to the rectilineal figure C and exceeding by a parallelogrammic figure similar to D. Let AB be bisected at E; let there be described on EB the parallelogram BF similar and similarly situated to D; and let GH be constructed at once equal to the sum of BF, C and similar and similarly situated to D. [VI. 25] Let KH correspond to FL and KG to FE. Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE. Let FL, FE be produced, let FLM be equal to KH, and FEN to KG, and let MN be completed; therefore MN is both equal and similar to GH. But GH is similar to EL; therefore MN is also similar to EL; [VI. 21]therefore EL is about the same diameter with MN. [VI. 26] Let their diameter FO be drawn, and let the figure be described. Since GH is equal to EL, C, while GH is equal to MN, therefore MN is also equal to EL, C. Let EL be subtracted from each; therefore the remainder, the gnomon XWV, is equal to C. Now, since AE is equal to EB, AN is also equal to NB [I. 36], that is, to LP [I. 43]. Let EO be added to each; therefore the whole AO is equal to the gnomon VWX. But the gnomon VWX is equal to C; therefore AO is also equal to C. Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [VI. 24].'"],["Rule","'ProofWordCount'",335],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν εὐθύγραμμον, ᾧ δεῖ ἴσον παρὰ τὴν ΑΒ παραβαλεῖν, τὸ Γ, ᾧ δὲ δεῖ ὅμοιον ὑπερβάλλειν, τὸ Δ: δεῖ δὴ παρὰ τὴν ΑΒ εὐθεῖαν τῷ Γ εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ὑπερβάλλον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ Δ. τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε, καὶ ἀναγεγράφθω ἀπὸ τῆς ΕΒ τῷ Δ ὅμοιον καὶ ὁμοίως κείμενον παραλληλόγραμμον τὸ ΒΖ, καὶ συναμφοτέροις μὲν τοῖς ΒΖ, Γ ἴσον, τῷ δὲ Δ ὅμοιον καὶ ὁμοίως κείμενον τὸ αὐτὸ συνεστάτω τὸ ΗΘ. ὁμόλογος δὲ ἔστω ἡ μὲν ΚΘ τῇ ΖΛ, ἡ δὲ ΚΗ τῇ ΖΕ. καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΗΘ τοῦ ΖΒ, μείζων ἄρα ἐστὶ καὶ ἡ μὲν ΚΘ τῆς ΖΛ, ἡ δὲ ΚΗ τῆς ΖΕ. ἐκβεβλήσθωσαν αἱ ΖΛ, ΖΕ, καὶ τῇ μὲν ΚΘ ἴση ἔστω ἡ ΖΛΜ, τῇ δὲ ΚΗ ἴση ἡ ΖΕΝ, καὶ συμπεπληρώσθω τὸ ΜΝ: τὸ ΜΝ ἄρα τῷ ΗΘ ἴσον τέ ἐστι καὶ ὅμοιον. ἀλλὰ τὸ ΗΘ τῷ ΕΛ ἐστιν ὅμοιον: καὶ τὸ ΜΝ ἄρα τῷ ΕΛ ὅμοιόν ἐστιν: περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὸ ΕΛ τῷ ΜΝ. ἤχθω αὐτῶν διάμετρος ἡ ΖΞ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ ἴσον ἐστὶ τὸ ΗΘ τοῖς ΕΛ, Γ, ἀλλὰ τὸ ΗΘ τῷ ΜΝ ἴσον ἐστίν, καὶ τὸ ΜΝ ἄρα τοῖς ΕΛ, Γ ἴσον ἐστίν. κοινὸν ἀφῃρήσθω τὸ ΕΛ: λοιπὸς ἄρα ὁ ΨΧΦ γνώμων τῷ Γ ἐστιν ἴσος. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, ἴσον ἐστὶ καὶ τὸ ΑΝ τῷ ΝΒ, τουτέστι τῷ ΛΟ. κοινὸν προσκείσθω τὸ ΕΞ: ὅλον ἄρα τὸ ΑΞ ἴσον ἐστὶ τῷ ΦΧΨ γνώμονι. ἀλλὰ ὁ ΦΧΨ γνώμων τῷ Γ ἴσος ἐστίν: καὶ τὸ ΑΞ ἄρα τῷ Γ ἴσον ἐστίν. παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΑΞ ὑπερβάλλον εἴδει παραλληλογράμμῳ τῷ ΠΟ ὁμοίῳ ὄντι τῷ Δ, ἐπεὶ καὶ τῷ ΕΛ ἐστιν ὅμοιον τὸ ΟΠ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",307]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",30]],["Association",["Rule","'VertexLabel'","'6.30'"],["Rule","'Text'","'To cut a given finite straight line in extreme and mean ratio.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην ἄκρον καὶ μέσον λόγον τεμεῖν.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",14]],["List",["Rule","'Book'",6],["Rule","'Theorem'",29]]]],["Rule","'Proof'","'Let AB be the given finite straight line; thus it is required to cut AB in extreme and mean ratio. On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC. [VI. 29] Now BC is a square; therefore AD is also a square. And, since BC is equal to CD, let CE be subtracted from each; therefore the remainder BF is equal to the remainder AD. But it is also equiangular with it; therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, as FE is to ED, so is AE to EB. But FE is equal to AB, and ED to AE. Therefore, as BA is to AE, so is AE to EB. And AB is greater than AE; therefore AE is also greater than EB. Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE.'"],["Rule","'ProofWordCount'",176],["Rule","'GreekProof'","'ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ: δεῖ δὴ τὴν ΑΒ εὐθεῖαν ἄκρον καὶ μέσον λόγον τεμεῖν. Ἀναγεγράφθω ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΒΓ, καὶ παραβεβλήσθω παρὰ τὴν ΑΓ τῇ ΒΓ ἴσον παραλληλόγραμμον τὸ ΓΔ ὑπερβάλλον εἴδει τῷ ΑΔ ὁμοίῳ τῷ ΒΓ. τετράγωνον δέ ἐστι τὸ ΒΓ: τετράγωνον ἄρα ἐστὶ καὶ τὸ ΑΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΒΓ τῷ ΓΔ, κοινὸν ἀφῃρήσθω τὸ ΓΕ: λοιπὸν ἄρα τὸ ΒΖ λοιπῷ τῷ ΑΔ ἐστιν ἴσον. ἔστι δὲ αὐτῷ καὶ ἰσογώνιον: τῶν ΒΖ, ΑΔ ἄρα ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: ἔστιν ἄρα ὡς ἡ ΖΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΑΕ πρὸς τὴν ΕΒ. ἴση δὲ ἡ μὲν ΖΕ τῇ ΑΒ, ἡ δὲ ΕΔ τῇ ΑΕ. ἔστιν ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΕ, οὕτως ἡ ΑΕ πρὸς τὴν ΕΒ. μείζων δὲ ἡ ΑΒ τῆς ΑΕ: μείζων ἄρα καὶ ἡ ΑΕ τῆς ΕΒ. ἡ ἄρα ΑΒ εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ε, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστι τὸ ΑΕ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",167]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",31]],["Association",["Rule","'VertexLabel'","'6.31'"],["Rule","'Text'","'In right - angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'ἐν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις.'"],["Rule","'GreekTextWordCount'",30],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",8]],["List",["Rule","'Book'",6],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC. Let AD be drawn perpendicular. Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8] And, since ABC is similar to ABD, therefore, as CB is to BA, so is AB to BD. [VI. Def. 1] And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [VI. 19] Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA. For the same reason also, as BC is to CD, so is the figure on BC to that on CA; so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC. But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC.'"],["Rule","'ProofWordCount'",223],["Rule","'GreekProof'","'ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν: λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις. ἤχθω κάθετος ἡ ΑΔ. ἐπεὶ οὖν ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΒΓ ἀπὸ τῆς πρὸς τῷ Α ὀρθῆς γωνίας ἐπὶ τὴν ΒΓ βάσιν κάθετος ἦκται ἡ ΑΔ, τὰ ΑΒΔ, ΑΔΓ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ τῷ ΑΒΓ καὶ ἀλλήλοις. καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓ τῷ ΑΒΔ, ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΔ. καὶ ἐπεὶ τρεῖς εὐθεῖαι ἀνάλογόν εἰσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον. ὡς ἄρα ἡ ΓΒ πρὸς τὴν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΓΒ εἶδος πρὸς τὸ ἀπὸ τῆς ΒΑ τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΒΓ πρὸς τὴν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΒΓ εἶδος πρὸς τὸ ἀπὸ τῆς ΓΑ. ὥστε καὶ ὡς ἡ ΒΓ πρὸς τὰς ΒΔ, ΔΓ, οὕτως τὸ ἀπὸ τῆς ΒΓ εἶδος πρὸς τὰ ἀπὸ τῶν ΒΑ, ΑΓ τὰ ὅμοια καὶ ὁμοίως ἀναγραφόμενα. ἴση δὲ ἡ ΒΓ ταῖς ΒΔ, ΔΓ: ἴσον ἄρα καὶ τὸ ἀπὸ τῆς ΒΓ εἶδος τοῖς ἀπὸ τῶν ΒΑ, ΑΓ εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις. ἐν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",257]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",32]],["Association",["Rule","'VertexLabel'","'6.32'"],["Rule","'Text'","'If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν τὰς δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ τῶν τριγώνων πλευραὶ ἐπ᾽ εὐθείας ἔσονται.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",14]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",6],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'Let ABC, DCE be two triangles having the two sides BA, AC proportional to the two sides DC, DE, so that, as AB is to AC, so is DC to DE, and AB parallel to DC, and AC to DE; I say that BC is in a straight line with CE. For, since AB is parallel to DC, and the straight line AC has fallen upon them, the alternate angles BAC, ACD are equal to one another. [I. 29] For the same reason the angle CDE is also equal to the angle ACD; so that the angle BAC is equal to the angle CDE. And, since ABC, DCE are two triangles having one angle, the angle at A, equal to one angle, the angle at D, and the sides about the equal angles proportional, so that, as BA is to AC, so is CD to DE, therefore the triangle ABC is equiangular with the triangle DCE; [VI. 6]therefore the angle ABC is equal to the angle DCE. But the angle ACD was also proved equal to the angle BAC; therefore the whole angle ACE is equal to the two angles ABC, BAC. Let the angle ACB be added to each; therefore the angles ACE, ACB are equal to the angles BAC, ACB, CBA. But the angles BAC, ABC, ACB are equal to two right angles; [I. 32] therefore the angles ACE, ACB are also equal to two right angles. Therefore with a straight line AC, and at the point C on it, the two straight lines BC, CE not lying on the same side make the adjacent angles ACE, ACB equal to two right angles; therefore BC is in a straight line with CE. [I. 14]'"],["Rule","'ProofWordCount'",287],["Rule","'GreekProof'","'ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΓΕ τὰς δύο πλευρὰς τὰς ΒΑ, ΑΓ ταῖς δυσὶ πλευραῖς ταῖς ΔΓ, ΔΕ ἀνάλογον ἔχοντα, ὡς μὲν τὴν ΑΒ πρὸς τὴν ΑΓ, οὕτως τὴν ΔΓ πρὸς τὴν ΔΕ, παράλληλον δὲ τὴν μὲν ΑΒ τῇ ΔΓ, τὴν δὲ ΑΓ τῇ ΔΕ: λέγω, ὅτι ἐπ᾽ εὐθείας ἐστὶν ἡ ΒΓ τῇ ΓΕ. ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΑΒ τῇ ΔΓ, καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΑΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ, ΑΓΔ ἴσαι ἀλλήλαις εἰσίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΓΔ ἴση ἐστίν. ὥστε καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση. καὶ ἐπεὶ δύο τρίγωνά ἐστι τὰ ΑΒΓ, ΔΓΕ μίαν γωνίαν τὴν πρὸς τῷ Α μιᾷ γωνίᾳ τῇ πρὸς τῷ Δ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, οὕτως τὴν ΓΔ πρὸς τὴν ΔΕ, ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΓΕ τριγώνῳ: ἴση ἄρα ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΓΕ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΒΑΓ ἴση: ὅλη ἄρα ἡ ὑπὸ ΑΓΕ δυσὶ ταῖς ὑπὸ ΑΒΓ, ΒΑΓ ἴση ἐστίν. κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ: αἱ ἄρα ὑπὸ ΑΓΕ, ΑΓΒ ταῖς ὑπὸ ΒΑΓ, ΑΓΒ, ΓΒΑ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΒΑΓ, ΑΒΓ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΑΓΕ, ΑΓΒ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. πρὸς δή τινι εὐθείᾳ τῇ ΑΓ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Γ δύο εὐθεῖαι αἱ ΒΓ, ΓΕ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΑΓΕ, ΑΓΒ δυσὶν ὀρθαῖς ἴσας ποιοῦσιν: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΒΓ τῇ ΓΕ. ἐὰν ἄρα δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν τὰς δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ τῶν τριγώνων πλευραὶ ἐπ᾽ εὐθείας ἔσονται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",303]]],["Rule",["Association",["Rule","'Book'",6],["Rule","'Theorem'",33]],["Association",["Rule","'VertexLabel'","'6.33'"],["Rule","'Text'","'In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centres or at the circumferences.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ἐν τοῖς ἴσοις κύκλοις αἱ γωνίαι τὸν αὐτὸν ἔχουσι λόγον ταῖς περιφερείαις, ἐφ᾽ ὧν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι.'"],["Rule","'GreekTextWordCount'",28],["Rule","'References'",["List",["List",["Rule","'Book'",3],["Rule","'Theorem'",27]],["List",["Rule","'Book'",5],["Rule","'Definition'",5]]]],["Rule","'Proof'","'Let ABC, DEF be equal circles, and let the angles BGC, EHF be angles at their centres G, H, and the angles BAC, EDF angles at the circumferences; I say that, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. For let any number of consecutive circumferences CK, KL be made equal to the circumference BC, and any number of consecutive circumferences FM, MN equal to the circumference EF; and let GK, GL, HM, HN be joined. Then, since the circumferences BC, CK, KL are equal to one another, the angles BGC, CGK, KGL are also equal to one another; [III. 27] therefore, whatever multiple the circumference BL is of BC, that multiple also is the angle BGL of the angle BGC. For the same reason also, whatever multiple the circumference NE is of EF, that multiple also is the angle NHE of the angle EHF. If then the circumference BL is equal to the circumference EN, the angle BGL is also equal to the angle EHN; [III. 27] if the circumference BL is greater than the circumference EN, the angle BGL is also greater than the angle EHN; and, if less, less. There being then four magnitudes, two circumferences BC, EF, and two angles BGC, EHF, there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL, and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN. And it has been proved that, if the circumference BL is in excess of the circumference EN, the angle BGL is also in excess of the angle EHN; if equal, equal; and if less, less. Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF. [V. Def. 5] But, as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF; for they are doubles respectively. Therefore also, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.'"],["Rule","'ProofWordCount'",370],["Rule","'GreekProof'","'ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ πρὸς μὲν τοῖς κέντροις αὐτῶν τοῖς Η, Θ γωνίαι ἔστωσαν αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ: λέγω, ὅτι ἐστὶν ὡς ἡ ΒΓ περιφέρεια πρὸς τὴν ΕΖ περιφέρειαν, οὕτως á¼¥ τε ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ καὶ ἡ ὑπὸ ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ. κείσθωσαν γὰρ τῇ μὲν ΒΓ περιφερείᾳ ἴσαι κατὰ τὸ ἑξῆς ὁσαιδηποτοῦν αἱ ΓΚ, ΚΛ, τῇ δὲ ΕΖ περιφερείᾳ ἴσαι ὁσαιδηποτοῦν αἱ ΖΜ, ΜΝ, καὶ ἐπεζεύχθωσαν αἱ ΗΚ, ΗΛ, ΘΜ, ΘΝ. ἐπεὶ οὖν ἴσαι εἰσὶν αἱ ΒΓ, ΓΚ, ΚΛ περιφέρειαι ἀλλήλαις, ἴσαι εἰσὶ καὶ αἱ ὑπὸ ΒΗΓ, ΓΗΚ, ΚΗΛ γωνίαι ἀλλήλαις: ὁσαπλασίων ἄρα ἐστὶν ἡ ΒΛ περιφέρεια τῆς ΒΓ, τοσαυταπλασίων ἐστὶ καὶ ἡ ὑπὸ ΒΗΛ γωνία τῆς ὑπὸ ΒΗΓ. διὰ τὰ αὐτὰ δὴ καὶ ὁσαπλασίων ἐστὶν ἡ ΝΕ περιφέρεια τῆς ΕΖ, τοσαυταπλασίων ἐστὶ καὶ ἡ ὑπὸ ΝΘΕ γωνία τῆς ὑπὸ ΕΘΖ. εἰ ἄρα ἴση ἐστὶν ἡ ΒΛ περιφέρεια τῇ ΕΝ περιφερείᾳ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΗΛ τῇ ὑπὸ ΕΘΝ, καὶ εἰ μείζων ἐστὶν ἡ ΒΛ περιφέρεια τῆς ΕΝ περιφερείας, μείζων ἐστὶ καὶ ἡ ὑπὸ ΒΗΛ γωνία τῆς ὑπὸ ΕΘΝ, καὶ εἰ ἐλάσσων, ἐλάσσων. τεσσάρων δὴ ὄντων μεγεθῶν, δύο μὲν περιφερειῶν τῶν ΒΓ, ΕΖ, δύο δὲ γωνιῶν τῶν ὑπὸ ΒΗΓ, ΕΘΖ, εἴληπται τῆς μὲν ΒΓ περιφερείας καὶ τῆς ὑπὸ ΒΗΓ γωνίας ἰσάκις πολλαπλασίων á¼¥ τε ΒΛ περιφέρεια καὶ ἡ ὑπὸ ΒΗΛ γωνία, τῆς δὲ ΕΖ περιφερείας καὶ τῆς ὑπὸ ΕΘΖ γωνίας á¼¥ τε ΕΝ περιφέρεια καὶ ἡ ὑπὸ ΕΘΝ γωνία. καὶ δέδεικται, ὅτι εἰ ὑπερέχει ἡ ΒΛ περιφέρεια τῆς ΕΝ περιφερείας, ὑπερέχει καὶ ἡ ὑπὸ ΒΗΛ γωνία τῆς ὑπὸ ΕΘΝ γωνίας, καὶ εἰ ἴση, ἴση, καὶ εἰ ἐλάσσων, ἐλάσσων. ἔστιν ἄρα, ὡς ἡ ΒΓ περιφέρεια πρὸς τὴν ΕΖ, οὕτως ἡ ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ. ἀλλ᾽ ὡς ἡ ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ, οὕτως ἡ ὑπὸ ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ: διπλασία γὰρ ἑκατέρα ἑκατέρας. καὶ ὡς ἄρα ἡ ΒΓ περιφέρεια πρὸς τὴν ΕΖ περιφέρειαν, οὕτως á¼¥ τε ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ καὶ ἡ ὑπὸ ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ. ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ γωνίαι τὸν αὐτὸν ἔχουσι λόγον ταῖς περιφερείαις, ἐφ᾽ ὧν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",378]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'7.1'"],["Rule","'Text'","'Two unequal numbers being set out, and the less being continually subtracted in turn from the greater, if the number which is left never measures the one before it until an unit is left, the original numbers will be prime to one another.'"],["Rule","'TextWordCount'",43],["Rule","'GreekText'","'δύο ἀριθμῶν ἀνίσων ἐκκειμένων, ἀνθυφαιρουμένου δὲ ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, ἐὰν ὁ λειπόμενος μηδέποτε καταμετρῇ τὸν πρὸ ἑαυτοῦ, ἕως οὗ λειφθῇ μονάς, οἱ ἐξ ἀρχῆς ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ἔσονται.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",12]]]],["Rule","'Proof'","'For, if AB, CD are not prime to one another, some number will measure them. Let a number measure them, and let it be E; let CD, measuring BF, leave FA less than itself, let AF, measuring DG, leave GC less than itself, and let GC, measuring FH, leave an unit HA. Since, then, E measures CD, and CD measures BF, therefore E also measures BF. But it also measures the whole BA; therefore it will also measure the remainder AF. But AF measures DG; therefore E also measures DG. But it also measures the whole DC therefore it will also measure the remainder CG. But CG measures FH; therefore E also measures FH. But it also measures the whole FA; therefore it will also measure the remainder, the unit AH, though it is a number: which is impossible. Therefore no number will measure the numbers AB, CD; therefore AB, CD are prime to one another. [VII. Def. 12]'"],["Rule","'ProofWordCount'",159],["Rule","'GreekProof'","'δύο γὰρ ἀνίσων ἀριθμῶν τῶν ΑΒ, ΓΔ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος ὁ λειπόμενος μηδέποτε καταμετρείτω τὸν πρὸ ἑαυτοῦ, ἕως οὗ λειφθῇ μονάς: λέγω, ὅτι οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους εἰσίν, τουτέστιν ὅτι τοὺς ΑΒ, ΓΔ μονὰς μόνη μετρεῖ. εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω, καὶ ἔστω ὁ Ε: καὶ ὁ μὲν ΓΔ τὸν ΒΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΖΑ, ὁ δὲ ΑΖ τὸν ΔΗ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΗΓ, ὁ δὲ ΗΓ τὸν ΖΘ μετρῶν λειπέτω μονάδα τὴν ΘΑ. ἐπεὶ οὖν ὁ Ε τὸν ΓΔ μετρεῖ, ὁ δὲ ΓΔ τὸν ΒΖ μετρεῖ καὶ ὁ Ε ἄρα τὸν ΒΖ μετρεῖ: μετρεῖ δὲ καὶ ὅλον τὸν ΒΑ: καὶ λοιπὸν ἄρα τὸν ΑΖ μετρήσει. ὁ δὲ ΑΖ τὸν ΔΗ μετρεῖ: καὶ ὁ Ε ἄρα τὸν ΔΗ μετρεῖ: μετρεῖ δὲ καὶ ὅλον τὸν ΔΓ: καὶ λοιπὸν ἄρα τὸν ΓΗ μετρήσει. ὁ δὲ ΓΗ τὸν ΖΘ μετρεῖ: καὶ ὁ Ε ἄρα τὸν ΖΘ μετρεῖ: μετρεῖ δὲ καὶ ὅλον τὸν ΖΑ: καὶ λοιπὴν ἄρα τὴν ΑΘ μονάδα μετρήσει ἀριθμὸς ὤν: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς ΑΒ, ΓΔ ἀριθμοὺς μετρήσει τις ἀριθμός: οἱ ΑΒ, ΓΔ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",202]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'7.2'"],["Rule","'Text'","'Given two numbers not prime to one another, to find their greatest common measure.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'δύο ἀριθμῶν δοθέντων μὴ πρώτων πρὸς ἀλλήλους τὸ μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let AB, CD be the two given numbers not prime to one another. Thus it is required to find the greatest common measure of AB, CD. If now CD measures AB —and it also measures itself— CD is a common measure of CD, AB. And it is manifest that it is also the greatest; for no greater number than CD will measure CD. But, if CD does not measure AB, then, the less of the numbers AB, CD being continually subtracted from the greater, some number will be left which will measure the one before it. For an unit will not be left; otherwise AB, CD will be prime to one another [VII. 1], which is contrary to the hypothesis. Therefore some number will be left which will measure the one before it. Now let CD, measuring BE, leave EA less than itself, let EA, measuring DF, leave FC less than itself, and let CF measure AE. Since then, CF measures AE, and AE measures DF, therefore CF will also measure DF. But it also measures itself; therefore it will also measure the whole CD. But CD measures BE; therefore CF also measures BE. But it also measures EA; therefore it will also measure the whole BA. But it also measures CD; therefore CF measures AB, CD. Therefore CF is a common measure of AB, CD. I say next that it is also the greatest. For, if CF is not the greatest common measure of AB, CD, some number which is greater than CF will measure the numbers AB, CD. Let such a number measure them, and let it be G. Now, since G measures CD, while CD measures BE, G also measures BE. But it also measures the whole BA; therefore it will also measure the remainder AE. But AE measures DF; therefore G will also measure DF. But it also measures the whole DC; therefore it will also measure the remainder CF, that is, the greater will measure the less: which is impossible. Therefore no number which is greater than CF will measure the numbers AB, CD; therefore CF is the greatest common measure of AB, CD.'"],["Rule","'ProofWordCount'",359],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες δύο ἀριθμοὶ μὴ πρῶτοι πρὸς ἀλλήλους οἱ ΑΒ, ΓΔ. δεῖ δὴ τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν. εἰ μὲν οὖν ὁ ΓΔ τὸν ΑΒ μετρεῖ, μετρεῖ δὲ καὶ ἑαυτόν, ὁ ΓΔ ἄρα τῶν ΓΔ, ΑΒ κοινὸν μέτρον ἐστίν. καὶ φανερόν, ὅτι καὶ μέγιστον: οὐδεὶς γὰρ μείζων τοῦ ΓΔ τὸν ΓΔ μετρήσει. εἰ δὲ οὐ μετρεῖ ὁ ΓΔ τὸν ΑΒ, τῶν ΑΒ, ΓΔ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος λειφθήσεταί τις ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ. μονὰς μὲν γὰρ οὐ λειφθήσεται: εἰ δὲ μή, ἔσονται οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους: ὅπερ οὐχ ὑπόκειται. λειφθήσεταί τις ἄρα ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ. καὶ ὁ μὲν ΓΔ τὸν ΒΕ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΕΑ, ὁ δὲ ΕΑ τὸν ΔΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΖΓ, ὁ δὲ ΓΖ τὸν ΑΕ μετρείτω. ἐπεὶ οὖν ὁ ΓΖ τὸν ΑΕ μετρεῖ, ὁ δὲ ΑΕ τὸν ΔΖ μετρεῖ, καὶ ὁ ΓΖ ἄρα τὸν ΔΖ μετρήσει: μετρεῖ δὲ καὶ ἑαυτόν: καὶ ὅλον ἄρα τὸν ΓΔ μετρήσει. ὁ δὲ ΓΔ τὸν ΒΕ μετρεῖ: καὶ ὁ ΓΖ ἄρα τὸν ΒΕ μετρεῖ: μετρεῖ δὲ καὶ τὸν ΕΑ: καὶ ὅλον ἄρα τὸν ΒΑ μετρήσει: μετρεῖ δὲ καὶ τὸν ΓΔ: ὁ ΓΖ ἄρα τοὺς ΑΒ, ΓΔ μετρεῖ. ὁ ΓΖ ἄρα τῶν ΑΒ, ΓΔ κοινὸν μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή ἐστιν ὁ ΓΖ τῶν ΑΒ, ΓΔ μέγιστον κοινὸν μέτρον, μετρήσει τις τοὺς ΑΒ, ΓΔ ἀριθμοὺς ἀριθμὸς μείζων ὢν τοῦ ΓΖ. μετρείτω, καὶ ἔστω ὁ Η. καὶ ἐπεὶ ὁ Η τὸν ΓΔ μετρεῖ, ὁ δὲ ΓΔ τὸν ΒΕ μετρεῖ, καὶ ὁ Η ἄρα τὸν ΒΕ μετρεῖ: μετρεῖ δὲ καὶ ὅλον τὸν ΒΑ: καὶ λοιπὸν ἄρα τὸν ΑΕ μετρήσει. ὁ δὲ ΑΕ τὸν ΔΖ μετρεῖ: καὶ ὁ Η ἄρα τὸν ΔΖ μετρήσει: μετρεῖ δὲ καὶ ὅλον τὸν ΔΓ: καὶ λοιπὸν ἄρα τὸν ΓΖ μετρήσει ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα τοὺς ΑΒ, ΓΔ ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ ΓΖ: ὁ ΓΖ ἄρα τῶν ΑΒ, ΓΔ μέγιστόν ἐστι κοινὸν μέτρον: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἀριθμὸς δύο ἀριθμοὺς μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",357]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'7.3'"],["Rule","'Text'","'Given three numbers not prime to one another, to find their greatest common measure.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'τριῶν ἀριθμῶν δοθέντων μὴ πρώτων πρὸς ἀλλήλους τὸ μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let A, B, C be the three given numbers not prime to one another; thus it is required to find the greatest common measure of A, B, C. For let the greatest common measure, D, of the two numbers A, B be taken; [VII. 2] then D either measures, or does not measure, C. First, let it measure it. But it measures A, B also; therefore D measures A, B, C; therefore D is a common measure of A, B, C. I say that it is also the greatest. For, if D is not the greatest common measure of A, B, C, some number which is greater than D will measure the numbers A, B, C. Let such a number measure them, and let it be E. Since then E measures A, B, C, it will also measure A, B; therefore it will also measure the greatest common measure of A, B. [VII. 2] But the greatest common measure of A, B is D; therefore E measures D, the greater the less: which is impossible. Therefore no number which is greater than D will measure the numbers A, B, C; therefore D is the greatest common measure of A, B, C.  Next, let D not measure C; I say first that C, D are not prime to one another. For, since A, B, C are not prime to one another, some number will measure them. Now that which measures A, B, C will also measure A, B, and will measure D, the greatest common measure of A, B. [VII. 2] But it measures C also; therefore some number will measure the numbers D, C; therefore D, C are not prime to one another. Let then their greatest common measure E be taken. [VII. 2] Then, since E measures D, and D measures A, B, therefore E also measures A, B. But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C. I say next that it is also the greatest. For, if E is not the greatest common measure of A, B, C, some number which is greater than E will measure the numbers A, B, C. Let such a number measure them, and let it be F. Now, since F measures A, B, C, it also measures A, B; therefore it will also measure the greatest common measure of A, B. [VII. 2] But the greatest common measure of A, B is D; therefore F measures D. And it measures C also; therefore F measures D, C; therefore it will also measure the greatest common measure of D, C. [VII. 2] But the greatest common measure of D, C is E; therefore F measures E, the greater the less: which is impossible. Therefore no number which is greater than E will measure the numbers A, B, C; therefore E is the greatest common measure of A, B, C.'"],["Rule","'ProofWordCount'",489],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ μὴ πρῶτοι πρὸς ἀλλήλους οἱ α, Β, Γ: δεῖ δὴ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν. εἰλήφθω γὰρ δύο τῶν Α, Β τὸ μέγιστον κοινὸν μέτρον ὁ Δ: ὁ δὴ Δ τὸν Γ ἤτοι μετρεῖ á¼¢ οὐ μετρεῖ. μετρείτω πρότερον: μετρεῖ δὲ καὶ τοὺς Α, Β: ὁ Δ ἄρα τοὺς Α, Β, Γ μετρεῖ: ὁ Δ ἄρα τῶν Α, Β, Γ κοινὸν μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή ἐστιν ὁ Δ τῶν Α, Β, Γ μέγιστον κοινὸν μέτρον, μετρήσει τις τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμὸς μείζων ὢν τοῦ Δ. μετρείτω, καὶ ἔστω ὁ Ε. ἐπεὶ οὖν ὁ Ε τοὺς Α, Β, Γ μετρεῖ, καὶ τοὺς Α, Β ἄρα μετρήσει: καὶ τὸ τῶν Α, Β ἄρα μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον κοινὸν μέτρον ἐστὶν ὁ Δ: ὁ Ε ἄρα τὸν Δ μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ Δ: ὁ Δ ἄρα τῶν Α, Β, Γ μέγιστόν ἐστι κοινὸν μέτρον. μὴ μετρείτω δὴ ὁ Δ τὸν Γ: λέγω πρῶτον, ὅτι οἱ Γ, Δ οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους. ἐπεὶ γὰρ οἱ Α, Β, Γ οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. ὁ δὴ τοὺς Α, Β, Γ μετρῶν καὶ τοὺς Α, Β μετρήσει, καὶ τὸ τῶν Α, Β μέγιστον κοινὸν μέτρον τὸν Δ μετρήσει: μετρεῖ δὲ καὶ τὸν Γ: τοὺς Δ, Γ ἄρα ἀριθμοὺς ἀριθμός τις μετρήσει: οἱ Δ, Γ ἄρα οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους. εἰλήφθω οὖν αὐτῶν τὸ μέγιστον κοινὸν μέτρον ὁ Ε. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ, ὁ δὲ Δ τοὺς Α, Β μετρεῖ, καὶ ὁ Ε ἄρα τοὺς Α, Β μετρεῖ: μετρεῖ δὲ καὶ τὸν Γ: ὁ Ε ἄρα τοὺς Α, Β, Γ μετρεῖ: ὁ Ε ἄρα τῶν Α, Β, Γ κοινόν ἐστι μέτρον. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή ἐστιν ὁ Ε τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον, μετρήσει τις τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμὸς μείζων ὢν τοῦ Ε. μετρείτω, καὶ ἔστω ὁ Ζ. καὶ ἐπεὶ ὁ Ζ τοὺς Α, Β, Γ μετρεῖ, καὶ τοὺς Α, Β μετρεῖ: καὶ τὸ τῶν Α, Β ἄρα μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον κοινὸν μέτρον ἐστὶν ὁ Δ: ὁ Ζ ἄρα τὸν Δ μετρεῖ: μετρεῖ δὲ καὶ τὸν Γ: ὁ Ζ ἄρα τοὺς Δ, Γ μετρεῖ: καὶ τὸ τῶν Δ, Γ ἄρα μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Δ, Γ μέγιστον κοινὸν μέτρον ἐστὶν ὁ Ε: ὁ Ζ ἄρα τὸν Ε μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ Ε: ὁ Ε ἄρα τῶν Α, Β, Γ μέγιστόν ἐστι κοινὸν μέτρον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",464]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'7.4'"],["Rule","'Text'","'Any number is either a part or parts of any number, the less of the greater.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'ἅπας ἀριθμὸς παντὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ μείζονος ἤτοι μέρος ἐστὶν á¼¢ μέρη.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let A, BC be two numbers, and let BC be the less; I say that BC is either a part, or parts, of A. For A, BC are either prime to one another or not. First, let A, BC be prime to one another. Then, if BC be divided into the units in it, each unit of those in BC will be some part of A; so that BC is parts of A. Next let A, BC not be prime to one another; then BC either measures, or does not measure, A. If now BC measures A, BC is a part of A. But, if not, let the greatest common measure D of A, BC be taken; [VII. 2] and let BC be divided into the numbers equal to D, namely BE, EF, FC. Now, since D measures A, D is a part of A. But D is equal to each of the numbers BE, EF, FC; therefore each of the numbers BE, EF, FC is also a part of A; so that BC is parts of A.'"],["Rule","'ProofWordCount'",178],["Rule","'GreekProof'","'ἔστωσαν δύο ἀριθμοὶ οἱ Α, ΒΓ, καὶ ἔστω ἐλάσσων ὁ ΒΓ: λέγω, ὅτι ὁ ΒΓ τοῦ Α ἤτοι μέρος ἐστὶν á¼¢ μέρη. οἱ Α, ΒΓ γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν á¼¢ οὔ. ἔστωσαν πρότερον οἱ Α, ΒΓ πρῶτοι πρὸς ἀλλήλους. διαιρεθέντος δὴ τοῦ ΒΓ εἰς τὰς ἐν αὐτῷ μονάδας ἔσται ἑκάστη μονὰς τῶν ἐν τῷ ΒΓ μέρος τι τοῦ Α: ὥστε μέρη ἐστὶν ὁ ΒΓ τοῦ Α. μὴ ἔστωσαν δὴ οἱ Α, ΒΓ πρῶτοι πρὸς ἀλλήλους: ὁ δὴ ΒΓ τὸν Α ἤτοι μετρεῖ á¼¢ οὐ μετρεῖ. εἰ μὲν οὖν ὁ ΒΓ τὸν Α μετρεῖ, μέρος ἐστὶν ὁ ΒΓ τοῦ Α. εἰ δὲ οὔ, εἰλήφθω τῶν Α, ΒΓ μέγιστον κοινὸν μέτρον ὁ Δ, καὶ διῃρήσθω ὁ ΒΓ εἰς τοὺς τῷ Δ ἴσους τοὺς ΒΕ, ΕΖ, ΖΓ. καὶ ἐπεὶ ὁ Δ τὸν Α μετρεῖ, μέρος ἐστὶν ὁ Δ τοῦ Α: ἴσος δὲ ὁ Δ ἑκάστῳ τῶν ΒΕ, ΕΖ, ΖΓ: καὶ ἕκαστος ἄρα τῶν ΒΕ, ΕΖ, ΖΓ τοῦ Α μέρος ἐστίν: ὥστε μέρη ἐστὶν ὁ ΒΓ τοῦ Α. ἅπας ἄρα ἀριθμὸς παντὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ μείζονος ἤτοι μέρος ἐστὶν á¼¢ μέρη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",183]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'7.5'"],["Rule","'Text'","'If a number be a part of a number, and another be the same part of another, the sum will also be the same part of the sum that the one is of the one.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'ἐὰν ἀριθμός ἀριθμοῦ μέρος ᾖ, καὶ ἕτερος ἑτέρου τὸ αὐτὸ μέρος ᾖ, καὶ συναμφότερος συναμφοτέρου τὸ αὐτὸ μέρος ἔσται, ὅπερ ὁ εἷς τοῦ ἑνός.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List"]],["Rule","'Proof'","'For let the number A be a part of BC, and another, D, the same part of another EF that A is of BC; I say that the sum of A, D is also the same part of the sum of BC, EF that A is of BC. For since, whatever part A is of BC, D is also the same part of EF, therefore, as many numbers as there are in BC equal to A, so many numbers are there also in EF equal to D. Let BC be divided into the numbers equal to A, namely BG, GC, and EF into the numbers equal to D, namely EH, HF; then the multitude of BG, GC will be equal to the multitude of EH, HF. And, since BG is equal to A, and EH to D, therefore BG, EH are also equal to A, D. For the same reason GC, HF are also equal to A, D. Therefore, as many numbers as there are in BC equal to A, so many are there also in BC, EF equal to A, D. Therefore, whatever multiple BC is of A, the same multiple also is the sum of BC, EF of the sum of A, D. Therefore, whatever part A is of BC, the same part also is the sum of A, D of the sum of BC, EF.'"],["Rule","'ProofWordCount'",228],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ Α ἀριθμοῦ τοῦ ΒΓ μέρος ἔστω, καὶ ἕτερος ὁ Δ ἑτέρου τοῦ ΕΖ τὸ αὐτὸ μέρος, ὅπερ ὁ Α τοῦ ΒΓ: λέγω, ὅτι καὶ συναμφότερος ὁ Α, Δ συναμφοτέρου τοῦ ΒΓ, ΕΖ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὁ Α τοῦ ΒΓ. ἐπεὶ γάρ, ὃ μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Δ τοῦ ΕΖ, ὅσοι ἄρα εἰσὶν ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ Α, τοσοῦτοί εἰσι καὶ ἐν τῷ ΕΖ ἀριθμοὶ ἴσοι τῷ Δ. διῃρήσθω ὁ μὲν ΒΓ εἰς τοὺς τῷ Α ἴσους τοὺς ΒΗ, ΗΓ, ὁ δὲ ΕΖ εἰς τοὺς τῷ Δ ἴσους τοὺς ΕΘ, ΘΖ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ. καὶ ἐπεὶ ἴσος ἐστὶν ὁ μὲν ΒΗ τῷ Α, ὁ δὲ ΕΘ τῷ Δ, καὶ οἱ ΒΗ, ΕΘ ἄρα τοῖς Α, Δ ἴσοι. διὰ τὰ αὐτὰ δὴ καὶ οἱ ΗΓ, ΘΖ τοῖς Α, Δ. ὅσοι ἄρα εἰσὶν ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ Α, τοσοῦτοί εἰσι καὶ ἐν τοῖς ΒΓ, ΕΖ ἴσοι τοῖς α, Δ. ὁσαπλασίων ἄρα ἐστὶν ὁ ΒΓ τοῦ Α, τοσαυταπλασίων ἐστὶ καὶ συναμφότερος ὁ ΒΓ, ΕΖ συναμφοτέρου τοῦ Α, Δ. ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ Α, Δ συναμφοτέρου τοῦ ΒΓ, ΕΖ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",215]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'7.6'"],["Rule","'Text'","'If a number be parts of a number, and another be the same parts of another, the sum will also be the same parts of the sum that the one is of the one.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἐὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, καὶ ἕτερος ἑτέρου τὰ αὐτὰ μέρη ᾖ, καὶ συναμφότερος συναμφοτέρου τὰ αὐτὰ μέρη ἔσται, ὅπερ ὁ εἷς τοῦ ἑνός.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'For let the number AB be parts of the number C, and another, DE, the same parts of another, F, that AB is of C; I say that the sum of AB, DE is also the same parts of the sum of C, F that AB is of C. For since, whatever parts AB is of C, DE is also the same parts of F, therefore, as many parts of C as there are in AB, so many parts of F are there also in DE. Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE; thus the multitude of AG, GB will be equal to the multitude of DH, HE. And since, whatever part AG is of C, the same part is DH of F also, therefore, whatever part AG is of C, the same part also is the sum of AG, DH of the sum of C, F. [VII. 5] For the same reason, whatever part GB is of C, the same part also is the sum of GB, HE of the sum of C, F. Therefore, whatever parts AB is of C, the same parts also is the sum of AB, DE of the sum of C, F.'"],["Rule","'ProofWordCount'",213],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ Γ μέρη ἔστω, καὶ ἕτερος ὁ ΔΕ ἑτέρου τοῦ Ζ τὰ αὐτὰ μέρη, ἅπερ ὁ ΑΒ τοῦ Γ: λέγω, ὅτι καὶ συναμφότερος ὁ ΑΒ, ΔΕ συναμφοτέρου τοῦ Γ, Ζ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὁ ΑΒ τοῦ Γ. ἐπεὶ γάρ, ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ αὐτὰ μέρη καὶ ὁ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μέρη τοῦ Γ, τοσαῦτά ἐστι καὶ ἐν τῷ ΔΕ μέρη τοῦ Ζ. διῃρήσθω ὁ μὲν ΑΒ εἰς τὰ τοῦ Γ μέρη τὰ ΑΗ, ΗΒ, ὁ δὲ ΔΕ εἰς τὰ τοῦ Ζ μέρη τὰ ΔΘ, ΘΕ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει τῶν ΔΘ, ΘΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΔΘ τοῦ Ζ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΗ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΑΗ, ΔΘ συναμφοτέρου τοῦ Γ, Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὃ μέρος ἐστὶν ὁ ΗΒ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΗΒ, ΘΕ συναμφοτέρου τοῦ Γ, Ζ. ἃ ἄρα μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ αὐτὰ μέρη ἐστὶ καὶ συναμφότερος ὁ ΑΒ, ΔΕ συναμφοτέρου τοῦ Γ, Ζ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",203]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'7.7'"],["Rule","'Text'","'If a number be that part of a number, which a number subtracted is of a number subtracted, the remainder will also be the same part of the remainder that the whole is of the whole.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν ἀριθμὸς ἀριθμοῦ μέρος ᾖ, ὅπερ ἀφαιρεθεὶς ἀφαιρεθέντος, καὶ ὁ λοιπὸς τοῦ λοιποῦ τὸ αὐτὸ μέρος ἔσται, ὅπερ ὁ ὅλος τοῦ ὅλου.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'For let the number AB be that part of the number CD which AE subtracted is of CF subtracted; I say that the remainder EB is also the same part of the remainder FD that the whole AB is of the whole CD. For, whatever part AE is of CF, the same part also let EB be of CG. Now since, whatever part AE is of CF, the same part also is EB of CG, therefore, whatever part AE is of CF, the same part also is AB of GF. [VII. 5] But, whatever part AE is of CF, the same part also, by hypothesis, is AB of CD; therefore, whatever part AB is of GF, the same part is it of CD also; therefore GF is equal to CD. Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD. Now since, whatever part AE is of CF, the same part also is EB of GC, while GC is equal to FD, therefore, whatever part AE is of CF, the same part also is EB of FD. But, whatever part AE is of CF, the same part also is AB of CD; therefore also the remainder EB is the same part of the remainder FD that the whole AB is of the whole CD.'"],["Rule","'ProofWordCount'",220],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ ΓΔ μέρος ἔστω, ὅπερ ἀφαιρεθεὶς ὁ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ: λέγω, ὅτι καὶ λοιπὸς ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ. ὃ γὰρ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἔστω καὶ ὁ ΕΒ τοῦ ΓΗ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΒ τοῦ ΓΗ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΗΖ. ὃ δὲ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ὑπόκειται καὶ ὁ ΑΒ τοῦ ΓΔ: ὃ ἄρα μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΗΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ τοῦ ΓΔ: ἴσος ἄρα ἐστὶν ὁ ΗΖ τῷ ΓΔ. κοινὸς ἀφῃρήσθω ὁ ΓΖ: λοιπὸς ἄρα ὁ ΗΓ λοιπῷ τῷ ΖΔ ἐστιν ἴσος. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΒ τοῦ ΗΓ, ἴσος δὲ ὁ ΗΓ τῷ ΖΔ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΒ τοῦ ΖΔ. ἀλλὰ ὃ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΓΔ: καὶ λοιπὸς ἄρα ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",221]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'7.8'"],["Rule","'Text'","'If a number be the same parts of a number that a number subtracted is of a number subtracted, the remainder will also be the same parts of the remainder that the whole is of the whole.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, ἅπερ ἀφαιρεθεὶς ἀφαιρεθέντος, καὶ ὁ λοιπὸς τοῦ λοιποῦ τὰ αὐτὰ μέρη ἔσται, ἅπερ ὁ ὅλος τοῦ ὅλου.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",7]]]],["Rule","'Proof'","'For let the number AB be the same parts of the number CD that AE subtracted is of CF subtracted; I say that the remainder EB is also the same parts of the remainder FD that the whole AB is of the whole CD. For let GH be made equal to AB. Therefore, whatever parts GH is of CD, the same parts also is AE of CF. Let GH be divided into the parts of CD, namely GK, KH, and AE into the parts of CF, namely AL, LE; thus the multitude of GK, KH will be equal to the multitude of AL, LE. Now since, whatever part GK is of CD, the same part also is AL of CF, while. CD is greater than CF, therefore GK is also greater than AL. Let GM be made equal to AL. Therefore, whatever part GK is of CD, the same part also is GM of CF; therefore also the remainder MK is the same part of the remainder FD that the whole GK is of the whole CD. [VII. 7] Again, since, whatever part KH is of CD, the same part also is EL of CF, while CD is greater than CF, therefore HK is also greater than EL. Let KN be made equal to EL. Therefore, whatever part KH is of CD, the same part also is KN of CF; therefore also the remainder NH is the same part of the remainder FD that the whole KH is of the whole CD. [VII. 7] But the remainder MK was also proved to be the same part of the remainder FD that the whole GK is of the whole CD; therefore also the sum of MK, NH is the same parts of DF that the whole HG is of the whole CD. But the sum of MK, NH is equal to EB, and HG is equal to BA; therefore the remainder EB is the same parts of the remainder FD that the whole AB is of the whole CD.'"],["Rule","'ProofWordCount'",338],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ ΓΔ μέρη ἔστω, ἅπερ ἀφαιρεθεὶς ὁ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ: λέγω, ὅτι καὶ λοιπὸς ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ. κείσθω γὰρ τῷ ΑΒ ἴσος ὁ ΗΘ. ἃ ἄρα μέρη ἐστὶν ὁ ΗΘ τοῦ ΓΔ, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ ΑΕ τοῦ ΓΖ. διῃρήσθω ὁ μὲν ΗΘ εἰς τὰ τοῦ ΓΔ μέρη τὰ ΗΚ, ΚΘ, ὁ δὲ ΑΕ εἰς τὰ τοῦ ΓΖ μέρη τὰ ΑΛ, ΛΕ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΗΚ, ΚΘ τῷ πλήθει τῶν ΑΛ, ΛΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΗΚ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΛ τοῦ ΓΖ, μείζων δὲ ὁ ΓΔ τοῦ ΓΖ, μείζων ἄρα καὶ ὁ ΗΚ τοῦ ΑΛ. κείσθω τῷ ΑΛ ἴσος ὁ ΗΜ. ὃ ἄρα μέρος ἐστὶν ὁ ΗΚ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΗΜ τοῦ ΓΖ: καὶ λοιπὸς ἄρα ὁ ΜΚ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΗΚ ὅλου τοῦ ΓΔ. πάλιν ἐπεί, ὃ μέρος ἐστὶν ὁ ΚΘ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΛ τοῦ ΓΖ, μείζων δὲ ὁ ΓΔ τοῦ ΓΖ, μείζων ἄρα καὶ ὁ ΘΚ τοῦ ΕΛ. κείσθω τῷ ΕΛ ἴσος ὁ ΚΝ. ὃ ἄρα μέρος ἐστὶν ὁ ΚΘ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΚΝ τοῦ ΓΖ: καὶ λοιπὸς ἄρα ὁ ΝΘ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΚΘ ὅλου τοῦ ΓΔ. ἐδείχθη δὲ καὶ λοιπὸς ὁ ΜΚ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ὤν, ὅπερ ὅλος ὁ ΗΚ ὅλου τοῦ ΓΔ: καὶ συναμφότερος ἄρα ὁ ΜΚ, ΝΘ τοῦ ΔΖ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὅλος ὁ ΘΗ ὅλου τοῦ ΓΔ. ἴσος δὲ συναμφότερος μὲν ὁ ΜΚ, ΝΘ τῷ ΕΒ, ὁ δὲ ΘΗ τῷ ΒΑ: καὶ λοιπὸς ἄρα ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",317]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'7.9'"],["Rule","'Text'","'If a number be a part of a number, and another be the same part of another, alternately also, whatever part or parts the first is of the third, the same part, or the same parts, will the second also be of the fourth.'"],["Rule","'TextWordCount'",44],["Rule","'GreekText'","'ἐὰν ἀριθμὸς ἀριθμοῦ μέρος ᾖ, καὶ ἕτερος ἑτέρου τὸ αὐτὸ μέρος ᾖ, καὶ ἐναλλάξ, ὃ μέρος ἐστὶν á¼¢ μέρη ὁ πρῶτος τοῦ τρίτου, τὸ αὐτὸ μέρος ἔσται á¼¢ τὰ αὐτὰ μέρη καὶ ὁ δεύτερος τοῦ τετάρτου.'"],["Rule","'GreekTextWordCount'",36],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",5]],["List",["Rule","'Book'",7],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'For let the number A be a part of the number BC, and another, D, the same part of another, EF, that A is of BC; I say that, alternately also, whatever part or parts A is of D, the same part or parts is BC of EF also. For since, whatever part A is of BC, the same part also is D of EF, therefore, as many numbers as there are in BC equal to A, so many also are there in EF equal to D. Let BC be divided into the numbers equal to A, namely BG, GC, and EF into those equal to D, namely EH, HF; thus the multitude of BG, GC will be equal to the multitude of EH, HF. Now, since the numbers BG, GC are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of BG, GC is equal to the multitude of EH, HF, therefore, whatever part or parts BG is of EH, the same part or the same parts is GC of HF also; so that, in addition, whatever part or parts BG is of EH, the same part also, or the same parts, is the sum BC of the sum EF. [VII. 5, 6] But BG is equal to A, and EH to D; therefore, whatever part or parts A is of D, the same part or the same parts is BC of EF also.'"],["Rule","'ProofWordCount'",244],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ Α ἀριθμοῦ τοῦ ΒΓ μέρος ἔστω, καὶ ἕτερος ὁ Δ ἑτέρου τοῦ ΕΖ τὸ αὐτὸ μέρος, ὅπερ ὁ Α τοῦ ΒΓ: λέγω, ὅτι καὶ ἐναλλάξ, ὃ μέρος ἐστὶν ὁ Α τοῦ Δ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΒΓ τοῦ ΕΖ á¼¢ μέρη. ἐπεὶ γὰρ ὃ μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Δ τοῦ ΕΖ, ὅσοι ἄρα εἰσὶν ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ Α, τοσοῦτοί εἰσι καὶ ἐν τῷ ΕΖ ἴσοι τῷ Δ. διῃρήσθω ὁ μὲν ΒΓ εἰς τοὺς τῷ Α ἴσους τοὺς ΒΗ, ΗΓ, ὁ δὲ ΕΖ εἰς τοὺς τῷ Δ ἴσους τοὺς ΕΘ, ΘΖ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΒΗ, ΗΓ ἀριθμοὶ ἀλλήλοις, εἰσὶ δὲ καὶ οἱ ΕΘ, ΘΖ ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ, ὃ ἄρα μέρος ἐστὶν ὁ ΒΗ τοῦ ΕΘ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΗΓ τοῦ ΘΖ á¼¢ τὰ αὐτὰ μέρη: ὥστε καὶ ὃ μέρος ἐστὶν ὁ ΒΗ τοῦ ΕΘ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΒΓ συναμφοτέρου τοῦ ΕΖ á¼¢ τὰ αὐτὰ μέρη. ἴσος δὲ ὁ μὲν ΒΗ τῷ Α, ὁ δὲ ΕΘ τῷ Δ: ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Δ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΒΓ τοῦ ΕΖ á¼¢ τὰ αὐτὰ μέρη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",239]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'7.10'"],["Rule","'Text'","'If a number be parts of a number, and another be the same parts of another, alternately also, whatever parts or part the first is of the third, the same parts or the same part will the second also be of the fourth.'"],["Rule","'TextWordCount'",43],["Rule","'GreekText'","'ἐὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, καὶ ἕτερος ἑτέρου τὰ αὐτὰ μέρη ᾖ, καὶ ἐναλλάξ, ἃ μέρη ἐστὶν ὁ πρῶτος τοῦ τρίτου á¼¢ μέρος, τὰ αὐτὰ μέρη ἔσται καὶ ὁ δεύτερος τοῦ τετάρτου á¼¢ τὸ αὐτὸ μέρος.'"],["Rule","'GreekTextWordCount'",36],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",5]],["List",["Rule","'Book'",7],["Rule","'Theorem'",6]],["List",["Rule","'Book'",7],["Rule","'Theorem'",9]]]],["Rule","'Proof'","'For let the number AB be parts of the number C, and another, DE, the same parts of another, F; I say that, alternately also, whatever parts or part AB is of DE, the same parts or the same part is C of F also. For since, whatever parts AB is of C, the same parts also is DE of F, therefore, as many parts of C as there are in AB, so many parts also of F are there in DE. Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE; thus the multitude of AG, GB will be equal to the multitude of DH, HE. Now since, whatever part AG is of C, the same part also is DH of F, alternately also, whatever part or parts AG is of DH, the same part or the same parts is C of F also. [VII. 9] For the same reason also, whatever part or parts GB is of HE, the same part or the same parts is C of F also; so that, in addition, whatever parts or part AB is of DE, the same parts also, or the same part, is C of F. [VII. 5, 6]'"],["Rule","'ProofWordCount'",211],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ Γ μέρη ἔστω, καὶ ἕτερος ὁ ΔΕ ἑτέρου τοῦ Ζ τὰ αὐτὰ μέρη: λέγω, ὅτι καὶ ἐναλλάξ, ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ ΔΕ á¼¢ μέρος, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ Γ τοῦ Ζ á¼¢ τὸ αὐτὸ μέρος. ἐπεὶ γάρ, ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μέρη τοῦ Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ μέρη τοῦ Ζ. διῃρήσθω ὁ μὲν ΑΒ εἰς τὰ τοῦ Γ μέρη τὰ ΑΗ, ΗΒ, ὁ δὲ ΔΕ εἰς τὰ τοῦ Ζ μέρη τὰ ΔΘ, ΘΕ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει τῶν ΔΘ, ΘΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΔΘ τοῦ Ζ, καὶ ἐναλλάξ, ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Ζ á¼¢ τὰ αὐτὰ μέρη. διὰ τὰ αὐτὰ δὴ καί, ὃ μέρος ἐστὶν ὁ ΗΒ τοῦ ΘΕ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Ζ á¼¢ τὰ αὐτὰ μέρη: ὥστε καί ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΗΒ τοῦ ΘΕ á¼¢ τὰ αὐτὰ μέρη: καὶ ὃ ἄρα μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΔΕ á¼¢ τὰ αὐτὰ μέρη: ἀλλ᾽ ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐδείχθη καὶ ὁ Γ τοῦ Ζ á¼¢ τὰ αὐτὰ μέρη, καὶ ἃ ἄρα μέρη ἐστὶν ὁ ΑΒ τοῦ ΔΕ á¼¢ μέρος, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ Γ τοῦ Ζ á¼¢ τὸ αὐτὸ μέρος: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",284]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'7.11'"],["Rule","'Text'","'If, as whole is to whole, so is a number subtracted to a number subtracted, the remainder will also be to the remainder as whole to whole.'"],["Rule","'TextWordCount'",27],["Rule","'GreekText'","'ἐὰν ᾖ ὡς ὅλος πρὸς ὅλον, οὕτως ἀφαιρεθεὶς πρὸς ἀφαιρεθέντα, καὶ ὁ λοιπὸς πρὸς τὸν λοιπὸν ἔσται, ὡς ὅλος πρὸς ὅλον.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",7]],["List",["Rule","'Book'",7],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'As the whole AB is to the whole CD, so let AE subtracted be to CF subtracted; I say that the remainder EB is also to the remainder FD as the whole AB to the whole CD. Since, as AB is to CD, so is AE to CF, whatever part or parts AB is of CD, the same part or the same parts is AE of CF also; [VII. Def. 20] Therefore also the remainder EB is the same part or parts of FD that AB is of CD. [VII. 7, 8] Therefore, as EB is to FD, so is AB to CD. [VII. Def. 20]'"],["Rule","'ProofWordCount'",106],["Rule","'GreekProof'","'ἔστω ὡς ὅλος ὁ ΑΒ πρὸς ὅλον τὸν ΓΔ, οὕτως ἀφαιρεθεὶς ὁ ΑΕ πρὸς ἀφαιρεθέντα τὸν ΓΖ: λέγω, ὅτι καὶ λοιπὸς ὁ ΕΒ πρὸς λοιπὸν τὸν ΖΔ ἐστιν, ὡς ὅλος ὁ ΑΒ πρὸς ὅλον τὸν ΓΔ. ἐπεί ἐστιν ὡς ὁ ΑΒ πρὸς τὸν ΓΔ, οὕτως ὁ ΑΕ πρὸς τὸν ΓΖ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΒ τοῦ ΓΔ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΕ τοῦ ΓΖ á¼¢ τὰ αὐτὰ μέρη. καὶ λοιπὸς ἄρα ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστὶν á¼¢ μέρη, ἅπερ ὁ ΑΒ τοῦ ΓΔ. ἔστιν ἄρα ὡς ὁ ΕΒ πρὸς τὸν ΖΔ, οὕτως ὁ ΑΒ πρὸς τὸν ΓΔ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",109]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'7.12'"],["Rule","'Text'","'If there be as many numbers as we please in proportion, then, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἀνάλογον, ἔσται ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",5]],["List",["Rule","'Book'",7],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'Let A, B, C, D be as many numbers as we please in proportion, so that, as A is to B, so is C to D; I say that, as A is to B, so are A, C to B, D. For since, as A is to B, so is C to D, whatever part or parts A is of B, the same part or parts is C of D also. [VII. Def. 20] Therefore also the sum of A, C is the same part or the same parts of the sum of B, D that A is of B. [VII. 5, 6] Therefore, as A is to B, so are A, C to B, D. [VII. Def. 20]'"],["Rule","'ProofWordCount'",119],["Rule","'GreekProof'","'ἔστωσαν ὁποσοιοῦν ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ: λέγω, ὅτι ἐστὶν ὡς ὁ Α πρὸς τὸν Β, οὕτως οἱ Α, Γ πρὸς τοὺς Β, Δ. ἐπεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ, ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Β á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Δ á¼¢ μέρη. καὶ συναμφότερος ἄρα ὁ Α, Γ συναμφοτέρου τοῦ Β, Δ τὸ αὐτὸ μέρος ἐστὶν á¼¢ τὰ αὐτὰ μέρη, ἅπερ ὁ Α τοῦ Β. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως οἱ Α, Γ πρὸς τοὺς Β, Δ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",116]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'7.13'"],["Rule","'Text'","'If four numbers be proportional, they will also be proportional alternately.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἐὰν τέσσαρες ἀριθμοὶ ἀνάλογον ὦσιν, καὶ ἐναλλὰξ ἀνάλογον ἔσονται.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",10]]]],["Rule","'Proof'","'Let the four numbers A, B, C, D be proportional, so that, as A is to B, so is C to D; I say that they will also be proportional alternately, so that, as A is to C, so will B be to D.  For since, as A is to B, so is C to D, therefore, whatever part or parts A is of B, the same part or the same parts is C of D also. [VII. Def. 20] Therefore, alternately, whatever part or parts A is of C, the same part or the same parts is B of D also. [VII. 10] Therefore, as A is to C, so is B to D. [VII. Def. 20]'"],["Rule","'ProofWordCount'",118],["Rule","'GreekProof'","'ἔστωσαν τέσσαρες ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ: λέγω, ὅτι καὶ ἐναλλὰξ ἀνάλογον ἔσονται, ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Β πρὸς τὸν Δ. ἐπεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ, ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Β á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Δ á¼¢ τὰ αὐτὰ μέρη. ἐναλλὰξ ἄρα, ὃ μέρος ἐστὶν ὁ Α τοῦ Γ á¼¢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Β τοῦ Δ á¼¢ τὰ αὐτὰ μέρη. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Β πρὸς τὸν Δ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",118]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'7.14'"],["Rule","'Text'","'If there be as many numbers as we please, and others equal to them in multitude, which taken two and two are in the same ratio, they will also be in the same ratio ex aequali.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ καὶ ἄλλοι αὐτοῖς ἴσοι τὸ πλῆθος σύνδυο λαμβανόμενοι καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσονται.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'Let there be as many numbers as we please A, B, C, and others equal to them in multitude D, E, F, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F; I say that, ex aequali, as A is to C, so also is D to F.  For, since, as A is to B, so is D to E, therefore, alternately, as A is to D, so is B to E. [VII. 13] Again, since, as B is to C, so is E to F, therefore, alternately, as B is to E, so is C to F. [VII. 13]  But, as B is to E, so is A to D; therefore also, as A is to D, so is C to F. Therefore, alternately, as A is to C, so is D to F.'"],["Rule","'ProofWordCount'",158],["Rule","'GreekProof'","'ἔστωσαν ὁποσοιοῦν ἀριθμοὶ οἱ Α, Β, Γ καὶ ἄλλοι αὐτοῖς ἴσοι τὸ πλῆθος σύνδυο λαμβανόμενοι ἐν τῷ αὐτῷ λόγῳ οἱ Δ, Ε, Ζ, ὡς μὲν ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε, ὡς δὲ ὁ Β πρὸς τὸν Γ, οὕτως ὁ Ε πρὸς τὸν Ζ: λέγω, ὅτι καὶ δι᾽ ἴσου ἐστὶν ὡς ὁ Î‘ πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ζ. ἐπεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, οὕτως ὁ Β πρὸς τὸν Ε. πάλιν, ἐπεί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, οὕτως ὁ Ε πρὸς τὸν Ζ, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Β πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Ζ. ὡς δὲ ὁ Β πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν δ: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Δ, οὕτως ὁ Γ πρὸς τὸν Ζ: ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ζ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",173]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'7.15'"],["Rule","'Text'","'If an unit measure any number, and another number measure any other number the same number of times, alternately also, the unit will measure the third number the same number of times that the second measures the fourth.'"],["Rule","'TextWordCount'",38],["Rule","'GreekText'","'ἐὰν μονὰς ἀριθμόν τινα μετρῇ, ἰσάκις δὲ ἕτερος ἀριθμὸς ἄλλον τινὰ ἀριθμὸν μετρῇ, καὶ ἐναλλὰξ ἰσάκις ἡ μονὰς τὸν τρίτον ἀριθμὸν μετρήσει καὶ ὁ δεύτερος τὸν τέταρτον.'"],["Rule","'GreekTextWordCount'",27],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'For let the unit A measure any number BC, and let another number D measure any other number EF the same number of times; I say that, alternately also, the unit A measures the number D the same number of times that BC measures EF. For, since the unit A measures the number BC the same number of times that D measures EF, therefore, as many units as there are in BC, so many numbers equal to D are there in EF also. Let BC be divided into the units in it, BG, GH, HC, and EF into the numbers EK, KL, LF equal to D. Thus the multitude of BG, GH, HC will be equal to the multitude of EK, KL, LF. And, since the units BG, GH, HC are equal to one another, and the numbers EK, KL, LF are also equal to one another, while the multitude of the units BG, GH, HC is equal to the multitude of the numbers EK, KL, LF, therefore, as the unit BG is to the number EK, so will the unit GH be to the number KL, and the unit HC to the number LF. Therefore also, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [VII. 12] therefore, as the unit BG is to the number EK, so is BC to EF. But the unit BG is equal to the unit A, and the number EK to the number D. Therefore, as the unit A is to the number D, so is BC to EF. Therefore the unit A measures the number D the same number of times that BC measures EF.'"],["Rule","'ProofWordCount'",285],["Rule","'GreekProof'","'μονὰς γὰρ ἡ Α ἀριθμόν τινα τὸν ΒΓ μετρείτω, ἰσάκις δὲ ἕτερος ἀριθμὸς ὁ Δ ἄλλον τινὰ ἀριθμὸν τὸν ΕΖ μετρείτω: λέγω, ὅτι καὶ ἐναλλὰξ ἰσάκις ἡ Α μονὰς τὸν Δ ἀριθμὸν μετρεῖ καὶ ὁ ΒΓ τὸν ΕΖ. ἐπεὶ γὰρ ἰσάκις ἡ Α μονὰς τὸν ΒΓ ἀριθμὸν μετρεῖ καὶ ὁ Δ τὸν ΕΖ, ὅσαι ἄρα εἰσὶν ἐν τῷ ΒΓ μονάδες, τοσοῦτοί εἰσι καὶ ἐν τῷ ΕΖ ἀριθμοὶ ἴσοι τῷ Δ. διῃρήσθω ὁ μὲν ΒΓ εἰς τὰς ἐν ἑαυτῷ μονάδας τὰς ΒΗ, ΗΘ, ΘΓ, ὁ δὲ ΕΖ εἰς τοὺς τῷ Δ ἴσους τοὺς ΕΚ, ΚΛ, ΛΖ. ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΘ, ΘΓ τῷ πλήθει τῶν ΕΚ, ΚΛ, ΛΖ. καὶ ἐπεὶ ἴσαι εἰσὶν αἱ ΒΗ, ΗΘ, ΘΓ μονάδες ἀλλήλαις, εἰσὶ δὲ καὶ οἱ ΕΚ, ΚΛ, ΛΖ ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΘ, ΘΓ μονάδων τῷ πλήθει τῶν ΕΚ, ΚΛ, ΛΖ ἀριθμῶν, ἔσται ἄρα ὡς ἡ ΒΗ μονὰς πρὸς τὸν ΕΚ ἀριθμόν, οὕτως ἡ ΗΘ μονὰς πρὸς τὸν ΚΛ ἀριθμὸν καὶ ἡ ΘΓ μονὰς πρὸς τὸν ΛΖ ἀριθμόν. ἔσται ἄρα καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους: ἔστιν ἄρα ὡς ἡ ΒΗ μονὰς πρὸς τὸν ΕΚ ἀριθμόν, οὕτως ὁ ΒΓ πρὸς τὸν ΕΖ. ἴση δὲ ἡ ΒΗ μονὰς τῇ Α μονάδι, ὁ δὲ ΕΚ ἀριθμὸς τῷ Δ ἀριθμῷ. ἔστιν ἄρα ὡς ἡ Α μονὰς πρὸς τὸν Δ ἀριθμόν, οὕτως ὁ ΒΓ πρὸς τὸν ΕΖ. ἰσάκις ἄρα ἡ Α μονὰς τὸν Δ ἀριθμὸν μετρεῖ καὶ ὁ ΒΓ τὸν ΕΖ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",256]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'7.16'"],["Rule","'Text'","'If two numbers by multiplying one another make certain numbers, the numbers so produced will be equal to one another.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινας, οἱ γενόμενοι ἐξ αὐτῶν ἴσοι ἀλλήλοις ἔσονται.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let A, B be two numbers, and let A by multiplying B make C, and B by multiplying A make D; I say that C is equal to D. For, since A by multiplying B has made C, therefore B measures C according to the units in A. But the unit E also measures the number A according to the units in it; therefore the unit E measures A the same number of times that B measures C.  Therefore, alternately, the unit E measures the number B the same number of times that A measures C. [VII. 15] Again, since B by multiplying A has made D, therefore A measures D according to the units in B. But the unit E also measures B according to the units in it; therefore the unit E measures the number B the same number of times that A measures D.  But the unit E measured the number B the same number of times that A measures C; therefore A measures each of the numbers C, D the same number of times. Therefore C is equal to D.'"],["Rule","'ProofWordCount'",184],["Rule","'GreekProof'","'ἔστωσαν δύο ἀριθμοὶ οἱ Α, Β, καὶ ὁ μὲν Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω, ὁ δὲ Β τὸν Α πολλαπλασιάσας τὸν Δ ποιείτω: λέγω, ὅτι ἴσος ἐστὶν ὁ Γ τῷ Δ. ἐπεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Α ἀριθμὸν κατὰ τὰς ἐν αὐτῷ μονάδας: ἰσάκις ἄρα ἡ Ε μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Γ. ἐναλλὰξ ἄρα ἰσάκις ἡ Ε μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Α τὸν Γ. πάλιν, ἐπεὶ ὁ Β τὸν Α πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ Α ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Β μονάδας. μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Β κατὰ τὰς ἐν αὐτῷ μονάδας: ἰσάκις ἄρα ἡ Ε μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Α τὸν Δ. ἰσάκις δὲ ἡ Ε μονὰς τὸν Β ἀριθμὸν ἐμέτρει καὶ ὁ Α τὸν Γ: ἰσάκις ἄρα ὁ Α ἑκάτερον τῶν Γ, Δ μετρεῖ. ἴσος ἄρα ἐστὶν ὁ Γ τῷ Δ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",180]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'7.17'"],["Rule","'Text'","'If a number by multiplying two numbers make certain numbers, the numbers so produced will have the same ratio as the numbers multiplied.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'ἐὰν ἀριθμὸς δύο ἀριθμοὺς πολλαπλασιάσας ποιῇ τινας, οἱ γενόμενοι ἐξ αὐτῶν τὸν αὐτὸν ἕξουσι λόγον τοῖς πολλαπλασιασθεῖσιν.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'For let the number A by multiplying the two numbers B, C make D, E; I say that, as B is to C, so is D to E. For, since A by multiplying B has made D, therefore B measures D according to the units in A. But the unit F also measures the number A according to the units in it; therefore the unit F measures the number A the same number of times that B measures D. Therefore, as the unit F is to the number A, so is B to D. [VII. Def. 20] For the same reason, as the unit F is to the number A, so also is C to E; therefore also, as B is to D, so is C to E. Therefore, alternately, as B is to C, so is D to E. [VII. 13]'"],["Rule","'ProofWordCount'",142],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ Α δύο ἀριθμοὺς τοὺς Β, Γ πολλαπλασιάσας τοὺς Δ, Ε ποιείτω: λέγω, ὅτι ἐστὶν ὡς ὁ Β πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ε. ἐπεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ Β ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεῖ δὲ καὶ ἡ Ζ μονὰς τὸν Α ἀριθμὸν κατὰ τὰς ἐν αὐτῷ μονάδας: ἰσάκις ἄρα ἡ Ζ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Δ. ἔστιν ἄρα ὡς ἡ Ζ μονὰς πρὸς τὸν Α ἀριθμόν, οὕτως ὁ Β πρὸς τὸν Δ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ Ζ μονὰς πρὸς τὸν Α ἀριθμόν, οὕτως ὁ Γ πρὸς τὸν Ε: καὶ ὡς ἄρα ὁ Β πρὸς τὸν Δ, οὕτως ὁ Γ πρὸς τὸν Ε. ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Β πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ε: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",146]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'7.18'"],["Rule","'Text'","'If two numbers by multiplying any number make certain numbers, the numbers so produced will have the same ratio as the multipliers.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ ἀριθμόν τινα πολλαπλασιάσαντες ποιῶσί τινας, οἱ γενόμενοι ἐξ αὐτῶν τὸν αὐτὸν ἕξουσι λόγον τοῖς πολλαπλασιάσασιν.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",16]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'For let two numbers A, B by multiplying any number C make D, E; I say that, as A is to B, so is D to E. For, since A by multiplying C has made D, therefore also C by multiplying A has made D. [VII. 16] For the same reason also C by multiplying B has made E. Therefore the number C by multiplying the two numbers A, B has made D, E. Therefore, as A is to B, so is D to E. [VII. 17]'"],["Rule","'ProofWordCount'",87],["Rule","'GreekProof'","'δύο γὰρ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν Γ πολλαπλασιάσαντες τοὺς Δ, Ε ποιείτωσαν: λέγω, ὅτι ἐστὶν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε. ἐπεὶ γὰρ ὁ Α τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν, καὶ ὁ Γ ἄρα τὸν Α πολλαπλασιάσας τὸν Δ πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Γ τὸν Β πολλαπλασιάσας τὸν Ε πεποίηκεν. ἀριθμὸς δὴ ὁ Γ δύο ἀριθμοὺς τοὺς Α, Β πολλαπλασιάσας τοὺς Δ, Ε πεποίηκεν. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν ε: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",94]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'7.19'"],["Rule","'Text'","'If four numbers be proportional, the number produced from the first and fourth will be equal to the number produced from the second and third; and, if the number produced from the first and fourth be equal to that produced from the second and third, the four numbers will be proportional.'"],["Rule","'TextWordCount'",51],["Rule","'GreekText'","'ἐὰν τέσσαρες ἀριθμοὶ ἀνάλογον ὦσιν, ὁ ἐκ πρώτου καὶ τετάρτου γενόμενος ἀριθμὸς ἴσος ἔσται τῷ ἐκ δευτέρου καὶ τρίτου γενομένῳ ἀριθμῷ: καὶ ἐὰν ὁ ἐκ πρώτου καὶ τετάρτου γενόμενος ἀριθμὸς ἴσος ᾖ τῷ ἐκ δευτέρου καὶ τρίτου, οἱ τέσσαρες ἀριθμοὶ ἀνάλογον ἔσονται.'"],["Rule","'GreekTextWordCount'",42],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",7],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let A, B, C, D be four numbers in proportion, so that, as A is to B, so is C to D; and let A by multiplying D make E, and let B by multiplying C make F; I say that E is equal to F. For let A by multiplying C make G. Since, then, A by multiplying C has made G, and by multiplying D has made E, the number A by multiplying the two numbers C, D has made G, E. Therefore, as C is to D, so is G to E. [VII. 17] But, as C is to D, so is A to B; therefore also, as A is to B, so is G to E. Again, since A by multiplying C has made G, but, further, B has also by multiplying C made F, the two numbers A, B by multiplying a certain number C have made G, F. Therefore, as A is to B, so is G to F. [VII. 18] But further, as A is to B, so is G to E also; therefore also, as G is to E, so is G to F. Therefore G has to each of the numbers E, F the same ratio; therefore E is equal to F. [cf. V. 9]  Again, let E be equal to F; I say that, as A is to B, so is C to D. For, with the same construction, since E is equal to F, therefore, as G is to E, so is G to F. [cf. V. 7] But, as G is to E, so is C to D, [VII. 17] and, as G is to F, so is A to B. [VII. 18]  Therefore also, as A is to B, so is C to D.'"],["Rule","'ProofWordCount'",297],["Rule","'GreekProof'","'ἔστωσαν τέσσαρες ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ, καὶ ὁ μὲν Α τὸν Δ πολλαπλασιάσας τὸν Ε ποιείτω, ὁ δὲ Β τὸν Γ πολλαπλασιάσας τὸν Ζ ποιείτω: λέγω, ὅτι ἴσος ἐστὶν ὁ Ε τῷ Ζ. ὁ γὰρ Α τὸν Γ πολλαπλασιάσας τὸν Η ποιείτω. ἐπεὶ οὖν ὁ Α τὸν Γ πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ Δ πολλαπλασιάσας τὸν Ε πεποίηκεν, ἀριθμὸς δὴ ὁ Α δύο ἀριθμοὺς τοὺς Γ, Δ πολλαπλασιάσας τοὺς Η, Ε πεποίηκεν. ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Η πρὸς τὸν Ε. ἀλλ᾽ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Β: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Ε. πάλιν, ἐπεὶ ὁ Α τὸν Γ πολλαπλασιάσας τὸν Η πεποίηκεν, ἀλλὰ μὴν καὶ ὁ Β τὸν Γ πολλαπλασιάσας τὸν Ζ πεποίηκεν, δύο δὴ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν Γ πολλαπλασιάσαντες τοὺς Η, Ζ πεποιήκασιν. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Ζ. ἀλλὰ μὴν καὶ ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Ε: καὶ ὡς ἄρα ὁ Η πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν Ζ. ὁ Η ἄρα πρὸς ἑκάτερον τῶν Ε, Ζ τὸν αὐτὸν ἔχει λόγον: ἴσος ἄρα ἐστὶν ὁ Ε τῷ Ζ. ἔστω δὴ πάλιν ἴσος ὁ Ε τῷ Ζ: λέγω, ὅτι ἐστὶν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴσος ἐστὶν ὁ Ε τῷ Ζ, ἔστιν ἄρα ὡς ὁ Η πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν Ζ. ἀλλ᾽ ὡς μὲν ὁ Η πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν δ, ὡς δὲ ὁ Η πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Β. καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",321]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'7.20'"],["Rule","'Text'","'The least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less.'"],["Rule","'TextWordCount'",33],["Rule","'GreekText'","'οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",4]],["List",["Rule","'Book'",7],["Rule","'Theorem'",12]],["List",["Rule","'Book'",7],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'For let CD, EF be the least numbers of those which have the same ratio with A, B; I say that CD measures A the same number of times that EF measures B. Now CD is not parts of A. For, if possible, let it be so; therefore EF is also the same parts of B that CD is of A. [VII. 13 and Def. 20] Therefore, as many parts of A as there are in CD, so many parts of B are there also in EF. Let CD be divided into the parts of A, namely CG, GD, and EF into the parts of B, namely EH, HF; thus the multitude of CG, GD will be equal to the multitude of EH, HF. Now, since the numbers CG, GD are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of CG, GD is equal to the multitude of EH, HF, therefore, as CG is to EH, so is GD to HF. Therefore also, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents. [VII. 12] Therefore, as CG is to EH, so is CD to EF. Therefore CG, EH are in the same ratio with CD, EF, being less than they: which is impossible, for by hypothesis CD, EF are the least numbers of those which have the same ratio with them. Therefore CD is not parts of A; therefore it is a part of it. [VII. 4] And EF is the same part of B that CD is of A; [VII. 13 and Def. 20] therefore CD measures A the same number of times that EF measures B.'"],["Rule","'ProofWordCount'",290],["Rule","'GreekProof'","'ἔστωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β οἱ ΓΔ, ΕΖ: λέγω, ὅτι ἰσάκις ὁ ΓΔ τὸν Α μετρεῖ καὶ ὁ ΕΖ τὸν Β. ὁ ΓΔ γὰρ τοῦ Α οὔκ ἐστι μέρη. εἰ γὰρ δυνατόν, ἔστω: καὶ ὁ ΕΖ ἄρα τοῦ Β τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὁ ΓΔ τοῦ Α. ὅσα ἄρα ἐστὶν ἐν τῷ ΓΔ μέρη τοῦ Α, τοσαῦτά ἐστι καὶ ἐν τῷ ΕΖ μέρη τοῦ Β. διῃρήσθω ὁ μὲν ΓΔ εἰς τὰ τοῦ Α μέρη τὰ ΓΗ, ΗΔ, ὁ δὲ ΕΖ εἰς τὰ τοῦ Β μέρη τὰ ΕΘ, ΘΖ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΓΗ, ΗΔ τῷ πλήθει τῶν ΕΘ, ΘΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΓΗ, ΗΔ ἀριθμοὶ ἀλλήλοις, εἰσὶ δὲ καὶ οἱ ΕΘ, ΘΖ ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῶν ΓΗ, ΗΔ τῷ πλήθει τῶν ΕΘ, ΘΖ, ἔστιν ἄρα ὡς ὁ ΓΗ πρὸς τὸν ΕΘ, οὕτως ὁ ΗΔ πρὸς τὸν ΘΖ. ἔσται ἄρα καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους. ἔστιν ἄρα ὡς ὁ ΓΗ πρὸς τὸν ΕΘ, οὕτως ὁ ΓΔ πρὸς τὸν ΕΖ: οἱ ΓΗ, ΕΘ ἄρα τοῖς ΓΔ, ΕΖ ἐν τῷ αὐτῷ λόγῳ εἰσὶν ἐλάσσονες ὄντες αὐτῶν: ὅπερ ἐστὶν ἀδύνατον: ὑπόκεινται γὰρ οἱ ΓΔ, ΕΖ ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. οὐκ ἄρα μέρη ἐστὶν ὁ ΓΔ τοῦ Α: μέρος ἄρα. καὶ ὁ ΕΖ τοῦ Β τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὁ ΓΔ τοῦ Α: ἰσάκις ἄρα ὁ ΓΔ τὸν Α μετρεῖ καὶ ὁ ΕΖ τὸν Β: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",256]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'7.21'"],["Rule","'Text'","'Numbers prime to one another are the least of those which have the same ratio with them.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'οἱ πρῶτοι πρὸς ἀλλήλους ἀριθμοὶ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",12]],["List",["Rule","'Book'",7],["Rule","'Theorem'",16]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let A, B be numbers prime to one another; I say that A, B are the least of those which have the same ratio with them. For, if not, there will be some numbers less than A, B which are in the same ratio with A, B. Let them be C, D. Since, then, the least numbers of those which have the same ratio measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent, [VII. 20] therefore C measures A the same number of times that D measures B. Now, as many times as C measures A, so many units let there be in E. Therefore D also measures B according to the units in E. And, since C measures A according to the units in E, therefore E also measures A according to the units in C. [VII. 16] For the same reason E also measures B according to the units in D. [VII. 16] Therefore E measures A, B which are prime to one another: which is impossible. [VII. Def. 12] Therefore there will be no numbers less than A, B which are in the same ratio with A, B. Therefore A, B are the least of those which have the same ratio with them.'"],["Rule","'ProofWordCount'",229],["Rule","'GreekProof'","'ἔστωσαν πρῶτοι πρὸς ἀλλήλους ἀριθμοὶ οἱ Α, Β: λέγω, ὅτι οἱ Α, Β ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. εἰ γὰρ μή, ἔσονταί τινες τῶν Α, Β ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β. ἔστωσαν οἱ Γ, Δ. ἐπεὶ οὖν οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάττων τὸν ἐλάττονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον, ἰσάκις ἄρα ὁ Γ τὸν Α μετρεῖ καὶ ὁ Δ τὸν Β. ὁσάκις δὴ ὁ Γ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. καὶ ὁ Δ ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας. καὶ ἐπεὶ ὁ Γ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, καὶ ὁ Ε ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας. διὰ τὰ αὐτὰ δὴ ὁ Ε καὶ τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας. ὁ Ε ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσονταί τινες τῶν Α, Β ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β. οἱ Α, Β ἄρα ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",208]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'7.22'"],["Rule","'Text'","'The least numbers of those which have the same ratio with them are prime to one another.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς πρῶτοι πρὸς ἀλλήλους εἰσίν.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",15]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'Let A, B be the least numbers of those which have the same ratio with them; I say that A, B are prime to one another. For, if they are not prime to one another, some number will measure them. Let some number measure them, and let it be C. And, as many times as C measures A, so many units let there be in D, and, as many times as C measures B, so many units let there be in E   Since C measures A according to the units in D, therefore C by multiplying D has made A. [VII. Def. 15] For the same reason also C by multiplying E has made B. Thus the number C by multiplying the two numbers D, E has made A, B; therefore, as D is to E, so is A to B; [VII. 17] therefore D, E are in the same ratio with A, B, being less than they: which is impossible. Therefore no number will measure the numbers A, B. Therefore A, B are prime to one another.'"],["Rule","'ProofWordCount'",178],["Rule","'GreekProof'","'ἔστωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς οἱ Α, Β: λέγω, ὅτι οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσι πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω, καὶ ἔστω ὁ Γ. καὶ ὁσάκις μὲν ὁ Γ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Δ, ὁσάκις δὲ ὁ Γ τὸν Β μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ ὁ Γ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας, ὁ Γ ἄρα τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Γ τὸν Ε πολλαπλασιάσας τὸν Β πεποίηκεν. ἀριθμὸς δὴ ὁ Γ δύο ἀριθμοὺς τοὺς Δ, Ε πολλαπλασιάσας τοὺς Α, Β πεποίηκεν: ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν Β: οἱ Δ, Ε ἄρα τοῖς Α, Β ἐν τῷ αὐτῷ λόγῳ εἰσὶν ἐλάσσονες ὄντες αὐτῶν: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Α, Β ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ α, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",164]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'7.23'"],["Rule","'Text'","'If two number be prime to one another, the number which measures the one of them will be prime to the remaining number.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ τὸν ἕνα αὐτῶν μετρῶν ἀριθμὸς πρὸς τὸν λοιπὸν πρῶτος ἔσται.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",12]]]],["Rule","'Proof'","'Let A, B be two numbers prime to one another, and let any number C measure A; I say that C, B are also prime to one another. For, if C, B are not prime to one another, some number will measure C, B. Let a number measure them, and let it be D. Since D measures C, and C measures A, therefore D also measures A. But it also measures B; therefore D measures A, B which are prime to one another: which is impossible. [VII. Def. 12] Therefore no number will measure the numbers C, B. Therefore C, B are prime to one another.'"],["Rule","'ProofWordCount'",106],["Rule","'GreekProof'","'ἔστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, τὸν δὲ Α μετρείτω τις ἀριθμὸς ὁ Γ: λέγω, ὅτι καὶ οἱ Γ, Β πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσιν οἱ Γ, Β πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς Γ, Β ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. ἐπεὶ ὁ Δ τὸν Γ μετρεῖ, ὁ δὲ Γ τὸν Α μετρεῖ, καὶ ὁ Δ ἄρα τὸν Α μετρεῖ. μετρεῖ δὲ καὶ τὸν Β: ὁ Δ ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Γ, Β ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ Γ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",106]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'7.24'"],["Rule","'Text'","'If two numbers be prime to any number, their product also will be prime to the same.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρός τινα ἀριθμὸν πρῶτοι ὦσιν, καὶ ὁ ἐξ αὐτῶν γενόμενος πρὸς τὸν αὐτὸν πρῶτος ἔσται.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",12]],["List",["Rule","'Book'",7],["Rule","'Definition'",15]],["List",["Rule","'Book'",7],["Rule","'Theorem'",16]],["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'For let the two numbers A, B be prime to any number C, and let A by multiplying B make D; I say that C, D are prime to one another. For, if C, D are not prime to one another, some number will measure C, D. Let a number measure them, and let it be E. Now, since C, A are prime to one another, and a certain number E measures C, therefore A, E are prime to one another. [VII. 23] As many times, then, as E measures D, so many units let there be in F; therefore F also measures D according to the units in E. [VII. 16] Therefore E by multiplying F has made D. [VII. Def. 15] But, further, A by multiplying B has also made D; therefore the product of E, F is equal to the product of A, B. But, if the product of the extremes be equal to that of the means, the four numbers are proportional; [VII. 19] therefore, as E is to A, so is B to F. But A, E are prime to one another, numbers which are prime to one another are also the least of those which have the same ratio, [VII. 21] and the least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures B. But it also measures C; therefore E measures B, C which are prime to one another: which is impossible. [VII. Def. 12] Therefore no number will measure the numbers C, D. Therefore C, D are prime to one another.'"],["Rule","'ProofWordCount'",297],["Rule","'GreekProof'","'δύο γὰρ ἀριθμοὶ οἱ Α, Β πρός τινα ἀριθμὸν τὸν Γ πρῶτοι ἔστωσαν, καὶ ὁ α τὸν Β πολλαπλασιάσας τὸν Δ ποιείτω: λέγω, ὅτι οἱ Γ, Δ πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσιν οἱ Γ, Δ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς Γ, Δ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Ε. καὶ ἐπεὶ οἱ Γ, Α πρῶτοι πρὸς ἀλλήλους εἰσίν, τὸν δὲ Γ μετρεῖ τις ἀριθμὸς ὁ Ε, οἱ Α, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. ὁσάκις δὴ ὁ Ε τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ζ: καὶ ὁ Ζ ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας. ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Δ πεποίηκεν: ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Ε, Ζ τῷ ἐκ τῶν Α, Β. ἐὰν δὲ ὁ ὑπὸ τῶν ἄκρων ἴσος ᾖ τῷ ὑπὸ τῶν μέσων, οἱ τέσσαρες ἀριθμοὶ ἀνάλογόν εἰσιν: ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Ζ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: ὁ Ε ἄρα τὸν Β μετρεῖ. μετρεῖ δὲ καὶ τὸν Γ: ὁ Ε ἄρα τοὺς Β, Γ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Γ, Δ ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ Γ, Δ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",260]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'7.25'"],["Rule","'Text'","'If two numbers be prime to one another, the product of one of them into itself will be prime to the remaining one.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ ἐκ τοῦ ἑνὸς αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτος ἔσται.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'Let A, B be two numbers prime to one another, and let A by multiplying itself make C: I say that B, C are prime to one another. For let D be made equal to A. Since A, B are prime to one another, and A is equal to D, therefore D, B are also prime to one another. Therefore each of the two numbers D, A is prime to B; therefore the product of D, A will also be prime to B. [VII. 24] But the number which is the product of D, A is C. Therefore C, B are prime to one another.'"],["Rule","'ProofWordCount'",105],["Rule","'GreekProof'","'ἔστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Γ ποιείτω: λέγω, ὅτι οἱ Β, Γ πρῶτοι πρὸς ἀλλήλους εἰσίν. κείσθω γὰρ τῷ Α ἴσος ὁ Δ. ἐπεὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, ἴσος δὲ ὁ Α τῷ Δ, καὶ οἱ Δ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. ἑκάτερος ἄρα τῶν Δ, Α πρὸς τὸν Β πρῶτός ἐστιν: καὶ ὁ ἐκ τῶν Δ, Α ἄρα γενόμενος πρὸς τὸν Β πρῶτος ἔσται. ὁ δὲ ἐκ τῶν Δ, Α γενόμενος ἀριθμός ἐστιν ὁ Γ. οἱ Γ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",101]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",26]],["Association",["Rule","'VertexLabel'","'7.26'"],["Rule","'Text'","'If two numbers be prime to two numbers, both to each, their products also will be prime to one another.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρὸς δύο ἀριθμοὺς ἀμφότεροι πρὸς ἑκάτερον πρῶτοι ὦσιν, καὶ οἱ ἐξ αὐτῶν γενόμενοι πρῶτοι πρὸς ἀλλήλους ἔσονται.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'For let the two numbers A, B be prime to the two numbers C, D; both to each, and let A by multiplying B make E, and let C by multiplying D make F; I say that E, F are prime to one another. For, since each of the numbers A, B is prime to C, therefore the product of A, B will also be prime to C. [VII. 24] But the product of A, B is E; therefore E, C are prime to one another. For the same reason E, D are also prime to one another. Therefore each of the numbers C, D is prime to E. Therefore the product of C, D will also be prime to E. [VII. 24] But the product of C, D is F. Therefore E, F are prime to one another.'"],["Rule","'ProofWordCount'",139],["Rule","'GreekProof'","'δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς δύο ἀριθμοὺς τοὺς Γ, Δ ἀμφότεροι πρὸς ἑκάτερον πρῶτοι ἔστωσαν, καὶ ὁ μὲν Α τὸν Β πολλαπλασιάσας τὸν Ε ποιείτω, ὁ δὲ Γ τὸν Δ πολλαπλασιάσας τὸν Ζ ποιείτω: λέγω, ὅτι οἱ Ε, Ζ πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ γὰρ ἑκάτερος τῶν Α, Β πρὸς τὸν Γ πρῶτός ἐστιν, καὶ ὁ ἐκ τῶν Α, Β ἄρα γενόμενος πρὸς τὸν Γ πρῶτος ἔσται. ὁ δὲ ἐκ τῶν Α, Β γενόμενός ἐστιν ὁ Ε: οἱ Ε, Γ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. διὰ τὰ αὐτὰ δὴ καὶ οἱ Δ, Ε πρῶτοι πρὸς ἀλλήλους εἰσίν. ἑκάτερος ἄρα τῶν Γ, Δ πρὸς τὸν Ε πρῶτός ἐστιν. καὶ ὁ ἐκ τῶν Γ, Δ ἄρα γενόμενος πρὸς τὸν Ε πρῶτος ἔσται. ὁ δὲ ἐκ τῶν Γ, Δ γενόμενός ἐστιν ὁ Ζ. οἱ Ε, Ζ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",143]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",27]],["Association",["Rule","'VertexLabel'","'7.27'"],["Rule","'Text'","'If two numbers be prime to one another, and each by multiplying itself make a certain number, the products will be prime to one another; and, if the original numbers by multiplying the products make certain numbers, the latter will also be prime to one another [and this is always the case with the extremes].'"],["Rule","'TextWordCount'",55],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ πολλαπλασιάσας ἑκάτερος ἑαυτὸν ποιῇ τινα, οἱ γενόμενοι ἐξ αὐτῶν πρῶτοι πρὸς ἀλλήλους ἔσονται, κἂν οἱ ἐξ ἀρχῆς τοὺς γενομένους πολλαπλασιάσαντες ποιῶσί τινας, κἀκεῖνοι πρῶτοι πρὸς ἀλλήλους ἔσονται καὶ ἀεὶ περὶ τοὺς ἄκρους τοῦτο συμβαίνει.'"],["Rule","'GreekTextWordCount'",42],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",25]],["List",["Rule","'Book'",7],["Rule","'Theorem'",26]]]],["Rule","'Proof'","'Let A, B be two numbers prime to one another, let A by multiplying itself make C, and by multiplying C make D, and let B by multiplying itself make E, and by multiplying E make F; I say that both C, E and D, F are prime to one another. For, since A, B are prime to one another, and A by multiplying itself has made C, therefore C, B are prime to one another. [VII. 25] Since then C, B are prime to one another, and B by multiplying itself has made E, therefore C, E are prime to one another. [id.] Again, since A, B are prime to one another, and B by multiplying itself has made E, therefore A, E are prime to one another. [id.] Since then the two numbers A, C are prime to the two numbers B, E, both to each, therefore also the product of A, C is prime to the product of B, E. [VII. 26] And the product of A, C is D, and the product of B, E is F. Therefore D, F are prime to one another.'"],["Rule","'ProofWordCount'",189],["Rule","'GreekProof'","'ἔστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, καὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Γ ποιείτω, τὸν δὲ Γ πολλαπλασιάσας τὸν Δ ποιείτω, ὁ δὲ Β ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε ποιείτω, τὸν δὲ Ε πολλαπλασιάσας τὸν Ζ ποιείτω: λέγω, ὅτι οἵ τε Γ, Ε καὶ οἱ Δ, Ζ πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ γὰρ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν, οἱ Γ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ οὖν οἱ Γ, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, οἱ Γ, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. πάλιν, ἐπεὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, οἱ Α, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ οὖν δύο ἀριθμοὶ οἱ Α, Γ πρὸς δύο ἀριθμοὺς τοὺς Β, Ε ἀμφότεροι πρὸς ἑκάτερον πρῶτοί εἰσιν, καὶ ὁ ἐκ τῶν Α, Γ ἄρα γενόμενος πρὸς τὸν ἐκ τῶν Β, Ε πρῶτός ἐστιν. καί ἐστιν ὁ μὲν ἐκ τῶν Α, Γ ὁ Δ, ὁ δὲ ἐκ τῶν Β, Ε ὁ Ζ. οἱ Δ, Ζ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",193]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",28]],["Association",["Rule","'VertexLabel'","'7.28'"],["Rule","'Text'","'If two numbers be prime to one another, the sum will also be prime to each of them; and, if the sum of two numbers be prime to any one of them, the original numbers will also be prime to one another.'"],["Rule","'TextWordCount'",42],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ συναμφότερος πρὸς ἑκάτερον αὐτῶν πρῶτος ἔσται: καὶ ἐὰν συναμφότερος πρὸς ἕνα τινὰ αὐτῶν πρῶτος ᾖ, καὶ οἱ ἐξ ἀρχῆς ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ἔσονται.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",12]]]],["Rule","'Proof'","'For let two numbers AB, BC prime to one another be added; I say that the sum AC is also prime to each of the numbers AB, BC. For, if CA, AB are not prime to one another, some number will measure CA, AB. Let a number measure them, and let it be D. Since then D measures CA, AB, therefore it will also measure the remainder BC. But it also measures BA; therefore D measures AB, BC which are prime to one another: which is impossible. [VII. Def. 12] Therefore no number will measure the numbers CA, AB; therefore CA, AB are prime to one another. For the same reason AC, CB are also prime to one another. Therefore CA is prime to each of the numbers AB, BC. Again, let CA, AB be prime to one another; I say that AB, BC are also prime to one another. For, if AB, BC are not prime to one another, some number will measure AB, BC. Let a number measure them, and let it be D. Now, since D measures each of the numbers AB, BC, it will also measure the whole CA. But it also measures AB; therefore D measures CA, AB which are prime to one another: which is impossible. [VII. Def. 12] Therefore no number will measure the numbers AB, BC. Therefore AB, BC are prime to one another.'"],["Rule","'ProofWordCount'",232],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ ΑΒ, ΒΓ: λέγω, ὅτι καὶ συναμφότερος ὁ ΑΓ πρὸς ἑκάτερον τῶν ΑΒ, ΒΓ πρῶτός ἐστιν. εἰ γὰρ μή εἰσιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς ΓΑ, ΑΒ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. ἐπεὶ οὖν ὁ Δ τοὺς ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸν ΒΓ μετρήσει. μετρεῖ δὲ καὶ τὸν ΒΑ: ὁ Δ ἄρα τοὺς ΑΒ, ΒΓ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς ΓΑ, ΑΒ ἀριθμοὺς ἀριθμός τις μετρήσει: οἱ ΓΑ, ΑΒ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. διὰ τὰ αὐτὰ δὴ καὶ οἱ ΑΓ, ΓΒ πρῶτοι πρὸς ἀλλήλους εἰσίν. ὁ ΓΑ ἄρα πρὸς ἑκάτερον τῶν ΑΒ, ΒΓ πρῶτός ἐστιν. ἔστωσαν δὴ πάλιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους: λέγω, ὅτι καὶ οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς ΑΒ, ΒΓ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. καὶ ἐπεὶ ὁ Δ ἑκάτερον τῶν ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸν ΓΑ μετρήσει. μετρεῖ δὲ καὶ τὸν ΑΒ: ὁ Δ ἄρα τοὺς ΓΑ, ΑΒ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς ΑΒ, ΒΓ ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ ΑΒ, ΒΓ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",210]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",29]],["Association",["Rule","'VertexLabel'","'7.29'"],["Rule","'Text'","'Any prime number is prime to any number which it does not measure.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἅπας πρῶτος ἀριθμὸς πρὸς ἅπαντα ἀριθμόν, ὃν μὴ μετρεῖ, πρῶτός ἐστιν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List"]],["Rule","'Proof'","'Let A be a prime number, and let it not measure B; I say that B, A are prime to one another. For, if B, A are not prime to one another, some number will measure them. Let C measure them. Since C measures B, and A does not measure B, therefore C is not the same with A. Now, since C measures B, A, therefore it also measures A which is prime, though it is not the same with it: which is impossible. Therefore no number will measure B, A. Therefore A, B are prime to one another.'"],["Rule","'ProofWordCount'",99],["Rule","'GreekProof'","'ἔστω πρῶτος ἀριθμὸς ὁ Α καὶ τὸν Β μὴ μετρείτω: λέγω, ὅτι οἱ Β, Α πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσιν οἱ Β, Α πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω ὁ Γ. ἐπεὶ ὁ Γ τὸν Β μετρεῖ, ὁ δὲ Α τὸν Β οὐ μετρεῖ, ὁ Γ ἄρα τῷ Α οὔκ ἐστιν ὁ αὐτός. καὶ ἐπεὶ ὁ Γ τοὺς Β, Α μετρεῖ, καὶ τὸν Α ἄρα μετρεῖ πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Β, Α μετρήσει τις ἀριθμός. οἱ Α, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",100]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",30]],["Association",["Rule","'VertexLabel'","'7.30'"],["Rule","'Text'","'If two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, τὸν δὲ γενόμενον ἐξ αὐτῶν μετρῇ τις πρῶτος ἀριθμός, καὶ ἕνα τῶν ἐξ ἀρχῆς μετρήσει.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",15]],["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",29]]]],["Rule","'Proof'","'For let the two numbers A, B by multiplying one another make C, and let any prime number D measure C; I say that D measures one of the numbers A, B. For let it not measure A. Now D is prime; therefore A, D are prime to one another. [VII. 29] And, as many times as D measures C, so many units let there be in E. Since then D measures C according to the units in E, therefore D by multiplying E has made C. [VII. Def. 15] Further, A by multiplying B has also made C; therefore the product of D, E is equal to the product of A, B. Therefore, as D is to A, so is B to E. [VII. 19] But D, A are prime to one another, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore D measures B. Similarly we can also show that, if D do not measure B, it will measure A. Therefore D measures one of the numbers A, B.'"],["Rule","'ProofWordCount'",207],["Rule","'GreekProof'","'δύο γὰρ ἀριθμοὶ οἱ Α, Β πολλαπλασιάσαντες ἀλλήλους τὸν Γ ποιείτωσαν, τὸν δὲ Γ μετρείτω τις πρῶτος ἀριθμὸς ὁ Δ: λέγω, ὅτι ὁ Δ ἕνα τῶν Α, Β μετρεῖ. τὸν γὰρ Α μὴ μετρείτω: καί ἐστι πρῶτος ὁ Δ: οἱ Α, Δ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ὁσάκις ὁ Δ τὸν Γ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν ὁ Δ τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, ὁ Δ ἄρα τὸν Ε πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν β πολλαπλασιάσας τὸν Γ πεποίηκεν: ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Δ, Ε τῷ ἐκ τῶν Α, Β. ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Ε. οἱ δὲ Δ, Α πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: ὁ Δ ἄρα τὸν Β μετρεῖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἐὰν τὸν Β μὴ μετρῇ, τὸν Α μετρήσει. ὁ Δ ἄρα ἕνα τῶν Α, Β μετρεῖ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",191]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",31]],["Association",["Rule","'VertexLabel'","'7.31'"],["Rule","'Text'","'Any composite number is measured by some prime number.'"],["Rule","'TextWordCount'",9],["Rule","'GreekText'","'ἅπας σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται.'"],["Rule","'GreekTextWordCount'",8],["Rule","'References'",["List"]],["Rule","'Proof'","'Let A be a composite number; I say that A is measured by some prime number. For, since A is composite,some number will measure it. Let a number measure it, and let it be B. Now, if B is prime, what was enjoined will have been done. But if it is composite, some number will measure it. Let a number measure it, and let it be C. Then, since C measures B, and B measures A, therefore C also measures A. And, if C is prime, what was enjoined will have been done. But if it is composite, some number will measure it. Thus, if the investigation be continued in this way, some prime number will be found which will measure the numberbefore it, which will also measure A. For, if it is not found, an infinite series of numbers will measure the number A, each of which is less than the other: which is impossible in numbers. Therefore some prime number will be found which willmeasure the one before it, which will also measure A. Therefore any composite number is measured by some prime number.'"],["Rule","'ProofWordCount'",187],["Rule","'GreekProof'","'ἔστω σύνθετος ἀριθμὸς ὁ Α: λέγω, ὅτι ὁ Α ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. ᾿επεὶ γὰρ σύνθετός ἐστιν ὁ Α, μετρήσει τις αὐτὸν ἀριθμός. μετρείτω, καὶ ἔστω ὁ Β. καὶ εἰ μὲν πρῶτός ἐστιν ὁ Β, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν ἀριθμός. μετρείτω, καὶ ἔστω ὁ Γ. καὶ ἐπεὶ ὁ Γ τὸν Β μετρεῖ, ὁ δὲ Β τὸν Α μετρεῖ, καὶ ὁ Γ ἄρα τὸν Α μετρεῖ. καὶ εἰ μὲν πρῶτός ἐστιν ὁ Γ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν ἀριθμός. τοιαύτης δὴ γινομένης ἐπισκέψεως ληφθήσεταί τις πρῶτος ἀριθμός, ὃς μετρήσει. εἰ γὰρ οὐ ληφθήσεται, μετρήσουσι τὸν Α ἀριθμὸν ἄπειροι ἀριθμοί, ὧν ἕτερος ἑτέρου ἐλάσσων ἐστίν: ὅπερ ἐστὶν ἀδύνατον ἐν ἀριθμοῖς. ληφθήσεταί τις ἄρα πρῶτος ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ, ὃς καὶ τὸν Α μετρήσει. ἅπας ἄρα σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",150]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",32]],["Association",["Rule","'VertexLabel'","'7.32'"],["Rule","'Text'","'Any number either is prime or is measured by some prime number.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'ἅπας ἀριθμὸς ἤτοι πρῶτός ἐστιν á¼¢ ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",31]]]],["Rule","'Proof'","'Let A be a number; I say that A either is prime or is measured by some prime number. If now A is prime, that which was enjoined will have been done. But if it is composite, some prime number will measure it. [VII. 31] Therefore any number either is prime or is measured by some prime number.'"],["Rule","'ProofWordCount'",58],["Rule","'GreekProof'","'ἔστω ἀριθμὸς ὁ Α: λέγω, ὅτι ὁ Α ἤτοι πρῶτός ἐστιν á¼¢ ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. εἰ μὲν οὖν πρῶτός ἐστιν ὁ Α, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν πρῶτος ἀριθμός. ἅπας ἄρα ἀριθμὸς ἤτοι πρῶτός ἐστιν á¼¢ ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",52]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",33]],["Association",["Rule","'VertexLabel'","'7.33'"],["Rule","'Text'","'Given as many numbers as we please, to find the least of those which have the same ratio with them.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ἀριθμῶν δοθέντων ὁποσωνοῦν εὑρεῖν τοὺς ἐλαχίστους τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",15]],["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",3]],["List",["Rule","'Book'",7],["Rule","'Theorem'",16]],["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'Let A, B, C be the given numbers, as many as we please; thus it is required to find the least ofthose which have the same ratio with A, B, C.  A, B, C are either prime to one another or not. Now, if A, B, C are prime to oneanother, they are the least of those which have the same ratio with them. [VII. 21] But, if not, let D the greatest common measure of A, B, C be taken, [VII. 3] and, as many times as D measures the numbers A, B, C  respectively, so many units let there be in the numbers E, F, G respectively. Therefore the numbers E, F, G measure the numbers A, B, C respectively according to the units in D. [VII. 16] Therefore E, F, G measure A, B, C the same number oftimes; therefore E, F, G are in the same ratio with A, B, C. [VII. Def. 20] I say next that they are the least that are in that ratio. For, if E, F, G are not the least of those which have the same ratio with A, B, C,there will be numbers less than E, F, G which are in the same ratio with A, B, C. Let them be H, K, L; therefore H measures A the same number of times that the numbers K, L measure the numbers B, C respectively. Now, as many times as H measures A, so many units let there be in M; therefore the numbers K, L also measure the numbers B, C respectively according to the units in M. And, since H measures A according to the units in M,therefore M also measures A according to the units in H. [VII. 16] For the same reason M also measures the numbers B, C according to the units in the numbers K, L respectively; Therefore M measures A, B, C. Now, since H measures A according to the units in M, therefore H by multiplying M has made A. [VII. Def. 15] For the same reason also E by multiplying D has made A. Therefore the product of E, D is equal to the product ofH, M. Therefore, as E is to H, so is M to D. [VII. 19] But E is greater than H; therefore M is also greater than D. And it measures A, B, C:which is impossible, for by hypothesis D is the greatest common measure of A, B, C. Therefore there cannot be any numbers less than E, F, G which are in the same ratio with A, B, C. Therefore E, F, G are the least of those which have thesame ratio with A, B, C.'"],["Rule","'ProofWordCount'",454],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες ὁποσοιοῦν ἀριθμοὶ οἱ Α, Β, Γ: δεῖ δὴ εὑρεῖν τοὺς ἐλαχίστους τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ. οἱ Α, Β, Γ γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν á¼¢ οὔ. εἰ μὲν οὖν οἱ Α, Β, Γ πρῶτοι πρὸς ἀλλήλους εἰσίν, ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. εἰ δὲ οὔ, εἰλήφθω τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον ὁ Δ, καὶ ὁσάκις ὁ Δ ἕκαστον τῶν Α, Β, Γ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν ἑκάστῳ τῶν Ε, Ζ, Η. καὶ ἕκαστος ἄρα τῶν Ε, Ζ, Η ἕκαστον τῶν Α, Β, Γ μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας. οἱ Ε, Ζ, Η ἄρα τοὺς Α, Β, Γ ἰσάκις μετροῦσιν: οἱ Ε, Ζ, Η ἄρα τοῖς Α, Β, Γ ἐν τῷ αὐτῷ λόγῳ εἰσίν. λέγω δή, ὅτι καὶ ἐλάχιστοι. εἰ γὰρ μή εἰσιν οἱ Ε, Ζ, Η ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ, ἔσονται τινες τῶν Ε, Ζ, Η ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β, Γ. ἔστωσαν οἱ Θ, Κ, Λ: ἰσάκις ἄρα ὁ Θ τὸν Α μετρεῖ καὶ ἑκάτερος τῶν Κ, Λ ἑκάτερον τῶν Β, Γ. ὁσάκις δὲ ὁ Θ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Μ: καὶ ἑκάτερος ἄρα τῶν Κ, Λ ἑκάτερον τῶν Β, Γ μετρεῖ κατὰ τὰς ἐν τῷ Μ μονάδας. καὶ ἐπεὶ ὁ Θ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Μ μονάδας, καὶ ὁ Μ ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Θ μονάδας. διὰ τὰ αὐτὰ δὴ ὁ Μ καὶ ἑκάτερον τῶν Β, Γ μετρεῖ κατὰ τὰς ἐν ἑκατέρῳ τῶν Κ, Λ μονάδας: ὁ Μ ἄρα τοὺς Α, Β, Γ μετρεῖ. καὶ ἐπεὶ ὁ Θ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Μ μονάδας, ὁ Θ ἄρα τὸν Μ πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Ε, Δ τῷ ἐκ τῶν Θ, Μ. ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Μ πρὸς τὸν Δ. μείζων δὲ ὁ Ε τοῦ Θ: μείζων ἄρα καὶ ὁ Μ τοῦ Δ. καὶ μετρεῖ τοὺς Α, Β, Γ: ὅπερ ἐστὶν ἀδύνατον: ὑπόκειται γὰρ ὁ Δ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον. οὐκ ἄρα ἔσονταί τινες τῶν Ε, Ζ, Η ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β, Γ. οἱ Ε, Ζ, Η ἄρα ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",410]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",34]],["Association",["Rule","'VertexLabel'","'7.34'"],["Rule","'Text'","'Given two numbers, to find the least number which they measure.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'δύο ἀριθμῶν δοθέντων εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν.'"],["Rule","'GreekTextWordCount'",8],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",15]],["List",["Rule","'Book'",7],["Rule","'Theorem'",16]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",33]]]],["Rule","'Proof'","'Let A, B be the two given numbers; thus it is required to find the least number which they measure. Now A, B are either prime to one another or not. First, let A, B be prime to one another, and let A by multiplying B make C; therefore also B by multiplying A has made C. [VII. 16] Therefore A, B measure C   I say next that it is also the least number they measure. For, if not, A, B will measure some number which is less than C. Let them measure D. Then, as many times as A measures D, so many units let there be in E, and, as many times as B measures D, so many units let there be in F; therefore A by multiplying E has made D, and B by multiplying F has made D; [VII. Def. 15] therefore the product of A, E is equal to the product of B, F. Therefore, as A is to B, so is F   E. [VII. 19] But A, B are prime, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore B measures E, as consequent consequent. And, since A by multiplying B, E has made C, D, therefore, as B is to E, so is C to D. [VII. 17] But B measures E; therefore C also measures D, the greater the less: which is impossible. Therefore A, B do not measure any number less than C; therefore C is the least that is measured by A, B. Next, let A, B not be prime to one another, and let F, E, the least numbers of those which have the same ratio with A, B, be taken; [VII. 33] therefore the product of A, E is equal to the product of B, F. [VII. 19] And let A by multiplying E make C; therefore also B by multiplying F has made C; therefore A, B measure C. I say next that it is also the least number that they measure. For, if not, A, B will measure some number which is less than C. Let them measure D. And, as many times as A measures D, so many units let there be in G, and, as many times as B measures D, so many units let there be in H. Therefore A by multiplying G has made D, and B by multiplying H has made D. Therefore the product of A, G is equal to the product of B, H; therefore, as A is to B, so is H to G. [VII. 19] But, as A is to B, so is F to E. Therefore also, as F is to E, so is H to G. But F, E are least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore E measures G. And, since A by multiplying E, G has made C, D, therefore, as E is to G, so is C to D. [VII. 17] But E measures G; therefore C also measures D, the greater the less: which is impossible. Therefore A, B will not measure any number which is less than C. Therefore C is the least that is measured by A, B.'"],["Rule","'ProofWordCount'",576],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες δύο ἀριθμοὶ οἱ Α, Β: δεῖ δὴ εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν. οἱ Α, Β γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν á¼¢ οὔ. ἔστωσαν πρότερον οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω: καὶ ὁ Β ἄρα τὸν Α πολλαπλασιάσας τὸν Γ πεποίηκεν. οἱ Α, Β ἄρα τὸν Γ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσί τινα ἀριθμὸν οἱ Α, Β ἐλάσσονα ὄντα τοῦ Γ. μετρείτωσαν τὸν Δ. καὶ ὁσάκις ὁ Α τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε, ὁσάκις δὲ ὁ Β τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ζ: ὁ μὲν Α ἄρα τὸν Ε πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ δὲ Β τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν: ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Α, Ε τῷ ἐκ τῶν Β, Ζ. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Ζ πρὸς τὸν Ε. οἱ δὲ Α, Β πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα: ὁ Β ἄρα τὸν Ε μετρεῖ, ὡς ἑπόμενος ἑπόμενον. καὶ ἐπεὶ ὁ Α τοὺς Β, Ε πολλαπλασιάσας τοὺς Γ, Δ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Β πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Δ. μετρεῖ δὲ ὁ Β τὸν Ε: μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β μετροῦσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Γ. ὁ Γ ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β μετρεῖται. μὴ ἔστωσαν δὴ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β οἱ Ζ, Ε: ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Α, Ε τῷ ἐκ τῶν Β, Ζ. καὶ ὁ Α τὸν Ε πολλαπλασιάσας τὸν Γ ποιείτω: καὶ ὁ Β ἄρα τὸν Ζ πολλαπλασιάσας τὸν Γ πεποίηκεν: οἱ Α, Β ἄρα τὸν Γ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσί τινα ἀριθμὸν οἱ Α, Β ἐλάσσονα ὄντα τοῦ Γ. μετρείτωσαν τὸν Δ. καὶ ὁσάκις μὲν ὁ Α τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Η, ὁσάκις δὲ ὁ Β τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Θ. ὁ μὲν Α ἄρα τὸν Η πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ δὲ Β τὸν Θ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Α, Η τῷ ἐκ τῶν Β, Θ: ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Θ πρὸς τὸν Η. ὡς δὲ ὁ Α πρὸς τὸν Β, οὕτως ὁ Ζ πρὸς τὸν Ε: καὶ ὡς ἄρα ὁ Ζ πρὸς τὸν Ε, οὕτως ὁ Θ πρὸς τὸν Η. οἱ δὲ Ζ, Ε ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα: ὁ Ε ἄρα τὸν Η μετρεῖ. καὶ ἐπεὶ ὁ Α τοὺς Ε, Η πολλαπλασιάσας τοὺς Γ, Δ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Η, οὕτως ὁ Γ πρὸς τὸν Δ. ὁ δὲ Ε τὸν Η μετρεῖ: καὶ ὁ Γ ἄρα τὸν Δ μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Γ. ὁ Γ ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β μετρεῖται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",546]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",35]],["Association",["Rule","'VertexLabel'","'7.35'"],["Rule","'Text'","'If two numbers measure any number, the least number measured by them will also measure the same.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ ἀριθμόν τινα μετρῶσιν, καὶ ὁ ἐλάχιστος ὑπ᾽ αὐτῶν μετρούμενος τὸν αὐτὸν μετρήσει.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List"]],["Rule","'Proof'","'For let the two numbers A, B measure any number CD, and let E be the least that they measure; I say that E also measures CD. For, if E does not measure CD, let E, measuring DF, leave CF less than itself. Now, since A, B measure E, and E measures DF, therefore A, B will also measure DF. But they also measure the whole CD; therefore they will also measure the remainder CF which is less than E: which is impossible. Therefore E cannot fail to measure CD; therefore it measures it.'"],["Rule","'ProofWordCount'",94],["Rule","'GreekProof'","'δύο γὰρ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν ΓΔ μετρείτωσαν, ἐλάχιστον δὲ τὸν Ε: λέγω, ὅτι καὶ ὁ Ε τὸν ΓΔ μετρεῖ. εἰ γὰρ οὐ μετρεῖ ὁ Ε τὸν ΓΔ, ὁ Ε τὸν ΔΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΓΖ. καὶ ἐπεὶ οἱ Α, Β τὸν Ε μετροῦσιν, ὁ δὲ Ε τὸν ΔΖ μετρεῖ, καὶ οἱ Α, Β ἄρα τὸν ΔΖ μετρήσουσιν. μετροῦσι δὲ καὶ ὅλον τὸν ΓΔ: καὶ λοιπὸν ἄρα τὸν ΓΖ μετρήσουσιν ἐλάσσονα ὄντα τοῦ Ε: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οὐ μετρεῖ ὁ Ε τὸν ΓΔ: μετρεῖ ἄρα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",95]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",36]],["Association",["Rule","'VertexLabel'","'7.36'"],["Rule","'Text'","'Given three numbers, to find the least number which they measure.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'τριῶν ἀριθμῶν δοθέντων εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν.'"],["Rule","'GreekTextWordCount'",8],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",34]],["List",["Rule","'Book'",7],["Rule","'Theorem'",35]]]],["Rule","'Proof'","'Let A, B, C be the three given numbers; thus it is required to find the least number which they measure. Let D, the least number measured by the two numbers A, B, be taken. [VII. 34] Then C either measures, or does not measure, D. First, let it measure it. But A, B also measure D; therefore A, B, C measure D. I say next that it is also the least that they measure. For, if not, A, B, C will measure some number which is less than D. Let them measure E. Since A, B, C measure E, therefore also A, B measure E. Therefore the least number measured by A, B will also measure E. [VII. 35] But D is the least number measured by A, B; therefore D will measure E, the greater the less: which is impossible. Therefore A, B, C will not measure any number which is less than D; therefore D is the least that A, B, C measure. Again, let C not measure D, and let E, the least number measured by C, D, be taken. [VII. 34] Since A, B measure D, and D measures E, therefore also A, B measure E. But C also measures E; therefore also A, B, C measure E. I say next that it is also the least that they measure. For, if not, A, B, C will measure some number which is less than E. Let them measure F. Since A, B, C measure F, therefore also A, B measure F; therefore the least number measured by A, B will also measure F. [VII. 35] But D is the least number measured by A, B; therefore D measures F. But C also measures F; therefore D, C measure F, so that the least number measured by D, C will also measure F. But E is the least number measured by C, D; therefore E measures F, the greater the less: which is impossible. Therefore A, B, C will not measure any number which is less than E. Therefore E is the least that is measured by A, B, C.'"],["Rule","'ProofWordCount'",354],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ οἱ Α, Β, Γ: δεῖ δὴ εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν. εἰλήφθω γὰρ ὑπὸ δύο τῶν Α, Β ἐλάχιστος μετρούμενος ὁ Δ. ὁ δὴ Γ τὸν Δ ἤτοι μετρεῖ á¼¢ οὐ μετρεῖ. μετρείτω πρότερον. μετροῦσι δὲ καὶ οἱ Α, Β τὸν Δ: οἱ Α, Β, Γ ἄρα τὸν Δ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσιν τινα ἀριθμὸν οἱ Α, Β, Γ ἐλάσσονα ὄντα τοῦ Δ. μετρείτωσαν τὸν Ε. ἐπεὶ οἱ Α, Β, Γ τὸν Ε μετροῦσιν, καὶ οἱ Α, Β ἄρα τὸν Ε μετροῦσιν. καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Α, Β μετρούμενος τὸν Ε μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Α, Β μετρούμενός ἐστιν ὁ Δ: ὁ Δ ἄρα τὸν Ε μετρήσει ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β, Γ μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Δ: οἱ Α, Β, Γ ἄρα ἐλάχιστον τὸν Δ μετροῦσιν. μὴ μετρείτω δὴ πάλιν ὁ Γ τὸν Δ, καὶ εἰλήφθω ὑπὸ τῶν Γ, Δ ἐλάχιστος μετρούμενος ἀριθμὸς ὁ Ε. ἐπεὶ οἱ Α, Β τὸν Δ μετροῦσιν, ὁ δὲ Δ τὸν Ε μετρεῖ, καὶ οἱ Α, Β ἄρα τὸν Ε μετροῦσιν. μετρεῖ δὲ καὶ ὁ Γ τὸν Ε: καὶ οἱ Α, Β, Γ ἄρα τὸν Ε μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσί τινα οἱ Α, Β, Γ ἐλάσσονα ὄντα τοῦ Ε. μετρείτωσαν τὸν Ζ. ἐπεὶ οἱ Α, Β, Γ τὸν Ζ μετροῦσιν, καὶ οἱ Α, Β ἄρα τὸν Ζ μετροῦσιν: καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Α, Β μετρούμενος τὸν Ζ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Α, Β μετρούμενός ἐστιν ὁ Δ: ὁ Δ ἄρα τὸν Ζ μετρεῖ. μετρεῖ δὲ καὶ ὁ Γ τὸν Ζ: οἱ Δ, Γ ἄρα τὸν Ζ μετροῦσιν: ὥστε καὶ ὁ ἐλάχιστος ὑπὸ τῶν Δ, Γ μετρούμενος τὸν Ζ μετρήσει. ὁ δὲ ἐλάχιστος ὑπὸ τῶν Γ, Δ μετρούμενός ἐστιν ὁ Ε: ὁ Ε ἄρα τὸν Ζ μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β, Γ μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Ε. ὁ Ε ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β, Γ μετρεῖται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",348]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",37]],["Association",["Rule","'VertexLabel'","'7.37'"],["Rule","'Text'","'If a number be measured by any number, the number which is measured will have a part called by the same name as the measuring number.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐὰν ἀριθμὸς ὑπό τινος ἀριθμοῦ μετρῆται, ὁ μετρούμενος ὁμώνυμον μέρος ἕξει τῷ μετροῦντι.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'For let the number A be measured by any number B; I say that A has a part called by the same name as B. For, as many times as B measures A, so many units let there be in C. Since B measures A according to the units in C, and the unit D also measures the number C according to the units in it, therefore the unit D measures the number C the same number of times as B measures A. Therefore, alternately, the unit D measures the number B the same number of times as C measures A; [VII. 15] therefore, whatever part the unit D is of the number B, the same part is C of A also. But the unit D is a part of the number B called by the same name as it; therefore C is also a part of A called by the same name as B, so that A has a part C which is called by the same name as B.'"],["Rule","'ProofWordCount'",171],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ Α ὑπό τινος ἀριθμοῦ τοῦ Β μετρείσθω: λέγω, ὅτι ὁ Α ὁμώνυμον μέρος ἔχει τῷ Β. ὁσάκις γὰρ ὁ Β τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Γ. ἐπεὶ ὁ Β τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας, μετρεῖ δὲ καὶ ἡ Δ μονὰς τὸν Γ ἀριθμὸν κατὰ τὰς ἐν αὐτῷ μονάδας, ἰσάκις ἄρα ἡ Δ μονὰς τὸν Γ ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Α. ἐναλλὰξ ἄρα ἰσάκις ἡ Δ μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Γ τὸν Α: ὃ ἄρα μέρος ἐστὶν ἡ Δ μονὰς τοῦ Β ἀριθμοῦ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Α. ἡ δὲ Δ μονὰς τοῦ Β ἀριθμοῦ μέρος ἐστὶν ὁμώνυμον αὐτῷ: καὶ ὁ Γ ἄρα τοῦ Α μέρος ἐστὶν ὁμώνυμον τῷ Β. ὥστε ὁ Α μέρος ἔχει τὸν Γ ὁμώνυμον ὄντα τῷ Β: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",142]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",38]],["Association",["Rule","'VertexLabel'","'7.38'"],["Rule","'Text'","'If a number have any part whatever, it will be measured by a number called by the same name as the part.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐὰν ἀριθμὸς μέρος ἔχῃ ὁτιοῦν, ὑπὸ ὁμωνύμου ἀριθμοῦ μετρηθήσεται τῷ μέρει.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'For let the number A have any part whatever, B, and let C be a number called by the same name as the part B; I say that C measures A. For, since B is a part of A called by the same name as C, and the unit D is also a part of C called by the same name as it, therefore, whatever part the unit D is of the number C, the same part is B of A also; therefore the unit D measures the number C the same number of times that B measures A. Therefore, alternately, the unit D measures the number B the same number of times that C measures A. [VII. 15] Therefore C measures A.'"],["Rule","'ProofWordCount'",123],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ Α μέρος ἐχέτω ὁτιοῦν τὸν Β, καὶ τῷ Β μέρει ὁμώνυμος ἔστω ἀριθμὸς ὁ Γ: λέγω, ὅτι ὁ Γ τὸν Α μετρεῖ. ἐπεὶ γὰρ ὁ Β τοῦ Α μέρος ἐστὶν ὁμώνυμον τῷ Γ, ἔστι δὲ καὶ ἡ Δ μονὰς τοῦ Γ μέρος ὁμώνυμον αὐτῷ, ὃ ἄρα μέρος ἐστὶν ἡ Δ μονὰς τοῦ Γ ἀριθμοῦ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Β τοῦ Α: ἰσάκις ἄρα ἡ Δ μονὰς τὸν Γ ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Α. ἐναλλὰξ ἄρα ἰσάκις ἡ Δ μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Γ τὸν Α. ὁ Γ ἄρα τὸν Α μετρεῖ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",104]]],["Rule",["Association",["Rule","'Book'",7],["Rule","'Theorem'",39]],["Association",["Rule","'VertexLabel'","'7.39'"],["Rule","'Text'","'To find the number which is the least that will have given parts.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἀριθμὸν εὑρεῖν, ὃς ἐλάχιστος ὢν ἕξει τὰ δοθέντα μέρη.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",36]],["List",["Rule","'Book'",7],["Rule","'Theorem'",37]],["List",["Rule","'Book'",7],["Rule","'Theorem'",38]]]],["Rule","'Proof'","'Let A, B, C be the given parts; thus it is required to find the number which is the least that will have the parts A, B, C. Let D, E, F be numbers called by the same name as the parts A, B, C, and let G, the least number measured by D, E, F, be taken. [VII. 36] Therefore G has parts called by the same name as D, E, F. [VII. 37] But A, B, C are parts called by the same name as D, E, F; therefore G has the parts A, B, C. I say next that it is also the least number that has. For, if not, there will be some number less than G which will have the parts A, B, C. Let it be H. Since H has the parts A, B, C, therefore H will be measured by numbers called by the same name as the parts A, B, C. [VII. 38] But D, E, F are numbers called by the same name as the parts A, B, C; therefore H is measured by D, E, F. And it is less than G: which is impossible. Therefore there will be no number less than G that will have the parts A, B, C.'"],["Rule","'ProofWordCount'",212],["Rule","'GreekProof'","'ἔστω τὰ δοθέντα μέρη τὰ Α, Β, Γ: δεῖ δὴ ἀριθμὸν εὑρεῖν, ὃς ἐλάχιστος ὢν ἕξει τὰ Α, Β, Γ μέρη. ἔστωσαν γὰρ τοῖς Α, Β, Γ μέρεσιν ὁμώνυμοι ἀριθμοὶ οἱ Δ, Ε, Ζ, καὶ εἰλήφθω ὑπὸ τῶν Δ, Ε, Ζ ἐλάχιστος μετρούμενος ἀριθμὸς ὁ Η. ὁ Η ἄρα ὁμώνυμα μέρη ἔχει τοῖς Δ, Ε, Ζ. τοῖς δὲ Δ, Ε, Ζ ὁμώνυμα μέρη ἐστὶ τὰ Α, Β, Γ: ὁ Η ἄρα ἔχει τὰ Α, Β, Γ μέρη. λέγω δή, ὅτι καὶ ἐλάχιστος ὤν. εἰ γὰρ μή, ἔσται τις τοῦ Η ἐλάσσων ἀριθμός, ὃς ἕξει τὰ Α, Β, Γ μέρη. ἔστω ὁ Θ. ἐπεὶ ὁ Θ ἔχει τὰ Α, Β, Γ μέρη, ὁ Θ ἄρα ὑπὸ ὁμωνύμων ἀριθμῶν μετρηθήσεται τοῖς Α, Β, Γ μέρεσιν. τοῖς δὲ Α, Β, Γ μέρεσιν ὁμώνυμοι ἀριθμοί εἰσιν οἱ Δ, Ε, Ζ: ὁ Θ ἄρα ὑπὸ τῶν Δ, Ε, Ζ μετρεῖται. καί ἐστιν ἐλάσσων τοῦ Η: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσται τις τοῦ Η ἐλάσσων ἀριθμός, ὃς ἕξει τὰ Α, Β, Γ μέρη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",171]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'8.1'"],["Rule","'Text'","'If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the numbers are the least of those which have the same ratio with them.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ δὲ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",14]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'Let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them A, D be prime to one another; I say that A, B, C, D are the least of those which have the same ratio with them. For, if not, let E, F, G, H be less than A, B, C, D, and in the same ratio with them. Now, since A, B, C, D are in the same ratio with E, F, G, H, and the multitude of the numbers A, B, C, D is equal to the multitude of the numbers E, F, G, H, therefore, ex aequali, as A is to D, so is E to H. [VII. 14]  But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20] Therefore A measures E, the greater the less: which is impossible. Therefore E, F, G, H which are less than A, B, C, D are not in the same ratio with them. Therefore A, B, C, D are the least of those which have the same ratio with them.'"],["Rule","'ProofWordCount'",221],["Rule","'GreekProof'","'ἔστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, δ, οἱ δὲ ἄκροι αὐτῶν οἱ Α, Δ πρῶτοι πρὸς ἀλλήλους ἔστωσαν: λέγω, ὅτι οἱ Α, Β, Γ, Δ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. εἰ γὰρ μή, ἔστωσαν ἐλάττονες τῶν Α, Β, Γ, Δ οἱ Ε, Ζ, Η, Θ ἐν τῷ αὐτῷ λόγῳ ὄντες αὐτοῖς. καὶ ἐπεὶ οἱ α, Β, Γ, Δ ἐν τῷ αὐτῷ λόγῳ εἰσὶ τοῖς Ε, Ζ, Η, Θ, καί ἐστιν ἴσον τὸ πλῆθος τῶν Α, Β, Γ, Δ τῷ πλήθει τῶν Ε, Ζ, Η, Θ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, ὁ Ε πρὸς τὸν Θ. οἱ δὲ Α, Δ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον. μετρεῖ ἄρα ὁ Α τὸν Ε ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Ε, Ζ, Η, Θ ἐλάσσονες ὄντες τῶν Α, Β, Γ, Δ ἐν τῷ αὐτῷ λόγῳ εἰσὶν αὐτοῖς. οἱ Α, Β, Γ, Δ ἄρα ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",199]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'8.2'"],["Rule","'Text'","'To find numbers in continued proportion, as many as may be prescribed, and the least that are in a given ratio.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'ἀριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον ἐλαχίστους, ὅσους ἂν ἐπιτάξῃ τις, ἐν τῷ δοθέντι λόγῳ.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",7],["Rule","'Theorem'",18]],["List",["Rule","'Book'",7],["Rule","'Theorem'",22]],["List",["Rule","'Book'",7],["Rule","'Theorem'",27]],["List",["Rule","'Book'",8],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let the ratio of A to B be the given ratio in least numbers; thus it is required to find numbers in continued proportion, as many as may be prescribed, and the least that are in the ratio of A to B. Let four be prescribed; let A by multiplying itself make C, and by multiplying B let it make D; let B by multiplying itself make E; further, let A by multiplying C, D, E make F, G, H, and let B by multiplying E make K. Now, since A by multiplying itself has made C, and by multiplying B has made D, therefore, as A is to B, so is C to D. [VII. 17] Again, since A by multiplying B has made D, and B by multiplying itself has made E, therefore the numbers A, B by multiplying B have made the numbers D, E respectively. Therefore, as A is to B, so is D to E. [VII. 18] But, as A is to B, so is C to D; therefore also, as C is to D, so is D to E. And, since A by multiplying C, D has made F, G, therefore, as C is to D, so is F to G. [VII. 17] But, as C is to D, so was A to B; therefore also, as A is to B, so is F to G. Again, since A by multiplying D, E has made G, H, therefore, as D is to E, so is G to H. [VII. 17] But, as D is to E, so is A to B. Therefore also, as A is to B, so is G to H. And, since A, B by multiplying E have made H, K, therefore, as A is to B, so is H to K. [VII. 18] But, as A is to B, so is F to G, and G to H. Therefore also, as F is to G, so is G to H, and H to K; therefore C, D, E, and F, G, H, K are proportional in the ratio of A to B. I say next that they are the least numbers that are so. For, since A, B are the least of those which have the same ratio with them, and the least of those which have the same ratio are prime to one another, [VII. 22] therefore A, B are prime to one another. And the numbers A, B by multiplying themselves respectively have made the numbers C, E, and by multiplying the numbers C, E respectively have made the numbers F, K; therefore C, E and F, K are prime to one another respectively. [VII. 27] But, if there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, they are the least of those which have the same ratio with them. [VIII. 1] Therefore C, D, E and F, G, H, K are the least of those which have the same ratio with A, B.'"],["Rule","'ProofWordCount'",508],["Rule","'GreekProof'","'ἔστω ὁ δοθεὶς λόγος ἐν ἐλαχίστοις ἀριθμοῖς ὁ τοῦ Α πρὸς τὸν Β: δεῖ δὴ ἀριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον ἐλαχίστους, ὅσους ἄν τις ἐπιτάξῃ, ἐν τῷ τοῦ Α πρὸς τὸν Β λόγῳ. Ἐπιτετάχθωσαν δὴ τέσσαρες, καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Γ ποιείτω, τὸν δὲ Β πολλαπλασιάσας τὸν Δ ποιείτω, καὶ ἔτι ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε ποιείτω, καὶ ἔτι ὁ Α τοὺς Γ, Δ, Ε πολλαπλασιάσας τοὺς Ζ, Η, Θ ποιείτω, ὁ δὲ Β τὸν Ε πολλαπλασιάσας τὸν Κ ποιείτω. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Γ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Δ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ. πάλιν, ἐπεὶ ὁ μὲν Α τὸν Β πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ δὲ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, ἑκάτερος ἄρα τῶν Α, Β τὸν Β πολλαπλασιάσας ἑκάτερον τῶν Δ, Ε πεποίηκεν. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε. ἀλλ᾽ ὡς ὁ Α πρὸς τὸν Β, ὁ Γ πρὸς τὸν Δ: καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Δ, ὁ Δ πρὸς τὸν Ε. καὶ ἐπεὶ ὁ Α τοὺς Γ, Δ πολλαπλασιάσας τοὺς Ζ, Η πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ζ πρὸς τὸν Η. ὡς δὲ ὁ Γ πρὸς τὸν Δ, οὕτως ἦν ὁ Α πρὸς τὸν Β: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, ὁ Ζ πρὸς τὸν Η. πάλιν, ἐπεὶ ὁ Α τοὺς Δ, Ε πολλαπλασιάσας τοὺς Η, Θ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ε, ὁ Η πρὸς τὸν Θ. ἀλλ᾽ ὡς ὁ Δ πρὸς τὸν Ε, ὁ Α πρὸς τὸν Β. καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Θ. καὶ ἐπεὶ οἱ Α, Β τὸν Ε πολλαπλασιάσαντες τοὺς Θ, Κ πεποιήκασιν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Θ πρὸς τὸν Κ. ἀλλ᾽ ὡς ὁ Α πρὸς τὸν Β, οὕτως ὅ τε Ζ πρὸς τὸν Η καὶ ὁ Η πρὸς τὸν Θ. καὶ ὡς ἄρα ὁ Ζ πρὸς τὸν Η, οὕτως ὅ τε Η πρὸς τὸν Θ καὶ ὁ Θ πρὸς τὸν Κ: οἱ Γ, Δ, Ε ἄρα καὶ οἱ Ζ, Η, Θ, Κ ἀνάλογόν εἰσιν ἐν τῷ τοῦ Α πρὸς τὸν Β λόγῳ. λέγω δή, ὅτι καὶ ἐλάχιστοι. ἐπεὶ γὰρ οἱ Α, Β ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, οἱ δὲ ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων πρῶτοι πρὸς ἀλλήλους εἰσίν, οἱ Α, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἑκάτερος μὲν τῶν Α, Β ἑαυτὸν πολλαπλασιάσας ἑκάτερον τῶν Γ, Ε πεποίηκεν, ἑκάτερον δὲ τῶν Γ, Ε πολλαπλασιάσας ἑκάτερον τῶν Ζ, Κ πεποίηκεν: οἱ Γ, Ε ἄρα καὶ οἱ Ζ, Κ πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐὰν δὲ ὦσιν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ δὲ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. οἱ Γ, Δ, Ε ἄρα καὶ οἱ Ζ, Η, Θ, Κ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστοι ὦσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, οἱ ἄκροι αὐτῶν τετράγωνοί εἰσιν, ἐὰν δὲ τέσσαρες, κύβοι.'"],["Rule","'GreekProofWordCount'",528]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'8.3'"],["Rule","'Text'","'If as many numbers as we please in continued proportion be the least of those which have the same ratio with them, the extremes of them are prime to one another.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, οἱ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους εἰσίν.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",22]],["List",["Rule","'Book'",7],["Rule","'Theorem'",27]],["List",["Rule","'Book'",7],["Rule","'Theorem'",33]],["List",["Rule","'Book'",8],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let as many numbers as we please, A, B, C, D, in continued proportion be the least of those which have the same ratio with them; I say that the extremes of them A, D are prime to one another. For let two numbers E, F, the least that are in the ratio of A, B, C, D, be taken, [VII. 33] then three others G, H, K with the same property; and others, more by one continually, [VIII. 2] until the multitude taken becomes equal to the multitude of the numbers A, B, C, D. Let them be taken, and let them be L, M, N, O. Now, since E, F are the least of those which have the same ratio with them, they are prime to one another. [VII. 22] And, since the numbers E, F by multiplying themselves respectively have made the numbers G, K, and by multiplying the numbers G, K respectively have made the numbers L, O, [VIII. 2] therefore both G, K and L, O are prime to one another. [VII. 27] And, since A, B, C, D are the least of those which have the same ratio with them, while L, M, N, O are the least that are in the same ratio with A, B, C, D, and the multitude of the numbers A, B, C, D is equal to the multitude of the numbers L, M, N, O, therefore the numbers A, B, C, D are equal to the numbers L, M, N, O respectively; therefore A is equal to L, and D to O. And L, O are prime to one another. Therefore A, D are also prime to one another.'"],["Rule","'ProofWordCount'",281],["Rule","'GreekProof'","'ἔστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς οἱ Α, Β, Γ, Δ: λέγω, ὅτι οἱ ἄκροι αὐτῶν οἱ Α, Δ πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰλήφθωσαν γὰρ δύο μὲν ἀριθμοὶ ἐλάχιστοι ἐν τῷ τῶν Α, Β, Γ, Δ λόγῳ οἱ Ε, Ζ, τρεῖς δὲ οἱ Η, Θ, Κ, καὶ ἑξῆς ἑνὶ πλείους, ἕως τὸ λαμβανόμενον πλῆθος ἴσον γένηται τῷ πλήθει τῶν Α, Β, Γ, Δ. εἰλήφθωσαν καὶ ἔστωσαν οἱ Λ, Μ, Ν, Ξ. καὶ ἐπεὶ οἱ Ε, Ζ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ ἑκάτερος τῶν Ε, Ζ ἑαυτὸν μὲν πολλαπλασιάσας ἑκάτερον τῶν Η, Κ πεποίηκεν, ἑκάτερον δὲ τῶν Η, Κ πολλαπλασιάσας ἑκάτερον τῶν Λ, Ξ πεποίηκεν, καὶ οἱ Η, Κ ἄρα καὶ οἱ Λ, Ξ πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ οἱ Α, Β, Γ, Δ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, εἰσὶ δὲ καὶ οἱ Λ, Μ, Ν, Ξ ἐλάχιστοι ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β, Γ, Δ, καί ἐστιν ἴσον τὸ πλῆθος τῶν Α, Β, Γ, Δ τῷ πλήθει τῶν Λ, Μ, Ν, Ξ, ἕκαστος ἄρα τῶν Α, Β, Γ, Δ ἑκάστῳ τῶν Λ, Μ, Ν, Ξ ἴσος ἐστίν: ἴσος ἄρα ἐστὶν ὁ μὲν Α τῷ Λ, ὁ δὲ Δ τῷ Ξ. καί εἰσιν οἱ Λ, Ξ πρῶτοι πρὸς ἀλλήλους. καὶ οἱ Α, Δ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",231]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'8.4'"],["Rule","'Text'","'Given as many ratios as we please in least numbers, to find numbers in continued proportion which are the least in the given ratios.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'λόγων δοθέντων ὁποσωνοῦν ἐν ἐλαχίστοις ἀριθμοῖς ἀριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον ἐλαχίστους ἐν τοῖς δοθεῖσι λόγοις.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",13]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",34]],["List",["Rule","'Book'",7],["Rule","'Theorem'",35]],["List",["Rule","'Book'",8],["Rule","'Definition'",20]]]],["Rule","'Proof'","'Let the given ratios in least numbers be that of A to B,that of C to D, and that of E to F; thus it is required to find numbers in continued proportion which are the least that are in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F. Let G, the least number measured by B, C, be taken. [VII. 34] And, as many times as B measures G, so many times also let A measure H, and, as many times as C measures G, so many times also let D measure K. Now E either measures or does not measure K. First, let it measure it. And, as many times as E measures K, so many times let F measure L also. Now, since A measures H the same number of times that B measures G,therefore, as A is to B, so is H to G. [VII. Def. 20, VII. 13] For the same reason also, as C is to D, so is G to K, and further, as E is to F, so is K to L; therefore H, G, K, L are continuously proportional in theratio of A to B, in the ratio of C to D, and in the ratio of E to F. I say next that they are also the least that have this property. For, if H, G, K, L are not the least numbers continuouslyproportional in the ratios of A to B, of C to D, and of E to F, let them be N, O, M, P. Then since, as A is to B, so is N to O, while A, B are least, and the least numbers measure those which have the sameratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; therefore B measures O. [VII. 20] For the same reasonC also measures O; therefore B, C measure O; therefore the least number measured by B, C will also measure O. [VII. 35] But G is the least number measured by B, C;therefore G measures O, the greater the less: which is impossible. Therefore there will be no numbers less than H, G, K, L which are continuously in the ratio of A to B, of C to D, and of E to F. Next, let E not measure K. Let M, the least number measured by E, K, be taken. And, as many times as K measures M, so many times let H, G measure N, O respectively, and, as many times as E measures M, so many times let F  measure P also. Since H measures N the same number of times that G measures O, therefore, as H is to G, so is N to O. [VII. 13 and Def. 20] But, as H is to G, so is A to B;therefore also, as A is to B, so is N to O. For the same reason also, as C is to D, so is O to M. Again, since E measures M the same number of times that F measures P,therefore, as E is to F, so is M to P; [VII. 13 and Def. 20] therefore N, O, M, P are continuously proportional in the ratios of A to B, of C to D, and of E to F. I say next that they are also the least that are in the ratios A: B, C: D, E: F. For, if not, there will be some numbers less than N, O, M, P continuously proportional in the ratios A: B, C: D, E: F. Let them be Q, R, S, T. Now since, as Q is to R, so is A to B, while A, B are least, and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent, [VII. 20] therefore B measures R. For the same reason C also measures R; therefore B, C measure R. Therefore the least number measured by B, C will also measure R. [VII. 35] But G is the least number measured by B, C;therefore G measures R. And, as G is to R, so is K to S: [VII. 13] therefore K also measures S. But E also measures S; therefore E, K measure S. Therefore the least number measured by E, K will also measure S. [VII. 35] But M is the least number measured by E, K; therefore M measures S, the greater the less: which is impossible. Therefore there will not be any numbers less than N, O, M, P continuously proportional in the ratios of A to B, of C to D, and of E to F; therefore N, O, M, P are the least numbers continuously proportional in the ratios A: B, C: D, E: F.'"],["Rule","'ProofWordCount'",836],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες λόγοι ἐν ἐλαχίστοις ἀριθμοῖς ὅ τε τοῦ Α πρὸς τὸν Β καὶ ὁ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι ὁ τοῦ Ε πρὸς τὸν Ζ: δεῖ δὴ ἀριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον ἐλαχίστους ἔν τε τῷ τοῦ Α πρὸς τὸν Β λόγῳ καὶ ἐν τῷ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι ἐν τῷ τοῦ Ε πρὸς τὸν Ζ. εἰλήφθω γὰρ ὁ ὑπὸ τῶν Β, Γ ἐλάχιστος μετρούμενος ἀριθμὸς ὁ Η. καὶ ὁσάκις μὲν ὁ Β τὸν Η μετρεῖ, τοσαυτάκις καὶ ὁ Α τὸν Θ μετρείτω, ὁσάκις δὲ ὁ Γ τὸν Η μετρεῖ, τοσαυτάκις καὶ ὁ Δ τὸν Κ μετρείτω. ὁ δὲ Ε τὸν Κ ἤτοι μετρεῖ á¼¢ οὐ μετρεῖ. μετρείτω πρότερον. καὶ ὁσάκις ὁ Ε τὸν Κ μετρεῖ, τοσαυτάκις καὶ ὁ Ζ τὸν Λ μετρείτω. καὶ ἐπεὶ ἰσάκις ὁ Α τὸν Θ μετρεῖ καὶ ὁ Β τὸν Η, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Θ πρὸς τὸν Η. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Η πρὸς τὸν Κ, καὶ ἔτι ὡς ὁ Ε πρὸς τὸν Ζ, οὕτως ὁ Κ πρὸς τὸν Λ: οἱ Θ, Η, Κ, Λ ἄρα ἑξῆς ἀνάλογόν εἰσιν ἔν τε τῷ τοῦ Α πρὸς τὸν Β καὶ ἐν τῷ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι ἐν τῷ τοῦ Ε πρὸς τὸν Ζ λόγῳ. λέγω δή, ὅτι καὶ ἐλάχιστοι. εἰ γὰρ μή εἰσιν οἱ Θ, Η, Κ, Λ ἑξῆς ἀνάλογον ἐλάχιστοι ἔν τε τοῖς τοῦ Α πρὸς τὸν Β καὶ τοῦ Γ πρὸς τὸν Δ καὶ ἐν τῷ τοῦ Ε πρὸς τὸν Ζ λόγοις, ἔστωσαν οἱ Ν, Ξ, Μ, Ο. καὶ ἐπεί ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Ν πρὸς τὸν Ξ, οἱ δὲ Α, Β ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον, ὁ Β ἄρα τὸν Ξ μετρεῖ. διὰ τὰ αὐτὰ δὴ καὶ ὁ Γ τὸν Ξ μετρεῖ: οἱ Β, Γ ἄρα τὸν Ξ μετροῦσιν: καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Β, Γ μετρούμενος τὸν Ξ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Β, Γ μετρεῖται ὁ Η: ὁ Η ἄρα τὸν Ξ μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσονταί τινες τῶν Θ, Η, Κ, Λ ἐλάσσονες ἀριθμοὶ ἑξῆς ἔν τε τῷ τοῦ Α πρὸς τὸν Β καὶ τῷ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι τῷ τοῦ Ε πρὸς τὸν Ζ λόγῳ. μὴ μετρείτω δὴ ὁ Ε τὸν Κ. καὶ εἰλήφθω ὑπὸ τῶν Ε, Κ ἐλάχιστος μετρούμενος ἀριθμὸς ὁ Μ. καὶ ὁσάκις μὲν ὁ Κ τὸν Μ μετρεῖ, τοσαυτάκις καὶ ἑκάτερος τῶν Θ, Η ἑκάτερον τῶν Ν, Ξ μετρείτω, ὁσάκις δὲ ὁ Ε τὸν Μ μετρεῖ, τοσαυτάκις καὶ ὁ Ζ τὸν Ο μετρείτω. ἐπεὶ ἰσάκις ὁ Θ τὸν Ν μετρεῖ καὶ ὁ Η τὸν Ξ, ἔστιν ἄρα ὡς ὁ Θ πρὸς τὸν Η, οὕτως ὁ Ν πρὸς τὸν Ξ. ὡς δὲ ὁ Θ πρὸς τὸν Η, οὕτως ὁ Α πρὸς τὸν Β: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, οὕτως ὁ Ν πρὸς τὸν ξ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ξ πρὸς τὸν Μ. πάλιν, ἐπεὶ ἰσάκις ὁ Ε τὸν Μ μετρεῖ καὶ ὁ Ζ τὸν Ο, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Ζ, οὕτως ὁ Μ πρὸς τὸν Ο: οἱ Ν, Ξ, Μ, Ο ἄρα ἑξῆς ἀνάλογόν εἰσιν ἐν τοῖς τοῦ τε Α πρὸς τὸν Β καὶ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι τοῦ Ε πρὸς τὸν Ζ λόγοις. λέγω δή, ὅτι καὶ ἐλάχιστοι ἐν τοῖς ΑΒ, ΓΔ, ΕΖ λόγοις. εἰ γὰρ μή, ἔσονταί τινες τῶν Ν, Ξ, Μ, Ο ἐλάσσονες ἀριθμοὶ ἑξῆς ἀνάλογον ἐν τοῖς ΑΒ, ΓΔ, ΕΖ λόγοις. ἔστωσαν οἱ Π, Ρ, Σ, Τ. καὶ ἐπεί ἐστιν ὡς ὁ Π πρὸς τὸν Ρ, οὕτως ὁ Α πρὸς τὸν Β, οἱ δὲ Α, Β ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον, ὁ Β ἄρα τὸν Ρ μετρεῖ. διὰ τὰ αὐτὰ δὴ καὶ ὁ Γ τὸν Ρ μετρεῖ: οἱ Β, Γ ἄρα τὸν Ρ μετροῦσιν. καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Β, Γ μετρούμενος τὸν Ρ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Β, Γ μετρούμενός ἐστιν ὁ Η: ὁ Η ἄρα τὸν Ρ μετρεῖ. καί ἐστιν ὡς ὁ Η πρὸς τὸν Ρ, οὕτως ὁ Κ πρὸς τὸν Σ: καὶ ὁ Κ ἄρα τὸν Σ μετρεῖ. μετρεῖ δὲ καὶ ὁ Ε τὸν Σ: οἱ Ε, Κ ἄρα τὸν Σ μετροῦσιν. καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Ε, Κ μετρούμενος τὸν Σ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Ε, Κ μετρούμενός ἐστιν ὁ Μ: ὁ Μ ἄρα τὸν Σ μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσονταί τινες τῶν Ν, Ξ, Μ, Ο ἐλάσσονες ἀριθμοὶ ἑξῆς ἀνάλογον ἔν τε τοῖς τοῦ Α πρὸς τὸν Β καὶ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι τοῦ Ε πρὸς τὸν Ζ λόγοις: οἱ Ν, Ξ, Μ, Ο ἄρα ἑξῆς ἀνάλογον ἐλάχιστοί εἰσιν ἐν τοῖς ΑΒ, ΓΔ, ΕΖ λόγοις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",844]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'8.5'"],["Rule","'Text'","'Plane numbers have to one another the ratio compounded of the ratios of their sides.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'οἱ ἐπίπεδοι ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχουσι τὸν συγκείμενον ἐκ τῶν πλευρῶν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",14]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",8],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let A, B be plane numbers, and let the numbers C, D be the sides of A, and E, F of B;I say that A has to B the ratio compounded of the ratios of the sides. For, the ratios being given which C has to E and D to F, let the least numbers G, H, K that are continuouslyin the ratios C: E, D: F be taken, so that, as C is to E, so is G to H, and, as D is to F, so is H to K. [VIII. 4] And let D by multiplying E make L. Now, since D by multiplying C has made A, and by multiplying E has made L, therefore, as C is to E, so is A to L. [VII. 17] But, as C is to E, so is G to H; therefore also, as G is to H, so is A to L. Again, since E by multiplying D has made L, and further by multiplying F has made B, therefore, as D is to F, so is L to B. [VII. 17] But, as D is to F, so is H to K; therefore also, as H is to K, so is L to B. But it was also proved that, as G is to H, so is A to L; therefore, ex aequali, as G is to K, so is A to B. [VII. 14]  But G has to K the ratio compounded of the ratios of thesides; therefore A also has to B the ratio compounded of the ratios of the sides.'"],["Rule","'ProofWordCount'",267],["Rule","'GreekProof'","'ἔστωσαν ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ τοῦ μὲν Α πλευραὶ ἔστωσαν οἱ Γ, Δ ἀριθμοί, τοῦ δὲ Β οἱ Ε, Ζ: λέγω, ὅτι ὁ Α πρὸς τὸν Β λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν. λόγων γὰρ δοθέντων τοῦ τε ὃν ἔχει ὁ Γ πρὸς τὸν Ε καὶ ὁ Δ πρὸς τὸν Ζ εἰλήφθωσαν ἀριθμοὶ ἑξῆς ἐλάχιστοι ἐν τοῖς ΓΕ, ΔΖ λόγοις, οἱ Η, Θ, Κ, ὥστε εἶναι ὡς μὲν τὸν Γ πρὸς τὸν Ε, οὕτως τὸν Η πρὸς τὸν θ, ὡς δὲ τὸν Δ πρὸς τὸν Ζ, οὕτως τὸν Θ πρὸς τὸν Κ. καὶ ὁ Δ τὸν Ε πολλαπλασιάσας τὸν Λ ποιείτω. καὶ ἐπεὶ ὁ Δ τὸν μὲν Γ πολλαπλασιάσας τὸν Α πεποίηκεν, τὸν δὲ Ε πολλαπλασιάσας τὸν Λ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν Λ. ὡς δὲ ὁ Γ πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν Θ: καὶ ὡς ἄρα ὁ Η πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Λ. πάλιν, ἐπεὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν Λ πεποίηκεν, ἀλλὰ μὴν καὶ τὸν Ζ πολλαπλασιάσας τὸν Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Λ πρὸς τὸν Β. ἀλλ᾽ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Θ πρὸς τὸν Κ: καὶ ὡς ἄρα ὁ Θ πρὸς τὸν Κ, οὕτως ὁ Λ πρὸς τὸν Β. ἐδείχθη δὲ καὶ ὡς ὁ Η πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Λ: δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Η πρὸς τὸν Κ, οὕτως ὁ Α πρὸς τὸν Β, ὁ δὲ Η πρὸς τὸν Κ λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν: καὶ ὁ Α ἄρα πρὸς τὸν Β λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",286]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'8.6'"],["Rule","'Text'","'If there be as many numbers as we please in continued proportion, and the first do not measure the second, neither will any other measure any other.'"],["Rule","'TextWordCount'",27],["Rule","'GreekText'","'ἐὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, ὁ δὲ πρῶτος τὸν δεύτερον μὴ μετρῇ, οὐδὲ ἄλλος οὐδεὶς οὐδένα μετρήσει.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",14]],["List",["Rule","'Book'",7],["Rule","'Theorem'",33]],["List",["Rule","'Book'",8],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'Let there be as many numbers as we please, A, B, C, D, E, in continued proportion, and let A not measure B; I say that neither will any other measure any other. Now it is manifest that A, B, C, D, E do not measure one another in order; for A does not even measure B. I say, then, that neither will any other measure any other. For, if possible, let A measure C. And, however many A, B, C are, let as many numbers F, G, H, the least of those which have the same ratio with A, B, C, be taken. [VII. 33] Now, since F, G, H are in the same ratio with A, B, C, and the multitude of the numbers A, B, C is equal to the multitude of the numbers F, G, H, therefore, ex aequali, as A is to C, so is F to H. [VII. 14] And since, as A is to B, so is F to G, while A does not measure B, therefore neither does F measure G; [VII. Def. 20] therefore F is not an unit, for the unit measures any number. Now F, H are prime to one another. [VIII. 3] And, as F is to H, so is A to C; therefore neither does A measure C. Similarly we can prove that neither will any other measure any other.'"],["Rule","'ProofWordCount'",233],["Rule","'GreekProof'","'ἔστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, δ, Ε, ὁ δὲ Α τὸν Β μὴ μετρείτω: λέγω, ὅτι οὐδὲ ἄλλος οὐδεὶς οὐδένα μετρήσει. ὅτι μὲν οὖν οἱ Α, Β, Γ, Δ, Ε ἑξῆς ἀλλήλους οὐ μετροῦσιν, φανερόν: οὐδὲ γὰρ ὁ Α τὸν Β μετρεῖ. λέγω δή, ὅτι οὐδὲ ἄλλος οὐδεὶς οὐδένα μετρήσει. εἰ γὰρ δυνατόν, μετρείτω ὁ Α τὸν Γ. καὶ ὅσοι εἰσὶν οἱ Α, Β, Γ, τοσοῦτοι εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ οἱ Ζ, Η, Θ. καὶ ἐπεὶ οἱ Ζ, Η, Θ ἐν τῷ αὐτῷ λόγῳ εἰσὶ τοῖς Α, Β, Γ, καί ἐστιν ἴσον τὸ πλῆθος τῶν Α, Β, Γ τῷ πλήθει τῶν Ζ, Η, Θ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Ζ πρὸς τὸν Θ. καὶ ἐπεί ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Ζ πρὸς τὸν Η, οὐ μετρεῖ δὲ ὁ Α τὸν Β, οὐ μετρεῖ ἄρα οὐδὲ ὁ Ζ τὸν Η: οὐκ ἄρα μονάς ἐστιν ὁ Ζ: ἡ γὰρ μονὰς πάντα ἀριθμὸν μετρεῖ. καί εἰσιν οἱ Ζ, Θ πρῶτοι πρὸς ἀλλήλους οὐδὲ ὁ Ζ ἄρα τὸν Θ μετρεῖ. καί ἐστιν ὡς ὁ Ζ πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Γ: οὐδὲ ὁ Α ἄρα τὸν Γ μετρεῖ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἄλλος οὐδεὶς οὐδένα μετρήσει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",223]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'8.7'"],["Rule","'Text'","'If there be as many numbers as we please in continued proportion, and the first measure the last, it will measure the second also.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'ἐὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, ὁ δὲ πρῶτος τὸν ἔσχατον μετρῇ, καὶ τὸν δεύτερον μετρήσει.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'Let there be as many numbers as we please, A, B, C, D, in continued proportion; and let A measure D; I say that A also measures B. For, if A does not measure B, neither will any other of the numbers measure any other. [VIII. 6] But A measures D. Therefore A also measures B.'"],["Rule","'ProofWordCount'",56],["Rule","'GreekProof'","'ἔστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, ὁ δὲ Α τὸν Δ μετρείτω: λέγω, ὅτι καὶ ὁ Α τὸν Β μετρεῖ. εἰ γὰρ οὐ μετρεῖ ὁ Α τὸν Β, οὐδὲ ἄλλος οὐδεὶς οὐδένα μετρήσει: μετρεῖ δὲ ὁ Α τὸν Δ. μετρεῖ ἄρα καὶ ὁ Α τὸν Β: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",53]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'8.8'"],["Rule","'Text'","'If between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many will also fall in continued proportion between the numbers which have the same ratio with the original numbers.'"],["Rule","'TextWordCount'",42],["Rule","'GreekText'","'ἐὰν δύο ἀριθμῶν μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν ἀριθμοί, ὅσοι εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται.'"],["Rule","'GreekTextWordCount'",35],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",14]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",33]],["List",["Rule","'Book'",8],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'Let the numbers C, D fall between the two numbers A, B in continued proportion with them, and let E be made in the same ratio to F as A is to B; I say that, as many numbers as have fallen between A, B in continued proportion, so many will also fall between E, F in continued proportion. For, as many as A, B, C, D are in multitude, let so many numbers G, H, K, L, the least of those which have the same ratio with A, C, D, B, be taken; [VII. 33] therefore the extremes of them G, L are prime to one another. [VIII. 3] Now, since A, C, D, B are in the same ratio with G, H, K, L, and the multitude of the numbers A, C, D, B is equal to the multitude of the numbers G, H, K, L, therefore, ex aequali, as A is to B, so is G to L. [VII. 14] But, as A is to B, so is E to F; therefore also, as G is to L, so is E to F. But G, L are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20] Therefore G measures E the same number of times as L measures F. Next, as many times as G measures E, so many times let H, K also measure M, N respectively; therefore G, H, K, L measure E, M, N, F the same number of times. Therefore G, H, K, L are in the same ratio with E, M, N, F. [VII. Def. 20] But G, H, K, L are in the same ratio with A, C, D, B; therefore A, C, D, B are also in the same ratio with E, M, N, F. But A, C, D, B are in continued proportion; therefore E, M, N, F are also in continued proportion. Therefore, as many numbers as have fallen between A, B in continued proportion with them, so many numbers have also fallen between E, F in continued proportion.'"],["Rule","'ProofWordCount'",377],["Rule","'GreekProof'","'δύο γὰρ ἀριθμῶν τῶν Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπιπτέτωσαν ἀριθμοὶ οἱ Γ, Δ, καὶ πεποιήσθω ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Ε πρὸς τὸν Ζ: λέγω, ὅτι ὅσοι εἰς τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς Ε, Ζ μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται. ὅσοι γάρ εἰσι τῷ πλήθει οἱ Α, Β, Γ, Δ, τοσοῦτοι εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Γ, Δ, Β οἱ Η, Θ, Κ, Λ: οἱ ἄρα ἄκροι αὐτῶν οἱ Η, Λ πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ οἱ Α, Γ, Δ, Β τοῖς Η, Θ, Κ, Λ ἐν τῷ αὐτῷ λόγῳ εἰσίν, καί ἐστιν ἴσον τὸ πλῆθος τῶν Α, Γ, Δ, Β τῷ πλήθει τῶν Η, Θ, Κ, Λ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Λ. ὡς δὲ ὁ Α πρὸς τὸν Β, οὕτως ὁ Ε πρὸς τὸν Ζ: καὶ ὡς ἄρα ὁ Η πρὸς τὸν Λ, οὕτως ὁ Ε πρὸς τὸν Ζ. οἱ δὲ Η, Λ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον. ἰσάκις ἄρα ὁ Η τὸν Ε μετρεῖ καὶ ὁ Λ τὸν Ζ. ὁσάκις δὴ ὁ Η τὸν Ε μετρεῖ, τοσαυτάκις καὶ ἑκάτερος τῶν Θ, Κ ἑκάτερον τῶν Μ, Ν μετρείτω: οἱ Η, Θ, Κ, Λ ἄρα τοὺς Ε, Μ, Ν, Ζ ἰσάκις μετροῦσιν. οἱ Η, Θ, Κ, Λ ἄρα τοῖς Ε, Μ, Ν, Ζ ἐν τῷ αὐτῷ λόγῳ εἰσίν. ἀλλὰ οἱ Η, Θ, Κ, Λ τοῖς Α, Γ, Δ, Β ἐν τῷ αὐτῷ λόγῳ εἰσίν: καὶ οἱ Α, Γ, Δ, Β ἄρα τοῖς Ε, Μ, Ν, Ζ ἐν τῷ αὐτῷ λόγῳ εἰσίν. οἱ δὲ Α, Γ, Δ, Β ἑξῆς ἀνάλογόν εἰσιν: καὶ οἱ Ε, Μ, Ν, Ζ ἄρα ἑξῆς ἀνάλογόν εἰσιν. ὅσοι ἄρα εἰς τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς Ε, Ζ μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",356]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'8.9'"],["Rule","'Text'","'If two numbers be prime to one another, and numbers fall between them in continued proportion, then, however many numbers fall between them in continued proportion, so many will also fall between each of them and an unit in continued proportion.'"],["Rule","'TextWordCount'",41],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν ἀριθμοί, ὅσοι εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ ἑκατέρου αὐτῶν καὶ μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται.'"],["Rule","'GreekTextWordCount'",39],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",8],["Rule","'Theorem'",1]],["List",["Rule","'Book'",8],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let A, B be two numbers prime to one another, and let C, D fall between them in continued proportion, and let the unit E be set out; I say that, as many numbers as fall between A, B in continued proportion, so many will also fall between either of the numbers A, B and the unit in continued proportion. For let two numbers F, G, the least that are in the ratio of A, C, D, B, be taken, three numbers H, K, L with the same property, and others more by one continually, until their multitude is equal to the multitude of A, C, D, B. [VIII. 2] Let them be taken, and let them be M, N, O, P. It is now manifest that F by multiplying itself has made H and by multiplying H has made M, while G by multiplying itself has made L and by multiplying L has made P. [VIII. 2] And, since M, N, O, P are the least of those which have the same ratio with F, G, and A, C, D, B are also the least of those which have the same ratio with F, G, [VIII. 1] while the multitude of the numbers M, N, O, P is equal to the multitude of the numbers A, C, D, B, therefore M, N, O, P are equal to A, C, D, B respectively; therefore M is equal to A, and P to B. Now, since F by multiplying itself has made H, therefore F measures H according to the units in F. But the unit E also measures F according to the units in it; therefore the unit E measures the number F the same number of times as F measures H. Therefore, as the unit E is to the number F, so is F to H. [VII. Def. 20] Again, since F by multiplying H has made M, therefore H measures M according to the units in F. But the unit E also measures the number F according to the units in it; therefore the unit E measures the number F the same number of times as H measures M. Therefore, as the unit E is to the number F, so is H to M. But it was also proved that, as the unit E is to the number F, so is F to H; therefore also, as the unit E is to the number F, so is F to H, and H to M. But M is equal to A; therefore, as the unit E is to the number F, so is F to H, and H to A. For the same reason also, as the unit E is to the number G, so is G to L and L to B. Therefore, as many numbers as have fallen between A, B in continued proportion, so many numbers also have fallen between each of the numbers A, B and the unit E in continued proportion.'"],["Rule","'ProofWordCount'",497],["Rule","'GreekProof'","'ἔστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β καὶ εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπιπτέτωσαν οἱ Γ, Δ, καὶ ἐκκείσθω ἡ Ε μονάς: λέγω, ὅτι ὅσοι εἰς τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί, τοσοῦτοι καὶ ἑκατέρου τῶν Α, Β καὶ τῆς μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται. εἰλήφθωσαν γὰρ δύο μὲν ἀριθμοὶ ἐλάχιστοι ἐν τῷ τῶν Α, Γ, Δ, Β λόγῳ ὄντες οἱ Ζ, Η, τρεῖς δὲ οἱ Θ, Κ, Λ, καὶ ἀεὶ ἑξῆς ἑνὶ πλείους, ἕως ἂν ἴσον γένηται τὸ πλῆθος αὐτῶν τῷ πλήθει τῶν Α, Γ, Δ, Β. εἰλήφθωσαν, καὶ ἔστωσαν οἱ Μ, Ν, Ξ, Ο. φανερὸν δή, ὅτι ὁ μὲν Ζ ἑαυτὸν πολλαπλασιάσας τὸν Θ πεποίηκεν, τὸν δὲ Θ πολλαπλασιάσας τὸν Μ πεποίηκεν, καὶ ὁ Η ἑαυτὸν μὲν πολλαπλασιάσας τὸν Λ πεποίηκεν, τὸν δὲ Λ πολλαπλασιάσας τὸν Ο πεποίηκεν. καὶ ἐπεὶ οἱ Μ, Ν, Ξ, Ο ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Ζ, Η, εἰσὶ δὲ καὶ οἱ Α, Γ, Δ, Β ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Ζ, Η, καί ἐστιν ἴσον τὸ πλῆθος τῶν Μ, Ν, Ξ, Ο τῷ πλήθει τῶν Α, Γ, Δ, Β, ἕκαστος ἄρα τῶν Μ, Ν, Ξ, Ο ἑκάστῳ τῶν Α, Γ, Δ, Β ἴσος ἐστίν: ἴσος ἄρα ἐστὶν ὁ μὲν Μ τῷ Α, ὁ δὲ Ο τῷ Β. καὶ ἐπεὶ ὁ Ζ ἑαυτὸν πολλαπλασιάσας τὸν Θ πεποίηκεν, ὁ Ζ ἄρα τὸν Θ μετρεῖ κατὰ τὰς ἐν τῷ Ζ μονάδας. μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Ζ κατὰ τὰς ἐν αὐτῷ μονάδας: ἰσάκις ἄρα ἡ Ε μονὰς τὸν Ζ ἀριθμὸν μετρεῖ καὶ ὁ Ζ τὸν Θ. ἔστιν ἄρα ὡς ἡ Ε μονὰς πρὸς τὸν Ζ ἀριθμόν, οὕτως ὁ Ζ πρὸς τὸν Θ. πάλιν, ἐπεὶ ὁ Ζ τὸν Θ πολλαπλασιάσας τὸν Μ πεποίηκεν, ὁ Θ ἄρα τὸν Μ μετρεῖ κατὰ τὰς ἐν τῷ Ζ μονάδας. μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Ζ ἀριθμὸν κατὰ τὰς ἐν αὐτῷ μονάδας: ἰσάκις ἄρα ἡ Ε μονὰς τὸν Ζ ἀριθμὸν μετρεῖ καὶ ὁ Θ τὸν Μ. ἔστιν ἄρα ὡς ἡ Ε μονὰς πρὸς τὸν Ζ ἀριθμόν, οὕτως ὁ Θ πρὸς τὸν Μ. ἐδείχθη δὲ καὶ ὡς ἡ Ε μονὰς πρὸς τὸν Ζ ἀριθμόν, οὕτως ὁ Ζ πρὸς τὸν Θ: καὶ ὡς ἄρα ἡ Ε μονὰς πρὸς τὸν Ζ ἀριθμόν, οὕτως ὁ Ζ πρὸς τὸν Θ καὶ ὁ Θ πρὸς τὸν Μ. ἴσος δὲ ὁ Μ τῷ Α: ἔστιν ἄρα ὡς ἡ Ε μονὰς πρὸς τὸν Ζ ἀριθμόν, οὕτως ὁ Ζ πρὸς τὸν Θ καὶ ὁ Θ πρὸς τὸν Α. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ Ε μονὰς πρὸς τὸν Η ἀριθμόν, οὕτως ὁ Η πρὸς τὸν Λ καὶ ὁ Λ πρὸς τὸν Β. ὅσοι ἄρα εἰς τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί, τοσοῦτοι καὶ ἑκατέρου τῶν Α, Β καὶ μονάδος τῆς Ε μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",474]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'8.10'"],["Rule","'Text'","'If numbers fall between each of two numbers and an unit in continued proportion, however many numbers fall between each of them and an unit in continued proportion, so many also will fall between the numbers themselves in continued proportion.'"],["Rule","'TextWordCount'",40],["Rule","'GreekText'","'ἐὰν δύο ἀριθμῶν ἑκατέρου καὶ μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν ἀριθμοί, ὅσοι ἑκατέρου αὐτῶν καὶ μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται.'"],["Rule","'GreekTextWordCount'",35],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",7],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'For let the numbers D, E and F, G respectively fall between the two numbers A, B and the unit C in continued proportion; I say that, as many numbers as have fallen between each of the numbers A, B and the unit C in continued proportion, so many numbers will also fall between A, B in continued proportion. For let D by multiplying F make H, and let the numbers D, F by multiplying H make K, L respectively. Now, since, as the unit C is to the number D, so is D to E, therefore the unit C measures the number D the same number of times as D measures E. [VII. Def. 20] But the unit C measures the number D according to the units in D; therefore the number D also measures E according to the units in D; therefore D by multiplying itself has made E. Again, since, as C is to the number D, so is E to A, therefore the unit C measures the number D the same number of times as E measures A. But the unit C measures the number D according to the units in D; therefore E also measures A according to the units in D; therefore D by multiplying E has made A. For the same reason also F by multiplying itself has made G, and by multiplying G has made B. And, since D by multiplying itself has made E and by multiplying F has made H, therefore, as D is to F, so is E to H. [VII. 17] For the same reason also, as D is to F, so is H to G. [VII. 18]  Therefore also, as E is to H, so is H to G. Again, since D by multiplying the numbers E, H has made A, K respectively, therefore, as E is to H, so is A to K. [VII. 17] But, as E is to H, so is D to F; therefore also, as D is to F, so is A to K. Again, since the numbers D, F by multiplying H have made K, L respectively, therefore, as D is to F, so is K to L. [VII. 18] But, as D is to F, so is A to K; therefore also, as A is to K, so is K to L. Further, since F by multiplying the numbers H, G has made L, B respectively, therefore, as H is to G, so is L to B. [VII. 17] But, as H is to G, so is D to F; therefore also, as D is to F, so is L to B. But it was also proved that, as D is to F, so is A to K and K to L; therefore also, as A is to K, so is K to L and L to B. Therefore A, K, L, B are in continued proportion. Therefore, as many numbers as fall between each of the numbers A, B and the unit C in continued proportion, so many also will fall between A, B in continued proportion.'"],["Rule","'ProofWordCount'",518],["Rule","'GreekProof'","'δύο γὰρ ἀριθμῶν τῶν Α, Β καὶ μονάδος τῆς Γ μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπιπτέτωσαν ἀριθμοὶ οἵ τε Δ, Ε καὶ οἱ Ζ, η: λέγω, ὅτι ὅσοι ἑκατέρου τῶν Α, Β καὶ μονάδος τῆς Γ μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται. ὁ Δ γὰρ τὸν Ζ πολλαπλασιάσας τὸν Θ ποιείτω, ἑκάτερος δὲ τῶν Δ, Ζ τὸν Θ πολλαπλασιάσας ἑκάτερον τῶν Κ, Λ ποιείτω. καὶ ἐπεί ἐστιν ὡς ἡ Γ μονὰς πρὸς τὸν Δ ἀριθμόν, οὕτως ὁ Δ πρὸς τὸν Ε, ἰσάκις ἄρα ἡ Γ μονὰς τὸν Δ ἀριθμὸν μετρεῖ καὶ ὁ Δ τὸν Ε. ἡ δὲ Γ μονὰς τὸν Δ ἀριθμὸν μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας: καὶ ὁ Δ ἄρα ἀριθμὸς τὸν Ε μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας: ὁ Δ ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν. πάλιν, ἐπεί ἐστιν ὡς ἡ Γ μονὰς πρὸς τὸν Δ ἀριθμὸν, οὕτως ὁ Ε πρὸς τὸν Α, ἰσάκις ἄρα ἡ Γ μονὰς τὸν Δ ἀριθμὸν μετρεῖ καὶ ὁ Ε τὸν Α. ἡ δὲ Γ μονὰς τὸν Δ ἀριθμὸν μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας: καὶ ὁ Ε ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας: ὁ Δ ἄρα τὸν Ε πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ μὲν Ζ ἑαυτὸν πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ Η πολλαπλασιάσας τὸν Β πεποίηκεν. καὶ ἐπεὶ ὁ Δ ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε πεποίηκεν, τὸν δὲ Ζ πολλαπλασιάσας τὸν Θ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Ε πρὸς τὸν Θ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Θ πρὸς τὸν Η. καὶ ὡς ἄρα ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Θ πρὸς τὸν Η. πάλιν, ἐπεὶ ὁ Δ ἑκάτερον τῶν Ε, Θ πολλαπλασιάσας ἑκάτερον τῶν Α, Κ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Κ. ἀλλ᾽ ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Δ πρὸς τὸν Ζ: καὶ ὡς ἄρα ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Κ. πάλιν, ἐπεὶ ἑκάτερος τῶν Δ, Ζ τὸν Θ πολλαπλασιάσας ἑκάτερον τῶν Κ, Λ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Κ πρὸς τὸν Λ. ἀλλ᾽ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Κ: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Κ, οὕτως ὁ Κ πρὸς τὸν Λ. ἔτι ἐπεὶ ὁ Ζ ἑκάτερον τῶν Θ, Η πολλαπλασιάσας ἑκάτερον τῶν Λ, Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Θ πρὸς τὸν Η, οὕτως ὁ Λ πρὸς τὸν Β. ὡς δὲ ὁ Θ πρὸς τὸν Η, οὕτως ὁ Δ πρὸς τὸν Ζ: καὶ ὡς ἄρα ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Λ πρὸς τὸν Β. ἐδείχθη δὲ καὶ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὅ τε Α πρὸς τὸν Κ καὶ ὁ Κ πρὸς τὸν Λ: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Κ, οὕτως ὁ Κ πρὸς τὸν Λ καὶ ὁ Λ πρὸς τὸν Β. οἱ Α, Κ, Λ, Β ἄρα κατὰ τὸ συνεχὲς ἑξῆς εἰσιν ἀνάλογον. ὅσοι ἄρα ἑκατέρου τῶν Α, Β καὶ τῆς Γ μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἐμπεσοῦνται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",545]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'8.11'"],["Rule","'Text'","'Between two square numbers there is one mean proportional number, and the square has to the square the ratio duplicate of that which the side has to the side.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'δύο τετραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καὶ ὁ τετράγωνος πρὸς τὸν τετράγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",9]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",7],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let A, B be square numbers, and let C be the side of A, and D of B; I say that between A, B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to D. For let C by multiplying D make E. Now, since A is a square and C is its side, therefore C by multiplying itself has made A. For the same reason also D by multiplying itself has made B. Since then C by multiplying the numbers C, D has made A, E respectively, therefore, as C is to D, so is A to E. [VII. 17] For the same reason also, as C is to D, so is E to B. [VII. 18]  Therefore also, as A is to E, so is E to B. Therefore between A, B there is one mean proportional number. I say next that A also has to B the ratio duplicate of that which C has to D. For, since A, E, B are three numbers in proportion, therefore A has to B the ratio duplicate of that which A has to E. [V. Def. 9] But, as A is to E, so is C to D. Therefore A has to B the ratio duplicate of that which the side C has to D.'"],["Rule","'ProofWordCount'",225],["Rule","'GreekProof'","'ἔστωσαν τετράγωνοι ἀριθμοὶ οἱ Α, Β, καὶ τοῦ μὲν Α πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ: λέγω, ὅτι τῶν Α, Β εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. ὁ Γ γὰρ τὸν Δ πολλαπλασιάσας τὸν Ε ποιείτω. καὶ ἐπεὶ τετράγωνός ἐστιν ὁ Α, πλευρὰ δὲ αὐτοῦ ἐστιν ὁ Γ, ὁ Γ ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Δ ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν. ἐπεὶ οὖν ὁ Γ ἑκάτερον τῶν Γ, Δ πολλαπλασιάσας ἑκάτερον τῶν Α, Ε πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Ε. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν Β. καὶ ὡς ἄρα ὁ Α πρὸς τὸν Ε, οὕτως ὁ Ε πρὸς τὸν Β. τῶν Α, Β ἄρα εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός. λέγω δή, ὅτι καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. ἐπεὶ γὰρ τρεῖς ἀριθμοὶ ἀνάλογόν εἰσιν οἱ Α, Ε, Β, ὁ Α ἄρα πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Α πρὸς τὸν Ε. ὡς δὲ ὁ Α πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Δ. ὁ Α ἄρα πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ἡ Γ πλευρὰ πρὸς τὴν Δ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",228]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'8.12'"],["Rule","'Text'","'Between two cube numbers there are two mean proportional numbers, and the cube has to the cube the ratio triplicate of that which the side has to the side.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'δύο κύβων ἀριθμῶν δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί, καὶ ὁ κύβος πρὸς τὸν κύβον τριπλασίονα λόγον ἔχει ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",10]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",7],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let A, B be cube numbers, and let C be the side of A, and D of B; I say that between A, B there are two mean proportional numbers, and A has to B the ratio triplicate of that which C has to D. For let C by multiplying itself make E, and by multiplying D let it make F; let D by multiplying itself make G, and let the numbers C, D by multiplying F make H, K respectively. Now, since A is a cube, and C its side, and C by multiplying itself has made E, therefore C by multiplying itself has made E and by multiplying E has made A. For the same reason also D by multiplying itself has made G and by multiplying G has made B. And, since C by multiplying the numbers C, D has made E, F respectively, therefore, as C is to D, so is E to F. [VII. 17] For the same reason also, as C is to D, so is F to G. [VII. 18]  Again, since C by multiplying the numbers E, F has made A, H respectively, therefore, as E is to F, so is A to H. [VII. 17] But, as E is to F, so is C to D. Therefore also, as C is to D, so is A to H. Again, since the numbers C, D by multiplying F have made H, K respectively, therefore, as C is to D, so is H to K. [VII. 18] Again, since D by multiplying each of the numbers F, G has made K, B respectively, therefore, as F is to G, so is K to B. [VII. 17] But, as F is to G, so is C to D; therefore also, as C is to D, so is A to H, H to K, and K to B. Therefore H, K are two mean proportionals between A, B. I say next that A also has to B the ratio triplicate of that which C has to D. For, since A, H, K, B are four numbers in proportion, therefore A has to B the ratio triplicate of that which A has to H. [V. Def. 10] But, as A is to H, so is C to D; therefore A also has to B the ratio triplicate of that which C has to D.'"],["Rule","'ProofWordCount'",398],["Rule","'GreekProof'","'ἔστωσαν κύβοι ἀριθμοὶ οἱ Α, Β καὶ τοῦ μὲν Α πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ: λέγω, ὅτι τῶν Α, Β δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί, καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. ὁ γὰρ Γ ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε ποιείτω, τὸν δὲ Δ πολλαπλασιάσας τὸν Ζ ποιείτω, ὁ δὲ Δ ἑαυτὸν πολλαπλασιάσας τὸν Η ποιείτω, ἑκάτερος δὲ τῶν Γ, Δ τὸν Ζ πολλαπλασιάσας ἑκάτερον τῶν Θ, Κ ποιείτω. καὶ ἐπεὶ κύβος ἐστὶν ὁ Α, πλευρὰ δὲ αὐτοῦ ὁ Γ, καὶ ὁ Γ ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, ὁ Γ ἄρα ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε πεποίηκεν, τὸν δὲ Ε πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Δ ἑαυτὸν μὲν πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ Η πολλαπλασιάσας τὸν Β πεποίηκεν. καὶ ἐπεὶ ὁ Γ ἑκάτερον τῶν Γ, Δ πολλαπλασιάσας ἑκάτερον τῶν Ε, Ζ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ζ πρὸς τὸν Η. πάλιν, ἐπεὶ ὁ Γ ἑκάτερον τῶν Ε, Ζ πολλαπλασιάσας ἑκάτερον τῶν Α, Θ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Θ. ὡς δὲ ὁ Ε πρὸς τὸν Ζ, οὕτως ὁ Γ πρὸς τὸν Δ: καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Θ. πάλιν, ἐπεὶ ἑκάτερος τῶν Γ, Δ τὸν Ζ πολλαπλασιάσας ἑκάτερον τῶν Θ, Κ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Θ πρὸς τὸν Κ. πάλιν, ἐπεὶ ὁ Δ ἑκάτερον τῶν Ζ, Η πολλαπλασιάσας ἑκάτερον τῶν Κ, Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ζ πρὸς τὸν Η, οὕτως ὁ Κ πρὸς τὸν Β. ὡς δὲ ὁ Ζ πρὸς τὸν Η, οὕτως ὁ Γ πρὸς τὸν Δ: καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Δ, οὕτως ὅ τε Α πρὸς τὸν Θ καὶ ὁ Θ πρὸς τὸν Κ καὶ ὁ Κ πρὸς τὸν Β. τῶν Α, Β ἄρα δύο μέσοι ἀνάλογόν εἰσιν οἱ Θ, Κ. λέγω δή, ὅτι καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. ἐπεὶ γὰρ τέσσαρες ἀριθμοὶ ἀνάλογόν εἰσιν οἱ Α, Θ, Κ, Β, ὁ Α ἄρα πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Α πρὸς τὸν Θ. ὡς δὲ ὁ Α πρὸς τὸν Θ, οὕτως ὁ Γ πρὸς τὸν Δ: καὶ ὁ Α ἄρα πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",419]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'8.13'"],["Rule","'Text'","'If there be as many numbers as we please in continued proportion, and each by multiplying itself make some number, the products will be proportional; and, if the original numbers by multiplying the products make certain numbers, the latter will also be proportional.'"],["Rule","'TextWordCount'",43],["Rule","'GreekText'","'ἐὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, καὶ πολλαπλασιάσας ἕκαστος ἑαυτὸν ποιῇ τινα, οἱ γενόμενοι ἐξ αὐτῶν ἀνάλογον ἔσονται: καὶ ἐὰν οἱ ἐξ ἀρχῆς τοὺς γενομένους πολλαπλασιάσαντες ποιῶσί τινας, καὶ αὐτοὶ ἀνάλογον ἔσονται καὶ ἀεὶ περὶ τοὺς ἄκρους τοῦτο συμβαίνει.'"],["Rule","'GreekTextWordCount'",39],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",14]]]],["Rule","'Proof'","'Let there be as many numbers as we please, A, B, C, in continued proportion, so that, as A is to B, so is B to C; let A, B, C by multiplying themselves make D, E, F, and by multiplying D, E, F let them make G, H, K; I say that D, E, F and G, H, K are in continued proportion. For let A by multiplying B make L, and let the numbers A, B by multiplying L make M. N respectively. And again let B by multiplying C make O, and let the numbers B, C by multiplying O make P, Q respectively. Then, in manner similar to the foregoing, we can prove that D, L, E and G, M, N, H are continuously proportional in the ratio of A to B, and further E, O, F and H, P, Q, K are continuously proportional in the ratio of B to C. Now, as A is to B, so is B to C; therefore D, L, E are also in the same ratio with E, O, F, and further G, M, N, H in the same ratio with H, P, Q, K. And the multitude of D, L, E is equal to the multitude of E, O, F, and that of G, M, N, H to that of H, P, Q, K; therefore, ex acquali, as D is to E, so is E to F, and, as G is to H, so is H to K. [VII. 14]'"],["Rule","'ProofWordCount'",252],["Rule","'GreekProof'","'ἔστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ Α, Β, Γ, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β πρὸς τὸν Γ, καὶ οἱ Α, Β, Γ ἑαυτοὺς μὲν πολλαπλασιάσαντες τοὺς Δ, Ε, Ζ ποιείτωσαν, τοὺς δὲ Δ, Ε, Ζ πολλαπλασιάσαντες τοὺς Η, Θ, Κ ποιείτωσαν: λέγω, ὅτι οἵ τε Δ, Ε, Ζ καὶ οἱ Η, Θ, Κ ἑξῆς ἀνάλογόν εἰσιν. ὁ μὲν γὰρ Α τὸν Β πολλαπλασιάσας τὸν Λ ποιείτω, ἑκάτερος δὲ τῶν Α, Β τὸν Λ πολλαπλασιάσας ἑκάτερον τῶν Μ, Ν ποιείτω. καὶ πάλιν ὁ μὲν Β τὸν Γ πολλαπλασιάσας τὸν Ξ ποιείτω, ἑκάτερος δὲ τῶν Β, Γ τὸν Ξ πολλαπλασιάσας ἑκάτερον τῶν Ο, Π ποιείτω. ὁμοίως δὴ τοῖς ἐπάνω δείξομεν, ὅτι οἱ Δ, Λ, Ε καὶ οἱ Η, Μ, Ν, Θ ἑξῆς εἰσιν ἀνάλογον ἐν τῷ τοῦ Α πρὸς τὸν Β λόγῳ, καὶ ἔτι οἱ Ε, Ξ, Ζ καὶ οἱ Θ, Ο, Π, Κ ἑξῆς εἰσιν ἀνάλογον ἐν τῷ τοῦ Β πρὸς τὸν Γ λόγῳ. καί ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β πρὸς τὸν Γ: καὶ οἱ Δ, Λ, Ε ἄρα τοῖς Ε, Ξ, Ζ ἐν τῷ αὐτῷ λόγῳ εἰσὶ καὶ ἔτι οἱ Η, Μ, Ν, Θ τοῖς Θ, Ο, Π, Κ. καί ἐστιν ἴσον τὸ μὲν τῶν Δ, Λ, Ε πλῆθος τῷ τῶν Ε, Ξ, Ζ πλήθει, τὸ δὲ τῶν Η, Μ, Ν, Θ τῷ τῶν Θ, Ο, Π, Κ: δι᾽ ἴσου ἄρα ἐστὶν ὡς μὲν ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Ε πρὸς τὸν Ζ, ὡς δὲ ὁ Η πρὸς τὸν Θ, οὕτως ὁ Θ πρὸς τὸν Κ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",261]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'8.14'"],["Rule","'Text'","'If a square measure a square, the side will also measure the side; and, if the side measure the side, the square will also measure the square.'"],["Rule","'TextWordCount'",27],["Rule","'GreekText'","'ἐὰν τετράγωνος τετράγωνον μετρῇ, καὶ ἡ πλευρὰ τὴν πλευρὰν μετρήσει: καὶ ἐὰν ἡ πλευρὰ τὴν πλευρὰν μετρῇ, καὶ ὁ τετράγωνος τὸν τετράγωνον μετρήσει.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",8],["Rule","'Theorem'",7]],["List",["Rule","'Book'",8],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'Let A, B be square numbers, let C, D be their sides, and let A measure B; I say that C also measures D. For let C by multiplying D make E; therefore A, E, B are continuously proportional in the ratio of C to D. [VIII. 11] And, since A, E, B are continuously proportional, and A measures B, therefore A also measures E. [VIII. 7] And, as A is to E, so is C to D; therefore also C measures D. [VII. Def. 20] Again, let C measure D; I say that A also measures B. For, with the same construction, we can in a similar manner prove that A, E, B are continuously proportional in the ratio of C to D. And since, as C is to D, so is A to E, and C measures D, therefore A also measures E. [VII. Def. 20] And A, E, B are continuously proportional; therefore A also measures B.'"],["Rule","'ProofWordCount'",160],["Rule","'GreekProof'","'ἔστωσαν τετράγωνοι ἀριθμοὶ οἱ α, Β, πλευραὶ δὲ αὐτῶν ἔστωσαν οἱ Γ, Δ, ὁ δὲ Α τὸν Β μετρείτω: λέγω, ὅτι καὶ ὁ Γ τὸν Δ μετρεῖ. ὁ Γ γὰρ τὸν Δ πολλαπλασιάσας τὸν Ε ποιείτω: οἱ Α, Ε, Β ἄρα ἑξῆς ἀνάλογόν εἰσιν ἐν τῷ τοῦ Γ πρὸς τὸν Δ λόγῳ. καὶ ἐπεὶ οἱ Α, Ε, Β ἑξῆς ἀνάλογόν εἰσιν, καὶ μετρεῖ ὁ Α τὸν Β, μετρεῖ ἄρα καὶ ὁ Α τὸν Ε. καί ἐστιν ὡς ὁ Α πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Δ: μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ. πάλιν δὴ ὁ Γ τὸν Δ μετρείτω: λέγω, ὅτι καὶ ὁ Α τὸν Β μετρεῖ. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι οἱ Α, Ε, Β ἑξῆς ἀνάλογόν εἰσιν ἐν τῷ τοῦ Γ πρὸς τὸν Δ λόγῳ. καὶ ἐπεί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Ε, μετρεῖ δὲ ὁ Γ τὸν Δ, μετρεῖ ἄρα καὶ ὁ Α τὸν Ε. καί εἰσιν οἱ Α, Ε, Β ἑξῆς ἀνάλογον: μετρεῖ ἄρα καὶ ὁ Α τὸν Β. ἐὰν ἄρα τετράγωνος τετράγωνον μετρῇ, καὶ ἡ πλευρὰ τὴν πλευρὰν μετρήσει: καὶ ἐὰν ἡ πλευρὰ τὴν πλευρὰν μετρῇ, καὶ ὁ τετράγωνος τὸν τετράγωνον μετρήσει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",202]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'8.15'"],["Rule","'Text'","'If a cube number measure a cube number, the side will also measure the side; and, if the side measure the side, the cube will also measure the cube.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'ἐὰν κύβος ἀριθμὸς κύβον ἀριθμὸν μετρῇ, καὶ ἡ πλευρὰ τὴν πλευρὰν μετρήσει: καὶ ἐὰν ἡ πλευρὰ τὴν πλευρὰν μετρῇ, καὶ ὁ κύβος τὸν κύβον μετρήσει.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",8],["Rule","'Theorem'",7]],["List",["Rule","'Book'",8],["Rule","'Theorem'",11]],["List",["Rule","'Book'",8],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'For let the cube number A measure the cube B, and let C be the side of A and D of B; I say that C measures D. For let C by multiplying itself make E, and let D by multiplying itself make G; further, let C by multiplying D make F, and let C, D by multiplying F make H, K respectively. Now it is manifest that E, F, G and A, H, K, B are continuously proportional in the ratio of C to D. [VIII. 11, 12] And, since A, H, K, B are continuously proportional, and A measures B, therefore it also measures H. [VIII. 7] And, as A is to H, so is C to D; therefore C also measures D. [VII. Def. 20] Next, let C measure D; I say that A will also measure B. For, with the same construction, we can prove in a similar manner that A, H, K, B are continuously proportional in the ratio of C to D. And, since C measures D, and, as C is to D, so is A to H, therefore A also measures H, [VII. Def. 20] so that A measures B also.'"],["Rule","'ProofWordCount'",198],["Rule","'GreekProof'","'κύβος γὰρ ἀριθμὸς ὁ Α κύβον τὸν Β μετρείτω, καὶ τοῦ μὲν Α πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ: λέγω, ὅτι ὁ Γ τὸν Δ μετρεῖ. ὁ Γ γὰρ ἑαυτὸν πολλαπλασιάσας τὸν Ε ποιείτω, ὁ δὲ Δ ἑαυτὸν πολλαπλασιάσας τὸν Η ποιείτω, καὶ ἔτι ὁ Γ τὸν Δ πολλαπλασιάσας τὸν Ζ ποιείτω, ἑκάτερος δὲ τῶν Γ, Δ τὸν Ζ πολλαπλασιάσας ἑκάτερον τῶν Θ, Κ ποιείτω. φανερὸν δή, ὅτι οἱ Ε, Ζ, Η καὶ οἱ Α, Θ, Κ, Β ἑξῆς ἀνάλογόν εἰσιν ἐν τῷ τοῦ Γ πρὸς τὸν Δ λόγῳ. καὶ ἐπεὶ οἱ Α, Θ, κ, Β ἑξῆς ἀνάλογόν εἰσιν, καὶ μετρεῖ ὁ Α τὸν Β, μετρεῖ ἄρα καὶ τὸν Θ. καί ἐστιν ὡς ὁ Α πρὸς τὸν Θ, οὕτως ὁ Γ πρὸς τὸν Δ: μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ. ἀλλὰ δὴ μετρείτω ὁ Γ τὸν Δ: λέγω, ὅτι καὶ ὁ Α τὸν Β μετρήσει. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δὴ δείξομεν, ὅτι οἱ Α, Θ, Κ, Β ἑξῆς ἀνάλογόν εἰσιν ἐν τῷ τοῦ Γ πρὸς τὸν Δ λόγῳ. καὶ ἐπεὶ ὁ Γ τὸν Δ μετρεῖ, καί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Θ, καὶ ὁ Α ἄρα τὸν Θ μετρεῖ: ὥστε καὶ τὸν Β μετρεῖ ὁ Α: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",211]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'8.16'"],["Rule","'Text'","'If a square number do not measure a square number, neither will the side measure the side; and, if the side do not measure the side, neither will the square measure the square.'"],["Rule","'TextWordCount'",33],["Rule","'GreekText'","'ἐὰν τετράγωνος ἀριθμὸς τετράγωνον ἀριθμὸν μὴ μετρῇ, οὐδὲ ἡ πλευρὰ τὴν πλευρὰν μετρήσει: κἂν ἡ πλευρὰ τὴν πλευρὰν μὴ μετρῇ, οὐδὲ ὁ τετράγωνος τὸν τετράγωνον μετρήσει.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",14]]]],["Rule","'Proof'","'Let A, B be square numbers, and let C, D be their sides; and let A not measure B; I say that neither does C measure D. For, if C measures D, A will also measure B. [VIII. 14] But A does not measure B; therefore neither will C measure D. Again, let C not measure D; I say that neither will A measure B. For, if A measures B, C will also measure D. [VIII. 14] But C does not measure D; therefore neither will A measure B.'"],["Rule","'ProofWordCount'",89],["Rule","'GreekProof'","'῎εστωσαν τετράγωνοι ἀριθμοὶ οἱ Α, Β, πλευραὶ δὲ αὐτῶν ἔστωσαν οἱ Γ, Δ, καὶ μὴ μετρείτω ὁ Α τὸν Β: λέγω, ὅτι οὐδὲ ὁ Γ τὸν Δ μετρεῖ. εἰ γὰρ μετρεῖ ὁ Γ τὸν Δ, μετρήσει καὶ ὁ Α τὸν Β. οὐ μετρεῖ δὲ ὁ Α τὸν Β: οὐδὲ ἄρα ὁ Γ τὸν Δ μετρήσει. μὴ μετρείτω δὴ πάλιν ὁ Γ τὸν Δ: λέγω, ὅτι οὐδὲ ὁ Α τὸν Β μετρήσει. εἰ γὰρ μετρεῖ ὁ Α τὸν Β, μετρήσει καὶ ὁ Γ τὸν Δ. οὐ μετρεῖ δὲ ὁ Γ τὸν Δ: οὐδ᾽ ἄρα ὁ Α τὸν Β μετρήσει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",103]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'8.17'"],["Rule","'Text'","'If a cube number do not measure a cube number, neither will the side measure the side; and, if the side do not measure the side, neither will the cube measure the cube.'"],["Rule","'TextWordCount'",33],["Rule","'GreekText'","'ἐὰν κύβος ἀριθμὸς κύβον ἀριθμὸν μὴ μετρῇ, οὐδὲ ἡ πλευρὰ τὴν πλευρὰν μετρήσει: κἂν ἡ πλευρὰ τὴν πλευρὰν μὴ μετρῇ, οὐδὲ ὁ κύβος τὸν κύβον μετρήσει.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'For let the cube number A not measure the cube number B, and let C be the side of A, and D of B; I say that C will not measure D. For if C measures D, A will also measure B. [VIII. 15] But A does not measure B; therefore neither does C measure D. Again, let C not measure D; I say that neither will A measure B. For, if A measures B, C will also measure D. [VIII. 15] But C does not measure D; therefore neither will A measure B.'"],["Rule","'ProofWordCount'",94],["Rule","'GreekProof'","'κύβος γὰρ ἀριθμὸς ὁ Α κύβον ἀριθμὸν τὸν Β μὴ μετρείτω, καὶ τοῦ μὲν Α πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ: λέγω, ὅτι ὁ Γ τὸν Δ οὐ μετρήσει. εἰ γὰρ μετρεῖ ὁ Γ τὸν Δ, καὶ ὁ Α τὸν Β μετρήσει. οὐ μετρεῖ δὲ ὁ Α τὸν Β: οὐδ᾽ ἄρα ὁ Γ τὸν Δ μετρεῖ. ἀλλὰ δὴ μὴ μετρείτω ὁ Γ τὸν Δ: λέγω, ὅτι οὐδὲ ὁ Α τὸν Β μετρήσει. εἰ γὰρ ὁ Α τὸν Β μετρεῖ, καὶ ὁ Γ τὸν Δ μετρήσει. οὐ μετρεῖ δὲ ὁ Γ τὸν Δ: οὐδ᾽ ἄρα ὁ Α τὸν Β μετρήσει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",107]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'8.18'"],["Rule","'Text'","'Between two similar plane numbers there is one mean proportional number; and the plane number has to the plane number the ratio duplicate of that which the corresponding side has to the corresponding side.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'δύο ὁμοίων ἐπιπέδων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός: καὶ ὁ ἐπίπεδος πρὸς τὸν ἐπίπεδον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",9]],["List",["Rule","'Book'",7],["Rule","'Definition'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",13]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'Let A, B be two similar plane numbers, and let the numbers C, D be the sides of A, and E, F of B. Now, since similar plane numbers are those which have their sides proportional, [VII. Def. 21] therefore, as C is to D, so is E to F. I say then that between A, B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to E, or D to F, that is, of that which the corresponding side has to the corresponding side. Now since, as C is to D, so is E to F, therefore, alternately, as C is to E, so is D to F. [VII. 13] And, since A is plane, and C, D are its sides, therefore D by multiplying C has made A. For the same reason also E by multiplying F has made B. Now let D by multiplying E make G. Then, since D by multiplying C has made A, and by multiplying E has made G, therefore, as C is to E, so is A to G. [VII. 17] But, as C is to E, so is D to F; therefore also, as D is to F, so is A to G. Again, since E by multiplying D has made G, and by multiplying F has made B, therefore, as D is to F, so is G to B. [VII. 17] But it was also proved that, as D is to F, so is A to G; therefore also, as A is to G, so is G to B. Therefore A, G, B are in continued proportion. Therefore between A, B there is one mean proportional number. I say next that A also has to B the ratio duplicate of that which the corresponding side has to the corresponding side, that is, of that which C has to E or D to F. For, since A, G, B are in continued proportion, A has to B the ratio duplicate of that which it has to G. [V. Def. 9] And, as A is to G, so is C to E, and so is D to F. Therefore A also has to B the ratio duplicate of that which C has to E or D to F.'"],["Rule","'ProofWordCount'",386],["Rule","'GreekProof'","'ἔστωσαν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ τοῦ μὲν Α πλευραὶ ἔστωσαν οἱ Γ, Δ ἀριθμοί, τοῦ δὲ Β οἱ Ε, Ζ. καὶ ἐπεὶ ὅμοιοι ἐπίπεδοί εἰσιν οἱ ἀνάλογον ἔχοντες τὰς πλευράς, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν Ζ. λέγω οὖν, ὅτι τῶν Α, Β εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Ε á¼¢ ὁ Δ πρὸς τὸν Ζ, τουτέστιν ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν. καὶ ἐπεί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν Ζ, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Γ πρὸς τὸν Ε, ὁ Δ πρὸς τὸν Ζ. καὶ ἐπεὶ ἐπίπεδός ἐστιν ὁ Α, πλευραὶ δὲ αὐτοῦ οἱ Γ, Δ, ὁ Δ ἄρα τὸν Γ πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε τὸν Ζ πολλαπλασιάσας τὸν Β πεποίηκεν. ὁ Δ δὴ τὸν Ε πολλαπλασιάσας τὸν Η ποιείτω. καὶ ἐπεὶ ὁ Δ τὸν μὲν Γ πολλαπλασιάσας τὸν Α πεποίηκεν, τὸν δὲ Ε πολλαπλασιάσας τὸν Η πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν Η. ἀλλ᾽ ὡς ὁ Γ πρὸς τὸν Ε, οὕτως ὁ Δ πρὸς τὸν Ζ: καὶ ὡς ἄρα ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Η. πάλιν, ἐπεὶ ὁ Ε τὸν μὲν Δ πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ Ζ πολλαπλασιάσας τὸν Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Η πρὸς τὸν Β. ἐδείχθη δὲ καὶ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Η: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Η, οὕτως ὁ Η πρὸς τὸν Β. οἱ Α, Η, Β ἄρα ἑξῆς ἀνάλογόν εἰσιν. τῶν Α, Β ἄρα εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός. λέγω δή, ὅτι καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἤπερ ὁ Γ πρὸς τὸν Ε á¼¢ ὁ Δ πρὸς τὸν Ζ. ἐπεὶ γὰρ οἱ Α, Η, Β ἑξῆς ἀνάλογόν εἰσιν, ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ πρὸς τὸν Η. καί ἐστιν ὡς ὁ Α πρὸς τὸν Η, οὕτως ὅ τε Γ πρὸς τὸν Ε καὶ ὁ Δ πρὸς τὸν Ζ. καὶ ὁ Α ἄρα πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Ε á¼¢ ὁ Δ πρὸς τὸν Ζ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",399]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'8.19'"],["Rule","'Text'","'Between two similar solid numbers there fall two mean proportional numbers; and the solid number has to the similar solid number the ratio triplicate of that which the corresponding side has to the corresponding side.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'δύο ὁμοίων στερεῶν ἀριθμῶν δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί: καὶ ὁ στερεὸς πρὸς τὸν ὅμοιον στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν.'"],["Rule","'GreekTextWordCount'",27],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",10]],["List",["Rule","'Book'",7],["Rule","'Definition'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",13]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",7],["Rule","'Theorem'",18]],["List",["Rule","'Book'",8],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let A, B be two similar solid numbers, and let C, D, E be the sides of A, and F, G, H of B. Now, since similar solid numbers are those which have their sides proportional, [VII. Def. 21] therefore, as C is to D, so is F to G, and, as D is to E, so is G to H.  I say that between A, B there fall two mean proportional numbers, and A has to B the ratio triplicate of that which C has to F, D to G, and also E to H. For let C by multiplying D make K, and let F by multiplying G make L. Now, since C, D are in the same ratio with F, G, and K is the product of C, D, and L the product of F, G, K, L are similar plane numbers; [VII. Def. 21] therefore between K, L there is one mean proportional number. [VIII. 18] Let it be M   Therefore M is the product of D, F, as was proved in the theorem preceding this. [VIII. 18] Now, since D by multiplying C has made K, and by multiplying F has made M, therefore, as C is to F, so is K to M. [VII. 17] But, as K is to M, so is M to L. Therefore K, M, L are continuously proportional in the ratio of C to F. And since, as C is to D, so is F to G, alternately therefore, as C is to F, so is D to G. [VII. 13] For the same reason also, as D is to G, so is E to H.  Therefore K, M, L are continuously proportional in the ratio of C to F, in the ratio of D to G, and also in the ratio of E to H. Next, let E, H by multiplying M make N, O respectively. Now, since A is a solid number, and C, D, E are its sides, therefore E by multiplying the product of C, D has made A. But the product of C, D is K; therefore E by multiplying K has made A. For the same reason also H by multiplying L has made B.  Now, since E by multiplying K has made A, and further also by multiplying M has made N, therefore, as K is to M, so is A to N. [VII. 17] But, as K is to M, so is C to F, D to G, and also E to H; therefore also, as C is to F, D to G, and E to H, so is A to N. Again, since E, H by multiplying M have made N, O respectively, therefore, as E is to H, so is N to O. [VII. 18] But, as E is to H, so is C to F and D to G; therefore also, as C is to F, D to G, and E to H, so is A to N and N to O. Again, since H by multiplying M has made O, and further also by multiplying L has made B, therefore, as M is to L, so is O to B. [VII. 17] But, as M is to L, so is C to F, D to G, and E to H. Therefore also, as C is to F, D to G, and E to H, so not only is O to B, but also A to N and N to O. Therefore A, N, O, B are continuously proportional in the aforesaid ratios of the sides. I say that A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, or D to G, and also E to H. For, since A, N, O, B are four numbers in continued proportion, therefore A has to B the ratio triplicate of that which A has to N. [V. Def. 10] But, as A is to N, so it was proved that C is to F, D to G, and also E to H. Therefore A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, D to G, and also E to H.'"],["Rule","'ProofWordCount'",734],["Rule","'GreekProof'","'ἔστωσαν δύο ὅμοιοι στερεοὶ οἱ Α, Β, καὶ τοῦ μὲν Α πλευραὶ ἔστωσαν οἱ Γ, Δ, Ε, τοῦ δὲ Β οἱ Ζ, Η, Θ. καὶ ἐπεὶ ὅμοιοι στερεοί εἰσιν οἱ ἀνάλογον ἔχοντες τὰς πλευράς, ἔστιν ἄρα ὡς μὲν ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ζ πρὸς τὸν Η, ὡς δὲ ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν Θ. λέγω, ὅτι τῶν Α, Β δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί, καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ. ὁ Γ γὰρ τὸν Δ πολλαπλασιάσας τὸν Κ ποιείτω, ὁ δὲ Ζ τὸν Η πολλαπλασιάσας τὸν Λ ποιείτω. καὶ ἐπεὶ οἱ Γ, Δ τοῖς Ζ, Η ἐν τῷ αὐτῷ λόγῳ εἰσίν, καὶ ἐκ μὲν τῶν Γ, Δ ἐστιν ὁ Κ, ἐκ δὲ τῶν Ζ, Η ὁ Λ, οἱ Κ, Λ ἄρα ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί: τῶν Κ, Λ ἄρα εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός. ἔστω ὁ Μ. ὁ Μ ἄρα ἐστὶν ὁ ἐκ τῶν Δ, Ζ, ὡς ἐν τῷ πρὸ τούτου θεωρήματι ἐδείχθη. καὶ ἐπεὶ ὁ Δ τὸν μὲν Γ πολλαπλασιάσας τὸν Κ πεποίηκεν, τὸν δὲ Ζ πολλαπλασιάσας τὸν Μ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Ζ, οὕτως ὁ Κ πρὸς τὸν Μ. ἀλλ᾽ ὡς ὁ Κ πρὸς τὸν Μ, ὁ Μ πρὸς τὸν Λ. οἱ Κ, Μ, Λ ἄρα ἑξῆς εἰσιν ἀνάλογον ἐν τῷ τοῦ Γ πρὸς τὸν Ζ λόγῳ. καὶ ἐπεί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ζ πρὸς τὸν Η, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Γ πρὸς τὸν Ζ, οὕτως ὁ Δ πρὸς τὸν Η. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Δ πρὸς τὸν Η, οὕτως ὁ Ε πρὸς τὸν Θ. οἱ Κ, Μ, Λ ἄρα ἑξῆς εἰσιν ἀνάλογον ἔν τε τῷ τοῦ Γ πρὸς τὸν Ζ λόγῳ καὶ τῷ τοῦ Δ πρὸς τὸν Η καὶ ἔτι τῷ τοῦ Ε πρὸς τὸν Θ. ἑκάτερος δὴ τῶν Ε, Θ τὸν Μ πολλαπλασιάσας ἑκάτερον τῶν Ν, Ξ ποιείτω. καὶ ἐπεὶ στερεός ἐστιν ὁ Α, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Γ, Δ, Ε, ὁ Ε ἄρα τὸν ἐκ τῶν Γ, Δ πολλαπλασιάσας τὸν Α πεποίηκεν. ὁ δὲ ἐκ τῶν Γ, Δ ἐστιν ὁ Κ: ὁ Ε ἄρα τὸν Κ πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Θ τὸν Λ πολλαπλασιάσας τὸν Β πεποίηκεν. καὶ ἐπεὶ ὁ Ε τὸν Κ πολλαπλασιάσας τὸν Α πεποίηκεν, ἀλλὰ μὴν καὶ τὸν Μ πολλαπλασιάσας τὸν Ν πεποίηκεν, ἔστιν ἄρα ὡς ὁ Κ πρὸς τὸν Μ, οὕτως ὁ Α πρὸς τὸν Ν. ὡς δὲ ὁ Κ πρὸς τὸν Μ, οὕτως ὅ τε Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ: καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Ν. πάλιν, ἐπεὶ ἑκάτερος τῶν Ε, Θ τὸν Μ πολλαπλασιάσας ἑκάτερον τῶν Ν, Ξ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Ν πρὸς τὸν Ξ. ἀλλ᾽ ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὅ τε Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η: καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ὁ Ε πρὸς τὸν Θ, οὕτως ὅ τε Α πρὸς τὸν Ν καὶ ὁ Ν πρὸς τὸν Ξ. πάλιν, ἐπεὶ ὁ Θ τὸν Μ πολλαπλασιάσας τὸν Ξ πεποίηκεν, ἀλλὰ μὴν καὶ τὸν Λ πολλαπλασιάσας τὸν Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Μ πρὸς τὸν Λ, οὕτως ὁ Ξ πρὸς τὸν Β. ἀλλ᾽ ὡς ὁ Μ πρὸς τὸν Λ, οὕτως ὅ τε Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ὁ Ε πρὸς τὸν Θ. καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ὁ Ε πρὸς τὸν Θ, οὕτως οὐ μόνον ὁ Ξ πρὸς τὸν Β, ἀλλὰ καὶ ὁ Α πρὸς τὸν ν καὶ ὁ Ν πρὸς τὸν Ξ. οἱ Α, Ν, Ξ, Β ἄρα ἑξῆς εἰσιν ἀνάλογον ἐν τοῖς εἰρημένοις τῶν πλευρῶν λόγοις. λέγω, ὅτι καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἤπερ ὁ Γ ἀριθμὸς πρὸς τὸν Ζ á¼¢ ὁ Δ πρὸς τὸν Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ. ἐπεὶ γὰρ τέσσαρες ἀριθμοὶ ἑξῆς ἀνάλογόν εἰσιν οἱ Α, Ν, Ξ, Β, ὁ Α ἄρα πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Α πρὸς τὸν Ν. ἀλλ᾽ ὡς ὁ Α πρὸς τὸν Ν, οὕτως ἐδείχθη ὅ τε Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ. καὶ ὁ Α ἄρα πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἤπερ ὁ Γ ἀριθμὸς πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",815]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'8.20'"],["Rule","'Text'","'If one mean proportional number fall between two numbers, the numbers will be similar plane numbers.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'ἐὰν δύο ἀριθμῶν εἷς μέσος ἀνάλογον ἐμπίπτῃ ἀριθμός, ὅμοιοι ἐπίπεδοι ἔσονται οἱ ἀριθμοί.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",13]],["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",33]]]],["Rule","'Proof'","'For let one mean proportional number C fall between the two numbers A, B;I say that A, B are similar plane numbers. Let D, E, the least numbers of those which have the same ratio with A, C, be taken; [VII. 33] therefore D measures A the same number of times that E measures C. [VII. 20] Now, as many times as D measures A, so many units let there be in F; therefore F by multiplying D has made A, so that A is plane, and D, F are its sides. Again, since D, E are the least of the numbers which havethe same ratio with C, B, therefore D measures C the same number of times that E measures B. [VII. 20] As many times, then, as E measures B, so many units let there be in G;therefore E measures B according to the units in G; therefore G by multiplying E has made B. Therefore B is plane, and E, G are its sides. Therefore A, B are plane numbers. I say next that they are also similar. For, since F by multiplying D has made A, and by multiplying E has made C, therefore, as D is to E, so is A to C, that is, C to B. [VII. 17] Again, since E by multiplying F, G has made C, B respectively,therefore, as F is to G, so is C to B. [VII. 17] But, as C is to B, so is D to E; therefore also, as D is to E, so is F to G. And alternately, as D is to F, so is E to G. [VII. 13] Therefore A, B are similar plane numbers; for their sidesare proportional.'"],["Rule","'ProofWordCount'",290],["Rule","'GreekProof'","'δύο γὰρ ἀριθμῶν τῶν Α, Β εἷς μέσος ἀνάλογον ἐμπιπτέτω ἀριθμὸς ὁ Γ: λέγω, ὅτι οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί. εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Γ οἱ Δ, Ε: ἰσάκις ἄρα ὁ Δ τὸν Α μετρεῖ καὶ ὁ Ε τὸν Γ. ὁσάκις δὴ ὁ Δ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ζ: ὁ Ζ ἄρα τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. ὥστε ὁ Α ἐπίπεδός ἐστιν, πλευραὶ δὲ αὐτοῦ οἱ Δ, Ζ. πάλιν, ἐπεὶ οἱ Δ, Ε ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Γ, Β, ἰσάκις ἄρα ὁ Δ τὸν Γ μετρεῖ καὶ ὁ Ε τὸν Β. ὁσάκις δὴ ὁ Ε τὸν Β μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Η. ὁ Ε ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Η μονάδας: ὁ Η ἄρα τὸν Ε πολλαπλασιάσας τὸν Β πεποίηκεν. ὁ Β ἄρα ἐπίπεδός ἐστι, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Ε, Η. οἱ Α, Β ἄρα ἐπίπεδοί εἰσιν ἀριθμοί. λέγω δή, ὅτι καὶ ὅμοιοι. ἐπεὶ γὰρ ὁ Ζ τὸν μὲν Δ πολλαπλασιάσας τὸν Α πεποίηκεν, τὸν δὲ Ε πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν Γ, τουτέστιν ὁ Γ πρὸς τὸν Β. πάλιν, ἐπεὶ ὁ Ε ἑκάτερον τῶν Ζ, Η πολλαπλασιάσας τοὺς Γ, Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ζ πρὸς τὸν Η, οὕτως ὁ Γ πρὸς τὸν Β. ὡς δὲ ὁ Γ πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε: καὶ ὡς ἄρα ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Ζ πρὸς τὸν Η. καὶ ἐναλλὰξ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Ε πρὸς τὸν Η. οἱ Α, Β ἄρα ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί: αἱ γὰρ πλευραὶ αὐτῶν ἀνάλογόν εἰσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",290]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'8.21'"],["Rule","'Text'","'If two mean proportional numbers fall between two numbers, the numbers are similar solid numbers.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἐὰν δύο ἀριθμῶν δύο μέσοι ἀνάλογον ἐμπίπτωσιν ἀριθμοί, ὅμοιοι στερεοί εἰσιν οἱ ἀριθμοί.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",14]],["List",["Rule","'Book'",7],["Rule","'Theorem'",18]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",33]],["List",["Rule","'Book'",8],["Rule","'Theorem'",2]],["List",["Rule","'Book'",8],["Rule","'Theorem'",3]],["List",["Rule","'Book'",8],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'For let two mean proportional numbers C, D fall between the two numbers A, B; I say that A, B are similar solid numbers. For let three numbers E, F, G, the least of those which have the same ratio with A, C, D, be taken; [VII. 33 or VIII. 2] therefore the extremes of them E, G are prime to one another. [VIII. 3] Now, since one mean proportional number F has fallen between E, G, therefore E, G are similar plane numbers. [VIII. 20] Let, then, H, K be the sides of E, and L, M of G. Therefore it is manifest from the theorem before this that E, F, G are continuously proportional in the ratio of H to L and that of K to M. Now, since E, F, G are the least of the numbers which have the same ratio with A, C, D, and the multitude of the numbers E, F, G is equal to the multitude of the numbers A, C, D, therefore, ex aequali, as E is to G, so is A to D. [VII. 14] But E, G are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio with them the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures A the same number of times that G measures D. Now, as many times as E measures A, so many units let there be in N. Therefore N by multiplying E has made A. But E is the product of H, K; therefore N by multiplying the product of H, K has made A. Therefore A is solid, and H, K, N are its sides. Again, since E, F, G are the least of the numbers which have the same ratio as C, D, B, therefore E measures C the same number of times that G measures B. Now, as many times as E measures C, so many units let there be in O. Therefore G measures B according to the units in O; therefore O by multiplying G has made B. But G is the product of L, M; therefore O by multiplying the product of L, M has made B. Therefore B is solid, and L, M, O are its sides; therefore A, B are solid. I say that they are also similar. For since N, O by multiplying E have made A, C, therefore, as N is to O, so is A to C, that is, E to F. [VII. 18] But, as E is to F, so is H to L and K to M; therefore also, as H is to L, so is K to M and N to O. And H, K, N are the sides of A, and O, L, M the sides of B. Therefore A, B are similar solid numbers.'"],["Rule","'ProofWordCount'",496],["Rule","'GreekProof'","'δύο γὰρ ἀριθμῶν τῶν Α, Β δύο μέσοι ἀνάλογον ἐμπιπτέτωσαν ἀριθμοὶ οἱ Γ, Δ: λέγω, ὅτι οἱ Α, Β ὅμοιοι στερεοί εἰσιν. εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Γ, Δ τρεῖς οἱ Ε, Ζ, Η: οἱ ἄρα ἄκροι αὐτῶν οἱ Ε, Η πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ τῶν Ε, Η εἷς μέσος ἀνάλογον ἐμπέπτωκεν ἀριθμὸς ὁ Ζ, οἱ Ε, Η ἄρα ἀριθμοὶ ὅμοιοι ἐπίπεδοί εἰσιν. ἔστωσαν οὖν τοῦ μὲν Ε πλευραὶ οἱ Θ, Κ, τοῦ δὲ Η οἱ Λ, Μ. φανερὸν ἄρα ἐστὶν ἐκ τοῦ πρὸ τούτου, ὅτι οἱ Ε, Ζ, Η ἑξῆς εἰσιν ἀνάλογον ἔν τε τῷ τοῦ Θ πρὸς τὸν Λ λόγῳ καὶ τῷ τοῦ Κ πρὸς τὸν Μ. καὶ ἐπεὶ οἱ Ε, Ζ, Η ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Γ, Δ, καί ἐστιν ἴσον τὸ πλῆθος τῶν Ε, Ζ, Η τῷ πλήθει τῶν Α, Γ, Δ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Ε πρὸς τὸν Η, οὕτως ὁ Α πρὸς τὸν Δ. οἱ δὲ Ε, Η πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: ἰσάκις ἄρα ὁ Ε τὸν Α μετρεῖ καὶ ὁ Η τὸν Δ. ὁσάκις δὴ ὁ Ε τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ν. ὁ Ν ἄρα τὸν Ε πολλαπλασιάσας τὸν Α πεποίηκεν. ὁ δὲ Ε ἐστιν ὁ ἐκ τῶν Θ, Κ: ὁ Ν ἄρα τὸν ἐκ τῶν Θ, Κ πολλαπλασιάσας τὸν Α πεποίηκεν. στερεὸς ἄρα ἐστὶν ὁ Α, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Θ, Κ, Ν. πάλιν, ἐπεὶ οἱ Ε, Ζ, Η ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Γ, Δ, Β, ἰσάκις ἄρα ὁ Ε τὸν Γ μετρεῖ καὶ ὁ Η τὸν Β. ὁσάκις δὴ ὁ Ε τὸν Γ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ξ. ὁ Η ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Ξ μονάδας: ὁ Ξ ἄρα τὸν Η πολλαπλασιάσας τὸν Β πεποίηκεν. ὁ δὲ Η ἐστιν ὁ ἐκ τῶν Λ, Μ: ὁ Ξ ἄρα τὸν ἐκ τῶν Λ, Μ πολλαπλασιάσας τὸν Β πεποίηκεν. στερεὸς ἄρα ἐστὶν ὁ Β, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Λ, Μ, Ξ: οἱ Α, Β ἄρα στερεοί εἰσιν. λέγω δή, ὅτι καὶ ὅμοιοι. ἐπεὶ γὰρ οἱ Ν, Ξ τὸν Ε πολλαπλασιάσαντες τοὺς Α, Γ πεποιήκασιν, ἔστιν ἄρα ὡς ὁ Ν πρὸς τὸν Ξ, ὁ Α πρὸς τὸν Γ, τουτέστιν ὁ Ε πρὸς τὸν Ζ. ἀλλ᾽ ὡς ὁ Ε πρὸς τὸν Ζ, ὁ Θ πρὸς τὸν Λ καὶ ὁ Κ πρὸς τὸν Μ: καὶ ὡς ἄρα ὁ Θ πρὸς τὸν Λ, οὕτως ὁ Κ πρὸς τὸν Μ καὶ ὁ Ν πρὸς τὸν Ξ. καί εἰσιν οἱ μὲν Θ, Κ, Ν πλευραὶ τοῦ Α, οἱ δὲ Ξ, Λ, Μ πλευραὶ τοῦ Β. οἱ Α, Β ἄρα ἀριθμοὶ ὅμοιοι στερεοί εἰσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",483]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'8.22'"],["Rule","'Text'","'If three numbers be in continued proportion, and the first be square, the third will also be square.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'ἐὰν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ δὲ πρῶτος τετράγωνος ᾖ, καὶ ὁ τρίτος τετράγωνος ἔσται.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let A, B, C be three numbers in continued proportion, and let A the first be square; I say that C the third is also square. For, since between A, C there is one mean proportional number, B, therefore A, C are similar plane numbers. [VIII. 20] But A is square; therefore C is also square.'"],["Rule","'ProofWordCount'",56],["Rule","'GreekProof'","'ἔστωσαν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, ὁ δὲ πρῶτος ὁ Α τετράγωνος ἔστω: λέγω, ὅτι καὶ ὁ τρίτος ὁ Γ τετράγωνός ἐστιν. ἐπεὶ γὰρ τῶν Α, Γ εἷς μέσος ἀνάλογόν ἐστιν ἀριθμὸς ὁ Β, οἱ Α, Γ ἄρα ὅμοιοι ἐπίπεδοί εἰσιν. τετράγωνος δὲ ὁ Α: τετράγωνος ἄρα καὶ ὁ Γ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",56]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'8.23'"],["Rule","'Text'","'If four numbers be in continued proportion, and the first be cube, the fourth will also be cube.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'ἐὰν τέσσαρες ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ δὲ πρῶτος κύβος ᾖ, καὶ ὁ τέταρτος κύβος ἔσται.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'Let A, B, C, D be four numbers in continued proportion, and let A be cube; I say that D is also cube. For, since between A, D there are two mean proportional numbers B, C, therefore A, D are similar solid numbers. [VIII. 21] But A is cube; therefore D is also cube.'"],["Rule","'ProofWordCount'",54],["Rule","'GreekProof'","'ἔστωσαν τέσσαρες ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, ὁ δὲ Α κύβος ἔστω: λέγω, ὅτι καὶ ὁ Δ κύβος ἐστίν. ἐπεὶ γὰρ τῶν Α, Δ δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοὶ οἱ Β, Γ, οἱ Α, Δ ἄρα ὅμοιοί εἰσι στερεοὶ ἀριθμοί. κύβος δὲ ὁ Α: κύβος ἄρα καὶ ὁ Δ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",55]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'8.24'"],["Rule","'Text'","'If two numbers have to one another the ratio which a square number has to a square number, and the first be square, the second will also be square.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχωσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ὁ δὲ πρῶτος τετράγωνος ᾖ, καὶ ὁ δεύτερος τετράγωνος ἔσται.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",8]],["List",["Rule","'Book'",8],["Rule","'Theorem'",18]],["List",["Rule","'Book'",8],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'For let the two numbers A, B have to one another the ratio which the square number C has to the square number D, and let A be square; I say that B is also square. For, since C, D are square, C, D are similar plane numbers. Therefore one mean proportional number falls between C, D. [VIII. 18] And, as C is to D, so is A to B; therefore one mean proportional number falls between A, B also. [VIII. 8] And A is square; therefore B is also square. [VIII. 22]'"],["Rule","'ProofWordCount'",93],["Rule","'GreekProof'","'δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς ἀλλήλους λόγον ἐχέτωσαν, ὃν τετράγωνος ἀριθμὸς ὁ Γ πρὸς τετράγωνον ἀριθμὸν τὸν Δ, ὁ δὲ Α τετράγωνος ἔστω: λέγω, ὅτι καὶ ὁ Β τετράγωνός ἐστιν. ἐπεὶ γὰρ οἱ Γ, Δ τετράγωνοί εἰσιν, οἱ Γ, Δ ἄρα ὅμοιοι ἐπίπεδοί εἰσιν. τῶν Γ, Δ ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. καί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, ὁ Α πρὸς τὸν Β: καὶ τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. καί ἐστιν ὁ Α τετράγωνος: καὶ ὁ Β ἄρα τετράγωνός ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",92]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'8.25'"],["Rule","'Text'","'If two numbers have to one another the ratio which a cube number has to a cube number, and the first be cube, the second will also be cube.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχωσιν, ὃν κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν, ὁ δὲ πρῶτος κύβος ᾖ, καὶ ὁ δεύτερος κύβος ἔσται.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",8]],["List",["Rule","'Book'",8],["Rule","'Theorem'",19]],["List",["Rule","'Book'",8],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'For let the two numbers A, B have to one another the ratio which the cube number C has to the cube number D, and let A be cube; I say that B is also cube. For, since C, D are cube, C, D are similar solid numbers. Therefore two mean proportional numbers fall between C, D. [VIII. 19] And, as many numbers as fall between C, D in continued proportion, so many will also fall between those which have the same ratio with them; [VIII. 8] so that two mean proportional numbers fall between A, B also. Let E, F so fall. Since, then, the four numbers A, E, F, B are in continued proportion, and A is cube, therefore B is also cube. [VIII. 23]'"],["Rule","'ProofWordCount'",127],["Rule","'GreekProof'","'δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς ἀλλήλους λόγον ἐχέτωσαν, ὃν κύβος ἀριθμὸς ὁ Γ πρὸς κύβον ἀριθμὸν τὸν Δ, κύβος δὲ ἔστω ὁ Α: λέγω δή, ὅτι καὶ ὁ Β κύβος ἐστίν. ἐπεὶ γὰρ οἱ Γ, Δ κύβοι εἰσίν, οἱ Γ, Δ ὅμοιοι στερεοί εἰσιν: τῶν Γ, Δ ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. ὅσοι δὲ εἰς τοὺς Γ, Δ μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτουσιν, τοσοῦτοι καὶ εἰς τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς: ὥστε καὶ τῶν Α, Β δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. ἐμπιπτέτωσαν οἱ Ε, Ζ. ἐπεὶ οὖν τέσσαρες ἀριθμοὶ οἱ Α, Ε, Ζ, Β ἑξῆς ἀνάλογόν εἰσιν, καί ἐστι κύβος ὁ Α, κύβος ἄρα καὶ ὁ Β: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",115]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",26]],["Association",["Rule","'VertexLabel'","'8.26'"],["Rule","'Text'","'Similar plane numbers have to one another the ratio which a square number has to a square number.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'οἱ ὅμοιοι ἐπίπεδοι ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",33]],["List",["Rule","'Book'",8],["Rule","'Theorem'",2]],["List",["Rule","'Book'",8],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let A, B be similar plane numbers; I say that A has to B the ratio which a square number has to a square number. For, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18] Let it so fall, and let it be C; and let D, E, F, the least numbers of those which have the same ratio with A, C, B, be taken; [VII. 33 or VIII. 2] therefore the extremes of them D, F are square. [VIII. 2] And since, as D is to F, so is A to B, and D, F are square, therefore A has to B the ratio which a square number has to a square number.'"],["Rule","'ProofWordCount'",123],["Rule","'GreekProof'","'ἔστωσαν ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β: λέγω, ὅτι ὁ Α πρὸς τὸν Β λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἐπεὶ γὰρ οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν, τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. ἐμπιπτέτω καὶ ἔστω ὁ Γ, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Γ, Β οἱ Δ, Ε, Ζ: οἱ ἄρα ἄκροι αὐτῶν οἱ Δ, Ζ τετράγωνοί εἰσιν. καὶ ἐπεί ἐστιν ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Β, καί εἰσιν οἱ Δ, Ζ τετράγωνοι, ὁ Α ἄρα πρὸς τὸν Β λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",108]]],["Rule",["Association",["Rule","'Book'",8],["Rule","'Theorem'",27]],["Association",["Rule","'VertexLabel'","'8.27'"],["Rule","'Text'","'Similar solid numbers have to one another the ratio which a cube number has to a cube number.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'οἱ ὅμοιοι στερεοὶ ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",33]],["List",["Rule","'Book'",8],["Rule","'Theorem'",2]],["List",["Rule","'Book'",8],["Rule","'Theorem'",19]]]],["Rule","'Proof'","'Let A, B be similar solid numbers; I say that A has to B the ratio which a cube number has to a cube number. For, since A, B are similar solid numbers, therefore two mean proportional numbers fall between A, B. [VIII. 19] Let C, D so fall, and let E, F, G, H, the least numbers of those which have the same ratio with A, C, D, B, and equal with them in multitude, be taken; [VII. 33 or VIII. 2] therefore the extremes of them E, H are cube. [VIII. 2] And, as E is to H, so is A to B; therefore A also has to B the ratio which a cube number has to a cube number.'"],["Rule","'ProofWordCount'",122],["Rule","'GreekProof'","'ἔστωσαν ὅμοιοι στερεοὶ ἀριθμοὶ οἱ Α, Β: λέγω, ὅτι ὁ Α πρὸς τὸν Β λόγον ἔχει, ὃν κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν. ἐπεὶ γὰρ οἱ Α, Β ὅμοιοι στερεοί εἰσιν, τῶν Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. ἐμπιπτέτωσαν οἱ Γ, Δ, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Γ, Δ, Β ἴσοι αὐτοῖς τὸ πλῆθος οἱ ε, Ζ, Η, Θ: οἱ ἄρα ἄκροι αὐτῶν οἱ Ε, Θ κύβοι εἰσίν. καί ἐστιν ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Β: καὶ ὁ Α ἄρα πρὸς τὸν Β λόγον ἔχει, ὃν κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",107]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'9.1'"],["Rule","'Text'","'If two similar plane numbers by multiplying one another make some number, the product will be square.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἐὰν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος τετράγωνος ἔσται.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",8],["Rule","'Theorem'",8]],["List",["Rule","'Book'",8],["Rule","'Theorem'",18]],["List",["Rule","'Book'",8],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'Let A, B be two similar plane numbers, and let A by multiplying B make C; I say that C is square. For let A by multiplying itself make D. Therefore D is square. Since then A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17] And, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18] But, if numbers fall between two numbers in continued proportion, as many as fall between them, so many also fall between those which have the same ratio; [VIII. 8] so that one mean proportional number falls between D, C also. And D is square; therefore C is also square. [VIII. 22]'"],["Rule","'ProofWordCount'",133],["Rule","'GreekProof'","'ἔστωσαν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω: λέγω, ὅτι ὁ Γ τετράγωνός ἐστιν. ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω. ὁ Δ ἄρα τετράγωνός ἐστιν. ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί, τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. ἐὰν δὲ δύο ἀριθμῶν μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν ἀριθμοί, ὅσοι εἰς αὐτοὺς ἐμπίπτουσι, τοσοῦτοι καὶ εἰς τοὺς τὸν αὐτὸν λόγον ἔχοντας: ὥστε καὶ τῶν Δ, Γ εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. καί ἐστι τετράγωνος ὁ Δ: τετράγωνος ἄρα καὶ ὁ Γ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",131]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'9.2'"],["Rule","'Text'","'If two numbers by multiplying one another make a square number, they are similar plane numbers.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσι τετράγωνον, ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",8],["Rule","'Theorem'",8]],["List",["Rule","'Book'",8],["Rule","'Theorem'",18]],["List",["Rule","'Book'",8],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let A, B be two numbers, and let A by multiplying B make the square number C; I say that A, B are similar plane numbers. For let A by multiplying itself make D; therefore D is square. Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17] And, since D is square, and C is so also, therefore D, C are similar plane numbers. Therefore one mean proportional number falls between D, C. [VIII. 18] And, as D is to C, so is A to B; therefore one mean proportional number falls between A, B also. [VIII. 8] But, if one mean proportional number fall between two numbers, they are similar plane numbers; [VIII. 20] therefore A, B are similar plane numbers.'"],["Rule","'ProofWordCount'",142],["Rule","'GreekProof'","'ἔστωσαν δύο ἀριθμοὶ οἱ Α, Β, καὶ ὁ Α τὸν Β πολλαπλασιάσας τετράγωνον τὸν Γ ποιείτω: λέγω, ὅτι οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί. ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω: ὁ Δ ἄρα τετράγωνός ἐστιν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ ὁ Δ τετράγωνός ἐστιν, ἀλλὰ καὶ ὁ Γ, οἱ Δ, Γ ἄρα ὅμοιοι ἐπίπεδοί εἰσιν. τῶν Δ, Γ ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει. καί ἐστιν ὡς ὁ Δ πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν Β: καὶ τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει. ἐὰν δὲ δύο ἀριθμῶν εἷς μέσος ἀνάλογον ἐμπίπτῃ, ὅμοιοι ἐπίπεδοί εἰσιν οἱ ἀριθμοί: οἱ ἄρα Α, Β ὅμοιοί εἰσιν ἐπίπεδοι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",139]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'9.3'"],["Rule","'Text'","'If a cube number by multiplying itself make some number, the product will be cube.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἐὰν κύβος ἀριθμὸς ἑαυτὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἔσται.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",8],["Rule","'Theorem'",8]],["List",["Rule","'Book'",8],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'For let the cube number A by multiplying itself make B; I say that B is cube. For let C, the side of A, be taken, and let C by multiplying itself make D. It is then manifest that C by multiplying D has made A. Now, since C by multiplying itself has made D, therefore C measures D according to the units in itself. But further the unit also measures C according to the units in it; therefore, as the unit is to C, so is C to D. [VII. Def. 20] Again, since C by multiplying D has made A, therefore D measures A according to the units in C. But the unit also measures C according to the units in it; therefore, as the unit is to C, so is D to A. But, as the unit is to C, so is C to D; therefore also, as the unit is to C, so is C to D, and D to A. Therefore between the unit and the number A two mean proportional numbers C, D have fallen in continued proportion. Again, since A by multiplying itself has made B, therefore A measures B according to the units in itself. But the unit also measures A according to the units in it; therefore, as the unit is to A, so is A to B. [VII. Def. 20] But between the unit and A two mean proportional numbers have fallen; therefore two mean proportional numbers will also fall between A, B. [VIII. 8] But, if two mean proportional numbers fall between two numbers, and the first be cube, the second will also be cube. [VIII. 23] And A is cube; therefore B is also cube.'"],["Rule","'ProofWordCount'",287],["Rule","'GreekProof'","'κύβος γὰρ ἀριθμὸς ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β ποιείτω: λέγω, ὅτι ὁ Β κύβος ἐστίν. εἰλήφθω γὰρ τοῦ Α πλευρὰ ὁ Γ, καὶ ὁ Γ ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω. φανερὸν δή ἐστιν, ὅτι ὁ Γ τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ ἐπεὶ ὁ Γ ἑαυτὸν πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ Γ ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας. ἀλλὰ μὴν καὶ ἡ μονὰς τὸν Γ μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας: ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Δ. πάλιν, ἐπεὶ ὁ Γ τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν, ὁ Δ ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Γ κατὰ τὰς ἐν αὐτῷ μονάδας: ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Δ πρὸς τὸν Α. ἀλλ᾽ ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Δ: καὶ ὡς ἄρα ἡ μονὰς πρὸς τὸν Γ, οὕτως ὁ Γ πρὸς τὸν Δ καὶ ὁ Δ πρὸς τὸν Α. τῆς ἄρα μονάδος καὶ τοῦ Α ἀριθμοῦ δύο μέσοι ἀνάλογον κατὰ τὸ συνεχὲς ἐμπεπτώκασιν ἀριθμοὶ οἱ Γ, Δ. πάλιν, ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας: ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς τὸν Β. τῆς δὲ μονάδος καὶ τοῦ Α δύο μέσοι ἀνάλογον ἐμπεπτώκασιν ἀριθμοί: καὶ τῶν Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπεσοῦνται ἀριθμοί. ἐὰν δὲ δύο ἀριθμῶν δύο μέσοι ἀνάλογον ἐμπίπτωσιν, ὁ δὲ πρῶτος κύβος ᾖ, καὶ ὁ δεύτερος κύβος ἔσται. καί ἐστιν ὁ Α κύβος: καὶ ὁ Β ἄρα κύβος ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",285]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'9.4'"],["Rule","'Text'","'If a cube number by multiplying a cube number make some number, the product will be cube.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἐὰν κύβος ἀριθμὸς κύβον ἀριθμὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἔσται.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",8],["Rule","'Theorem'",8]],["List",["Rule","'Book'",8],["Rule","'Theorem'",19]],["List",["Rule","'Book'",8],["Rule","'Theorem'",23]],["List",["Rule","'Book'",9],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'For let the cube number A by multiplying the cube number B make C; I say that C is cube. For let A by multiplying itself make D; therefore D is cube. [IX. 3] And, since A by multiplying itself has made D, and by multiplying B has made C therefore, as A is to B, so is D to C. [VII. 17] And, since A, B are cube numbers, A, B are similar solid numbers. Therefore two mean proportional numbers fall between A, B; [VIII. 19] so that two mean proportional numbers will fall between D, C also. [VIII. 8] And D is cube; therefore C is also cube [VIII. 23]'"],["Rule","'ProofWordCount'",112],["Rule","'GreekProof'","'κύβος γὰρ ἀριθμὸς ὁ Α κύβον ἀριθμὸν τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω: λέγω, ὅτι ὁ Γ κύβος ἐστίν. ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω: ὁ Δ ἄρα κύβος ἐστίν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Α, Β κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν οἱ Α, Β. τῶν Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί: ὥστε καὶ τῶν Δ, Γ δύο μέσοι ἀνάλογον ἐμπεσοῦνται ἀριθμοί. καί ἐστι κύβος ὁ Δ: κύβος ἄρα καὶ ὁ Γ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",108]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'9.5'"],["Rule","'Text'","'If a cube number by multiplying any number make a cube number, the multiplied number will also be cube.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἐὰν κύβος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας κύβον ποιῇ, καὶ ὁ πολλαπλασιασθεὶς κύβος ἔσται.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",17]],["List",["Rule","'Book'",8],["Rule","'Theorem'",8]],["List",["Rule","'Book'",8],["Rule","'Theorem'",19]],["List",["Rule","'Book'",8],["Rule","'Theorem'",23]],["List",["Rule","'Book'",9],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'For let the cube number A by multiplying any number B make the cube number C; I say that B is cube. For let A by multiplying itself make D; therefore D is cube. [IX. 3] Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17] And since D, C are cube, they are similar solid numbers. Therefore two mean proportional numbers fall between D, C. [VIII. 19] And, as D is to C, so is A to B; therefore two mean proportional numbers fall between A, B also. [VIII. 8] And A is cube; therefore B is also cube. [VIII. 23]'"],["Rule","'ProofWordCount'",121],["Rule","'GreekProof'","'κύβος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλαπλασιάσας κύβον τὸν Γ ποιείτω: λέγω, ὅτι ὁ Β κύβος ἐστίν. ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω: κύβος ἄρα ἐστίν ὁ Δ. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Δ, Γ κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν. τῶν Δ, Γ ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. καί ἐστιν ὡς ὁ Δ πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν Β: καὶ τῶν α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. καί ἐστι κύβος ὁ Α: κύβος ἄρα ἐστὶ καὶ ὁ Β: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",120]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'9.6'"],["Rule","'Text'","'If a number by multiplying itself make a cube number, it will itself also be cube.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'ἐὰν ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ αὐτὸς κύβος ἔσται.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",8],["Rule","'Theorem'",8]],["List",["Rule","'Book'",8],["Rule","'Theorem'",19]],["List",["Rule","'Book'",8],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'For let the number A by multiplying itself make the cube number B; I say that A is also cube. For let A by multiplying B make C. Since, then, A by multiplying itself has made B, and by multiplying B has made C, therefore C is cube. And, since A by multiplying itself has made B, therefore A measures B according to the units in itself. But the unit also measures A according to the units in it. Therefore, as the unit is to A, so is A to B. [VII. Def. 20] And, since A by multiplying B has made C, therefore B measures C according to the units in A. But the unit also measures A according to the units in it. Therefore, as the unit is to A, so is B to C. [VII. Def. 20] But, as the unit is to A, so is A to B; therefore also, as A is to B, so is B to C. And, since B, C are cube, they are similar solid numbers. Therefore there are two mean proportional numbers between B, C. [VIII. 19] And, as B is to C, so is A to B. Therefore there are two mean proportional numbers between A, B also. [VIII. 8] And B is cube; therefore A is also cube. [cf. VIII. 23]'"],["Rule","'ProofWordCount'",223],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ Α ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β ποιείτω: λέγω, ὅτι καὶ ὁ Α κύβος ἐστίν. ῾ο γὰρ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω. ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Γ ἄρα κύβος ἐστίν. καὶ ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας. ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β. καὶ ἐπεὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας. ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Γ. ἀλλ᾽ ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Β, Γ κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν. τῶν Β, Γ ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί. καί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β. καὶ τῶν Α, Β ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί. καί ἐστι κύβος ὁ Β: κύβος ἄρα ἐστὶ καὶ ὁ Α: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",226]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'9.7'"],["Rule","'Text'","'If a composite number by multiplying any number make some number, the product will be solid.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'ἐὰν σύνθετος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος στερεὸς ἔσται.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",13]],["List",["Rule","'Book'",7],["Rule","'Definition'",15]]]],["Rule","'Proof'","'For let the composite number A by multiplying any number B make C; I say that C is solid. For, since A is composite, it will be measured by some number. [VII. Def. 13] Let it be measured by D; and, as many times as D measures A, so many units let there be in E. Since then D measures A according to the units in E, therefore E by multiplying D has made A. [VII. Def. 15] And, since A by multiplying B has made C, and A is the product of D, E, therefore the product of D, E by multiplying B has made C. Therefore C is solid, and D, E, B are its sides.'"],["Rule","'ProofWordCount'",118],["Rule","'GreekProof'","'σύνθετος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω: λέγω, ὅτι ὁ Γ στερεός ἐστιν. ἐπεὶ γὰρ ὁ Α σύνθετός ἐστιν, ὑπὸ ἀριθμοῦ τινος μετρηθήσεται. μετρείσθω ὑπὸ τοῦ Δ, καὶ ὁσάκις ὁ Δ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν ὁ Δ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, ὁ Ε ἄρα τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ ἐπεὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ δὲ Α ἐστιν ὁ ἐκ τῶν Δ, Ε, ὁ ἄρα ἐκ τῶν Δ, Ε τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. ὁ Γ ἄρα στερεός ἐστιν, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Δ, Ε, Β: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",115]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'9.8'"],["Rule","'Text'","'If as many numbers as we please beginning from an unit be in continued proportion, the third from the unit will be square, as will also those which successively leave out one; the fourth will be cube, as will also all those which leave out two; and the seventh will be at once cube and square, as will also those which leave out five.'"],["Rule","'TextWordCount'",64],["Rule","'GreekText'","'ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ μὲν τρίτος ἀπὸ τῆς μονάδος τετράγωνος ἔσται καὶ οἱ ἕνα διαλείποντες, ὁ δὲ τέταρτος κύβος καὶ οἱ δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος κύβος ἅμα καὶ τετράγωνος καὶ οἱ πέντε διαλείποντες.'"],["Rule","'GreekTextWordCount'",40],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",8],["Rule","'Theorem'",22]],["List",["Rule","'Book'",8],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion; I say that B, the third from the unit, is square, as are also all those which leave out one; C, the fourth, is cube, as are also all those which leave out two; and F, the seventh, is at once cube and square, as are also all those which leave out five. For since, as the unit is to A, so is A to B, therefore the unit measures the number A the same number of times that A measures B. [VII. Def. 20] But the unit measures the number A according to the units in it; therefore A also measures B according to the units in A. Therefore A by multiplying itself has made B; therefore B is square. And, since B, C, D are in continued proportion, and B is square, therefore D is also square. [VIII. 22] For the same reason F is also square. Similarly we can prove that all those which leave out one are square. I say next that C, the fourth from the unit, is cube, as are also all those which leave out two. For since, as the unit is to A, so is B to C, therefore the unit measures the number A the same number of times that B measures C. But the unit measures the number A according to the units in A; therefore B also measures C according to the units in A. Therefore A by multiplying B has made C. Since then A by multiplying itself has made B, and by multiplying B has made C, therefore C is cube. And, since C, D, E, F are in continued proportion, and C is cube, therefore F is also cube. [VIII. 23] But it was also proved square; therefore the seventh from the unit is both cube and square. Similarly we can prove that all the numbers which leave out five are also both cube and square.'"],["Rule","'ProofWordCount'",343],["Rule","'GreekProof'","'ἔστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, Ε, Ζ: λέγω, ὅτι ὁ μὲν τρίτος ἀπὸ τῆς μονάδος ὁ Β τετράγωνός ἐστι καὶ οἱ ἕνα διαλείποντες πάντες, ὁ δὲ τέταρτος ὁ Γ κύβος καὶ οἱ δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος ὁ Ζ κύβος ἅμα καὶ τετράγωνος καὶ οἱ πέντε διαλείποντες πάντες. ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας: καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν: τετράγωνος ἄρα ἐστὶν ὁ Β. καὶ ἐπεὶ οἱ Β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Β τετράγωνός ἐστιν, καὶ ὁ Δ ἄρα τετράγωνός ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ζ τετράγωνός ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ ἕνα διαλείποντες πάντες τετράγωνοί εἰσιν. λέγω δή, ὅτι καὶ ὁ τέταρτος ἀπὸ τῆς μονάδος ὁ Γ κύβος ἐστὶ καὶ οἱ δύο διαλείποντες πάντες. ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Γ, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Γ. ἡ δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας: καὶ ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας: ὁ Α ἄρα τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, κύβος ἄρα ἐστὶν ὁ Γ. καὶ ἐπεὶ οἱ Γ, Δ, Ε, Ζ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Γ κύβος ἐστίν, καὶ ὁ Ζ ἄρα κύβος ἐστίν. ἐδείχθη δὲ καὶ τετράγωνος: ὁ ἄρα ἕβδομος ἀπὸ τῆς μονάδος κύβος τέ ἐστι καὶ τετράγωνος. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ πέντε διαλείποντες πάντες κύβοι τέ εἰσι καὶ τετράγωνοι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",318]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'9.9'"],["Rule","'Text'","'If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be square, all the rest will also be square.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἑξῆς κατὰ τὸ συνεχὲς ἀριθμοὶ ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα τετράγωνος ᾖ, καὶ οἱ λοιποὶ πάντες τετράγωνοι ἔσονται.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",22]],["List",["Rule","'Book'",8],["Rule","'Theorem'",23]],["List",["Rule","'Book'",9],["Rule","'Theorem'",3]],["List",["Rule","'Book'",9],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'And, if the number after the unit be cube, all the rest will also be cube. Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, be square; I say that all the rest will also be square. Now it has been proved that B, the third from the unit, is square, as are also all those which leave out one; [IX. 8] I say that all the rest are also square. For, since A, B, C are in continued proportion, and A is square, therefore C is also square. [VIII. 22] Again, since B, C, D are in continued proportion, and B is square, D is also square. [VIII. 22] Similarly we can prove that all the rest are also square. Next, let A be cube; I say that all the rest are also cube. Now it has been proved that C, the fourth from the unit, is cube, as also are all those which leave out two; [IX. 8] I say that all the rest are also cube. For, since, as the unit is to A, so is A to B, therefore the unit measures A the same number of times as A measures B. But the unit measures A according to the units in it; therefore A also measures B according to the units in itself; therefore A by multiplying itself has made B. And A is cube. But, if a cube number by multiplying itself make some number, the product is cube. [IX. 3] Therefore B is also cube. And, since the four numbers A, B, C, D are in continued proportion, and A is cube, D also is cube. [VIII. 23] For the same reason E is also cube, and similarly all the rest are cube.'"],["Rule","'ProofWordCount'",312],["Rule","'GreekProof'","'καὶ ἐὰν ὁ μετὰ τὴν μονάδα κύβος ᾖ, καὶ οἱ λοιποὶ πάντες κύβοι ἔσονται. ἔστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α τετράγωνος ἔστω: λέγω, ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοι ἔσονται. ὅτι μὲν οὖν ὁ τρίτος ἀπὸ τῆς μονάδος ὁ Β τετράγωνός ἐστι καὶ οἱ ἕνα διαλείποντες πάντες, δέδεικται: λέγω δή, ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοί εἰσιν. ἐπεὶ γὰρ οἱ Α, Β, Γ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α τετράγωνος, καὶ ὁ Γ ἄρα τετράγωνός ἐστιν. πάλιν, ἐπεὶ καὶ οἱ β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Β τετράγωνος, καὶ ὁ Δ ἄρα τετράγωνός ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοί εἰσιν. ἀλλὰ δὴ ἔστω ὁ Α κύβος: λέγω, ὅτι καὶ οἱ λοιποὶ πάντες κύβοι εἰσίν. ὅτι μὲν οὖν ὁ τέταρτος ἀπὸ τῆς μονάδος ὁ Γ κύβος ἐστὶ καὶ οἱ δύο διαλείποντες πάντες, δέδεικται: λέγω δή, ὅτι καὶ οἱ λοιποὶ πάντες κύβοι εἰσίν. ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β, ἰσάκις ἄρα ἡ μονὰς τὸν Α μετρεῖ καὶ ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας: καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας: ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν. καί ἐστιν ὁ Α κύβος. ἐὰν δὲ κύβος ἀριθμὸς ἑαυτὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἐστίν: καὶ ὁ Β ἄρα κύβος ἐστίν. καὶ ἐπεὶ τέσσαρες ἀριθμοὶ οἱ Α, Β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α κύβος, καὶ ὁ Δ ἄρα κύβος ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε κύβος ἐστίν, καὶ ὁμοίως οἱ λοιποὶ πάντες κύβοι εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",287]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'9.10'"],["Rule","'Text'","'If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be not square, neither will any other be square except the third from the unit and all those which leave out one.'"],["Rule","'TextWordCount'",43],["Rule","'GreekText'","'ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα μὴ ᾖ τετράγωνος, οὐδ᾽ ἄλλος οὐδεὶς τετράγωνος ἔσται χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων πάντων.'"],["Rule","'GreekTextWordCount'",33],["Rule","'References'",["List",["List",["Rule","'Book'",8],["Rule","'Theorem'",25]],["List",["Rule","'Book'",8],["Rule","'Theorem'",26]],["List",["Rule","'Book'",9],["Rule","'Theorem'",6]],["List",["Rule","'Book'",9],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'And, if the number after the unit be not cube, neither will any other be cube except the fourth from the unit and all those which leave out two. Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, not be square; I say that neither will any other be square except the third from the unit <and those which leave out one>. For, if possible, let C be square. But B is also square; [IX. 8] [therefore B, C have to one another the ratio which a square number has to a square number]. And, as B is to C, so is A to B; therefore A, B have to one another the ratio which a square number has to a square number; [so that A, B are similar plane numbers]. [VIII. 26, converse] And B is square; therefore A is also square: which is contrary to the hypothesis. Therefore C is not square. Similarly we can prove that neither is any other of the numbers square except the third from the unit and those which leave out one. Next, let A not be cube. I say that neither will any other be cube except the fourth from the unit and those which leave out two. For, if possible, let D be cube. Now C is also cube; for it is fourth from the unit. [IX. 8] And, as C is to D, so is B to C; therefore B also has to C the ratio which a cube has to a cube. And C is cube; therefore B is also cube. [VIII. 25] And since, as the unit is to A, so is A to B, and the unit measures A according to the units in it, therefore A also measures B according to the units in itself; therefore A by multiplying itself has made the cube number B. But, if a number by multiplying itself make a cube number, it is also itself cube. [IX. 6] Therefore A is also cube: which is contrary to the hypothesis. Therefore D is not cube. Similarly we can prove that neither is any other of the numbers cube except the fourth from the unit and those which leave out two.'"],["Rule","'ProofWordCount'",386],["Rule","'GreekProof'","'καὶ ἐὰν ὁ μετὰ τὴν μονάδα κύβος μὴ ᾖ, οὐδὲ ἄλλος οὐδεὶς κύβος ἔσται χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων πάντων. ἔστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α μὴ ἔστω τετράγωνος: λέγω, ὅτι οὐδὲ ἄλλος οὐδεὶς τετράγωνος ἔσται χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων. εἰ γὰρ δυνατόν, ἔστω ὁ Γ τετράγωνος. ἔστι δὲ καὶ ὁ Β τετράγωνος: οἱ Β, Γ ἄρα πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β: οἱ Α, Β ἄρα πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ὥστε οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν. καί ἐστι τετράγωνος ὁ Β: τετράγωνος ἄρα ἐστὶ καὶ ὁ Α: ὅπερ οὐχ ὑπέκειτο. οὐκ ἄρα ὁ Γ τετράγωνός ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλος οὐδεὶς τετράγωνός ἐστι χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων. ἀλλὰ δὴ μὴ ἔστω ὁ Α κύβος. λέγω, ὅτι οὐδ᾽ ἄλλος οὐδεὶς κύβος ἔσται χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων. εἰ γὰρ δυνατόν, ἔστω ὁ Δ κύβος. ἔστι δὲ καὶ ὁ Γ κύβος: τέταρτος γάρ ἐστιν ἀπὸ τῆς μονάδος. καί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, ὁ Β πρὸς τὸν Γ: καὶ ὁ Β ἄρα πρὸς τὸν Γ λόγον ἔχει, ὃν κύβος πρὸς κύβον. καί ἐστιν ὁ Γ κύβος: καὶ ὁ Β ἄρα κύβος ἐστίν. καὶ ἐπεί ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς τὸν Β, ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας, καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας: ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β πεποίηκεν. ἐὰν δὲ ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ αὐτὸς κύβος ἔσται. κύβος ἄρα καὶ ὁ Α: ὅπερ οὐχ ὑπόκειται. οὐκ ἄρα ὁ Δ κύβος ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλος οὐδεὶς κύβος ἐστὶ χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",342]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'9.11'"],["Rule","'Text'","'If as many numbers as we please beginning from an unit be in continued proportion, the less measures the greater according to some one of the numbers which have place among the proportional numbers.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ ἐλάττων τὸν μείζονα μετρεῖ κατά τινα τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον ἀριθμοῖς.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let there be as many numbers as we please, B, C, D, E, beginning from the unit A and in continued proportion; I say that B, the least of the numbers B, C, D, E, measures E according to some one of the numbers C, D. For since, as the unit A is to B, so is D to E, therefore the unit A measures the number B the same number of times as D measures E; therefore, alternately, the unit A measures D the same number of times as B measures E. [VII. 15] But the unit A measures D according to the units in it; therefore B also measures E according to the units in D; so that B the less measures E the greater according to some number of those which have place among the proportional numbers.'"],["Rule","'ProofWordCount'",140],["Rule","'GreekProof'","'ἔστωσαν ἀπὸ μονάδος τῆς Α ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Β, Γ, Δ, Ε: λέγω, ὅτι τῶν Β, Γ, Δ, Ε ὁ ἐλάχιστος ὁ Β τὸν Ε μετρεῖ κατά τινα τῶν Γ, Δ. ἐπεὶ γάρ ἐστιν ὡς ἡ Α μονὰς πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε, ἰσάκις ἄρα ἡ Α μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Δ τὸν Ε: ἐναλλὰξ ἄρα ἰσάκις ἡ Α μονὰς τὸν Δ μετρεῖ καὶ ὁ Β τὸν Ε. ἡ δὲ Α μονὰς τὸν Δ μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας: καὶ ὁ Β ἄρα τὸν Ε μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας: ὥστε ὁ ἐλάσσων ὁ Β τὸν μείζονα τὸν Ε μετρεῖ κατά τινα ἀριθμὸν τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον ἀριθμοῖς. Πόρισμα Καὶ φανερόν, ὅτι ἣν ἔχει τάξιν ὁ μετρῶν ἀπὸ μονάδος, τὴν αὐτὴν ἔχει καὶ ὁ καθ᾽ ὃν μετρεῖ ἀπὸ τοῦ μετρουμένου ἐπὶ τὸ πρὸ αὐτοῦ. _ ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",151]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'9.12'"],["Rule","'Text'","'If as many numbers as we please beginning from an unit be in continued proportion, by however many prime numbers the last is measured, the next to the unit will also be measured by the same.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὑφ᾽ ὅσων ἂν ὁ ἔσχατος πρώτων ἀριθμῶν μετρῆται, ὑπὸ τῶν αὐτῶν καὶ ὁ παρὰ τὴν μονάδα μετρηθήσεται.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",14]],["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",29]],["List",["Rule","'Book'",9],["Rule","'Theorem'",8]],["List",["Rule","'Book'",9],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'Let there be as many numbers as we please, A, B, C, D, beginning from an unit, and in continued proportion; I say that, by however many prime numbers D is measured, A will also be measured by the same. For let D be measured by any prime number E; I say that E measures A. For suppose it does not; now E is prime, and any prime number is prime to any which it does not measure; [VII. 29] therefore E, A are prime to one another. And, since E measures D, let it measure it according to F, therefore E by multiplying F has made D. Again, since A measures D according to the units in C, [IX. 11 and Por.] therefore A by multiplying C has made D. But, further, E has also by multiplying F made D; therefore the product of A, C is equal to the product of E, F. Therefore, as A is to E, so is F to C. [VII. 19] But A, E are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures C. Let it measure it according to G; therefore E by multiplying G has made C. But, further, by the theorem before this, A has also by multiplying B made C. [IX. 11 and Por.] Therefore the product of A, B is equal to the product of E, G. Therefore, as A is to E, so is G to B. [VII. 19] But A, E are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent: [VII. 20] therefore E measures B. Let it measure it according to H; therefore E by multiplying H has made B. But further A has also by multiplying itself made B; [IX. 8] therefore the product of E, H is equal to the square on A. Therefore, as E is to A, so is A to H. [VII. 19] But A, E are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures A, as antecedent antecedent. But, again, it also does not measure it: which is impossible. Therefore E, A are not prime to one another. Therefore they are composite to one another. But numbers composite to one another are measured by some number. [VII. Def. 14] And, since E is by hypothesis prime, and the prime is not measured by any number other than itself, therefore E measures A, E, so that E measures A. [But it also measures D; therefore E measures A, D.] Similarly we can prove that, by however many prime numbers D is measured, A will also be measured by the same.'"],["Rule","'ProofWordCount'",512],["Rule","'GreekProof'","'ἔστωσαν ἀπὸ μονάδος ὁποσοιδηποτοῦν ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ: λέγω, ὅτι ὑφ᾽ ὅσων ἂν ὁ Δ πρώτων ἀριθμῶν μετρῆται, ὑπὸ τῶν αὐτῶν καὶ ὁ Α μετρηθήσεται. μετρείσθω γὰρ ὁ Δ ὑπό τινος πρώτου ἀριθμοῦ τοῦ Ε: λέγω, ὅτι ὁ ε τὸν Α μετρεῖ. μὴ γάρ: καί ἐστιν ὁ Ε πρῶτος, ἅπας δὲ πρῶτος ἀριθμὸς πρὸς ἅπαντα, ὃν μὴ μετρεῖ, πρῶτός ἐστιν: οἱ Ε, Α ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ: ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. πάλιν, ἐπεὶ ὁ Α τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας, ὁ Α ἄρα τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Ε τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν: ὁ ἄρα ἐκ τῶν Α, Γ ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Ζ. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Ε, ὁ Ζ πρὸς τὸν Γ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: μετρεῖ ἄρα ὁ Ε τὸν Γ. μετρείτω αὐτὸν κατὰ τὸν Η: ὁ Ε ἄρα τὸν Η πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν διὰ τὸ πρὸ τούτου καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. ὁ ἄρα ἐκ τῶν Α, Β ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Η. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Ε, ὁ Η πρὸς τὸν Β. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: μετρεῖ ἄρα ὁ Ε τὸν Β. μετρείτω αὐτὸν κατὰ τὸν Θ: ὁ Ε ἄρα τὸν Θ πολλαπλασιάσας τὸν Β πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν: ὁ ἄρα ἐκ τῶν Ε, Θ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Α. ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Α, ὁ Α πρὸς τὸν Θ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: μετρεῖ ἄρα ὁ Ε τὸν Α ὡς ἡγούμενος ἡγούμενον. ἀλλὰ μὴν καὶ οὐ μετρεῖ: ὅπερ ἀδύνατον. οὐκ ἄρα οἱ Ε, Α πρῶτοι πρὸς ἀλλήλους εἰσίν. σύνθετοι ἄρα. οἱ δὲ σύνθετοι ὑπὸ πρώτου ἀριθμοῦ τινος μετροῦνται. καὶ ἐπεὶ ὁ Ε πρῶτος ὑπόκειται, ὁ δὲ πρῶτος ὑπὸ ἑτέρου ἀριθμοῦ οὐ μετρεῖται á¼¢ ὑφ᾽ ἑαυτοῦ, ὁ Ε ἄρα τοὺς Α, Ε μετρεῖ: ὥστε ὁ Ε τὸν Α μετρεῖ. μετρεῖ δὲ καὶ τὸν Δ: ὁ Ε ἄρα τοὺς Α, Δ μετρεῖ. ὁμοίως δὴ δείξομεν, ὅτι ὑφ᾽ ὅσων ἂν ὁ Δ πρώτων ἀριθμῶν μετρῆται, ὑπὸ τῶν αὐτῶν καὶ ὁ Α μετρηθήσεται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",461]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'9.13'"],["Rule","'Text'","'If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be prime, the greatest will not be measured by any except those which have a place among the proportional numbers.'"],["Rule","'TextWordCount'",41],["Rule","'GreekText'","'ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα πρῶτος ᾖ, ὁ μέγιστος ὑπ᾽ οὐδενὸς ἄλλου μετρηθήσεται παρὲξ τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον ἀριθμοῖς.'"],["Rule","'GreekTextWordCount'",29],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",7],["Rule","'Theorem'",31]],["List",["Rule","'Book'",9],["Rule","'Theorem'",8]],["List",["Rule","'Book'",9],["Rule","'Theorem'",11]],["List",["Rule","'Book'",9],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'Let there be as many numbers as we please, A, B, C, D, beginning from an unit and in continued proportion, and let A, the number after the unit, be prime; I say that D, the greatest of them, will not be measured by any other number except A, B, C. For, if possible, let it be measured by E, and let E not be the same with any of the numbers A, B, C. It is then manifest that E is not prime. For, if E is prime and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible. Therefore E is not prime. Therefore it is composite. But any composite number is measured by some prime number; [VII. 31] therefore E is measured by some prime number. I say next that it will not be measured by any other prime except A. For, if E is measured by another, and E measures D, that other will also measure D; so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible. Therefore A measures E. And, since E measures D, let it measure it according to F. I say that F is not the same with any of the numbers A, B, C. For, if F is the same with one of the numbers A, B, C, and measures D according to E, therefore one of the numbers A, B, C also measures D according to E. But one of the numbers A, B, C measures D according to some one of the numbers A, B, C; [IX. 11] therefore E is also the same with one of the numbers A, B, C: which is contrary to the hypothesis. Therefore F is not the same as any one of the numbers A, B, C. Similarly we can prove that F is measured by A, by proving again that F is not prime. For, if it is, and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible; therefore F is not prime. Therefore it is composite. But any composite number is measured by some prime number; [VII. 31] therefore F is measured by some prime number. I say next that it will not be measured by any other prime except A. For, if any other prime number measures F, and F measures D, that other will also measure D; so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible. Therefore A measures F. And, since E measures D according to F, therefore E by multiplying F has made D. But, further, A has also by multiplying C made D; [IX. 11] therefore the product of A, C is equal to the product of E, F. Therefore, proportionally, as A is to E, so is F to C. [VII. 19] But A measures E; therefore F also measures C. Let it measure it according to G. Similarly, then, we can prove that G is not the same with any of the numbers A, B, and that it is measured by A. And, since F measures C according to G therefore F by multiplying G has made C. But, further, A has also by multiplying B made C; [IX. 11] therefore the product of A, B is equal to the product of F, G. Therefore, proportionally, as A is to F, so is G to B. [VII. 19] But A measures F; therefore G also measures B. Let it measure it according to H. Similarly then we can prove that H is not the same with A. And, since G measures B according to H, therefore G by multiplying H has made B. But further A has also by multiplying itself made B; [IX. 8] therefore the product of H, G is equal to the square on A. Therefore, as H is to A, so is A to G. [VII. 19] But A measures G; therefore H also measures A, which is prime, though it is not the same with it: which is absurd. Therefore D the greatest will not be measured by any other number except A, B, C.'"],["Rule","'ProofWordCount'",730],["Rule","'GreekProof'","'ἔστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α πρῶτος ἔστω: λέγω, ὅτι ὁ μέγιστος αὐτῶν ὁ Δ ὑπ᾽ οὐδενὸς ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ. εἰ γὰρ δυνατόν, μετρείσθω ὑπὸ τοῦ Ε, καὶ ὁ Ε μηδενὶ τῶν Α, Β, Γ ἔστω ὁ αὐτός. φανερὸν δή, ὅτι ὁ Ε πρῶτος οὔκ ἐστιν. εἰ γὰρ ὁ Ε πρῶτός ἐστι καὶ μετρεῖ τὸν Δ, καὶ τὸν Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ὁ Ε πρῶτός ἐστιν. σύνθετος ἄρα. πᾶς δὲ σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται: ὁ Ε ἄρα ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. λέγω δή, ὅτι ὑπ᾽ οὐδενὸς ἄλλου πρώτου μετρηθήσεται πλὴν τοῦ Α. εἰ γὰρ ὑφ᾽ ἑτέρου μετρεῖται ὁ Ε, ὁ δὲ Ε τὸν Δ μετρεῖ, κἀκεῖνος ἄρα τὸν Δ μετρήσει: ὥστε καὶ τὸν Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός: ὅπερ ἐστὶν ἀδύνατον. ὁ Α ἄρα τὸν Ε μετρεῖ. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ. λέγω, ὅτι ὁ Ζ οὐδενὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. εἰ γὰρ ὁ Ζ ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτὸς καὶ μετρεῖ τὸν Δ κατὰ τὸν Ε, καὶ εἷς ἄρα τῶν Α, Β, Γ τὸν Δ μετρεῖ κατὰ τὸν Ε. ἀλλὰ εἷς τῶν Α, Β, Γ τὸν Δ μετρεῖ κατά τινα τῶν Α, Β, Γ: καὶ ὁ Ε ἄρα ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός: ὅπερ οὐχ ὑπόκειται. οὐκ ἄρα ὁ Ζ ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. ὁμοίως δὴ δείξομεν, ὅτι μετρεῖται ὁ Ζ ὑπὸ τοῦ Α, δεικνύντες πάλιν, ὅτι ὁ Ζ οὔκ ἐστι πρῶτος. εἰ γάρ, καὶ μετρεῖ τὸν Δ, καὶ τὸν α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα πρῶτός ἐστιν ὁ Ζ: σύνθετος ἄρα. ἅπας δὲ σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται: ὁ Ζ ἄρα ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. λέγω δή, ὅτι ὑφ᾽ ἑτέρου πρώτου οὐ μετρηθήσεται πλὴν τοῦ α. εἰ γὰρ ἕτερός τις πρῶτος τὸν Ζ μετρεῖ, ὁ δὲ Ζ τὸν Δ μετρεῖ, κἀκεῖνος ἄρα τὸν Δ μετρήσει: ὥστε καὶ τὸν Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός: ὅπερ ἐστὶν ἀδύνατον. ὁ Α ἄρα τὸν Ζ μετρεῖ. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ κατὰ τὸν Ζ, ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν: ὁ ἄρα ἐκ τῶν Α, Γ ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Ζ. ἀνάλογον ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Ε, οὕτως ὁ Ζ πρὸς τὸν Γ. ὁ δὲ Α τὸν Ε μετρεῖ: καὶ ὁ Ζ ἄρα τὸν Γ μετρεῖ. μετρείτω αὐτὸν κατὰ τὸν Η. ὁμοίως δὴ δείξομεν, ὅτι ὁ Η οὐδενὶ τῶν Α, Β ἐστιν ὁ αὐτός, καὶ ὅτι μετρεῖται ὑπὸ τοῦ Α. καὶ ἐπεὶ ὁ Ζ τὸν Γ μετρεῖ κατὰ τὸν Η, ὁ Ζ ἄρα τὸν Η πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν: ὁ ἄρα ἐκ τῶν Α, Β ἴσος ἐστὶ τῷ ἐκ τῶν Ζ, Η. ἀνάλογον ἄρα ὡς ὁ Α πρὸς τὸν Ζ, ὁ Η πρὸς τὸν Β. μετρεῖ δὲ ὁ Α τὸν Ζ: μετρεῖ ἄρα καὶ ὁ Η τὸν Β. μετρείτω αὐτὸν κατὰ τὸν Θ. ὁμοίως δὴ δείξομεν, ὅτι ὁ Θ τῷ Α οὐκ ἔστιν ὁ αὐτός. καὶ ἐπεὶ ὁ Η τὸν Β μετρεῖ κατὰ τὸν Θ, ὁ Η ἄρα τὸν Θ πολλαπλασιάσας τὸν Β πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν: ὁ ἄρα ὑπὸ Θ, Η ἴσος ἐστὶ τῷ ἀπὸ τοῦ Α τετραγώνῳ. ἔστιν ἄρα ὡς ὁ Θ πρὸς τὸν Α, ὁ Α πρὸς τὸν Η. μετρεῖ δὲ ὁ Α τὸν Η: μετρεῖ ἄρα καὶ ὁ Θ τὸν Α πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός: ὅπερ ἄτοπον. οὐκ ἄρα ὁ μέγιστος ὁ Δ ὑπὸ ἑτέρου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Α, Β, Γ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",653]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'9.14'"],["Rule","'Text'","'If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it.'"],["Rule","'TextWordCount'",27],["Rule","'GreekText'","'ἐὰν ἐλάχιστος ἀριθμὸς ὑπὸ πρώτων ἀριθμῶν μετρῆται, ὑπ᾽ οὐδενὸς ἄλλου πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν ἐξ ἀρχῆς μετρούντων.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",30]]]],["Rule","'Proof'","'For let the number A be the least that is measured by the prime numbers B, C, D; I say that A will not be measured by any other prime number except B, C, D. For, if possible, let it be measured by the prime number E, and let E not be the same with any one of the numbers B, C, D. Now, since E measures A, let it measure it according to F; therefore E by multiplying F has made A. And A is measured by the prime numbers B, C, D. But, if two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers; [VII. 30] therefore B, C, D will measure one of the numbers E, F. Now they will not measure E; for E is prime and not the same with any one of the numbers B, C, D. Therefore they will measure F, which is less than A: which is impossible, for A is by hypothesis the least number measured by B, C, D. Therefore no prime number will measure A except B, C, D.'"],["Rule","'ProofWordCount'",195],["Rule","'GreekProof'","'ἐλάχιστος γὰρ ἀριθμὸς ὁ Α ὑπὸ πρώτων ἀριθμῶν τῶν Β, Γ, Δ μετρείσθω: λέγω, ὅτι ὁ Α ὑπ᾽ οὐδενὸς ἄλλου πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Β, Γ, Δ. εἰ γὰρ δυνατόν, μετρείσθω ὑπὸ πρώτου τοῦ Ε, καὶ ὁ Ε μηδενὶ τῶν Β, Γ, Δ ἔστω ὁ αὐτός. καὶ ἐπεὶ ὁ Ε τὸν Α μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ: ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ μετρεῖται ὁ Α ὑπὸ πρώτων ἀριθμῶν τῶν Β, Γ, Δ. ἐὰν δὲ δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, τὸν δὲ γενόμενον ἐξ αὐτῶν μετρῇ τις πρῶτος ἀριθμός, καὶ ἕνα τῶν ἐξ ἀρχῆς μετρήσει: οἱ Β, Γ, Δ ἄρα ἕνα τῶν Ε, Ζ μετρήσουσιν. τὸν μὲν οὖν Ε οὐ μετρήσουσιν: ὁ γὰρ Ε πρῶτός ἐστι καὶ οὐδενὶ τῶν Β, Γ, Δ ὁ αὐτός. τὸν Ζ ἄρα μετροῦσιν ἐλάσσονα ὄντα τοῦ Α: ὅπερ ἀδύνατον. ὁ γὰρ Α ὑπόκειται ἐλάχιστος ὑπὸ τῶν Β, Γ, Δ μετρούμενος. οὐκ ἄρα τὸν Α μετρήσει πρῶτος ἀριθμὸς παρὲξ τῶν Β, Γ, Δ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",168]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'9.15'"],["Rule","'Text'","'If three numbers in continued proportion be the least of those which have the same ratio with them, any two whatever added together will be prime to the remaining number.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'ἐὰν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, δύο ὁποιοιοῦν συντεθέντες πρὸς τὸν λοιπὸν πρῶτοί εἰσιν.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",3]],["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",7],["Rule","'Theorem'",22]],["List",["Rule","'Book'",7],["Rule","'Theorem'",24]],["List",["Rule","'Book'",7],["Rule","'Theorem'",25]],["List",["Rule","'Book'",7],["Rule","'Theorem'",28]],["List",["Rule","'Book'",8],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let A, B, C, three numbers in continued proportion, be the least of those which have the same ratio with them; I say that any two of the numbers A, B, C whatever added together are prime to the remaining number, namely A, B to C; B, C to A; and further A, C to B. For let two numbers DE, EF, the least of those which have the same ratio with A, B, C, be taken. [VIII. 2] It is then manifest that DE by multiplying itself has made A, and by multiplying EF has made B, and, further, EF by multiplying itself has made C. [VIII. 2] Now, since DE, EF are least, they are prime to one another. [VII. 22] But, if two numbers be prime to one another, their sum is also prime to each; [VII. 28] therefore DF is also prime to each of the numbers DE, EF. But further DE is also prime to EF; therefore DF, DE are prime to EF. But, if two numbers be prime to any number, their product is also prime to the other; [VII. 24] so that the product of FD, DE is prime to EF; hence the product of FD, DE is also prime to the square on EF. [VII. 25] But the product of FD, DE is the square on DE together with the product of DE, EF; [II. 3] therefore the square on DE together with the product of DE, EF is prime to the square on EF. And the square on DE is A, the product of DE, EF is B, and the square on EF is C; therefore A, B added together are prime to C. Similarly we can prove that B, C added together are prime to A. I say next that A, C added together are also prime to B. For, since DF is prime to each of the numbers DE, EF, the square on DF is also prime to the product of DE, EF. [VII. 24, 25] But the squares on DE, EF together with twice the product of DE, EF are equal to the square on DF; [II. 4] therefore the squares on DE, EF together with twice the product of DE, EF are prime to the product of DE, EF. Separando, the squares on DE, EF together with once the product of DE, EF are prime to the product of DE, EF. Therefore, separando again, the squares on DE, EF are prime to the product of DE, EF. And the square on DE is A, the product of DE, EF is B, and the square on EF is C. Therefore A, C added together are prime to B.'"],["Rule","'ProofWordCount'",450],["Rule","'GreekProof'","'ἔστωσαν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς οἱ Α, Β, Γ: λέγω, ὅτι τῶν Α, Β, Γ δύο ὁποιοιοῦν συντεθέντες πρὸς τὸν λοιπὸν πρῶτοί εἰσιν, οἱ μὲν Α, Β πρὸς τὸν Γ, οἱ δὲ Β, Γ πρὸς τὸν Α καὶ ἔτι οἱ Α, Γ πρὸς τὸν Β. εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ δύο οἱ ΔΕ, ΕΖ. φανερὸν δή, ὅτι ὁ μὲν ΔΕ ἑαυτὸν πολλαπλασιάσας τὸν Α πεποίηκεν, τὸν δὲ ΕΖ πολλαπλασιάσας τὸν Β πεποίηκεν, καὶ ἔτι ὁ ΕΖ ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν. καὶ ἐπεὶ οἱ ΔΕ, ΕΖ ἐλάχιστοί εἰσιν, πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐὰν δὲ δύο ἀριθμοί πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ συναμφότερος πρὸς ἑκάτερον πρῶτός ἐστιν: καὶ ὁ ΔΖ ἄρα πρὸς ἑκάτερον τῶν ΔΕ, ΕΖ πρῶτός ἐστιν. ἀλλὰ μὴν καὶ ὁ ΔΕ πρὸς τὸν ΕΖ πρῶτός ἐστιν: οἱ ΔΖ, ΔΕ ἄρα πρὸς τὸν ΕΖ πρῶτοί εἰσιν. ἐὰν δὲ δύο ἀριθμοὶ πρός τινα ἀριθμὸν πρῶτοι ὦσιν, καὶ ὁ ἐξ αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτός ἐστιν: ὥστε ὁ ἐκ τῶν ΖΔ, ΔΕ πρὸς τὸν ΕΖ πρῶτός ἐστιν: ὥστε καὶ ὁ ἐκ τῶν ΖΔ, ΔΕ πρὸς τὸν ἀπὸ τοῦ ΕΖ πρῶτός ἐστιν. ἐὰν γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ ἐκ τοῦ ἑνὸς αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτός ἐστιν. ἀλλ᾽ ὁ ἐκ τῶν ΖΔ, ΔΕ ὁ ἀπὸ τοῦ ΔΕ ἐστι μετὰ τοῦ ἐκ τῶν ΔΕ, ΕΖ: ὁ ἄρα ἀπὸ τοῦ ΔΕ μετὰ τοῦ ἐκ τῶν ΔΕ, ΕΖ πρὸς τὸν ἀπὸ τοῦ ΕΖ πρῶτός ἐστιν. καί ἐστιν ὁ μὲν ἀπὸ τοῦ ΔΕ ὁ Α, ὁ δὲ ἐκ τῶν ΔΕ, ΕΖ ὁ Β, ὁ δὲ ἀπὸ τοῦ ΕΖ ὁ Γ: οἱ Α, Β ἄρα συντεθέντες πρὸς τὸν Γ πρῶτοί εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ Β, Γ πρὸς τὸν Α πρῶτοί εἰσιν. λέγω δή, ὅτι καὶ οἱ Α, Γ πρὸς τὸν Β πρῶτοί εἰσιν. ἐπεὶ γὰρ ὁ ΔΖ πρὸς ἑκάτερον τῶν ΔΕ, ΕΖ πρῶτός ἐστιν, καὶ ὁ ἀπὸ τοῦ ΔΖ πρὸς τὸν ἐκ τῶν ΔΕ, ΕΖ πρῶτός ἐστιν. ἀλλὰ τῷ ἀπὸ τοῦ ΔΖ ἴσοι εἰσὶν οἱ ἀπὸ τῶν ΔΕ, ΕΖ μετὰ τοῦ δὶς ἐκ τῶν ΔΕ, ΕΖ: καὶ οἱ ἀπὸ τῶν ΔΕ, ΕΖ ἄρα μετὰ τοῦ δὶς ὑπὸ τῶν ΔΕ, ΕΖ πρὸς τὸν ὑπὸ τῶν ΔΕ, ΕΖ πρῶτοί εἰσι. διελόντι οἱ ἀπὸ τῶν ΔΕ, ΕΖ μετὰ τοῦ ἅπαξ ὑπὸ ΔΕ, ΕΖ πρὸς τὸν ὑπὸ ΔΕ, ΕΖ πρῶτοί εἰσιν. ἔτι διελόντι οἱ ἀπὸ τῶν ΔΕ, ΕΖ ἄρα πρὸς τὸν ὑπὸ ΔΕ, ΕΖ πρῶτοί εἰσιν. καί ἐστιν ὁ μὲν ἀπὸ τοῦ ΔΕ ὁ Α, ὁ δὲ ὑπὸ τῶν ΔΕ, ΕΖ ὁ Β, ὁ δὲ ἀπὸ τοῦ ΕΖ ὁ Γ. οἱ Α, Γ ἄρα συντεθέντες πρὸς τὸν Β πρῶτοί εἰσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",445]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'9.16'"],["Rule","'Text'","'If two numbers be prime to one another, the second will not be to any other number as the first is to the second.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, οὐκ ἔσται ὡς ὁ πρῶτος πρὸς τὸν δεύτερον, οὕτως ὁ δεύτερος πρὸς ἄλλον τινά.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'For let the two numbers A, B be prime to one another; I say that B is not to any other number as A is to B. For, if possible, as A is to B, so let B be to C. Now A, B are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore A measures B as antecedent antecedent. But it also measures itself; therefore A measures A, B which are prime to one another: which is absurd. Therefore B will not be to C, as A is to B.'"],["Rule","'ProofWordCount'",117],["Rule","'GreekProof'","'δύο γὰρ ἀριθμοὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους ἔστωσαν: λέγω, ὅτι οὐκ ἔστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β πρὸς ἄλλον τινά. εἰ γὰρ δυνατόν, ἔστω ὡς ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Γ. οἱ δὲ Α, Β πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: μετρεῖ ἄρα ὁ Α τὸν Β ὡς ἡγούμενος ἡγούμενον. μετρεῖ δὲ καὶ ἑαυτόν: ὁ Α ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἄτοπον. οὐκ ἄρα ἔσται ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β πρὸς τὸν Γ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",116]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'9.17'"],["Rule","'Text'","'If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the last will not be to any other number as the first to the second.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ δὲ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, οὐκ ἔσται ὡς ὁ πρῶτος πρὸς τὸν δεύτερον, οὕτως ὁ ἔσχατος πρὸς ἄλλον τινά.'"],["Rule","'GreekTextWordCount'",28],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",13]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'For let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them, A, D, be prime to one another; I say that D is not to any other number as A is to B. For, if possible, as A is to B, so let D be to E; therefore, alternately, as A is to D, so is B to E. [VII. 13] But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent. [VII. 20] Therefore A measures B. And, as A is to B, so is B to C. Therefore B also measures C; so that A also measures C. And since, as B is to C, so is C to D, and B measures C, therefore C also measures D. But A measured C; so that A also measures D. But it also measures itself; therefore A measures A, D which are prime to one another: which is impossible. Therefore D will not be to any other number as A is to B.'"],["Rule","'ProofWordCount'",201],["Rule","'GreekProof'","'ἔστωσαν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, οἱ δὲ ἄκροι αὐτῶν οἱ Α, Δ πρῶτοι πρὸς ἀλλήλους ἔστωσαν: λέγω, ὅτι οὐκ ἔστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς ἄλλον τινά. εἰ γὰρ δυνατόν, ἔστω ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε: ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, ὁ Β πρὸς τὸν Ε. οἱ δὲ Α, Δ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον. μετρεῖ ἄρα ὁ Α τὸν Β. καί ἐστιν ὡς ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Γ. καὶ ὁ Β ἄρα τὸν Γ μετρεῖ: ὥστε καὶ ὁ Α τὸν Γ μετρεῖ. καὶ ἐπεί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Δ, μετρεῖ δὲ ὁ Β τὸν Γ, μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ. ἀλλ᾽ ὁ Α τὸν Γ ἐμέτρει: ὥστε ὁ Α καὶ τὸν Δ μετρεῖ. μετρεῖ δὲ καὶ ἑαυτόν. ὁ Α ἄρα τοὺς α, Δ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσται ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς ἄλλον τινά: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",208]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'9.18'"],["Rule","'Text'","'Given two numbers, to investigate whether it is possible to find a third proportional to them.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'δύο ἀριθμῶν δοθέντων ἐπισκέψασθαι, εἰ δυνατόν ἐστιν αὐτοῖς τρίτον ἀνάλογον προσευρεῖν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",9],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'Let A, B be the given two numbers, and let it be required to investigate whether it is possible to find a third proportional to them. Now A, B are either prime to one another or not. And, if they are prime to one another, it has been proved that it is impossible to find a third proportional to them. [IX. 16] Next, let A, B not be prime to one another, and let B by multiplying itself make C. Then A either measures C or does not measure it. First, let it measure it according to D; therefore A by multiplying D has made C. But, further, B has also by multiplying itself made C; therefore the product of A, D is equal to the square on B. Therefore, as A is to B, so is B to D; [VII. 19] therefore a third proportional number D has been found to A, B. Next, let A not measure C; I say that it is impossible to find a third proportional number to A, B. For, if possible, let D, such third proportional, have been found. Therefore the product of A, D is equal to the square on B. But the square on B is C; therefore the product of A, D is equal to C. Hence A by multiplying D has made C; therefore A measures C according to D. But, by hypothesis, it also does not measure it: which is absurd. Therefore it is not possible to find a third proportional number to A, B when A does not measure C.'"],["Rule","'ProofWordCount'",263],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες δύο ἀριθμοὶ οἱ Α, Β, καὶ δέον ἔστω ἐπισκέψασθαι, εἰ δυνατόν ἐστιν αὐτοῖς τρίτον ἀνάλογον προσευρεῖν. οἱ δὴ Α, Β ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν á¼¢ οὔ. καὶ εἰ πρῶτοι πρὸς ἀλλήλους εἰσίν, δέδεικται, ὅτι ἀδύνατόν ἐστιν αὐτοῖς τρίτον ἀνάλογον προσευρεῖν. ἀλλὰ δὴ μὴ ἔστωσαν οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Γ ποιείτω: ὁ Α δὴ τὸν Γ ἤτοι μετρεῖ á¼¢ οὐ μετρεῖ. μετρείτω πρότερον κατὰ τὸν Δ: ὁ Α ἄρα τὸν Δ πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν: ὁ ἄρα ἐκ τῶν Α, Δ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Β. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Δ: τοῖς Α, Β ἄρα τρίτος ἀριθμὸς ἀνάλογον προσηύρηται ὁ Δ. ἀλλὰ δὴ μὴ μετρείτω ὁ Α τὸν Γ: λέγω, ὅτι τοῖς Α, Β ἀδύνατόν ἐστι τρίτον ἀνάλογον προσευρεῖν ἀριθμόν. εἰ γὰρ δυνατόν, προσηυρήσθω ὁ Δ. ὁ ἄρα ἐκ τῶν Α, Δ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Β. ὁ δὲ ἀπὸ τοῦ Β ἐστιν ὁ Γ: ὁ ἄρα ἐκ τῶν Α, Δ ἴσος ἐστὶ τῷ Γ. ὥστε ὁ Α τὸν Δ πολλαπλασιάσας τὸν Γ πεποίηκεν: ὁ Α ἄρα τὸν Γ μετρεῖ κατὰ τὸν Δ. ἀλλὰ μὴν ὑπόκειται καὶ μὴ μετρῶν: ὅπερ ἄτοπον. οὐκ ἄρα δυνατόν ἐστι τοῖς Α, Β τρίτον ἀνάλογον προσευρεῖν ἀριθμόν, ὅταν ὁ Α τὸν Γ μὴ μετρῇ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",233]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'9.19'"],["Rule","'Text'","'Given three numbers, to investigate when it is possible to find a fourth proportional to them.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'τριῶν ἀριθμῶν δοθέντων ἐπισκέψασθαι, πότε δυνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",14]],["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]],["List",["Rule","'Book'",9],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'Let A, B, C be the given three numbers, and let it be required to investigate when it is possible to find a fourth proportional to them. Now either they are not in continued proportion, and the extremes of them are prime to one another; or they are in continued proportion, and the extremes of them are not prime to one another; or they are not in continued proportion, nor are the extremes of them prime to one another; or they are in continued proportion, and the extremes of them are prime to one another. If then A, B, C are in continued proportion, and the extremes of them A, C are prime to one another, it has been proved that it is impossible to find a fourth proportional number to them. [IX. 17] <*>Next, let A, B, C not be in continued proportion, the extremes being again prime to one another; I say that in this case also it is impossible to find a fourth proportional to them. For, if possible, let D have been found, so that, as A is to B, so is C to D, and let it be contrived that, as B is to C, so is D to E. Now, since, as A is to B, so is C to D, and, as B is to C, so is D to E, therefore, ex aequali, as A is to C, so is C to E. [VII. 14] But A, C are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio, the antecedent the antecedent and the consequent the consequent. [VII. 20] Therefore A measures C as antecedent antecedent. But it also measures itself; therefore A measures A, C which are prime to one another: which is impossible. Therefore it is not possible to find a fourth proportional to A, B, C.<*> Next, let A, B, C be again in continued proportion, but let A, C not be prime to one another. I say that it is possible to find a fourth proportional to them. For let B by multiplying C make D; therefore A either measures D or does not measure it. First, let it measure it according to E; therefore A by multiplying E has made D. But, further, B has also by multiplying C made D; therefore the product of A, E is equal to the product of B, C; therefore, proportionally, as A is to B, so is C to E; [VII. 19] therefore E has been found a fourth proportional to A, B, C. Next, let A not measure D; I say that it is impossible to find a fourth proportional number to A, B, C. For, if possible, let E have been found; therefore the product of A, E is equal to the product of B, C. [VII. 19] But the product of B, C is D; therefore the product of A, E is also equal to D. Therefore A by multiplying E has made D; therefore A measures D according to E, so that A measures D. But it also does not measure it: which is absurd. Therefore it is not possible to find a fourth proportional number to A, B, C when A does not measure D. Next, let A, B, C not be in continued proportion, nor the extremes prime to one another. And let B by multiplying C make D. Similarly then it can be proved that, if A measures D, it is possible to find a fourth proportional to them, but, if it does not measure it, impossible.'"],["Rule","'ProofWordCount'",603],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ οἱ Α, Β, Γ, καὶ δέον ἔστω ἐπισκέψασθαι, πότε δυνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν. ἤτοι οὖν οὔκ εἰσιν ἑξῆς ἀνάλογον, καὶ οἱ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους εἰσίν, á¼¢ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ ἄκροι αὐτῶν οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους, á¼¢ οὔτε ἑξῆς εἰσιν ἀνάλογον, οὔτε οἱ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους εἰσίν, á¼¢ καὶ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ μὲν οὖν οἱ Α, Β, Γ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ ἄκροι αὐτῶν οἱ Α, Γ πρῶτοι πρὸς ἀλλήλους εἰσίν, δέδεικται, ὅτι ἀδύνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν ἀριθμόν. μὴ ἔστωσαν δὴ οἱ Α, Β, Γ ἑξῆς ἀνάλογον τῶν ἄκρων πάλιν ὄντων πρώτων πρὸς ἀλλήλους. λέγω, ὅτι καὶ οὕτως ἀδύνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν. εἰ γὰρ δυνατόν, προσευρήσθω ὁ Δ, ὥστε εἶναι ὡς τὸν Α πρὸς τὸν Β, τὸν Γ πρὸς τὸν Δ, καὶ γεγονέτω ὡς ὁ Β πρὸς τὸν Γ, ὁ Δ πρὸς τὸν Ε. καὶ ἐπεί ἐστιν ὡς μὲν ὁ Α πρὸς τὸν Β, ὁ Γ πρὸς τὸν Δ, ὡς δὲ ὁ Β πρὸς τὸν Γ, ὁ Δ πρὸς τὸν Ε, δι᾽ ἴσου ἄρα ὡς ὁ Α πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Ε. οἱ δὲ Α, Γ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον. μετρεῖ ἄρα ὁ Α τὸν Γ ὡς ἡγούμενος ἡγούμενον. μετρεῖ δὲ καὶ ἑαυτόν: ὁ Α ἄρα τοὺς Α, Γ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοῖς Α, Β, Γ δυνατόν ἐστι τέταρτον ἀνάλογον προσευρεῖν. ἀλλὰ δὴ πάλιν ἔστωσαν οἱ Α, Β, Γ ἑξῆς ἀνάλογον, οἱ δὲ Α, Γ μὴ ἔστωσαν πρῶτοι πρὸς ἀλλήλους. λέγω, ὅτι δυνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν. ὁ γὰρ Β τὸν Γ πολλαπλασιάσας τὸν Δ ποιείτω: ὁ Α ἄρα τὸν Δ ἤτοι μετρεῖ á¼¢ οὐ μετρεῖ. μετρείτω αὐτὸν πρότερον κατὰ τὸν Ε: ὁ Α ἄρα τὸν Ε πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Β τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν: ὁ ἄρα ἐκ τῶν Α, Ε ἴσος ἐστὶ τῷ ἐκ τῶν Β, Γ. ἀνάλογον ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Β, ὁ Γ πρὸς τὸν Ε: τοῖς Α, Β, Γ ἄρα τέταρτος ἀνάλογον προσηύρηται ὁ Ε. ἀλλὰ δὴ μὴ μετρείτω ὁ Α τὸν Δ: λέγω, ὅτι ἀδύνατόν ἐστι τοῖς Α, Β, Γ τέταρτον ἀνάλογον προσευρεῖν ἀριθμόν. εἰ γὰρ δυνατόν, προσευρήσθω ὁ Ε: ὁ ἄρα ἐκ τῶν Α, Ε ἴσος ἐστὶ τῷ ἐκ τῶν Β, Γ. ἀλλὰ ὁ ἐκ τῶν Β, Γ ἐστιν ὁ Δ: καὶ ὁ ἐκ τῶν Α, Ε ἄρα ἴσος ἐστὶ τῷ Δ. ὁ Α ἄρα τὸν Ε πολλαπλασιάσας τὸν Δ πεποίηκεν: ὁ Α ἄρα τὸν Δ μετρεῖ κατὰ τὸν Ε: ὥστε μετρεῖ ὁ Α τὸν Δ. ἀλλὰ καὶ οὐ μετρεῖ: ὅπερ ἄτοπον. οὐκ ἄρα δυνατόν ἐστι τοῖς Α, Β, Γ τέταρτον ἀνάλογον προσευρεῖν ἀριθμόν, ὅταν ὁ Α τὸν Δ μὴ μετρῇ. ἀλλὰ δὴ οἱ Α, Β, Γ μήτε ἑξῆς ἔστωσαν ἀνάλογον μήτε οἱ ἄκροι πρῶτοι πρὸς ἀλλήλους. καὶ ὁ Β τὸν Γ πολλαπλασιάσας τὸν Δ ποιείτω. ὁμοίως δὴ δειχθήσεται, ὅτι εἰ μὲν μετρεῖ ὁ Α τὸν Δ, δυνατόν ἐστιν αὐτοῖς ἀνάλογον προσευρεῖν, εἰ δὲ οὐ μετρεῖ, ἀδύνατον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",536]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'9.20'"],["Rule","'Text'","'Prime numbers are more than any assigned multitude of prime numbers.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'οἱ πρῶτοι ἀριθμοὶ πλείους εἰσὶ παντὸς τοῦ προτεθέντος πλήθους πρώτων ἀριθμῶν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",31]]]],["Rule","'Proof'","'Let A, B, C be the assigned prime numbers; I say that there are more prime numbers than A, B, C. For let the least number measured by A, B, C be taken, and let it be DE; let the unit DF be added to DE. Then EF is either prime or not. First, let it be prime; then the prime numbers A, B, C, EF have been found which are more than A, B, C. Next, let EF not be prime; therefore it is measured by some prime number. [VII. 31] Let it be measured by the prime number G. I say that G is not the same with any of the numbers A, B, C. For, if possible, let it be so. Now A, B, C measure DE; therefore G also will measure DE. But it also measures EF. Therefore G, being a number, will measure the remainder, the unit DF: which is absurd. Therefore G is not the same with any one of the numbers A, B, C. And by hypothesis it is prime. Therefore the prime numbers A, B, C, G have been found which are more than the assigned multitude of A, B, C.'"],["Rule","'ProofWordCount'",199],["Rule","'GreekProof'","'ἔστωσαν οἱ προτεθέντες πρῶτοι ἀριθμοὶ οἱ Α, Β, Γ: λέγω, ὅτι τῶν Α, Β, Γ πλείους εἰσὶ πρῶτοι ἀριθμοί. εἰλήφθω γὰρ ὁ ὑπὸ τῶν Α, Β, Γ ἐλάχιστος μετρούμενος καὶ ἔστω ὁ ΔΕ, καὶ προσκείσθω τῷ ΔΕ μονὰς ἡ ΔΖ. ὁ δὴ ΕΖ ἤτοι πρῶτός ἐστιν á¼¢ οὔ. ἔστω πρότερον πρῶτος: εὑρημένοι ἄρα εἰσὶ πρῶτοι ἀριθμοὶ οἱ Α, Β, Γ, ΕΖ πλείους τῶν Α, Β, Γ. ἀλλὰ δὴ μὴ ἔστω ὁ ΕΖ πρῶτος: ὑπὸ πρώτου ἄρα τινὸς ἀριθμοῦ μετρεῖται. μετρείσθω ὑπὸ πρώτου τοῦ Η: λέγω, ὅτι ὁ Η οὐδενὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. εἰ γὰρ δυνατόν, ἔστω. οἱ δὲ Α, Β, Γ τὸν ΔΕ μετροῦσιν: καὶ ὁ Η ἄρα τὸν ΔΕ μετρήσει. μετρεῖ δὲ καὶ τὸν ΕΖ: καὶ λοιπὴν τὴν ΔΖ μονάδα μετρήσει ὁ Η ἀριθμὸς ὤν: ὅπερ ἄτοπον. οὐκ ἄρα ὁ Η ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. καὶ ὑπόκειται πρῶτος. εὑρημένοι ἄρα εἰσὶ πρῶτοι ἀριθμοὶ πλείους τοῦ προτεθέντος πλήθους τῶν Α, Β, Γ οἱ Α, Β, Γ, Η: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",168]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'9.21'"],["Rule","'Text'","'If as many even numbers as we please be added together, the whole is even.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἐὰν ἄρτιοι ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, ὁ ὅλος ἄρτιός ἐστιν.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",6]]]],["Rule","'Proof'","'For let as many even numbers as we please, AB, BC, CD, DE, be added together; I say that the whole AE is even. For, since each of the numbers AB, BC, CD, DE is even, it has a half part; [VII. Def. 6] so that the whole AE also has a half part. But an even number is that which is divisible into two equal parts; therefore AE is even.'"],["Rule","'ProofWordCount'",71],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ ἄρτιοι ἀριθμοὶ ὁποσοιοῦν οἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ: λέγω, ὅτι ὅλος ὁ ΑΕ ἄρτιός ἐστιν. ἐπεὶ γὰρ ἕκαστος τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ ἄρτιός ἐστιν, ἔχει μέρος ἥμισυ: ὥστε καὶ ὅλος ὁ ΑΕ ἔχει μέρος ἥμισυ. ἄρτιος δὲ ἀριθμός ἐστιν ὁ δίχα διαιρούμενος: ἄρτιος ἄρα ἐστὶν ὁ ΑΕ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",53]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'9.22'"],["Rule","'Text'","'If as many odd numbers as we please be added together, and their multitude be even, the whole will be even.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'ἐὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ πλῆθος αὐτῶν ἄρτιον ᾖ, ὁ ὅλος ἄρτιος ἔσται.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",7]],["List",["Rule","'Book'",9],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'For let as many odd numbers as we please, AB, BC, CD, DE, even in multitude, be added together; I say that the whole AE is even. For, since each of the numbers AB, BC, CD, DE is odd, if an unit be subtracted from each, each of the remainders will be even; [VII. Def. 7] so that the sum of them will be even. [IX. 21] But the multitude of the units is also even. Therefore the whole AE is also even. [IX. 21]'"],["Rule","'ProofWordCount'",85],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ περισσοὶ ἀριθμοὶ ὁσοιδηποτοῦν ἄρτιοι τὸ πλῆθος οἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ: λέγω, ὅτι ὅλος ὁ ΑΕ ἄρτιός ἐστιν. ἐπεὶ γὰρ ἕκαστος τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ περιττός ἐστιν, ἀφαιρεθείσης μονάδος ἀφ᾽ ἑκάστου ἕκαστος τῶν λοιπῶν ἄρτιος ἔσται: ὥστε καὶ ὁ συγκείμενος ἐξ αὐτῶν ἄρτιος ἔσται. ἔστι δὲ καὶ τὸ πλῆθος τῶν μονάδων ἄρτιον. καὶ ὅλος ἄρα ὁ ΑΕ ἄρτιός ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",66]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'9.23'"],["Rule","'Text'","'If as many odd numbers as we please be added together, and their multitude be odd, the whole will also be odd.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ πλῆθος αὐτῶν περισσὸν ᾖ, καὶ ὁ ὅλος περισσὸς ἔσται.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",7]],["List",["Rule","'Book'",9],["Rule","'Theorem'",21]],["List",["Rule","'Book'",9],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'For let as many odd numbers as we please, AB, BC, CD, the multitude of which is odd, be added together; I say that the whole AD is also odd. Let the unit DE be subtracted from CD; therefore the remainder CE is even. [VII. Def. 7] But CA is also even; [IX. 22] therefore the whole AE is also even. [IX. 21] And DE is an unit. Therefore AD is odd. [VII. Def. 7]'"],["Rule","'ProofWordCount'",75],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ ὁποσοιοῦν περισσοὶ ἀριθμοί, ὧν τὸ πλῆθος περισσὸν ἔστω, οἱ ΑΒ, ΒΓ, ΓΔ: λέγω, ὅτι καὶ ὅλος ὁ ΑΔ περισσός ἐστιν. ἀφῃρήσθω ἀπὸ τοῦ ΓΔ μονὰς ἡ ΔΕ: λοιπὸς ἄρα ὁ ΓΕ ἄρτιός ἐστιν. ἔστι δὲ καὶ ὁ ΓΑ ἄρτιος: καὶ ὅλος ἄρα ὁ ΑΕ ἄρτιός ἐστιν. καί ἐστι μονὰς ἡ ΔΕ. περισσὸς ἄρα ἐστὶν ὁ ΑΔ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",61]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'9.24'"],["Rule","'Text'","'If from an even number an even number be subtracted, the remainder will be even.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἐὰν ἀπὸ ἀρτίου ἀριθμοῦ ἄρτιος ἀφαιρεθῇ, ὁ λοιπὸς ἄρτιος ἔσται.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",6]]]],["Rule","'Proof'","'For from the even number AB let the even number BC be subtracted: I say that the remainder CA is even. For, since AB is even, it has a half part. [VII. Def. 6] For the same reason BC also has a half part; so that the remainder [CA also has a half part, and]AC is therefore even.'"],["Rule","'ProofWordCount'",59],["Rule","'GreekProof'","'ἀπὸ γὰρ ἀρτίου τοῦ ΑΒ ἄρτιος ἀφῃρήσθω ὁ ΒΓ: λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν. ἐπεὶ γὰρ ὁ ΑΒ ἄρτιός ἐστιν, ἔχει μέρος ἥμισυ. διὰ τὰ αὐτὰ δὴ καὶ ὁ ΒΓ ἔχει μέρος ἥμισυ: ὥστε καὶ λοιπὸς ὁ ΓΑ ἔχει μέρος ἥμισυ ἄρτιος ἄρα ἐστὶν ὁ ΑΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",52]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'9.25'"],["Rule","'Text'","'If from an even number an odd number be subtracted, the remainder will be odd.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἐὰν ἀπὸ ἀρτίου ἀριθμοῦ περισσὸς ἀφαιρεθῇ, ὁ λοιπὸς περισσὸς ἔσται.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",7]],["List",["Rule","'Book'",9],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'For from the even number AB let the odd number BC be subtracted; I say that the remainder CA is odd. For let the unit CD be subtracted from BC; therefore DB is even. [VII. Def. 7] But AB is also even; therefore the remainder AD is also even. [IX. 24] And CD is an unit; therefore CA is odd. [VII. Def. 7]'"],["Rule","'ProofWordCount'",63],["Rule","'GreekProof'","'ἀπὸ γὰρ ἀρτίου τοῦ ΑΒ περισσὸς ἀφῃρήσθω ὁ ΒΓ: λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ περισσός ἐστιν. ἀφῃρήσθω γὰρ ἀπὸ τοῦ ΒΓ μονὰς ἡ ΓΔ: ὁ ΔΒ ἄρα ἄρτιός ἐστιν. ἔστι δὲ καὶ ὁ ΑΒ ἄρτιος: καὶ λοιπὸς ἄρα ὁ ΑΔ ἄρτιός ἐστιν. καί ἐστι μονὰς ἡ ΓΔ: ὁ ΓΑ ἄρα περισσός ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",56]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",26]],["Association",["Rule","'VertexLabel'","'9.26'"],["Rule","'Text'","'If from an odd number an odd number be subtracted, the remainder will be even.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἐὰν ἀπὸ περισσοῦ ἀριθμοῦ περισσὸς ἀφαιρεθῇ, ὁ λοιπὸς ἄρτιος ἔσται.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",7]],["List",["Rule","'Book'",9],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'For from the odd number AB let the odd number BC be subtracted; I say that the remainder CA is even. For, since AB is odd, let the unit BD be subtracted; therefore the remainder AD is even. [VII. Def. 7] For the same reason CD is also even; [VII. Def. 7] so that the remainder CA is also even. [IX. 24]'"],["Rule","'ProofWordCount'",62],["Rule","'GreekProof'","' Ἀπὸ γὰρ περισσοῦ τοῦ ΑΒ περισσὸς ἀφῃρήσθω ὁ ΒΓ:  λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν. Ἐπεὶ γὰρ ὁ ΑΒ περισσός ἐστιν, ἀφῃρήσθω μονὰς ἡ ΒΔ: λοιπὸς ἄρα ὁ ΑΔ ἄρτιός ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ὁ ΓΔ ἄρτιός ἐστιν:  ὥστε καὶ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",52]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",27]],["Association",["Rule","'VertexLabel'","'9.27'"],["Rule","'Text'","'If from an odd number an even number be subtracted, the remainder will be odd.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἐὰν ἀπὸ περισσοῦ ἀριθμοῦ ἄρτιος ἀφαιρεθῇ, ὁ λοιπὸς περισσὸς ἔσται.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",7]],["List",["Rule","'Book'",9],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'For from the odd number AB let the even number BC be subtracted; I say that the remainder CA is odd. Let the unit AD be subtracted; therefore DB is even. [VII. Def. 7] But BC is also even; therefore the remainder CD is even. [IX. 24] Therefore CA is odd. [VII. Def. 7]'"],["Rule","'ProofWordCount'",54],["Rule","'GreekProof'","'ἀπὸ γὰρ περισσοῦ τοῦ ΑΒ ἄρτιος ἀφῃρήσθω ὁ ΒΓ: λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ περισσός ἐστιν. ἀφῃρήσθω γὰρ μονὰς ἡ ΑΔ: ὁ ΔΒ ἄρα ἄρτιός ἐστιν. ἔστι δὲ καὶ ὁ ΒΓ ἄρτιος: καὶ λοιπὸς ἄρα ὁ ΓΔ ἄρτιός ἐστιν. περισσὸς ἄρα ὁ ΓΑ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",47]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",28]],["Association",["Rule","'VertexLabel'","'9.28'"],["Rule","'Text'","'If an odd number by multiplying an even number make some number, the product will be even.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἐὰν περισσὸς ἀριθμὸς ἄρτιον πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος ἄρτιος ἔσται.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",15]],["List",["Rule","'Book'",9],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'For let the odd number A by multiplying the even number B make C; I say that C is even. For, since A by multiplying B has made C, therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15] And B is even; therefore C is made up of even numbers. But, if as many even numbers as we please be added together, the whole is even. [IX. 21] Therefore C is even.'"],["Rule","'ProofWordCount'",84],["Rule","'GreekProof'","'περισσὸς γὰρ ἀριθμὸς ὁ Α ἄρτιον τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω: λέγω, ὅτι ὁ Γ ἄρτιός ἐστιν. ᾿επεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Γ ἄρα σύγκειται ἐκ τοσούτων ἴσων τῷ Β, ὅσαι εἰσὶν ἐν τῷ Α μονάδες. καί ἐστιν ὁ Β ἄρτιος: ὁ Γ ἄρα σύγκειται ἐξ ἀρτίων. ἐὰν δὲ ἄρτιοι ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, ὁ ὅλος ἄρτιός ἐστιν. ἄρτιος ἄρα ἐστὶν ὁ Γ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",73]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",29]],["Association",["Rule","'VertexLabel'","'9.29'"],["Rule","'Text'","'If an odd number by multiplying an odd number make some number, the product will be odd.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἐὰν περισσὸς ἀριθμὸς περισσὸν ἀριθμὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος περισσὸς ἔσται.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",15]],["List",["Rule","'Book'",9],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'For let the odd number A by multiplying the odd number B make C; I say that C is odd. For, since A by multiplying B has made C, therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15] And each of the numbers A, B is odd; therefore C is made up of odd numbers the multitude of which is odd. Thus C is odd. [IX. 23]'"],["Rule","'ProofWordCount'",79],["Rule","'GreekProof'","'περισσὸς γὰρ ἀριθμὸς ὁ Α περισσὸν τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω: λέγω, ὅτι ὁ Γ περισσός ἐστιν. ἐπεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Γ ἄρα σύγκειται ἐκ τοσούτων ἴσων τῷ Β, ὅσαι εἰσὶν ἐν τῷ Α μονάδες. καί ἐστιν ἑκάτερος τῶν Α, Β περισσός: ὁ Γ ἄρα σύγκειται ἐκ περισσῶν ἀριθμῶν, ὧν τὸ πλῆθος περισσόν ἐστιν. ὥστε ὁ Γ περισσός ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",70]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",30]],["Association",["Rule","'VertexLabel'","'9.30'"],["Rule","'Text'","'If an odd number measure an even number, it will also measure the half of it.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'ἐὰν περισσὸς ἀριθμὸς ἄρτιον ἀριθμὸν μετρῇ, καὶ τὸν ἥμισυν αὐτοῦ μετρήσει.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",9],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'For let the odd number A measure the even number B; I say that it will also measure the half of it. For, since A measures B, let it measure it according to C; I say that C is not odd. For, if possible, let it be so. Then, since A measures B according to C, therefore A by multiplying C has made B. Therefore B is made up of odd numbers the multitude of which is odd. Therefore B is odd: [IX. 23] which is absurd, for by hypothesis it is even. Therefore C is not odd; therefore C is even. Thus A measures B an even number of times. For this reason then it also measures the half of it.'"],["Rule","'ProofWordCount'",122],["Rule","'GreekProof'","'περισσὸς γὰρ ἀριθμὸς ὁ Α ἄρτιον τὸν Β μετρείτω: λέγω, ὅτι καὶ τὸν ἥμισυν αὐτοῦ μετρήσει. ἐπεὶ γὰρ ὁ Α τὸν Β μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Γ: λέγω, ὅτι ὁ Γ οὐκ ἔστι περισσός. εἰ γὰρ δυνατόν, ἔστω. καὶ ἐπεὶ ὁ Α τὸν Β μετρεῖ κατὰ τὸν Γ, ὁ Α ἄρα τὸν Γ πολλαπλασιάσας τὸν Β πεποίηκεν. ὁ Β ἄρα σύγκειται ἐκ περισσῶν ἀριθμῶν, ὧν τὸ πλῆθος περισσόν ἐστιν. ὁ Β ἄρα περισσός ἐστιν: ὅπερ ἄτοπον: ὑπόκειται γὰρ ἄρτιος. οὐκ ἄρα ὁ Γ περισσός ἐστιν: ἄρτιος ἄρα ἐστὶν ὁ Γ. ὥστε ὁ Α τὸν Β μετρεῖ ἀρτιάκις. διὰ δὴ τοῦτο καὶ τὸν ἥμισυν αὐτοῦ μετρήσει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",109]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",31]],["Association",["Rule","'VertexLabel'","'9.31'"],["Rule","'Text'","'If an odd number be prime to any number, it will also be prime to the double of it.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἐὰν περισσὸς ἀριθμὸς πρός τινα ἀριθμὸν πρῶτος ᾖ, καὶ πρὸς τὸν διπλασίονα αὐτοῦ πρῶτος ἔσται.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",9],["Rule","'Theorem'",30]]]],["Rule","'Proof'","'For let the odd number A be prime to any number B, and let C be double of B; I say that A is prime to C. For, if they are not prime to one another, some number will measure them. Let a number measure them, and let it be D. Now A is odd; therefore D is also odd. And since D which is odd measures C, and C is even, therefore [D] will measure the half of C also. [IX. 30] But B is half of C; therefore D measures B. But it also measures A; therefore D measures A, B which are prime to one another: which is impossible. Therefore A cannot but be prime to C. Therefore A, C are prime to one another.'"],["Rule","'ProofWordCount'",128],["Rule","'GreekProof'","'περισσὸς γὰρ ἀριθμὸς ὁ Α πρός τινα ἀριθμὸν τὸν Β πρῶτος ἔστω, τοῦ δὲ Β διπλασίων ἔστω ὁ Γ: λέγω, ὅτι ὁ Α καὶ πρὸς τὸν Γ πρῶτός ἐστιν. εἰ γὰρ μή εἰσιν οἱ Α, Γ πρῶτοι, μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. καί ἐστιν ὁ Α περισσός: περισσὸς ἄρα καὶ ὁ Δ. καὶ ἐπεὶ ὁ Δ περισσὸς ὢν τὸν Γ μετρεῖ, καί ἐστιν ὁ Γ ἄρτιος, καὶ τὸν ἥμισυν ἄρα τοῦ Γ μετρήσει ὁ Δ. τοῦ δὲ Γ ἥμισύ ἐστιν ὁ Β: ὁ Δ ἄρα τὸν Β μετρεῖ. μετρεῖ δὲ καὶ τὸν Α. ὁ Δ ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ὁ Α πρὸς τὸν Γ πρῶτος οὔκ ἐστιν. οἱ Α, Γ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",132]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",32]],["Association",["Rule","'VertexLabel'","'9.32'"],["Rule","'Text'","'Each of the numbers which are continually doubled beginning from a dyad is even-times even only.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'τῶν ἀπὸ δυάδος διπλασιαζομένων ἀριθμῶν ἕκαστος ἀρτιάκις ἄρτιός ἐστι μόνον.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",8]],["List",["Rule","'Book'",9],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'For let as many numbers as we please, B, C, D, have been continually doubled beginning from the dyad A; I say that B, C, D are eventimes even only. Now that each of the numbers B, C, D is even-times even is manifest; for it is doubled from a dyad. I say that it is also even-times even only. For let an unit be set out. Since then as many numbers as we please beginning from an unit are in continued proportion, and the number A after the unit is prime, therefore D, the greatest of the numbers A, B, C, D, will not be measured by any other number except A, B, C. [IX. 13] And each of the numbers A, B, C is even; therefore D is even-times even only. [VII. Def. 8] Similarly we can prove that each of the numbers B, C is even-times even only.'"],["Rule","'ProofWordCount'",151],["Rule","'GreekProof'","'ἀπὸ γὰρ δυάδος τῆς Α δεδιπλασιάσθωσαν ὁσοιδηποτοῦν ἀριθμοὶ οἱ Β, Γ, Δ: λέγω, ὅτι οἱ Β, Γ, Δ ἀρτιάκις ἄρτιοί εἰσι μόνον. ὅτι μὲν οὖν ἕκαστος τῶν Β, Γ, Δ ἀρτιάκις ἄρτιός ἐστιν, φανερόν: ἀπὸ γὰρ δυάδος ἐστὶ διπλασιασθείς. λέγω, ὅτι καὶ μόνον. ἐκκείσθω γὰρ μονάς. ἐπεὶ οὖν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ μετὰ τὴν μονάδα ὁ Α πρῶτός ἐστιν, ὁ μέγιστος τῶν Α, Β, Γ, Δ ὁ Δ ὑπ᾽ οὐδενὸς ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ. καί ἐστιν ἕκαστος τῶν Α, Β, Γ ἄρτιος: ὁ Δ ἄρα ἀρτιάκις ἄρτιός ἐστι μόνον. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκάτερος τῶν Β, Γ ἀρτιάκις ἄρτιός ἐστι μόνον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",114]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",33]],["Association",["Rule","'VertexLabel'","'9.33'"],["Rule","'Text'","'If a number have its half odd, it is even-times odd only.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'ἐὰν ἀριθμὸς τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις περισσός ἐστι μόνον.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",8]],["List",["Rule","'Book'",7],["Rule","'Definition'",9]]]],["Rule","'Proof'","'For let the number A have its half odd; I say that A is even-times odd only. Now that it is even-times odd is manifest; for the half of it, being odd, measures it an even number of times. [VII. Def. 9] I say next that it is also even-times odd only. For, if A is even-times even also, it will be measured by an even number according to an even number; [VII. Def. 8] so that the half of it will also be measured by an even number though it is odd: which is absurd. Therefore A is even-times odd only.'"],["Rule","'ProofWordCount'",102],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ Α τὸν ἥμισυν ἐχέτω περισσόν: λέγω, ὅτι ὁ Α ἀρτιάκις περισσός ἐστι μόνον. ὅτι μὲν οὖν ἀρτιάκις περισσός ἐστιν, φανερόν: ὁ γὰρ ἥμισυς αὐτοῦ περισσὸς ὢν μετρεῖ αὐτὸν ἀρτιάκις. λέγω δή, ὅτι καὶ μόνον. εἰ γὰρ ἔσται ὁ Α καὶ ἀρτιάκις ἄρτιος, μετρηθήσεται ὑπὸ ἀρτίου κατὰ ἄρτιον ἀριθμόν: ὥστε καὶ ὁ ἥμισυς αὐτοῦ μετρηθήσεται ὑπὸ ἀρτίου ἀριθμοῦ περισσὸς ὤν: ὅπερ ἐστὶν ἄτοπον. ὁ Α ἄρα ἀρτιάκις περισσός ἐστι μόνον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",75]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",34]],["Association",["Rule","'VertexLabel'","'9.34'"],["Rule","'Text'","'If a number neither be one of those which are continually doubled from a dyad, nor have its half odd, it is both even-times even and even-times odd.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐὰν ἀριθμὸς μήτε τῶν ἀπὸ δυάδος διπλασιαζομένων ᾖ μήτε τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις τε ἄρτιός ἐστι καὶ ἀρτιάκις περισσός.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",8]]]],["Rule","'Proof'","'For let the number A neither be one of those doubled from a dyad, nor have its half odd; I say that A is both even-times even and even-times odd. Now that A is even-times even is manifest; for it has not its half odd. [VII. Def. 8] I say next that it is also even-times odd. For, if we bisect A, then bisect its half, and do this continually, we shall come upon some odd number which will measure A according to an even number. For, if not, we shall come upon a dyad, and A will be among those which are doubled from a dyad: which is contrary to the hypothesis. Thus A is even-times odd. But it was also proved even-times even. Therefore A is both even-times even and even-times odd.'"],["Rule","'ProofWordCount'",134],["Rule","'GreekProof'","'ἀριθμὸς γὰρ ὁ Α μήτε τῶν ἀπὸ δυάδος διπλασιαζομένων ἔστω μήτε τὸν ἥμισυν ἐχέτω περισσόν: λέγω, ὅτι ὁ Α ἀρτιάκις τέ ἐστιν ἄρτιος καὶ ἀρτιάκις περισσός. ὅτι μὲν οὖν ὁ Α ἀρτιάκις ἐστὶν ἄρτιος, φανερόν: τὸν γὰρ ἥμισυν οὐκ ἔχει περισσόν. λέγω δή, ὅτι καὶ ἀρτιάκις περισσός ἐστιν. ἐὰν γὰρ τὸν Α τέμνωμεν δίχα καὶ τὸν ἥμισυν αὐτοῦ δίχα καὶ τοῦτο ἀεὶ ποιῶμεν, καταντήσομεν εἴς τινα ἀριθμὸν περισσόν, ὃς μετρήσει τὸν Α κατὰ ἄρτιον ἀριθμόν. εἰ γὰρ οὔ, καταντήσομεν εἰς δυάδα, καὶ ἔσται ὁ Α τῶν ἀπὸ δυάδος διπλασιαζομένων: ὅπερ οὐχ ὑπόκειται. ὥστε ὁ Α ἀρτιάκις περισσός ἐστιν. ἐδείχθη δὲ καὶ ἀρτιάκις ἄρτιος. ὁ Α ἄρα ἀρτιάκις τε ἄρτιός ἐστι καὶ ἀρτιάκις περισσός: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",116]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",35]],["Association",["Rule","'VertexLabel'","'9.35'"],["Rule","'Text'","'If as many numbers as we please be in continued proportion, and there be subtracted from the second and the last numbers equal to the first, then, as the excess of the second is to the first, so will the excess of the last be to all those before it.'"],["Rule","'TextWordCount'",50],["Rule","'GreekText'","'ἐὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, ἀφαιρεθῶσι δὲ ἀπό τε τοῦ δευτέρου καὶ τοῦ ἐσχάτου ἴσοι τῷ πρώτῳ, ἔσται ὡς ἡ τοῦ δευτέρου ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας.'"],["Rule","'GreekTextWordCount'",37],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Theorem'",11]],["List",["Rule","'Book'",7],["Rule","'Theorem'",12]],["List",["Rule","'Book'",7],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'Let there be as many numbers as we please in continued proportion, A, BC, D, EF, beginning from A as least, and let there be subtracted from BC and EF the numbers BG, FH, each equal to A; I say that, as GC is to A, so is EH to A, BC, D. For let FK be made equal to BC, and FL equal to D. Then, since FK is equal to BC, and of these the part FH is equal to the part BG, therefore the remainder HK is equal to the remainder GC. And since, as EF is to D, so is D to BC, and BC to A, while D is equal to FL, BC to FK, and A to FH, therefore, as EF is to FL, so is LF to FK, and FK to FH.  Separando, as EL is to LF, so is LK to FK, and KH to FH. [VII. 11, 13] Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [VII. 12] therefore, as KH is to FH, so are EL, LK. KH to LF, FK, HF. But KH is equal to CG, FH to A, and LF, FK, HF to D, BC, A; therefore, as CG is to A, so is EH to D, BC, A. Therefore, as the excess of the second is to the first, so is the excess of the last to all those before it.'"],["Rule","'ProofWordCount'",249],["Rule","'GreekProof'","'ἔστωσαν ὁποσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, ΒΓ, Δ, ΕΖ ἀρχόμενοι ἀπὸ ἐλαχίστου τοῦ Α, καὶ ἀφῃρήσθω ἀπὸ τοῦ ΒΓ καὶ τοῦ ΕΖ τῷ Α ἴσος ἑκάτερος τῶν ΒΗ, ΖΘ: λέγω, ὅτι ἐστὶν ὡς ὁ ΗΓ πρὸς τὸν Α, οὕτως ὁ ΕΘ πρὸς τοὺς Α, ΒΓ, Δ. κείσθω γὰρ τῷ μὲν ΒΓ ἴσος ὁ ΖΚ, τῷ δὲ Δ ἴσος ὁ ΖΛ. καὶ ἐπεὶ ὁ ΖΚ τῷ ΒΓ ἴσος ἐστίν, ὧν ὁ ΖΘ τῷ ΒΗ ἴσος ἐστίν, λοιπὸς ἄρα ὁ ΘΚ λοιπῷ τῷ ΗΓ ἐστιν ἴσος. καὶ ἐπεί ἐστιν ὡς ὁ ΕΖ πρὸς τὸν Δ, οὕτως ὁ Δ πρὸς τὸν ΒΓ καὶ ὁ ΒΓ πρὸς τὸν Α, ἴσος δὲ ὁ μὲν Δ τῷ ΖΛ, ὁ δὲ ΒΓ τῷ ΖΚ, ὁ δὲ Α τῷ ΖΘ, ἔστιν ἄρα ὡς ὁ ΕΖ πρὸς τὸν ΖΛ, οὕτως ὁ ΛΖ πρὸς τὸν ΖΚ καὶ ὁ ΖΚ πρὸς τὸν ΖΘ. διελόντι, ὡς ὁ ΕΛ πρὸς τὸν ΛΖ, οὕτως ὁ ΛΚ πρὸς τὸν ΖΚ καὶ ὁ ΚΘ πρὸς τὸν ΖΘ. ἔστιν ἄρα καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους: ἔστιν ἄρα ὡς ὁ ΚΘ πρὸς τὸν ΖΘ, οὕτως οἱ ΕΛ, ΛΚ, ΚΘ πρὸς τοὺς ΛΖ, ΖΚ, ΘΖ. ἴσος δὲ ὁ μὲν ΚΘ τῷ ΓΗ, ὁ δὲ ΖΘ τῷ Α, οἱ δὲ ΛΖ, ΖΚ, ΘΖ τοῖς Δ, ΒΓ, Α: ἔστιν ἄρα ὡς ὁ ΓΗ πρὸς τὸν Α, οὕτως ὁ ΕΘ πρὸς τοὺς Δ, ΒΓ, Α. ἔστιν ἄρα ὡς ἡ τοῦ δευτέρου ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",259]]],["Rule",["Association",["Rule","'Book'",9],["Rule","'Theorem'",36]],["Association",["Rule","'VertexLabel'","'9.36'"],["Rule","'Text'","'If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect.'"],["Rule","'TextWordCount'",41],["Rule","'GreekText'","'ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἐκτεθῶσιν ἐν τῇ διπλασίονι ἀναλογίᾳ, ἕως οὗ ὁ σύμπας συντεθεὶς πρῶτος γένηται, καὶ ὁ σύμπας ἐπὶ τὸν ἔσχατον πολλαπλασιασθεὶς ποιῇ τινα, ὁ γενόμενος τέλειος ἔσται.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["List",["List",["Rule","'Book'",7],["Rule","'Definition'",20]],["List",["Rule","'Book'",7],["Rule","'Definition'",22]],["List",["Rule","'Book'",7],["Rule","'Theorem'",14]],["List",["Rule","'Book'",7],["Rule","'Theorem'",19]],["List",["Rule","'Book'",7],["Rule","'Theorem'",20]],["List",["Rule","'Book'",7],["Rule","'Theorem'",21]],["List",["Rule","'Book'",7],["Rule","'Theorem'",29]],["List",["Rule","'Book'",9],["Rule","'Theorem'",13]],["List",["Rule","'Book'",9],["Rule","'Theorem'",35]]]],["Rule","'Proof'","'For let as many numbers as we please, A, B, C, D, beginning from an unit be set out in double proportion, until the sum of all becomes prime, let E be equal to the sum, and let E by multiplying D make FG; I say that FG is perfect. For, however many A, B, C, D are in multitude, let so many E, HK, L, M be taken in double proportion beginning from E; therefore, ex aequali, as A is to D, so is E to M. [VII. 14] Therefore the product of E, D is equal to the product of A, M. [VII. 19] And the product of E, D is FG; therefore the product of A, M is also FG. Therefore A by multiplying M has made FG; therefore M measures FG according to the units in A. And A is a dyad; therefore FG is double of M. But M, L, HK, E are continuously double of each other; therefore E, HK, L, M, FG are continuously proportional in double proportion. Now let there be subtracted from the second HK and the last FG the numbers HN, FO, each equal to the first E; therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. [IX. 35] Therefore, as NK is to E, so is OG to M, L, KH, E. And NK is equal to E; therefore OG is also equal to M, L, HK, E. But FO is also equal to E, and E is equal to A, B, C, D and the unit. Therefore the whole FG is equal to E, HK, L, M and A, B, C, D and the unit; and it is measured by them. I say also that FG will not be measured by any other number except A, B, C, D, E, HK, L, M and the unit. For, if possible, let some number P measure FG, and let P not be the same with any of the numbers A, B, C, D, E, HK, L, M. And, as many times as P measures FG, so many units let there be in Q; therefore Q by multiplying P has made FG. But, further, E has also by multiplying D made FG; therefore, as E is to Q, so is P to D. [VII. 19] And, since A, B, C, D are continuously proportional beginning from an unit, therefore D will not be measured by any other number except A, B, C. [IX. 13] And, by hypothesis, P is not the same with any of the numbers A, B, C; therefore P will not measure D. But, as P is to D, so is E to Q; therefore neither does E measure Q. [VII. Def. 20] And E is prime; and any prime number is prime to any number which it does not measure. [VII. 29] Therefore E, Q are prime to one another. But primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] and, as E is to Q, so is P to D; therefore E measures P the same number of times that Q measures D. But D is not measured by any other number except A, B, C; therefore Q is the same with one of the numbers A, B, C. Let it be the same with B. And, however many B, C, D are in multitude, let so many E, HK, L be taken beginning from E. Now E, HK, L are in the same ratio with B, C, D; therefore, ex aequali, as B is to D, so is E to L. [VII. 14] Therefore the product of B, L is equal to the product of D, E. [VII. 19] But the product of D, E is equal to the product of Q, P; therefore the product of Q, P is also equal to the product of B, L. Therefore, as Q is to B, so is L to P. [VII. 19] And Q is the same with B; therefore L is also the same with P; which is impossible, for by hypothesis P is not the same with any of the numbers set out. Therefore no number will measure FG except A, B, C, D, E, HK, L, M and the unit. And FG was proved equal to A, B, C, D, E, HK, L, M and the unit; and a perfect number is that which is equal to its own parts; [VII. Def. 22] therefore FG is perfect.'"],["Rule","'ProofWordCount'",782],["Rule","'GreekProof'","'ἀπὸ γὰρ μονάδος ἐκκείσθωσαν ὁσοιδηποτοῦν ἀριθμοὶ ἐν τῇ διπλασίονι ἀναλογίᾳ, ἕως οὗ ὁ σύμπας συντεθεὶς πρῶτος γένηται, οἱ Α, Β, Γ, Δ, καὶ τῷ σύμπαντι ἴσος ἔστω ὁ Ε, καὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν ΖΗ ποιείτω. λέγω, ὅτι ὁ ΖΗ τέλειός ἐστιν. ὅσοι γάρ εἰσιν οἱ Α, Β, Γ, Δ τῷ πλήθει, τοσοῦτοι ἀπὸ τοῦ Ε εἰλήφθωσαν ἐν τῇ διπλασίονι ἀναλογίᾳ οἱ Ε, ΘΚ, Λ, Μ: δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν Μ. ὁ ἄρα ἐκ τῶν Ε, Δ ἴσος ἐστὶ τῷ ἐκ τῶν Α, Μ. καί ἐστιν ὁ ἐκ τῶν Ε, Δ ὁ ΖΗ: καὶ ὁ ἐκ τῶν Α, Μ ἄρα ἐστὶν ὁ ΖΗ. ὁ Α ἄρα τὸν Μ πολλαπλασιάσας τὸν ΖΗ πεποίηκεν: ὁ Μ ἄρα τὸν ΖΗ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. καί ἐστι δυὰς ὁ Α: διπλάσιος ἄρα ἐστὶν ὁ ΖΗ τοῦ Μ. εἰσὶ δὲ καὶ οἱ Μ, Λ, ΘΚ, Ε ἑξῆς διπλάσιοι ἀλλήλων: οἱ Ε, ΘΚ, Λ, Μ, ΖΗ ἄρα ἑξῆς ἀνάλογόν εἰσιν ἐν τῇ διπλασίονι ἀναλογίᾳ. ἀφῃρήσθω δὴ ἀπὸ τοῦ δευτέρου τοῦ ΘΚ καὶ τοῦ ἐσχάτου τοῦ ΖΗ τῷ πρώτῳ τῷ Ε ἴσος ἑκάτερος τῶν ΘΝ, ΖΞ: ἔστιν ἄρα ὡς ἡ τοῦ δευτέρου ἀριθμοῦ ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας. ἔστιν ἄρα ὡς ὁ ΝΚ πρὸς τὸν Ε, οὕτως ὁ ΞΗ πρὸς τοὺς Μ, Λ, ΚΘ, Ε. καί ἐστιν ὁ ΝΚ ἴσος τῷ Ε: καὶ ὁ ΞΗ ἄρα ἴσος ἐστὶ τοῖς Μ, Λ, ΘΚ, Ε. ἔστι δὲ καὶ ὁ ΖΞ τῷ Ε ἴσος, ὁ δὲ Ε τοῖς Α, Β, Γ, Δ καὶ τῇ μονάδι. ὅλος ἄρα ὁ ΖΗ ἴσος ἐστὶ τοῖς τε Ε, ΘΚ, Λ, Μ καὶ τοῖς Α, Β, Γ, Δ καὶ τῇ μονάδι: καὶ μετρεῖται ὑπ᾽ αὐτῶν. λέγω, ὅτι καὶ ὁ ΖΗ ὑπ᾽ οὐδενὸς ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ καὶ τῆς μονάδος. εἰ γὰρ δυνατόν, μετρείτω τις τὸν ΖΗ ὁ Ο, καὶ ὁ Ο μηδενὶ τῶν Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ ἔστω ὁ αὐτός. καὶ ὁσάκις ὁ Ο τὸν ΖΗ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Π: ὁ Π ἄρα τὸν Ο πολλαπλασιάσας τὸν ΖΗ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν ΖΗ πεποίηκεν: ἔστιν ἄρα ὡς ὁ ε πρὸς τὸν Π, ὁ Ο πρὸς τὸν Δ. καὶ ἐπεὶ ἀπὸ μονάδος ἑξῆς ἀνάλογόν εἰσιν οἱ Α, Β, Γ, Δ, ὁ Δ ἄρα ὑπ᾽ οὐδενὸς ἄλλου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Α, Β, Γ. καὶ ὑπόκειται ὁ Ο οὐδενὶ τῶν Α, Β, Γ ὁ αὐτός: οὐκ ἄρα μετρήσει ὁ Ο τὸν Δ. ἀλλ᾽ ὡς ὁ Ο πρὸς τὸν Δ, ὁ Ε πρὸς τὸν Π: οὐδὲ ὁ Ε ἄρα τὸν Π μετρεῖ. καί ἐστιν ὁ Ε πρῶτος: πᾶς δὲ πρῶτος ἀριθμὸς πρὸς ἅπαντα, ὃν μὴ μετρεῖ, πρῶτος ἐστιν. οἱ Ε, Π ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: καί ἐστιν ὡς ὁ Ε πρὸς τὸν Π, ὁ Ο πρὸς τὸν Δ: ἰσάκις ἄρα ὁ Ε τὸν Ο μετρεῖ καὶ ὁ Π τὸν Δ: ἰσάκις ἄρα ὁ Ε τὸν Ο μετρεῖ καὶ ὁ Π τὸν Δ. ὁ δὲ Δ ὑπ᾽ οὐδενὸς ἄλλου μετρεῖται παρὲξ τῶν Α, Β, Γ: ὁ Π ἄρα ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. ἔστω τῷ Β ὁ αὐτός. καὶ ὅσοι εἰσὶν οἱ Β, Γ, Δ τῷ πλήθει τοσοῦτοι εἰλήφθωσαν ἀπὸ τοῦ Ε οἱ Ε, ΘΚ, Λ. καί εἰσιν οἱ Ε, ΘΚ, Λ τοῖς Β, Γ, Δ ἐν τῷ αὐτῷ λόγῳ: δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Β πρὸς τὸν Δ, ὁ Ε πρὸς τὸν Λ. ὁ ἄρα ἐκ τῶν Β, Λ ἴσος ἐστὶ τῷ ἐκ τῶν Δ, Ε: ἀλλ᾽ ὁ ἐκ τῶν Δ, Ε ἴσος ἐστὶ τῷ ἐκ τῶν π, Ο: καὶ ὁ ἐκ τῶν Π, Ο ἄρα ἴσος ἐστὶ τῷ ἐκ τῶν Β, Λ. ἔστιν ἄρα ὡς ὁ Π πρὸς τὸν Β, ὁ Λ πρὸς τὸν Ο. καί ἐστιν ὁ Π τῷ Β ὁ αὐτός: καὶ ὁ Λ ἄρα τῷ Ο ἐστιν ὁ αὐτός: ὅπερ ἀδύνατον: ὁ γὰρ Ο ὑπόκειται μηδενὶ τῶν ἐκκειμένων ὁ αὐτός. οὐκ ἄρα τὸν ΖΗ μετρήσει τις ἀριθμὸς παρὲξ τῶν α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ καὶ τῆς μονάδος. καὶ ἐδείχθη ὁ ΖΗ τοῖς Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ καὶ τῇ μονάδι ἴσος. τέλειος δὲ ἀριθμός ἐστιν ὁ τοῖς ἑαυτοῦ μέρεσιν ἴσος ὤν: τέλειος ἄρα ἐστὶν ὁ ΖΗ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",755]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'10.1'"],["Rule","'Text'","'Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the lesser magnitude set out.'"],["Rule","'TextWordCount'",54],["Rule","'GreekText'","'δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος ἀφαιρεθῇ μεῖζον á¼¢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον á¼¢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους.'"],["Rule","'GreekTextWordCount'",34],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",4]]]],["Rule","'Proof'","'Let AB, C be two unequal magnitudes of which AB is the greater: I say that, if from AB there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C. For C if multiplied will sometime be greater than AB. [cf. v. Def. 4] Let it be multiplied, and let DE be a multiple of C, and greater than. AB; let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE. Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, GE. Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA. And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK. But DF is equal to C; therefore C is also greater than AK. Therefore AK is less than C. Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C.'"],["Rule","'ProofWordCount'",273],["Rule","'GreekProof'","'ἔστω δύο μεγέθη ἄνισα τὰ ΑΒ, Γ, ὧν μεῖζον τὸ ΑΒ: λέγω, ὅτι, ἐὰν ἀπὸ τοῦ ΑΒ ἀφαιρεθῇ μεῖζον á¼¢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον á¼¢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ Γ μεγέθους. τὸ Γ γὰρ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ ΑΒ μεῖζον. πεπολλαπλασιάσθω, καὶ ἔστω τὸ ΔΕ τοῦ μὲν Γ πολλαπλάσιον, τοῦ δὲ ΑΒ μεῖζον, καὶ διῃρήσθω τὸ ΔΕ εἰς τὰ τῷ Γ ἴσα τὰ ΔΖ, ΖΗ, ΗΕ, καὶ ἀφῃρήσθω ἀπὸ μὲν τοῦ ΑΒ μεῖζον á¼¢ τὸ ἥμισυ τὸ ΒΘ, ἀπὸ δὲ τοῦ ΑΘ μεῖζον á¼¢ τὸ ἥμισυ τὸ ΘΚ, καὶ τοῦτο ἀεὶ γιγνέσθω, ἕως ἂν αἱ ἐν τῷ ΑΒ διαιρέσεις ἰσοπληθεῖς γένωνται ταῖς ἐν τῷ ΔΕ διαιρέσεσιν. ἔστωσαν οὖν αἱ ΑΚ, ΚΘ, ΘΒ διαιρέσεις ἰσοπληθεῖς οὖσαι ταῖς ΔΖ, ΖΗ, ΗΕ: καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΔΕ τοῦ ΑΒ, καὶ ἀφῄρηται ἀπὸ μὲν τοῦ ΔΕ ἔλασσον τοῦ ἡμίσεος τὸ ΕΗ, ἀπὸ δὲ τοῦ ΑΒ μεῖζον á¼¢ τὸ ἥμισυ τὸ ΒΘ, λοιπὸν ἄρα τὸ ΗΔ λοιποῦ τοῦ ΘΑ μεῖζόν ἐστιν. καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΗΔ τοῦ ΘΑ, καὶ ἀφῄρηται τοῦ μὲν ΗΔ ἥμισυ τὸ ΗΖ, τοῦ δὲ ΘΑ μεῖζον á¼¢ τὸ ἥμισυ τὸ ΘΚ, λοιπὸν ἄρα τὸ ΔΖ λοιποῦ τοῦ ΑΚ μεῖζόν ἐστιν. ἴσον δὲ τὸ ΔΖ τῷ Γ: καὶ τὸ Γ ἄρα τοῦ ΑΚ μεῖζόν ἐστιν. ἔλασσον ἄρα τὸ ΑΚ τοῦ Γ. καταλείπεται ἄρα ἀπὸ τοῦ ΑΒ μεγέθους τὸ ΑΚ μέγεθος ἔλασσον ὂν τοῦ ἐκκειμένου ἐλάσσονος μεγέθους τοῦ Γ: ὅπερ ἔδει δεῖξαι. _ ὁμοίως δὲ δειχθήσεται, κἂν ἡμίση ᾖ τὰ ἀφαιρούμενα.'"],["Rule","'GreekProofWordCount'",250]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'10.2'"],["Rule","'Text'","'If, when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν δύο μεγεθῶν ἐκκειμένων ἀνίσων ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος τὸ καταλειπόμενον μηδέποτε καταμετρῇ τὸ πρὸ ἑαυτοῦ, ἀσύμμετρα ἔσται τὰ μεγέθη.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List"]],["Rule","'Proof'","'For, there being two unequal magnitudes AB, CD, and AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it; I say that the magnitudes AB, CD are incommensurable. For, if they are commensurable, some magnitude will measure them. Let a magnitude measure them, if possible, and let it be E; let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave AG less than itself, and let this process be repeated continually, until there is left some magnitude which is less than E. Suppose this done, and let there be left AG less than E. Then, since E measures AB, while AB measures DF, therefore E will also measure FD. But it measures the whole CD also; therefore it will also measure the remainder CF. But CF measures BG; therefore E also measures BG. But it measures the whole AB also; therefore it will also measure the remainder AG, the greater the less: which is impossible. Therefore no magnitude will measure the magnitudes AB, CD; therefore the magnitudes AB, CD are incommensurable.'"],["Rule","'ProofWordCount'",192],["Rule","'GreekProof'","'δύο γὰρ μεγεθῶν ὄντων ἀνίσων τῶν ΑΒ, ΓΔ καὶ ἐλάσσονος τοῦ ΑΒ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος τὸ περιλειπόμενον μηδέποτε καταμετρείτω τὸ πρὸ ἑαυτοῦ: λέγω, ὅτι ἀσύμμετρά ἐστι τὰ ΑΒ, ΓΔ μεγέθη. εἰ γάρ ἐστι σύμμετρα, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Ε: καὶ τὸ μὲν ΑΒ τὸ ΖΔ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΓΖ, τὸ δὲ ΓΖ τὸ ΒΗ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΑΗ, καὶ τοῦτο ἀεὶ γινέσθω, ἕως οὗ λειφθῇ τι μέγεθος, ὅ ἐστιν ἔλασσον τοῦ Ε. γεγονέτω, καὶ λελείφθω τὸ ΑΗ ἔλασσον τοῦ Ε. ἐπεὶ οὖν τὸ Ε τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΔΖ μετρεῖ, καὶ τὸ Ε ἄρα τὸ ΖΔ μετρήσει. μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ: καὶ λοιπὸν ἄρα τὸ ΓΖ μετρήσει. ἀλλὰ τὸ ΓΖ τὸ ΒΗ μετρεῖ: καὶ τὸ Ε ἄρα τὸ ΒΗ μετρεῖ. μετρεῖ δὲ καὶ ὅλον τὸ ΑΒ: καὶ λοιπὸν ἄρα τὸ ΑΗ μετρήσει, τὸ μεῖζον τὸ ἔλασσον. ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΑΒ, ΓΔ μεγέθη μετρήσει τι μέγεθος: ἀσύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΓΔ μεγέθη. ἐὰν ἄρα δύο μεγεθῶν ἀνίσων, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",182]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'10.3'"],["Rule","'Text'","'Given two commensurable magnitudes, to find their greatest common measure.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'δύο μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",2]]]],["Rule","'Proof'","'Let the two given commensurable magnitudes be AB, CD of which AB is the less; thus it is required to find the greatest common measure of AB, CD. Now the magnitude AB either measures CD or it does not. If then it measures it—and it measures itself also— AB is a common measure of AB, CD. And it is manifest that it is also the greatest; for a greater magnitude than the magnitude AB will not measure AB. Next, let AB not measure CD. Then, if the less be continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB, CD are not incommensurable; [cf. X. 2] let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE. Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB. But it measures itself also; therefore AF will also measure the whole AB. But AB measures DE; therefore AF will also measure ED. But it measures CE also; therefore it also measures the whole CD. Therefore AF is a common measure of AB, CD. I say next that it is also the greatest. For, if not, there will be some magnitude greater than AF which will measure AB, CD. Let it be G. Since then G measures AB, while AB measures ED, therefore G will also measure ED. But it measures the whole CD also; therefore G will also measure the remainder CE. But CE measures FB; therefore G will also measure FB. But it measures the whole AB also, and it will therefore measure the remainder AF, the greater the less: which is impossible. Therefore no magnitude greater than AF will measure AB, CD; therefore AF is the greatest common measure of AB, CD. Therefore the greatest common measure of the two given commensurable magnitudes AB, CD has been found.'"],["Rule","'ProofWordCount'",327],["Rule","'GreekProof'","'ἔστω τὰ δοθέντα δύο μεγέθη σύμμετρα τὰ ΑΒ, ΓΔ, ὧν ἔλασσον τὸ ΑΒ: δεῖ δὴ τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν. τὸ ΑΒ γὰρ μέγεθος ἤτοι μετρεῖ τὸ ΓΔ á¼¢ οὔ. εἰ μὲν οὖν μετρεῖ, μετρεῖ δὲ καὶ ἑαυτό, τὸ ΑΒ ἄρα τῶν ΑΒ, ΓΔ κοινὸν μέτρον ἐστίν: καὶ φανερόν, ὅτι καὶ μέγιστον. μεῖζον γὰρ τοῦ ΑΒ μεγέθους τὸ ΑΒ οὐ μετρήσει. μὴ μετρείτω δὴ τὸ ΑΒ τὸ ΓΔ. καὶ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, τὸ περιλειπόμενον μετρήσει ποτὲ τὸ πρὸ ἑαυτοῦ διὰ τὸ μὴ εἶναι ἀσύμμετρα τὰ ΑΒ, ΓΔ: καὶ τὸ μὲν ΑΒ τὸ ΕΔ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΕΓ, τὸ δὲ ΕΓ τὸ ΖΒ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΑΖ, τὸ δὲ ΑΖ τὸ ΓΕ μετρείτω. ἐπεὶ οὖν τὸ ΑΖ τὸ ΓΕ μετρεῖ, ἀλλὰ τὸ ΓΕ τὸ ΖΒ μετρεῖ, καὶ τὸ ΑΖ ἄρα τὸ ΖΒ μετρήσει. μετρεῖ δὲ καὶ ἑαυτό: καὶ ὅλον ἄρα τὸ ΑΒ μετρήσει τὸ ΑΖ. ἀλλὰ τὸ ΑΒ τὸ ΔΕ μετρεῖ: καὶ τὸ ΑΖ ἄρα τὸ ΕΔ μετρήσει. μετρεῖ δὲ καὶ τὸ ΓΕ: καὶ ὅλον ἄρα τὸ ΓΔ μετρεῖ: τὸ ΑΖ ἄρα τῶν ΑΒ, ΓΔ κοινὸν μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή, ἔσται τι μέγεθος μεῖζον τοῦ ΑΖ, ὃ μετρήσει τὰ ΑΒ, ΓΔ. ἔστω τὸ Η. ἐπεὶ οὖν τὸ Η τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΕΔ μετρεῖ, καὶ τὸ Η ἄρα τὸ ΕΔ μετρήσει. μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ: καὶ λοιπὸν ἄρα τὸ ΓΕ μετρήσει τὸ Η. ἀλλὰ τὸ ΓΕ τὸ ΖΒ μετρεῖ: καὶ τὸ Η ἄρα τὸ ΖΒ μετρήσει. μετρεῖ δὲ καὶ ὅλον τὸ ΑΒ, καὶ λοιπὸν τὸ ΑΖ μετρήσει, τὸ μεῖζον τὸ ἔλασσον: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα μεῖζόν τι μέγεθος τοῦ ΑΖ τὰ ΑΒ, ΓΔ μετρήσει: τὸ ΑΖ ἄρα τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν μέτρον ἐστίν. δύο ἄρα μεγεθῶν συμμέτρων δοθέντων τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν μέτρον ηὕρηται: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν μέγεθος δύο μεγέθη μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει.'"],["Rule","'GreekProofWordCount'",331]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'10.4'"],["Rule","'Text'","'Given three commensurable magnitudes, to find their greatest common measure.'"],["Rule","'TextWordCount'",10],["Rule","'GreekText'","'τριῶν μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'Let A, B, C be the three given commensurable magnitudes; thus it is required to find the greatest common measure of A, B, C. Let the greatest common measure of the two magnitudes A, B be taken, and let it be D; [X. 3] then D either measures C, or does not measure it. First, let it measure it. Since then D measures C, while it also measures A, B, therefore D is a common measure of A, B, C. And it is manifest that it is also the greatest; for a greater magnitude than the magnitude D does not measure A, B. Next, let D not measure C. I say first that C, D are commensurable. For, since A, B, C are commensurable, some magnitude will measure them, and this will of course measure A, B also; so that it will also measure the greatest common measure of A, B, namely D. [X. 3] But it also measures C; so that the said magnitude will measure C, D; therefore C, D are commensurable. Now let their greatest common measure be taken, and let it be E. [X. 3] Since then E measures D, while D measures A, B, therefore E will also measure A, B. But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C. I say next that it is also the greatest. For, if possible, let there be some magnitude F greater than E, and let it measure A, B, C. Now, since F measures A, B, C, it will also measure A, B, and will measure the greatest common measure of A, B. [X. 3] But the greatest common measure of A, B is D; therefore F measures D. But it measures C also; therefore F measures C, D; therefore F will also measure the greatest common measure of C, D. [X. 3] But that is E; therefore F will measure E, the greater the less: which is impossible. Therefore no magnitude greater than the magnitude E will measure A, B, C; therefore E is the greatest common measure of A, B, C if D do not measure C, and, if it measure it, D is itself the greatest common measure. Therefore the greatest common measure of the three given commensurable magnitudes has been found.'"],["Rule","'ProofWordCount'",390],["Rule","'GreekProof'","'ἔστω τὰ δοθέντα τρία μεγέθη σύμμετρα τὰ Α, Β, Γ: δεῖ δὴ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν. εἰλήφθω γὰρ δύο τῶν Α, Β τὸ μέγιστον κοινὸν μέτρον, καὶ ἔστω τὸ Δ: τὸ δὴ Δ τὸ Γ ἤτοι μετρεῖ á¼¢ οὔ μετρεῖ. μετρείτω πρότερον. ἐπεὶ οὖν τὸ Δ τὸ Γ μετρεῖ, μετρεῖ δὲ καὶ τὰ Α, Β, τὸ Δ ἄρα τὰ Α, Β, Γ μετρεῖ: τὸ Δ ἄρα τῶν Α, Β, Γ κοινὸν μέτρον ἐστίν. καὶ φανερόν, ὅτι καὶ μέγιστον: μεῖζον γὰρ τοῦ Δ μεγέθους τὰ Α, Β οὐ μετρεῖ. μὴ μετρείτω δὴ τὸ Δ τὸ Γ. λέγω πρῶτον, ὅτι σύμμετρά ἐστι τὰ Γ, Δ. ἐπεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, Γ, μετρήσει τι αὐτὰ μέγεθος, ὃ δηλαδὴ καὶ τὰ Α, Β μετρήσει: ὥστε καὶ τὸ τῶν Α, Β μέγιστον κοινὸν μέτρον τὸ Δ μετρήσει. μετρεῖ δὲ καὶ τὸ Γ: ὥστε τὸ εἰρημένον μέγεθος μετρήσει τὰ Γ, Δ: σύμμετρα ἄρα ἐστὶ τὰ Γ, Δ. εἰλήφθω οὖν αὐτῶν τὸ μέγιστον κοινὸν μέτρον, καὶ ἔστω τὸ Ε. ἐπεὶ οὖν τὸ Ε τὸ Δ μετρεῖ, ἀλλὰ τὸ Δ τὰ Α, Β μετρεῖ, καὶ τὸ Ε ἄρα τὰ Α, Β μετρήσει. μετρεῖ δὲ καὶ τὸ Γ. τὸ Ε ἄρα τὰ Α, Β, Γ μετρεῖ: τὸ Ε ἄρα τῶν Α, Β, Γ κοινόν ἐστι μέτρον. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ δυνατόν, ἔστω τι τοῦ Ε μεῖζον μέγεθος τὸ Ζ, καὶ μετρείτω τὰ Α, Β, Γ. καὶ ἐπεὶ τὸ Ζ τὰ Α, Β, Γ μετρεῖ, καὶ τὰ Α, Β ἄρα μετρήσει καὶ τὸ τῶν Α, Β μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον κοινὸν μέτρον ἐστὶ τὸ Δ: τὸ Ζ ἄρα τὸ Δ μετρεῖ. μετρεῖ δὲ καὶ τὸ Γ: τὸ Ζ ἄρα τὰ Γ, Δ μετρεῖ: καὶ τὸ τῶν Γ, Δ ἄρα μέγιστον κοινὸν μέτρον μετρήσει τὸ Ζ. ἔστι δὲ τὸ ε: τὸ Ζ ἄρα τὸ Ε μετρήσει, τὸ μεῖζον τὸ ἔλασσον: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα μεῖζόν τι τοῦ Ε μεγέθους μέγεθος τὰ Α, Β, Γ μετρεῖ: τὸ Ε ἄρα τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον ἐστίν, ἐὰν μὴ μετρῇ τὸ Δ τὸ Γ, ἐὰν δὲ μετρῇ, αὐτὸ τὸ Δ. τριῶν ἄρα μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον κοινὸν μέτρον ηὕρηται ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν μέγεθος τρία μεγέθη μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει. ὁμοίως δὴ καὶ ἐπὶ πλειόνων τὸ μέγιστον κοινὸν μέτρον ληφθήσεται, καὶ τὸ πόρισμα προχωρήσει. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",404]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'10.5'"],["Rule","'Text'","'Commensurable magnitudes have to one another the ratio which a number has to a number.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'τὰ σύμμετρα μεγέθη πρὸς ἄλληλα λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",7],["Rule","'Definition'",20]]]],["Rule","'Proof'","'Let A, B be commensurable magnitudes; I say that A has to B the ratio which a number has to a number. For, since A, B are commensurable, some magnitude will measure them. Let it measure them, and let it be C. And, as many times as C measures A, so many units let there be in D; and, as many times as C measures B, so many units let there be in E. Since then C measures A according to the units in D, while the unit also measures D according to the units in it, therefore the unit measures the number D the same number of times as the magnitude C measures A; therefore, as C is to A, so is the unit to D; [VII. Def. 20] therefore, inversely, as A is to C, so is D to the unit. [cf. V. 7] Again, since C measures B according to the units in E, while the unit also measures E according to the units in it, therefore the unit measures E the same number of times as C measures B; therefore, as C is to B, so is the unit to E. But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to B, so is the number D to E. [V. 22]  Therefore the commensurable magnitudes A, B have to one another the ratio which the number D has to the number E.'"],["Rule","'ProofWordCount'",250],["Rule","'GreekProof'","'ἔστω σύμμετρα μεγέθη τὰ Α, Β: λέγω, ὅτι τὸ Α πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐπεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, καὶ ἔστω τὸ Γ. καὶ ὁσάκις τὸ Γ τὸ Α μετρεῖ τοσαῦται μονάδες ἔστωσαν ἐν τῷ Δ, ὁσάκις δὲ τὸ Γ τὸ Β μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν τὸ Γ τὸ Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας, μετρεῖ δὲ καὶ ἡ μονὰς τὸν Δ κατὰ τὰς ἐν αὐτῷ μονάδας, ἰσάκις ἄρα ἡ μονὰς τὸν Δ μετρεῖ ἀριθμὸν καὶ τὸ Γ μέγεθος τὸ Α: ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ: ἀνάπαλιν ἄρα, ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὴν μονάδα. πάλιν ἐπεὶ τὸ Γ τὸ Β μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, μετρεῖ δὲ καὶ ἡ μονὰς τὸν Ε κατὰ τὰς ἐν αὐτῷ μονάδας, ἰσάκις ἄρα ἡ μονὰς τὸν Ε μετρεῖ καὶ τὸ Γ τὸ Β: ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Β, οὕτως ἡ μονὰς πρὸς τὸν Ε. ἐδείχθη δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, ὁ Δ πρὸς τὴν μονάδα: δι᾽ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Δ ἀριθμὸς πρὸς τὸν Ε. τὰ ἄρα σύμμετρα μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον ἔχει, ὃν ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε: ὅπερ ἔδει δεῖξαι. ἐὰν δύο μεγέθη πρὸς ἄλληλα λόγον ἔχῃ, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρα ἔσται τὰ μεγέθη. δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον ἐχέτω, ὃν ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε: λέγω, ὅτι σύμμετρά ἐστι τὰ Α, Β μεγέθη. ὅσαι γάρ εἰσιν ἐν τῷ Δ μονάδες, εἰς τοσαῦτα ἴσα διῃρήσθω τὸ Α, καὶ ἑνὶ αὐτῶν ἴσον ἔστω τὸ Γ: ὅσαι δέ εἰσιν ἐν τῷ Ε μονάδες, ἐκ τοσούτων μεγεθῶν ἴσων τῷ Γ συγκείσθω τὸ Ζ. ἐπεὶ οὖν, ὅσαι εἰσὶν ἐν τῷ Δ μονάδες, τοσαῦτά εἰσι καὶ ἐν τῷ Α μεγέθη ἴσα τῷ Γ, ὃ ἄρα μέρος ἐστὶν ἡ μονὰς τοῦ Δ, τὸ αὐτὸ μέρος ἐστὶ καὶ τὸ Γ τοῦ Α: ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ. μετρεῖ δὲ ἡ μονὰς τὸν Δ ἀριθμόν: μετρεῖ ἄρα καὶ τὸ Γ τὸ Α. καὶ ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ ἀριθμόν, ἀνάπαλιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ ἀριθμὸς πρὸς τὴν μονάδα. πάλιν ἐπεί, ὅσαι εἰσὶν ἐν τῷ Ε μονάδες, τοσαῦτά εἰσι καὶ ἐν τῷ Ζ ἴσα τῷ Γ, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ, οὕτως ἡ μονὰς πρὸς τὸν Ε ἀριθμόν. ἐδείχθη δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὴν μονάδα: δι᾽ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Ζ, οὕτως ὁ Δ πρὸς τὸν Ε. ἀλλ᾽ ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ἐστὶ τὸ Α πρὸς τὸ Β: καὶ ὡς ἄρα τὸ Α πρὸς τὸ Β, οὕτως καὶ πρὸς τὸ Ζ. τὸ Α ἄρα πρὸς ἑκάτερον τῶν Β, Ζ τὸν αὐτὸν ἔχει λόγον: ἴσον ἄρα ἐστὶ τὸ Β τῷ Ζ. μετρεῖ δὲ τὸ Γ τὸ Ζ: μετρεῖ ἄρα καὶ τὸ Β. ἀλλὰ μὴν καὶ τὸ Α: τὸ Γ ἄρα τὰ Α, Β μετρεῖ. σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Β. ἐὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν ὦσι δύο ἀριθμοί, ὡς οἱ Δ, Ε, καὶ εὐθεῖα, ὡς ἡ Α, δύνατόν ἐστι ποιῆσαι ὡς ὁ Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν, οὕτως τὴν εὐθεῖαν πρὸς εὐθεῖαν. ἐὰν δὲ καὶ τῶν Α, Ζ μέση ἀνάλογον ληφθῇ, ὡς ἡ Β, ἔσται ὡς ἡ Α πρὸς τὴν Ζ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Β, τουτέστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον. ἀλλ᾽ ὡς ἡ Α πρὸς τὴν Ζ, οὕτως ἐστὶν ὁ Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν: γέγονεν ἄρα καὶ ὡς ὁ Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν, οὕτως τὸ ἀπὸ τῆς Α εὐθείας πρὸς τὸ ἀπὸ τῆς Β εὐθείας: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",683]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'10.6'"],["Rule","'Text'","'If two magnitudes have to one another the ratio which a number has to a number, the magnitudes will be commensurable.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'ἐὰν δύο μεγέθη πρὸς ἄλληλα λόγον ἔχῃ, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρα ἔσται τὰ μεγέθη.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",7],["Rule","'Definition'",20]]]],["Rule","'Proof'","'For let the two magnitudes A, B have to one another the ratio which the number D has to the number E; I say that the magnitudes A, B are commensurable. For let A be divided into as many equal parts as there are units in D, and let C be equal to one of them; and let F be made up of as many magnitudes equal to C asthere are units in E. Since then there are in A as many magnitudes equal to C as there are units in D, whatever part the unit is of D, the same part is C of A also; therefore, as C is to A, so is the unit to D. [VII. Def. 20] But the unit measures the number D; therefore C also measures A. And since, as C is to A, so is the unit to D, therefore, inversely, as A is to C, so is the number D to the unit. [cf. V. 7] Again, since there are in F as many magnitudes equal to C as there are units in E, therefore, as C is to F, so is the unit to E. [VII. Def. 20] But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to F, so is D to E. [v. 22] But, as D is to E, so is A to B; therefore also, as A is to B, so is it to F also. [V. 11] Therefore A has the same ratio to each of the magnitudes B, F;therefore B is equal to F. [V. 9] But C measures F; therefore it measures B also. Further it measures A also; therefore C measures A, B. Therefore A is commensurable with B.'"],["Rule","'ProofWordCount'",302],["Rule","'GreekProof'","'Δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον ἐχέτω, ὃν ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε: λέγω, ὅτι σύμμετρά ἐστι τὰ Α, Β μεγέθη. Ὅσαι γάρ εἰσιν ἐν τῷ Δ μονάδες, εἰς τοσαῦτα ἴσα διῃρήσθω τὸ Α, καὶ ἑνὶ αὐτῶν ἴσον ἔστω τὸ Γ: ὅσαι δέ εἰσιν ἐν τῷ Ε μονάδες, ἐκ τοσούτων μεγεθῶν ἴσων τῷ Γ συγκείσθω τὸ Ζ. Ἐπεὶ οὖν, ὅσαι εἰσὶν ἐν τῷ Δ μονάδες, τοσαῦτά εἰσι καὶ ἐν τῷ Α μεγέθη ἴσα τῷ Γ, ὃ ἄρα μέρος ἐστὶν ἡ μονὰς τοῦ Δ, τὸ αὐτὸ μέρος ἐστὶ καὶ τὸ Γ τοῦ Α: ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ. μετρεῖ δὲ ἡ μονὰς τὸν Δ ἀριθμόν: μετρεῖ ἄρα καὶ τὸ Γ τὸ Α.  καὶ ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ [ἀριθμόν], ἀνάπαλιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ ἀριθμὸς πρὸς τὴν μονάδα. πάλιν ἐπεί, ὅσαι εἰσὶν ἐν τῷ Ε μονάδες, τοσαῦτά εἰσι καὶ ἐν τῷ Ζ ἴσα τῷ Γ, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ, οὕτως ἡ μονὰς πρὸς τὸν Ε [ἀριθμόν]. ἐδείχθη δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὴν μονάδα: δι᾽ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Ζ, οὕτως ὁ Δ πρὸς τὸν Ε. ἀλλ᾽ ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ἐστὶ τὸ Α πρὸς τὸ Β: καὶ ὡς ἄρα τὸ Α πρὸς τὸ Β, οὕτως καὶ πρὸς τὸ Ζ. τὸ Α ἄρα πρὸς ἑκάτερον τῶν Β, Ζ τὸν αὐτὸν ἔχει λόγον: ἴσον ἄρα ἐστὶ τὸ Β τῷ Ζ. μετρεῖ δὲ τὸ Γ τὸ Ζ: μετρεῖ ἄρα καὶ τὸ Β. ἀλλὰ μὴν καὶ τὸ Α:  τὸ Γ ἄρα τὰ Α, Β μετρεῖ. σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Β. Ἐὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",306]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'10.7'"],["Rule","'Text'","'Incommensurable magnitudes have not to one another the ratio which a number has to a number.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'τὰ ἀσύμμετρα μεγέθη πρὸς ἄλληλα λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'Let A, B be incommensurable magnitudes; I say that A has not to B the ratio which a number has to a number. For, if A has to B the ratio which a number has to a number, A will be commensurable with B. [X. 6] But it is not; therefore A has not to B the ratio which a number has to a number.'"],["Rule","'ProofWordCount'",65],["Rule","'GreekProof'","'ἔστω ἀσύμμετρα μεγέθη τὰ Α, Β: λέγω, ὅτι τὸ Α πρὸς τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. εἰ γὰρ ἔχει τὸ Α πρὸς τὸ Β λόγον, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρον ἔσται τὸ Α τῷ Β. οὐκ ἔστι δέ: οὐκ ἄρα τὸ Α πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. τὰ ἄρα ἀσύμμετρα μεγέθη πρὸς ἄλληλα λόγον οὐκ ἔχει, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",67]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'10.8'"],["Rule","'Text'","'If two magnitudes have not to one another the ratio which a number has to a number, the magnitudes will be incommensurable.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐὰν δύο μεγέθη πρὸς ἄλληλα λόγον μὴ ἔχῃ, ὃν ἀριθμὸς πρὸς ἀριθμόν, ἀσύμμετρα ἔσται τὰ μεγέθη.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'For let the two magnitudes A, B not have to one another the ratio which a number has to a number; I say that the magnitudes A, B are incommensurable. For, if they are commensurable, A will have to B the ratio which a number has to a number. [X. 5] But it has not; therefore the magnitudes A, B are incommensurable.'"],["Rule","'ProofWordCount'",62],["Rule","'GreekProof'","'δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον μὴ ἐχέτω, ὃν ἀριθμὸς πρὸς ἀριθμόν: λέγω, ὅτι ἀσύμμετρά ἐστι τὰ Α, Β μεγέθη. εἰ γὰρ ἔσται σύμμετρα, τὸ Α πρὸς τὸ Β λόγον ἕξει, ὃν ἀριθμὸς πρὸς ἀριθμόν. οὐκ ἔχει δέ. ἀσύμμετρα ἄρα ἐστὶ τὰ Α, Β μεγέθη. ἐὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",57]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'10.9'"],["Rule","'Text'","'The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either.'"],["Rule","'TextWordCount'",102],["Rule","'GreekText'","'τὰ ἀπὸ τῶν μήκει συμμέτρων εὐθειῶν τετράγωνα πρὸς ἄλληλα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ τὰ τετράγωνα τὰ πρὸς ἄλληλα λόγον ἔχοντα, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ τὰς πλευρὰς ἕξει μήκει συμμέτρους.'"],["Rule","'GreekTextWordCount'",37],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",20]],["List",["Rule","'Book'",8],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'For let A, B be commensurable in length; I say that the square on A has to the square on B the ratio which a square number has to a square number. For, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [X. 5] Let it have to it the ratio which C has to D. Since then, as A is to B, so is C to D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, for similar figures are in the duplicate ratio of their corresponding sides; [VI. 20] and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, for between two square numbers there is one mean proportional number, and the square number has to the square number the ratio duplicate of that which the side has to the side; [VIII. 11] therefore also, as the square on A is to the square on B, so is the square on C to the square on D. Next, as the square on A is to the square on B, so let the square on C be to the square on D; I say that A is commensurable in length with B. For since, as the square on A is to the square on B, so is the square on C to the square on D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, therefore also, as A is to B, so is C to D. Therefore A has to B the ratio which the number C has to the number D; therefore A is commensurable in length with B. [X. 6] Next, let A be incommensurable in length with B; I say that the square on A has not to the square on B the ratio which a square number has to a square number. For, if the square on A has to the square on B the ratio which a square number has to a square number, A will be commensurable with B. But it is not; therefore the square on A has not to the square on B the ratio which a square number has to a square number. Again, let the square on A not have to the square on B the ratio which a square number has to a square number; I say that A is incommensurable in length with B. For, if A is commensurable with B, the square on A will have to the square on B the ratio which a square number has to a square number. But it has not; therefore A is not commensurable in length with B.'"],["Rule","'ProofWordCount'",503],["Rule","'GreekProof'","'τὰ δὲ ἀπὸ τῶν μήκει ἀσυμμέτρων εὐθειῶν τετράγωνα πρὸς ἄλληλα λόγον οὐκ ἔχει, ὅνπερ τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ τὰ τετράγωνα τὰ πρὸς ἄλληλα λόγον μὴ ἔχοντα, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὰς πλευρὰς ἕξει μήκει συμμέτρους. ἔστωσαν γὰρ αἱ Α, Β μήκει σύμμετροι: λέγω, ὅτι τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἐπεὶ γὰρ σύμμετρός ἐστιν ἡ Α τῇ Β μήκει, ἡ Α ἄρα πρὸς τὴν Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ Γ πρὸς τὸν Δ. ἐπεὶ οὖν ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ Γ πρὸς τὸν Δ, ἀλλὰ τοῦ μὲν τῆς Α πρὸς τὴν Β λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τῆς Α τετραγώνου πρὸς τὸ ἀπὸ τῆς Β τετράγωνον: τὰ γὰρ ὅμοια σχήματα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν: τοῦ δὲ τοῦ Γ ἀριθμοῦ πρὸς τὸν Δ ἀριθμὸν λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τοῦ Γ τετραγώνου πρὸς τὸν ἀπὸ τοῦ Δ τετράγωνον: δύο γὰρ τετραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καὶ ὁ τετράγωνος πρὸς τὸν τετράγωνον ἀριθμὸν διπλασίονα λόγον ἔχει, ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν: ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος ἀριθμὸς πρὸς τὸν ἀπὸ τοῦ Δ ἀριθμοῦ τετράγωνον ἀριθμόν. ἀλλὰ δὴ ἔστω ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος πρὸς τὸν ἀπὸ τοῦ Δ τετράγωνον: λέγω, ὅτι σύμμετρός ἐστιν ἡ Α τῇ Β μήκει. ἐπεὶ γάρ ἐστιν ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος πρὸς τὸν ἀπὸ τοῦ Δ τετράγωνον, ἀλλ᾽ ὁ μὲν τοῦ ἀπὸ τῆς Α τετραγώνου πρὸς τὸ ἀπὸ τῆς Β τετράγωνον λόγος διπλασίων ἐστὶ τοῦ τῆς Α πρὸς τὴν Β λόγου, ὁ δὲ τοῦ ἀπὸ τοῦ Γ ἀριθμοῦ τετραγώνου ἀριθμοῦ πρὸς τὸν ἀπὸ τοῦ Δ ἀριθμοῦ τετράγωνον ἀριθμὸν λόγος διπλασίων ἐστὶ τοῦ τοῦ Γ ἀριθμοῦ πρὸς τὸν Δ ἀριθμὸν λόγου, ἔστιν ἄρα καὶ ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ Γ ἀριθμὸς πρὸς τὸν Δ ἀριθμόν. ἡ Α ἄρα πρὸς τὴν Β, λόγον ἔχει, ὃν ἀριθμὸς ὁ Γ πρὸς ἀριθμὸν τὸν Δ: σύμμετρος ἄρα ἐστὶν ἡ Α τῇ Β μήκει. ἀλλὰ δὴ ἀσύμμετρος ἔστω ἡ Α τῇ Β μήκει: λέγω, ὅτι τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. εἰ γὰρ ἔχει τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, σύμμετρος ἔσται ἡ Α τῇ Β. οὐκ ἔστι δέ: οὐκ ἄρα τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. πάλιν δὴ τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: λέγω, ὅτι ἀσύμμετρός ἐστιν ἡ Α τῇ Β μήκει. εἰ γάρ ἐστι σύμμετρος ἡ Α τῇ Β, ἕξει τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Β λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. οὐκ ἔχει δέ: οὐκ ἄρα σύμμετρός ἐστιν ἡ Α τῇ Β μήκει. τὰ ἄρα ἀπὸ τῶν μήκει συμμέτρων, καὶ τὰ ἑξῆς. Πόρισμα καὶ φανερὸν ἐκ τῶν δεδειγμένων ἔσται, ὅτι αἱ μήκει σύμμετροι πάντως καὶ δυνάμει, αἱ δὲ δυνάμει οὐ πάντως καὶ μήκει [εἴπερ τὰ ἀπὸ τῶν μήκει συμμέτρων εὐθειῶν τετράγωνα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, τὰ δὲ λόγον ἔχοντα, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρά ἐστιν. ὥστε αἱ μήκει σύμμετροι εὐθεῖαι οὐ μόνον εἰσὶ μήκει σύμμετροι, ἀλλὰ καὶ δυνάμει. πάλιν ἐπεί, ὅσα τετράγωνα πρὸς ἄλληλα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, μήκει ἐδείχθη σύμμετρα καὶ δυνάμει ὄντα σύμμετρα τῷ τὰ τετράγωνα λόγον ἔχειν, ὃν ἀριθμὸς πρὸς ἀριθμόν, ὅσα ἄρα τετράγωνα λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ἀλλὰ ἁπλῶς, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρα μὲν ἔσται αὐτὰ τὰ τετράγωνα δυνάμει, οὐκέτι δὲ καὶ μήκει: ὥστε τὰ μὲν μήκει σύμμετρα πάντως καὶ δυνάμει, τὰ δὲ δυνάμει οὐ πάντως καὶ μήκει, εἰ μὴ καὶ λόγον ἔχοιεν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. λέγω δή, ὅτι καὶ αἱ μήκει ἀσύμμετροι οὐ πάντως καὶ δυνάμει, ἐπειδήπερ αἱ δυνάμει σύμμετροι δύνανται λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ διὰ τοῦτο δυνάμει οὖσαι σύμμετροι μήκει εἰσὶν ἀσύμμετροι. ὥστε οὐχ αἱ τῷ μήκει ἀσύμμετροι πάντως καὶ δυνάμει, ἀλλὰ δύνανται μήκει οὖσαι ἀσύμμετροι δυνάμει εἶναι καὶ ἀσύμμετροι καὶ σύμμετροι. αἱ δὲ δυνάμει ἀσύμμετροι πάντως καὶ μήκει ἀσύμμετροι: εἰ γὰρ εἰσι μήκει σύμμετροι, ἔσονται καὶ δυνάμει σύμμετροι. ὑπόκεινται δὲ καὶ ἀσύμμετροι: ὅπερ ἄτοπον. αἱ ἄρα δυνάμει ἀσύμμετροι πάντως καὶ μήκει]. λῆμμα δέδεικται ἐν τοῖς ἀριθμητικοῖς, ὅτι οἱ ὅμοιοι ἐπίπεδοι ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ὅτι, ἐὰν δύο ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχωσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ὅμοιοί εἰσιν ἐπίπεδοι. καὶ δῆλον ἐκ τούτων, ὅτι οἱ μὴ ὅμοιοι ἐπίπεδοι ἀριθμοί, τουτέστιν οἱ μὴ ἀνάλογον ἔχοντες τὰς πλευράς, πρὸς ἀλλήλους λόγον οὐκ ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. εἰ γὰρ ἕξουσιν, ὅμοιοι ἐπίπεδοι ἔσονται: ὅπερ οὐχ ὑπόκειται. οἱ ἄρα μὴ ὅμοιοι ἐπίπεδοι πρὸς ἀλλήλους λόγον οὐκ ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.'"],["Rule","'GreekProofWordCount'",860]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'10.10'"],["Rule","'Text'","'To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'τῇ προτεθείσῃ εὐθείᾳ προσευρεῖν δύο εὐθείας ἀσυμμέτρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ δυνάμει.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'Let A be the assigned straight line; thus it is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A. Let two numbers B, C be set out which have not to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers; and let it be contrived that, as B is to C, so is the square on A to the square on D  —for we have learnt how to do this— [X. 6] therefore the square on A is commensurable with the square on D. [X. 6] And, since B has not to C the ratio which a square number has to a square number, therefore neither has the square on A to the square on D the ratio which a square number has to a square number; therefore A is incommensurable in length with D. [X. 9] Let E be taken a mean proportional between A, D; therefore, as A is to D, so is the square on A to the square on E. [V. Def. 9] But A is incommensurable in length with D; therefore the square on A is also incommensurable with the square on E; [X. 11] therefore A is incommensurable in square with E. Therefore two straight lines D, E have been found incommensurable, D in length only, and E in square and of course in length also, with the assigned straight line A.]'"],["Rule","'ProofWordCount'",251],["Rule","'GreekProof'","'ἔστω ἡ προτεθεῖσα εὐθεῖα ἡ Α: δεῖ δὴ τῇ Α προσευρεῖν δύο εὐθείας ἀσυμμέτρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ δυνάμει. Ἐκκείσθωσαν γὰρ δύο ἀριθμοὶ οἱ Β, Γ πρὸς ἀλλήλους λόγον μὴ ἔχοντες, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, τουτέστι μὴ ὅμοιοι ἐπίπεδοι, καὶ γεγονέτω ὡς ὁ Β πρὸς τὸν Γ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Δ τετράγωνον: ἐμάθομεν γάρ: σύμμετρον ἄρα τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς Δ. καὶ ἐπεὶ ὁ Β πρὸς τὸν Γ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾽ ἄρα τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Δ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ Δ μήκει. εἰλήφθω τῶν Α, Δ μέση ἀνάλογον ἡ Ε: ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Δ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Ε. ἀσύμμετρος δέ ἐστιν ἡ Α τῇ Δ μήκει: ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς Α τετράγωνον τῷ ἀπὸ τῆς Ε τετραγώνῳ: ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ Ε δυνάμει. τῇ ἄρα προτεθείσῃ εὐθείᾳ τῇ Α προσεύρηνται δύο εὐθεῖαι ἀσύμμετροι αἱ Δ, Ε, μήκει μὲν μόνον ἡ Δ, δυνάμει δὲ καὶ μήκει δηλαδὴ ἡ Ε ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",205]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'10.11'"],["Rule","'Text'","'If four magnitudes be proportional, and the first be commensurable with the second, the third will also be commensurable with the fourth; and, if the first be incommensurable with the second, the third will also be incommensurable with the fourth.'"],["Rule","'TextWordCount'",40],["Rule","'GreekText'","'ἐὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ πρῶτον τῷ δευτέρῳ σύμμετρον ᾖ, καὶ τὸ τρίτον τῷ τετάρτῳ σύμμετρον ἔσται: κἂν τὸ πρῶτον τῷ δευτέρῳ ἀσύμμετρον ᾖ, καὶ τὸ τρίτον τῷ τετάρτῳ ἀσύμμετρον ἔσται.'"],["Rule","'GreekTextWordCount'",33],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",7]],["List",["Rule","'Book'",10],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let A, B, C, D be four magnitudes in proportion, so that, as A is to B, so is C to D, and let A be commensurable with B; I say that C will also be commensurable with D. For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [X. 5] And, as A is to B, so is C to D; therefore C also has to D the ratio which a number has to a number; therefore C is commensurable with D. [X. 6] Next, let A be incommensurable with B; I say that C will also be incommensurable with D. For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [X. 7] And, as A is to B, so is C to D; therefore neither has C to D the ratio which a number has to a number; therefore C is incommensurable with D. [X. 8]'"],["Rule","'ProofWordCount'",171],["Rule","'GreekProof'","'ἔστωσαν τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, τὸ Α δὲ τῷ Β σύμμετρον ἔστω: λέγω, ὅτι καὶ τὸ Γ τῷ Δ σύμμετρον ἔσται. ἐπεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Β, τὸ Α ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: καὶ τὸ Γ ἄρα πρὸς τὸ Δ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν: σύμμετρον ἄρα ἐστὶ τὸ Γ τῷ Δ. ἀλλὰ δὴ τὸ Α τῷ Β ἀσύμμετρον ἔστω: λέγω, ὅτι καὶ τὸ Γ τῷ Δ ἀσύμμετρον ἔσται. ἐπεὶ γὰρ ἀσύμμετρόν ἐστι τὸ Α τῷ Β, τὸ Α ἄρα πρὸς τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: οὐδὲ τὸ Γ ἄρα πρὸς τὸ Δ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν: ἀσύμμετρον ἄρα ἐστὶ τὸ Γ τῷ Δ. ἐὰν ἄρα τέσσαρα μεγέθη, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",170]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'10.12'"],["Rule","'Text'","'Magnitudes commensurable with the same magnitude are commensurable with one another also.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'τὰ τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ σύμμετρα.'"],["Rule","'GreekTextWordCount'",9],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",8],["Rule","'Theorem'",4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]]]],["Rule","'Proof'","'For let each of the magnitudes A, B be commensurable with C; I say that A is also commensurable with B. For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [X. 5] Let it have the ratio which D has to E. Again, since C is commensurable with B, therefore C has to B the ratio which a number has to a number. [X. 5] Let it have the ratio which F has to G. And, given any number of ratios we please, namely the ratio which D has to E and that which F has to G, let the numbers H, K, L be taken continuously in the given ratios; [cf. VIII. 4] so that, as D is to E, so is H to K, and, as F is to G, so is K to L.  Since, then, as A is to C, so is D to E, while, as D is to E, so is H to K, therefore also, as A is to C, so is H to K. [V. 11] Again, since, as C is to B, so is F to G, while, as F is to G, so is K to L, therefore also, as C is to B, so is K to L. [V. 11] But also, as A is to C, so is H to K; therefore, ex aequali, as A is to B, so is H to L. [V. 22] Therefore A has to B the ratio which a number has to a number; therefore A is commensurable with B. [X. 6]'"],["Rule","'ProofWordCount'",272],["Rule","'GreekProof'","'ἑκάτερον γὰρ τῶν Α, Β τῷ Γ ἔστω σύμμετρον. λέγω, ὅτι καὶ τὸ Α τῷ Β ἐστι σύμμετρον. ἐπεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Γ, τὸ Α ἄρα πρὸς τὸ Γ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ Δ πρὸς τὸν Ε. πάλιν, ἐπεὶ σύμμετρόν ἐστι τὸ Γ τῷ Β, τὸ Γ ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ Ζ πρὸς τὸν Η. καὶ λόγων δοθέντων ὁποσωνοῦν τοῦ τε, ὃν ἔχει ὁ Δ πρὸς τὸν Ε, καὶ ὁ Ζ πρὸς τὸν Η εἰλήφθωσαν ἀριθμοὶ ἑξῆς ἐν τοῖς δοθεῖσι λόγοις οἱ Θ, Κ, Λ: ὥστε εἶναι ὡς μὲν τὸν Δ πρὸς τὸν Ε, οὕτως τὸν Θ πρὸς τὸν Κ, ὡς δὲ τὸν Ζ πρὸς τὸν Η, οὕτως τὸν Κ πρὸς τὸν Λ. ἐπεὶ οὖν ἐστιν ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὸν Ε, ἀλλ᾽ ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Θ πρὸς τὸν Κ, ἔστιν ἄρα καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ. πάλιν, ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Β, οὕτως ὁ Ζ πρὸς τὸν Η, ἀλλ᾽ ὡς ὁ Ζ πρὸς τὸν Η, οὕτως ὁ Κ πρὸς τὸν Λ, καὶ ὡς ἄρα τὸ Γ πρὸς τὸ Β, οὕτως ὁ Κ πρὸς τὸν Λ. ἔστι δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ: δι᾽ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Θ πρὸς τὸν Λ. τὸ Α ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς ὁ Θ πρὸς ἀριθμὸν τὸν Λ: σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Β. τὰ ἄρα τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ σύμμετρα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",285]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'10.13'"],["Rule","'Text'","'If two magnitudes be commensurable, and the one of them be incommensurable with any magnitude, the remaining one will also be incommensurable with the same.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ἐὰν ᾖ δύο μεγέθη σύμμετρα, τὸ δὲ ἕτερον αὐτῶν μεγέθει τινὶ ἀσύμμετρον ᾖ, καὶ τὸ λοιπὸν τῷ αὐτῷ ἀσύμμετρον ἔσται.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'Let A, B be two commensurable magnitudes, and let one of them, A, be incommensurable with any other magnitude C; I say that the remaining one, B, will also be incommensurable with C. For, if B is commensurable with C, while A is also commensurable with B, A is also commensurable with C. [X. 12] But it is also incommensurable with it: which is impossible. Therefore B is not commensurable with C; therefore it is incommensurable with it.'"],["Rule","'ProofWordCount'",78],["Rule","'GreekProof'","'ἔστω δύο μεγέθη σύμμετρα τὰ Α, Β, τὸ δὲ ἕτερον αὐτῶν τὸ Α ἄλλῳ τινὶ τῷ Γ ἀσύμμετρον ἔστω: λέγω, ὅτι καὶ τὸ λοιπὸν τὸ Β τῷ Γ ἀσύμμετρόν ἐστιν. εἰ γάρ ἐστι σύμμετρον τὸ Β τῷ Γ, ἀλλὰ καὶ τὸ Α τῷ Β σύμμετρόν ἐστιν, καὶ τὸ Α ἄρα τῷ Γ σύμμετρόν ἐστιν. ἀλλὰ καὶ ἀσύμμετρον: ὅπερ ἀδύνατον. οὐκ ἄρα σύμμετρόν ἐστι τὸ Β τῷ Γ: ἀσύμμετρον ἄρα. ἐὰν ἄρα ᾖ δύο μεγέθη σύμμετρα, καὶ τὰ ἑξῆς. λῆμμα δύο δοθεισῶν εὐθειῶν ἀνίσων εὑρεῖν, τίνι μεῖζον δύναται ἡ μείζων τῆς ἐλάσσονος. ἔστωσαν αἱ δοθεῖσαι δύο ἄνισοι εὐθεῖαι αἱ ΑΒ, Γ, ὧν μείζων ἔστω ἡ ΑΒ: δεῖ δὴ εὑρεῖν, τίνι μεῖζον δύναται ἡ ΑΒ τῆς Γ. γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ εἰς αὐτὸ ἐνηρμόσθω τῇ Γ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΒ. φανερὸν δή, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΑΔΒ γωνία, καὶ ὅτι ἡ ΑΒ τῆς ΑΔ, τουτέστι τῆς Γ, μεῖζον δύναται τῇ ΔΒ. ὁμοίως δὲ καὶ δύο δοθεισῶν εὐθειῶν ἡ δυναμένη αὐτὰς εὑρίσκεται οὕτως. ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ, καὶ δέον ἔστω εὑρεῖν τὴν δυναμένην αὐτάς. κείσθωσαν γάρ, ὥστε ὀρθὴν γωνίαν περιέχειν τὴν ὑπὸ ΑΔ, ΔΒ, καὶ ἐπεζεύχθω ἡ ΑΒ: φανερὸν πάλιν, ὅτι ἡ τὰς ΑΔ, ΔΒ δυναμένη ἐστὶν ἡ ΑΒ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",211]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'10.14'"],["Rule","'Text'","'If four straight lines be proportional, and the square on the first be greater than the square on the second by the square on a straight line commensurable with the first, the square on the third will also be greater than the square on the fourth by  the square on a straight line commensurable with the third. And, if the square on the first be greater than the square on the second by the square on a straight line incommensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line in-commensurable with the third.'"],["Rule","'TextWordCount'",109],["Rule","'GreekText'","'ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, δύνηται δὲ ἡ πρώτη τῆς δευτέρας μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει, καὶ ἡ τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει.'"],["Rule","'GreekTextWordCount'",29],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",17]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",6],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'Let A, B, C, D be four straight lines in proportion, so that, as A is to B, so is C to D; and let the square on A be greater than the square on B by the square on E, andlet the square on C be greater than the square on D by the square on F; I say that, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C isalso incommensurable with F. For since, as A is to B, so is C to D, therefore also, as the square on A is to the square on B, so is the square on C to the square on D. [VI. 22] But the squares on E, B are equal to the square on A,and the squares on D, F are equal to the square on C. Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D; therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D; [V. 17]therefore also, as E is to B, so is F to D; [VI. 22] therefore, inversely, as B is to E, so is D to F. But, as A is to B, so also is C to D; therefore, ex aequali, as A is to E, so is C to F. [V. 22] Therefore, if A is commensurable with E, C is also commensurablewith F, and, if A is incommensurable with E, C is also incommensurable with F. [X. 11]'"],["Rule","'ProofWordCount'",280],["Rule","'GreekProof'","'καὶ ἐὰν ἡ πρώτη τῆς δευτέρας μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, καὶ ἡ τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, Δ, ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, καὶ ἡ Α μὲν τῆς β μεῖζον δυνάσθω τῷ ἀπὸ τῆς Ε, ἡ δὲ Γ τῆς Δ μεῖζον δυνάσθω τῷ ἀπὸ τῆς Ζ: λέγω, ὅτι, εἴτε σύμμετρός ἐστιν ἡ Α τῇ Ε, σύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός ἐστιν ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ. ἐπεὶ γάρ ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Β, οὕτως τὸ ἀπὸ τῆς Γ πρὸς τὸ ἀπὸ τῆς Δ. ἀλλὰ τῷ μὲν ἀπὸ τῆς Α ἴσα ἐστὶ τὰ ἀπὸ τῶν Ε, Β, τῷ δὲ ἀπὸ τῆς Γ ἴσα ἐστὶ τὰ ἀπὸ τῶν Δ, Ζ. ἔστιν ἄρα ὡς τὰ ἀπὸ τῶν Ε, Β πρὸς τὸ ἀπὸ τῆς Β, οὕτως τὰ ἀπὸ τῶν Δ, Ζ πρὸς τὸ ἀπὸ τῆς Δ: διελόντι ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς Β, οὕτως τὸ ἀπὸ τῆς Ζ πρὸς τὸ ἀπὸ τῆς Δ: ἔστιν ἄρα καὶ ὡς ἡ Ε πρὸς τὴν Β, οὕτως ἡ Ζ πρὸς τὴν Δ: ἀνάπαλιν ἄρα ἐστὶν ὡς ἡ Β πρὸς τὴν Ε, οὕτως ἡ Δ πρὸς τὴν Ζ. ἔστι δὲ καὶ ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ: δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ Α πρὸς τὴν Ε, οὕτως ἡ Γ πρὸς τὴν Ζ. εἴτε οὖν σύμμετρός ἐστιν ἡ Α τῇ Ε, σύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός ἐστιν ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ. ἐὰν ἄρα, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",305]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'10.15'"],["Rule","'Text'","'If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἐὰν δύο μεγέθη σύμμετρα συντεθῇ, καὶ τὸ ὅλον ἑκατέρῳ αὐτῶν σύμμετρον ἔσται: κἂν τὸ ὅλον ἑνὶ αὐτῶν σύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη σύμμετρα ἔσται.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",1]]]],["Rule","'Proof'","'For let the two commensurable magnitudes AB, BC be added together; I say that the whole AC is also commensurable with each of the magnitudes AB, BC. For, since AB, BC are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC, it will also measure the whole AC. But it measures AB, BC also; therefore D measures AB, BC, AC; therefore AC is commensurable with each of the magnitudes AB, BC. [X. Def. 1] Next, let AC be commensurable with AB; I say that AB, BC are also commensurable. For, since AC, AB are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures CA, AB, it will also measure the remainder BC. But it measures AB also; therefore D will measure AB, BC; therefore AB, BC are commensurable. [X. Def. 1]'"],["Rule","'ProofWordCount'",153],["Rule","'GreekProof'","'Συγκείσθω γὰρ δύο μεγέθη σύμμετρα τὰ ΑΒ, ΒΓ: λέγω, ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἐστι σύμμετρον. ἐπεὶ γὰρ σύμμετρά ἐστι τὰ ΑΒ, ΒΓ, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον τὸ ΑΓ μετρήσει. μετρεῖ δὲ καὶ τὰ ΑΒ, ΒΓ. τὸ Δ ἄρα τὰ ΑΒ, ΒΓ, ΑΓ μετρεῖ: σύμμετρον ἄρα ἐστὶ τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ. ἀλλὰ δὴ τὸ ΑΓ ἔστω σύμμετρον τῷ ΑΒ: λέγω δή, ὅτι καὶ τὰ ΑΒ, ΒΓ σύμμετρά ἐστιν. ἐπεὶ γὰρ σύμμετρά ἐστι τὰ ΑΓ, ΑΒ, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει. μετρεῖ δὲ καὶ τὸ ΑΒ: τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρήσει: σύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΒΓ. ἐὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",144]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'10.16'"],["Rule","'Text'","'If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἐὰν δύο μεγέθη ἀσύμμετρα συντεθῇ, καὶ τὸ ὅλον ἑκατέρῳ αὐτῶν ἀσύμμετρον ἔσται: κἂν τὸ ὅλον ἑνὶ αὐτῶν ἀσύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη ἀσύμμετρα ἔσται.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",1]]]],["Rule","'Proof'","'For let the two incommensurable magnitudes AB, BC be added together; I say that the whole AC is also incommensurable with each of the magnitudes AB, BC. For, if CA, AB are not incommensurable, some magnitude will measure them. Let it measure them, if possible, and let it be D. Since then D measures CA, AB, therefore it will also measure the remainder BC. But it measures AB also; therefore D measures AB, BC. Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure CA, AB; therefore CA, AB are incommensurable. [X. Def. 1] Similarly we can prove that AC, CB are also incommensurable. Therefore AC is incommensurable with each of the magnitudes AB, BC. Next, let AC be incommensurable with one of the magnitudes AB, BC. First, let it be incommensurable with AB; I say that AB, BC are also incommensurable. For, if they are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC. therefore it will also measure the whole AC. But it measures AB also; therefore D measures CA, AB. Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [X. Def. 1]'"],["Rule","'ProofWordCount'",225],["Rule","'GreekProof'","'Συγκείσθω γὰρ δύο μεγέθη ἀσύμμετρα τὰ ΑΒ, ΒΓ: λέγω, ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἀσύμμετρόν ἐστιν. εἰ γὰρ μή ἐστιν ἀσύμμετρα τὰ ΓΑ, ΑΒ, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει. μετρεῖ δὲ καὶ τὸ ΑΒ: τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρεῖ. σύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΒΓ: ὑπέκειντο δὲ καὶ ἀσύμμετρα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΓΑ, ΑΒ μετρήσει τι μέγεθος: ἀσύμμετρα ἄρα ἐστὶ τὰ ΓΑ, ΑΒ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τὰ ΑΓ, ΓΒ ἀσύμμετρά ἐστιν. τὸ ΑΓ ἄρα ἑκατέρῳ τῶν ΑΒ, ΒΓ ἀσύμμετρόν ἐστιν. ἀλλὰ δὴ τὸ ΑΓ ἑνὶ τῶν ΑΒ, ΒΓ ἀσύμμετρον ἔστω. ἔστω δὴ πρότερον τῷ ΑΒ: λέγω, ὅτι καὶ τὰ ΑΒ, ΒΓ ἀσύμμετρά ἐστιν. εἰ γὰρ ἔσται σύμμετρα, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸ ΑΓ μετρήσει. μετρεῖ δὲ καὶ τὸ ΑΒ: τὸ Δ ἄρα τὰ ΓΑ, ΑΒ μετρεῖ. σύμμετρα ἄρα ἐστὶ τὰ ΓΑ, ΑΒ: ὑπέκειτο δὲ καὶ ἀσύμμετρα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΑΒ, ΒΓ μετρήσει τι μέγεθος: ἀσύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΒΓ. ἐὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς. λῆμμα ἐὰν παρά τινα εὐθεῖαν παραβληθῇ παραλληλόγραμμον ἐλλεῖπον εἴδει τετραγώνῳ, τὸ παραβληθὲν ἴσον ἐστὶ τῷ ὑπὸ τῶν ἐκ τῆς παραβολῆς γενομένων τμημάτων τῆς εὐθείας. παρὰ γὰρ εὐθεῖαν τὴν ΑΒ παραβεβλήσθω παραλληλόγραμμον τὸ ΑΔ ἐλλεῖπον εἴδει τετραγώνῳ τῷ ΔΒ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΔ τῷ ὑπὸ τῶν ΑΓ, ΓΒ. καί ἐστιν αὐτόθεν φανερόν: ἐπεὶ γὰρ τετράγωνόν ἐστι τὸ ΔΒ, ἴση ἐστὶν ἡ ΔΓ τῇ ΓΒ, καί ἐστι τὸ ΑΔ τὸ ὑπὸ τῶν ΑΓ, ΓΔ, τουτέστι τὸ ὑπὸ τῶν ΑΓ, ΓΒ. ἐὰν ἄρα παρά τινα εὐθεῖαν, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",295]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'10.17'"],["Rule","'Text'","'If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater. And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with  the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length.'"],["Rule","'TextWordCount'",135],["Rule","'GreekText'","'ἐὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς σύμμετρα αὐτὴν διαιρῇ μήκει, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει.'"],["Rule","'GreekTextWordCount'",38],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let A, BC be two unequal straight lines, of which BC isthe greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficientby a square figure. Let this be the rectangle BD, DC, [cf. Lemma]and let BD be commensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC. For let BC be bisected at the point E, and let EF be made equal to DE. Therefore the remainder DC is equal to BF. And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D,therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [II. 5] And the same is true of their quadruples; therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the squareon EC. But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE, for DF is double of DE. And the square on BC is equal to four times the square on EC, for again BC is double of CE. Therefore the squares on A, DF are equal to the square on BC, so that the square on BC is greater than the square on A bythe square on DF. It is to be proved that BC is also commensurable with DF. Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15] But CD is commensurable in length with CD, BF, forCD is equal to BF. [X. 6] Therefore BC is also commensurable in length with BF, CD, [X. 12] so that BC is also commensurable in length with the remainder FD; [X. 15]therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC. Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth partof the square on A and deficient by a square figure, and let it be the rectangle BD, DC. It is to be proved that BD is commensurable in length with DC. With the same construction, we can prove similarly thatthe square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line commensurable with BC. Therefore BC is commensurable in length with FD,so that BC is also commensurable in length with the remainder, the sum of BF, DC. [X. 15] But the sum of BF, DC is commensurable with DC, [X. 6] so that BC is also commensurable in length with CD; [X. 12] and therefore, separando, BD is commensurable in lengthwith DC. [X. 15]'"],["Rule","'ProofWordCount'",543],["Rule","'GreekProof'","'καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διαιρεῖ μήκει. ἔστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ ΒΓ, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος τῆς Α, τουτέστι τῷ ἀπὸ τῆς ἡμισείας τῆς Α, ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔ, ΔΓ, σύμμετρος δὲ ἔστω ἡ ΒΔ τῇ ΔΓ μήκει: λέγω, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. τετμήσθω γὰρ ἡ ΒΓ δίχα κατὰ τὸ Ε σημεῖον, καὶ κείσθω τῇ ΔΕ ἴση ἡ ΕΖ. λοιπὴ ἄρα ἡ ΔΓ ἴση ἐστὶ τῇ ΒΖ. καὶ ἐπεὶ εὐθεῖα ἡ ΒΓ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Ε, εἰς δὲ ἄνισα κατὰ τὸ Δ, τὸ ἄρα ὑπὸ ΒΔ, ΔΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΔ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΓ τετραγώνῳ: καὶ τὰ τετραπλάσια: τὸ ἄρα τετράκις ὑπὸ τῶν ΒΔ, ΔΓ μετὰ τοῦ τετραπλασίου τοῦ ἀπὸ τῆς ΔΕ ἴσον ἐστὶ τῷ τετράκις ἀπὸ τῆς ΕΓ τετραγώνῳ. ἀλλὰ τῷ μέν τετραπλασίῳ τοῦ ὑπὸ τῶν ΒΔ, ΔΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς Α τετράγωνον, τῷ δὲ τετραπλασίῳ τοῦ ἀπὸ τῆς ΔΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΔΖ τετράγωνον: διπλασίων γάρ ἐστιν ἡ ΔΖ τῆς ΔΕ. τῷ δὲ τετραπλασίῳ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΓ τετράγωνον: διπλασίων γάρ ἐστι πάλιν ἡ ΒΓ τῆς ΓΕ. τὰ ἄρα ἀπὸ τῶν Α, ΔΖ τετράγωνα ἴσα ἐστὶ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ: ὥστε τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς Α μεῖζόν ἐστι τῷ ἀπὸ τῆς ΔΖ: ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῇ ΔΖ. δεικτέον, ὅτι καὶ σύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ. ἐπεὶ γὰρ σύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, σύμμετρος ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει. ἀλλὰ ἡ ΓΔ ταῖς ΓΔ, ΒΖ ἐστι σύμμετρος μήκει: ἴση γάρ ἐστιν ἡ ΓΔ τῇ ΒΖ. καὶ ἡ ΒΓ ἄρα σύμμετρός ἐστι ταῖς ΒΖ, ΓΔ μήκει: ὥστε καὶ λοιπῇ τῇ ΖΔ σύμμετρός ἐστιν ἡ ΒΓ μήκει: ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. ἀλλὰ δὴ ἡ ΒΓ τῆς Α μεῖζον δυνάσθω τῷ ἀπὸ συμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι σύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. δύναται δὲ ἡ ΒΓ τῆς Α μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ. σύμμετρος ἄρα ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει: ὥστε καὶ λοιπῇ συναμφοτέρῳ τῇ ΒΖ, ΔΓ σύμμετρός ἐστιν ἡ ΒΓ μήκει. ἀλλὰ συναμφότερος ἡ ΒΖ, ΔΓ σύμμετρός ἐστι τῇ ΔΓ μήκει. ὥστε καὶ ἡ ΒΓ τῇ ΓΔ σύμμετρός ἐστι μήκει: καὶ διελόντι ἄρα ἡ ΒΔ τῇ ΔΓ ἐστι σύμμετρος μήκει. ἐὰν ἄρα ὦσι δύο εὐθεῖαι ἄνισοι, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",483]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'10.18'"],["Rule","'Text'","'If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are incommensurable, the square on the greater will be greater than the square on the less by the square on a straight line incommensurable with the greater. And, if the square on the greater be greater than the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it divides it into parts which are incommensurable. Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure.'"],["Rule","'TextWordCount'",170],["Rule","'GreekText'","'ἐὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, καὶ εἰς ἀσύμμετρα αὐτὴν διαιρῇ μήκει, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.'"],["Rule","'GreekTextWordCount'",37],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'Let this be the rectangle BD, DC, [cf. Lemma before X. 17] and let BD be incommensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD. It is to be proved that BC is incommensurable in length with DF. Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16] But DC is commensurable with the sum of BF, DC; [X. 6] therefore BC is also incommensurable with the sum of BF, DC; [X. 13] so that BC is also incommensurable in length with the remainder FD. [X. 16] And the square on BC is greater than the square on A by the square on FD; therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC, and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure. Let this be the rectangle BD, DC. It is to be proved that BD is incommensurable in length with DC. For, with the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC; therefore BC is incommensurable in length with FD. so that BC is also incommensurable with the remainder, the sum of BF, DC. [X. 16] But the sum of BF, DC is commensurable in length with DC; [X. 6] therefore BC is also incommensurable in length with DC, [X. 13] so that, separando, BD is also incommensurable in length with DC. [X. 16]'"],["Rule","'ProofWordCount'",361],["Rule","'GreekProof'","'καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διαιρεῖ μήκει. ῎εστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ ΒΓ, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔΓ, ἀσύμμετρος δὲ ἔστω ἡ ΒΔ τῇ ΔΓ μήκει: λέγω, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. τῶν γὰρ αὐτῶν κατασκευασθέντων τῷ πρότερον ὁμοίως δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. δεικτέον οὖν, ὅτι ἀσύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ μήκει. ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει. ἀλλὰ ἡ ΔΓ σύμμετρός ἐστι συναμφοτέραις ταῖς ΒΖ, ΔΓ: καὶ ἡ ΒΓ ἄρα ἀσύμμετρός ἐστι συναμφοτέραις ταῖς ΒΖ, ΔΓ. ὥστε καὶ λοιπῇ τῇ ΖΔ ἀσύμμετρός ἐστιν ἡ ΒΓ μήκει. καὶ ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ: ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. δυνάσθω δὴ πάλιν ἡ ΒΓ τῆς Α μεῖζον τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. ἀλλὰ ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει: ὥστε καὶ λοιπῇ συναμφοτέρῳ τῇ ΒΖ, ΔΓ ἀσύμμετρός ἐστιν ἡ ΒΓ. ἀλλὰ συναμφότερος ἡ ΒΖ, ΔΓ τῇ ΔΓ σύμμετρός ἐστι μήκει: καὶ ἡ ΒΓ ἄρα τῇ ΔΓ ἀσύμμετρός ἐστι μήκει: ὥστε καὶ διελόντι ἡ ΒΔ τῇ ΔΓ ἀσύμμετρός ἐστι μήκει. ἐὰν ἄρα ὦσι δύο εὐθεῖαι, καὶ τὰ ἑξῆς. λῆμμα ἐπεὶ δέδεικται, ὅτι αἱ μήκει σύμμετροι πάντως καὶ δυνάμει εἰσὶ σύμμετροι, αἱ δὲ δυνάμει οὐ πάντως καὶ μήκει, ἀλλὰ δὴ δύνανται μήκει καὶ σύμμετροι εἶναι καὶ ἀσύμμετροι, φανερόν, ὅτι, ἐὰν τῇ ἐκκειμένῃ ῥητῇ σύμμετρός τις ᾖ μήκει, λέγεται ῥητὴ καὶ σύμμετρος αὐτῇ οὐ μόνον μήκει, ἀλλὰ καὶ δυνάμει, ἐπεὶ αἱ μήκει σύμμετροι πάντως καὶ δυνάμει. ἐὰν δὲ τῇ ἐκκειμένῃ ῥητῇ σύμμετρός τις ᾖ δυνάμει, εἰ μὲν καὶ μήκει, λέγεται καὶ οὕτως ῥητὴ καὶ σύμμετρος αὐτῇ μήκει καὶ δυνάμει: εἰ δὲ τῇ ἐκκειμένῃ πάλιν ῥητῇ σύμμετρός τις οὖσα δυνάμει μήκει αὐτῇ ᾖ ἀσύμμετρος, λέγεται καὶ οὕτως ῥητὴ δυνάμει μόνον σύμμετρος.'"],["Rule","'GreekProofWordCount'",413]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'10.19'"],["Rule","'Text'","'The rectangle contained by rational straight lines commensurable in length is rational.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'τὸ ὑπὸ ῥητῶν μήκει συμμέτρων κατά τινα τῶν προειρημένων τρόπων εὐθειῶν περιεχόμενον ὀρθογώνιον ῥητόν ἐστιν.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in length; I say that AC is rational. For on AB let the square AD be described; therefore AD is rational. [X. Def. 4] And, since AB is commensurable in length with BC, while AB is equal to BD, therefore BD is commensurable in length with BC. And, as BD is to BC, so is DA to AC. [VI. 1] Therefore DA is commensurable with AC. [X. 11] But DA is rational; therefore AC is also rational. [X. Def. 4]'"],["Rule","'ProofWordCount'",95],["Rule","'GreekProof'","'ὑπὸ γὰρ ῥητῶν μήκει συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ ὀρθογώνιον περιεχέσθω τὸ ΑΓ: λέγω, ὅτι ῥητόν ἐστι τὸ ΑΓ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ: ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, ἴση δέ ἐστιν ἡ ΑΒ τῇ ΒΔ, σύμμετρος ἄρα ἐστὶν ἡ ΒΔ τῇ ΒΓ μήκει. καί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΒΓ, οὕτως τὸ ΔΑ πρὸς τὸ ΑΓ. σύμμετρον ἄρα ἐστὶ τὸ ΔΑ τῷ ΑΓ. ῥητὸν δὲ τὸ ΔΑ: ῥητὸν ἄρα ἐστὶ καὶ τὸ ΑΓ. τὸ ἄρα ὑπὸ ῥητῶν μήκει συμμέτρων, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",96]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'10.20'"],["Rule","'Text'","'If a rational area be applied to a rational straight line, it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν ῥητὸν παρὰ ῥητὴν παραβληθῇ, πλάτος ποιεῖ ῥητὴν καὶ σύμμετρον τῇ, παρ᾽ ἣν παράκειται, μήκει.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'For let the rational area AC be applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth; I say that BC is rational and commensurable in length with BA. For on AB let the square AD be described; therefore AD is rational. [X. Def. 4] But AC is also rational; therefore DA is commensurable with AC. And, as DA is to AC, so is DB to BC. [VI. 1] Therefore DB is also commensurable with BC; [X. 11] and DB is equal to BA; therefore AB is also commensurable with BC. But AB is rational; therefore BC is also rational and commensurable in length with AB.'"],["Rule","'ProofWordCount'",115],["Rule","'GreekProof'","'ῥητὸν γὰρ τὸ ΑΓ παρὰ ῥητὴν κατά τινα πάλιν τῶν προειρημένων τρόπων τὴν ΑΒ παραβεβλήσθω πλάτος ποιοῦν τὴν ΒΓ: λέγω, ὅτι ῥητή ἐστιν ἡ ΒΓ καὶ σύμμετρος τῇ ΒΑ μήκει. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ: ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. ῥητὸν δὲ καὶ τὸ ΑΓ: σύμμετρον ἄρα ἐστὶ τὸ ΔΑ τῷ ΑΓ. καί ἐστιν ὡς τὸ ΔΑ πρὸς τὸ ΑΓ, οὕτως ἡ ΔΒ πρὸς τὴν ΒΓ. σύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΒ τῇ ΒΓ: ἴση δὲ ἡ ΔΒ τῇ ΒΑ: σύμμετρος ἄρα καὶ ἡ ΑΒ τῇ ΒΓ. ῥητὴ δέ ἐστιν ἡ ΑΒ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΓ καὶ σύμμετρος τῇ ΑΒ μήκει. ἐὰν ἄρα ῥητὸν παρὰ ῥητὴν παραβληθῇ, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",115]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'10.21'"],["Rule","'Text'","'The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called medial.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'τὸ ὑπὸ ῥητῶν δυνάμει μόνον συμμέτρων εὐθειῶν περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν, καλείσθω δὲ μέση.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only; I say that AC is irrational, and the side of the square equal to it is irrational; and let the latter be called medial. For on AB let the square AD be described; therefore AD is rational. [X. Def. 4] And, since AB is incommensurable in length with BC, for by hypothesis they are commensurable in square only, while AB is equal to BD, therefore DB is also incommensurable in length with BC. And, as DB is to BC, so is AD to AC; [VI. 1] therefore DA is incommensurable with AC. [X. 11] But DA is rational; therefore AC is irrational, so that the side of the square equal to AC is also irrational. [X. Def. 4] And let the latter be called medial.'"],["Rule","'ProofWordCount'",143],["Rule","'GreekProof'","'ὑπὸ γὰρ ῥητῶν δυνάμει μόνον συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ ὀρθογώνιον περιεχέσθω τὸ ΑΓ: λέγω, ὅτι ἄλογόν ἐστι τὸ ΑΓ, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν, καλείσθω δὲ μέση. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ: ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει: δυνάμει γὰρ μόνον ὑπόκεινται σύμμετροι: ἴση δὲ ἡ ΑΒ τῇ ΒΔ, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΒ τῇ ΒΓ μήκει. καί ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΓ, οὕτως τὸ ΑΔ πρὸς τὸ ΑΓ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΔΑ τῷ ΑΓ. ῥητὸν δὲ τὸ ΔΑ: ἄλογον ἄρα ἐστὶ τὸ ΑΓ: ὥστε καὶ ἡ δυναμένη τὸ ΑΓ τουτέστιν ἡ ἴσον αὐτῷ τετράγωνον δυναμένη ἄλογός ἐστιν, καλείσθω δὲ μέση: ὅπερ ἔδει δεῖξαι. λῆμμα ἐὰν ὦσι δύο εὐθεῖαι, ἔστιν ὡς ἡ πρώτη πρὸς τὴν δευτέραν, οὕτως τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ὑπὸ τῶν δύο εὐθειῶν. ἔστωσαν δύο εὐθεῖαι αἱ ΖΕ, ΕΗ. λέγω, ὅτι ἐστὶν ὡς ἡ ΖΕ πρὸς τὴν ΕΗ, οὕτως τὸ ἀπὸ τῆς ΖΕ πρὸς τὸ ὑπὸ τῶν ΖΕ, ΕΗ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΖΕ τετράγωνον τὸ ΔΖ, καὶ συμπεπληρώσθω τὸ ΗΔ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΖΕ πρὸς τὴν ΕΗ, οὕτως τὸ ΖΔ πρὸς τὸ ΔΗ, καί ἐστι τὸ μὲν ΖΔ τὸ ἀπὸ τῆς ΖΕ, τὸ δὲ ΔΗ τὸ ὑπὸ τῶν ΔΕ, ΕΗ, τουτέστι τὸ ὑπὸ τῶν ΖΕ, ΕΗ, ἔστιν ἄρα ὡς ἡ ΖΕ τὴν ΕΗ, οὕτως τὸ ἀπὸ τῆς ΖΕ πρὸς τὸ ὑπὸ τῶν ΖΕ, ΕΗ. ὁμοίως δὲ καὶ ὡς τὸ ὑπὸ τῶν ΗΕ, ΕΖ πρὸς τὸ ἀπὸ τῆς ΕΖ, τουτέστιν ὡς τὸ ΗΔ πρὸς τὸ ΖΔ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΖ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",268]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'10.22'"],["Rule","'Text'","'The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'τὸ ἀπὸ μέσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ῥητὴν καὶ ἀσύμμετρον τῇ, παρ᾽ ἣν παράκειται, μήκει.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",14]],["List",["Rule","'Book'",6],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]]]],["Rule","'Proof'","'Let A be medial and CB rational, and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth; I say that CD is rational and incommensurable in length with CB. For, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines commensurable in square only. [X. 21] Let the square on it be equal to GF. But the square on it is also equal to BD; therefore BD is equal to GF. But it is also equiangular with it; and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, proportionally, as BC is to EG, so is EF to CD. Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD. [VI. 22] But the square on CB is commensurable with the square on EG, for each of these straight lines is rational; therefore the square on EF is also commensurable with the square on CD. [X. 11] But the square on EF is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is rational. And, since EF is incommensurable in length with EG, for they are commensurable in square only, and, as EF is to EG, so is the square on EF to the rectangle FE, EG, [Lemma]therefore the square on EF is incommensurable with the rectangle FE, EG. [X. 11] But the square on CD is commensurable with the square on EF, for the straight lines are rational in square; and the rectangle DC, CB is commensurable with the rectangle FE, EG, for they are equal to the square on A; therefore the square on CD is also incommensurable with the rectangle DC, CB. [X. 13] But, as the square on CD is to the rectangle DC, CB, so is DC to CB; [Lemma]therefore DC is incommensurable in length with CB. [X. 11] Therefore CD is rational and incommensurable in length with CB.'"],["Rule","'ProofWordCount'",351],["Rule","'GreekProof'","'ἔστω μέση μὲν ἡ Α, ῥητὴ δὲ ἡ ΓΒ, καὶ τῷ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω χωρίον ὀρθογώνιον τὸ ΒΔ πλάτος ποιοῦν τὴν ΓΔ: λέγω, ὅτι ῥητή ἐστιν ἡ ΓΔ καὶ ἀσύμμετρος τῇ ΓΒ μήκει. ἐπεὶ γὰρ μέση ἐστὶν ἡ Α, δύναται χωρίον περιεχόμενον ὑπὸ ῥητῶν δυνάμει μόνον συμμέτρων. δυνάσθω τὸ ΗΖ. δύναται δὲ καὶ τὸ ΒΔ: ἴσον ἄρα ἐστὶ τὸ ΒΔ τῷ ΗΖ. ἔστι δὲ αὐτῷ καὶ ἰσογώνιον: τῶν δὲ ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΓ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΖ πρὸς τὴν ΓΔ. ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς ΒΓ πρὸς τὸ ἀπὸ τῆς ΕΗ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΓΔ. σύμμετρον δέ ἐστι τὸ ἀπὸ τῆς ΓΒ τῷ ἀπὸ τῆς ΕΗ: ῥητὴ γάρ ἐστιν ἑκατέρα αὐτῶν: σύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΓΔ. ῥητὸν δέ ἐστι τὸ ἀπὸ τῆς ΕΖ: ῥητὸν ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΓΔ: ῥητὴ ἄρα ἐστὶν ἡ ΓΔ. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΕΖ τῇ ΕΗ μήκει: δυνάμει γὰρ μόνον εἰσὶ σύμμετροι: ὡς δὲ ἡ ΕΖ πρὸς τὴν ΕΗ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ὑπὸ τῶν ΖΕ, ΕΗ, ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΖ τῷ ὑπὸ τῶν ΖΕ, ΕΗ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΕΖ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΓΔ: ῥηταὶ γάρ εἰσι δυνάμει: τῷ δὲ ὑπὸ τῶν ΖΕ, ΕΗ σύμμετρόν ἐστι τὸ ὑπὸ τῶν ΑΓ, ΓΒ: ἴσα γάρ ἐστι τῷ ἀπὸ τῆς Α: ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΓΔ τῷ ὑπὸ τῶν ΔΓ, ΓΒ. ὡς δὲ τὸ ἀπὸ τῆς ΓΔ πρὸς τὸ ὑπὸ τῶν ΔΓ, ΓΒ, οὕτως ἐστὶν ἡ ΔΓ πρὸς τὴν ΓΒ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΓ τῇ ΓΒ μήκει. ῥητὴ ἄρα ἐστὶν ἡ ΓΔ καὶ ἀσύμμετρος τῇ ΓΒ μήκει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",306]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'10.23'"],["Rule","'Text'","'A straight line commensurable with a medial straight line is medial.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἡ τῇ μέσῃ σύμμετρος μέση ἐστίν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'Let A be medial, and let B be commensurable with A; I say that B is also medial. For let a rational straight line CD be set out, and to CD let the rectangular area CE equal to the square on A be applied, producing ED as breadth; therefore ED is rational and incommensurable in length with CD. [X. 22] And let the rectangular area CF equal to the square on B be applied to CD, producing DF as breadth. Since then A is commensurable with B, the square on A is also commensurable with the square on B. But EC is equal to the square on A, and CF is equal to the square on B; therefore EC is commensurable with CF. And, as EC is to CF, so is ED to DF; [VI. 1] therefore ED is commensurable in length with DF. [X. 11] But ED is rational and incommensurable in length with DC; therefore DF is also rational [X. Def. 3] and incommensurable in length with DC. [X. 13] Therefore CD, DF are rational and commensurable in square only. But the straight line the square on which is equal to the rectangle contained by rational straight lines commensurable in square only is medial; [X. 21] therefore the side of the square equal to the rectangle CD, DF is medial. And B is the side of the square equal to the rectangle CD, DF; therefore B is medial.'"],["Rule","'ProofWordCount'",240],["Rule","'GreekProof'","'ἔστω μέση ἡ Α, καὶ τῇ Α σύμμετρος ἔστω ἡ Β: λέγω, ὅτι καὶ ἡ Β μέση ἐστίν. Ἐκκείσθω γὰρ ῥητὴ ἡ ΓΔ, καὶ τῷ μὲν ἀπὸ τῆς Α ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω χωρίον ὀρθογώνιον τὸ ΓΕ πλάτος ποιοῦν τὴν ΕΔ: ῥητὴ ἄρα ἐστὶν ἡ ΕΔ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. τῷ δὲ ἀπὸ τῆς Β ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω χωρίον ὀρθογώνιον τὸ ΓΖ πλάτος ποιοῦν τὴν ΔΖ. ἐπεὶ οὖν σύμμετρός ἐστιν ἡ Α τῇ Β, σύμμετρόν ἐστι καὶ τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς Β. ἀλλὰ τῷ μὲν ἀπὸ τῆς Α ἴσον ἐστὶ τὸ ΕΓ, τῷ δὲ ἀπὸ τῆς Β ἴσον ἐστὶ τὸ ΓΖ: σύμμετρον ἄρα ἐστὶ τὸ ΕΓ τῷ ΓΖ. καί ἐστιν ὡς τὸ ΕΓ πρὸς τὸ ΓΖ, οὕτως ἡ ΕΔ πρὸς τὴν ΔΖ: σύμμετρος ἄρα ἐστὶν ἡ ΕΔ τῇ ΔΖ μήκει. ῥητὴ δέ ἐστιν ἡ ΕΔ καὶ ἀσύμμετρος τῇ ΔΓ μήκει: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΔΖ καὶ ἀσύμμετρος τῇ ΔΓ μήκει: αἱ ΓΔ, ΔΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. ἡ δὲ τὸ ὑπὸ ῥητῶν δυνάμει μόνον συμμέτρων δυναμένη μέση ἐστίν. ἡ ἄρα τὸ ὑπὸ τῶν ΓΔ, ΔΖ δυναμένη μέση ἐστίν: καὶ δύναται τὸ ὑπὸ τῶν ΓΔ, ΔΖ ἡ Β: μέση ἄρα ἐστὶν ἡ Β. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι τὸ τῷ μέσῳ χωρίῳ σύμμετρον μέσον ἐστίν. δύνανται γὰρ αὐτὰ εὐθεῖαι, αἵ εἰσι δυνάμει σύμμετροι, ὧν ἡ ἑτέρα μέση: ὥστε καὶ ἡ λοιπὴ μέση ἐστίν. ὡσαύτως δὲ τοῖς ἐπὶ τῶν ῥητῶν εἰρημένοις καὶ ἐπὶ τῶν μέσων ἐξακολουθεῖ, τὴν τῇ μέσῃ μήκει σύμμετρον λέγεσθαι μέσην καὶ σύμμετρον αὐτῇ μὴ μόνον μήκει, ἀλλὰ καὶ δυνάμει, ἐπειδήπερ καθόλου αἱ μήκει σύμμετροι πάντως καὶ δυνάμει. ἐὰν δὲ τῇ μέσῃ σύμμετρός τις ᾖ δυνάμει, εἰ μὲν καὶ μήκει, λέγονται καὶ οὕτως μέσαι καὶ σύμμετροι μήκει καὶ δυνάμει, εἰ δὲ δυνάμει μόνον, λέγονται μέσαι δυνάμει μόνον σύμμετροι.'"],["Rule","'GreekProofWordCount'",299]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'10.24'"],["Rule","'Text'","'The rectangle contained by medial straight lines commensurable in length is medial.'"],["Rule","'TextWordCount'",12],["Rule","'GreekText'","'τὸ ὑπὸ μέσων μήκει συμμέτρων εὐθειῶν κατά τινα τῶν εἰρημένων τρόπων περιεχόμενον ὀρθογώνιον μέσον ἐστίν.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in length; I say that AC is medial. For on AB let the square AD be described; therefore AD is medial. And, since AB is commensurable in length with BC, while AB is equal to BD, therefore DB is also commensurable in length with BC; so that DA is also commensurable with AC. [VI. 1, X. 11] But DA is medial; therefore AC is also medial. [X. 23]'"],["Rule","'ProofWordCount'",85],["Rule","'GreekProof'","'ὑπὸ γὰρ μέσων μήκει συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ περιεχέσθω ὀρθογώνιον τὸ ΑΓ: λέγω, ὅτι τὸ ΑΓ μέσον ἐστίν. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ: μέσον ἄρα ἐστὶ τὸ ΑΔ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, ἴση δὲ ἡ ΑΒ τῇ ΒΔ, σύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΒ τῇ ΒΓ μήκει: ὥστε καὶ τὸ ΔΑ τῷ ΑΓ σύμμετρόν ἐστιν. μέσον δὲ τὸ ΔΑ: μέσον ἄρα καὶ τὸ ΑΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",76]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'10.25'"],["Rule","'Text'","'The rectangle contained by medial straight lines commensurable in square only is either rational or medial.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'τὸ ὑπὸ μέσων δυνάμει μόνον συμμέτρων εὐθειῶν περιεχόμενον ὀρθογώνιον ἤτοι ῥητὸν á¼¢ μέσον ἐστίν.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in square only; I say that AC is either rational or medial. For on AB, BC let the squares AD, BE be described; therefore each of the squares AD, BE is medial. Let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to AD, producing FH as breadth, to HM let there be applied the rectangular parallelogram MK equal to AC, producing HK as breadth, and further to KN let there be similarly applied NL equal to BE, producing KL as breadth; therefore FH, HK, KL are in a straight line. Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL, therefore each of the rectangles GH, NL is also medial. And they are applied to the rational straight line FG; therefore each of the straight lines FH, KL is rational and incommensurable in length with FG. [X. 22] And, since AD is commensurable with BE, therefore GH is also commensurable with NL. And, as GH is to NL, so is FH to KL; [VI. 1] therefore FH is commensurable in length with KL. [X. 11] Therefore FH, KL are rational straight lines commensurable in length; therefore the rectangle FH, KL is rational. [X. 19] And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO. But, as DB is to BC, so is DA to AC, [VI. 1] and, as AB is to BO, so is AC to CO; [id.]therefore, as DA is to AC, so is AC to CO. But AD is equal to GH, AC to MK and CO to NL; therefore, as GH is to MK, so is MK to NL; therefore also, as FH is to HK, so is HK to KL; [VI. 1, V. 11] therefore the rectangle FH, KL is equal to the square on HK. [VI. 17] But the rectangle FH, KL is rational; therefore the square on HK is also rational. Therefore HK is rational. And, if it is commensurable in length with FG, HN is rational; [X. 19] but, if it is incommensurable in length with FG, KH, HM are rational straight lines commensurable in square only, and therefore HN is medial. [X. 21] Therefore HN is either rational or medial. But HN is equal to AC; therefore AC is either rational or medial.'"],["Rule","'ProofWordCount'",423],["Rule","'GreekProof'","'ὑπὸ γὰρ μέσων δυνάμει μόνον συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ ὀρθογώνιον περιεχέσθω τὸ ΑΓ: λέγω, ὅτι τὸ ΑΓ ἤτοι ῥητὸν á¼¢ μέσον ἐστίν. Ἀναγεγράφθω γὰρ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα τὰ ΑΔ, ΒΕ: μέσον ἄρα ἐστὶν ἑκάτερον τῶν ΑΔ, ΒΕ. καὶ ἐκκείσθω ῥητὴ ἡ ΖΗ, καὶ τῷ μὲν ΑΔ ἴσον παρὰ τὴν ΖΗ παραβεβλήσθω ὀρθογώνιον παραλληλόγραμμον τὸ ΗΘ πλάτος ποιοῦν τὴν ΖΘ, τῷ δὲ ΑΓ ἴσον παρὰ τὴν ΘΜ παραβεβλήσθω ὀρθογώνιον παραλληλόγραμμον τὸ ΜΚ πλάτος ποιοῦν τὴν ΘΚ, καὶ ἔτι τῷ ΒΕ ἴσον ὁμοίως παρὰ τὴν ΚΝ παραβεβλήσθω τὸ ΝΛ πλάτος ποιοῦν τὴν ΚΛ: ἐπ᾽ εὐθείας ἄρα εἰσὶν αἱ ΖΘ, ΘΚ, ΚΛ. ἐπεὶ οὖν μέσον ἐστὶν ἑκάτερον τῶν ΑΔ, ΒΕ, καί ἐστιν ἴσον τὸ μὲν ΑΔ τῷ ΗΘ, τὸ δὲ ΒΕ τῷ ΝΛ, μέσον ἄρα καὶ ἑκάτερον τῶν ΗΘ, ΝΛ. καὶ παρὰ ῥητὴν τὴν ΖΗ παράκειται: ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΖΘ, ΚΛ καὶ ἀσύμμετρος τῇ ΖΗ μήκει. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ΑΔ τῷ ΒΕ, σύμμετρον ἄρα ἐστὶ καὶ τὸ ΗΘ τῷ ΝΛ. καί ἐστιν ὡς τὸ ΗΘ πρὸς τὸ ΝΛ, οὕτως ἡ ΖΘ πρὸς τὴν ΚΛ: σύμμετρος ἄρα ἐστὶν ἡ ΖΘ τῇ ΚΛ μήκει. αἱ ΖΘ, ΚΛ ἄρα ῥηταί εἰσι μήκει σύμμετροι: ῥητὸν ἄρα ἐστὶ τὸ ὑπὸ τῶν ΖΘ, ΚΛ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΔΒ τῇ ΒΑ, ἡ δὲ ΞΒ τῇ ΒΓ, ἔστιν ἄρα ὡς ἡ ΔΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΞ. ἀλλ᾽ ὡς μὲν ἡ ΔΒ πρὸς τὴν ΒΓ, οὕτως τὸ ΔΑ πρὸς τὸ ΑΓ: ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΞ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΞ: ἔστιν ἄρα ὡς τὸ ΔΑ πρὸς τὸ ΑΓ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΞ. ἴσον δέ ἐστι τὸ μὲν ΑΔ τῷ ΗΘ, τὸ δὲ ΑΓ τῷ ΜΚ, τὸ δὲ ΓΞ τῷ ΝΛ: ἔστιν ἄρα ὡς τὸ ΗΘ πρὸς τὸ ΜΚ, οὕτως τὸ ΜΚ πρὸς τὸ ΝΛ: ἔστιν ἄρα καὶ ὡς ἡ ΖΘ πρὸς τὴν ΘΚ, οὕτως ἡ ΘΚ πρὸς τὴν ΚΛ: τὸ ἄρα ὑπὸ τῶν ΖΘ, ΚΛ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΘΚ. ῥητὸν δὲ τὸ ὑπὸ τῶν ΖΘ, ΚΛ: ῥητὸν ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΘΚ: ῥητὴ ἄρα ἐστὶν ἡ ΘΚ. καὶ εἰ μὲν σύμμετρός ἐστι τῇ ΖΗ μήκει, ῥητόν ἐστι τὸ ΘΝ: εἰ δὲ ἀσύμμετρός ἐστι τῇ ΖΗ μήκει, αἱ ΚΘ, ΘΜ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: μέσον ἄρα τὸ ΘΝ. τὸ ΘΝ ἄρα ἤτοι ῥητὸν á¼¢ μέσον ἐστίν. ἴσον δὲ τὸ ΘΝ τῷ ΑΓ: τὸ ΑΓ ἄρα ἤτοι ῥητὸν á¼¢ μέσον ἐστίν. τὸ ἄρα ὑπὸ μέσων δυνάμει μόνον συμμέτρων, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",415]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",26]],["Association",["Rule","'VertexLabel'","'10.26'"],["Rule","'Text'","'A medial area does not exceed a medial area by a rational area.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'μέσον μέσου οὐχ ὑπερέχει ῥητῷ.'"],["Rule","'GreekTextWordCount'",5],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'For, if possible, let the medial area AB exceed the medial area AC by the rational area DB, and let a rational straight line EF be set out; to EF let there be applied the rectangular parallelogram FH equal to AB, producing EH as breadth, and let the rectangle FG equal to AC be subtracted; therefore the remainder BD is equal to the remainder KH. But DB is rational; therefore KH is also rational. Since, then, each of the rectangles AB, AC is medial, and AB is equal to FH, and AC to FG, therefore each of the rectangles FH, FG is also medial. And they are applied to the rational straight line EF; therefore each of the straight lines HE, EG is rational and incommensurable in length with EF. [X. 22] And, since [DB is rational and is equal to KH, therefore]KH is [also]rational; and it is applied to the rational straight line EF; therefore GH is rational and commensurable in length with EF. [X. 20] But EG is also rational, and is incommensurable in length with EF; therefore EG is incommensurable in length with GH. [X. 13] And, as EG is to GH, so is the square on EG to the rectangle EG, GH; therefore the square on EG is incommensurable with the rectangle EG, GH. [X. 11] But the squares on EG, GH are commensurable with the square on EG, for both are rational; and twice the rectangle EG, GH is commensurable with the rectangle EG, GH, for it is double of it; [X. 6] therefore the squares on EG, GH are incommensurable with twice the rectangle EG, GH; [X. 13] therefore also the sum of the squares on EG, GH and twice the rectangle EG, GH, that is, the square on EH [II. 4], is incommensurable with the squares on EG, GH. [X. 16] But the squares on EG, GH are rational; therefore the square on EH is irrational. [X. Def. 4] Therefore EH is irrational. But it is also rational: which is impossible.'"],["Rule","'ProofWordCount'",340],["Rule","'GreekProof'","'εἰ γὰρ δυνατόν, μέσον τὸ ΑΒ μέσου τοῦ ΑΓ ὑπερεχέτω ῥητῷ τῷ ΔΒ, καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τῷ ΑΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω παραλληλόγραμμον ὀρθογώνιον τὸ ΖΘ πλάτος ποιοῦν τὴν ΕΘ, τῷ δὲ ΑΓ ἴσον ἀφῃρήσθω τὸ ΖΗ: λοιπὸν ἄρα τὸ ΒΔ λοιπῷ τῷ ΚΘ ἐστιν ἴσον. ῥητὸν δέ ἐστι τὸ ΔΒ: ῥητὸν ἄρα ἐστὶ καὶ τὸ ΚΘ. ἐπεὶ οὖν μέσον ἐστὶν ἑκάτερον τῶν ΑΒ, ΑΓ, καί ἐστι τὸ μὲν ΑΒ τῷ ΖΘ ἴσον, τὸ δὲ ΑΓ τῷ ΖΗ, μέσον ἄρα καὶ ἑκάτερον τῶν ΖΘ, ΖΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται: ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΘΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ ῥητόν ἐστι τὸ ΔΒ καί ἐστιν ἴσον τῷ ΚΘ, ῥητὸν ἄρα ἐστὶ καὶ τὸ ΚΘ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται: ῥητὴ ἄρα ἐστὶν ἡ ΗΘ καὶ σύμμετρος τῇ ΕΖ μήκει. ἀλλὰ καὶ ἡ ΕΗ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΕΖ μήκει: ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΗ τῇ ΗΘ μήκει. καί ἐστιν ὡς ἡ ΕΗ πρὸς τὴν ΗΘ, οὕτως τὸ ἀπὸ τῆς ΕΗ πρὸς τὸ ὑπὸ τῶν ΕΗ, ΗΘ: ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΗ τῷ ὑπὸ τῶν ΕΗ, ΗΘ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΕΗ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΕΗ, ΗΘ τετράγωνα: ῥητὰ γὰρ ἀμφότερα: τῷ δὲ ὑπὸ τῶν ΕΗ, ΗΘ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΕΗ, ΗΘ: διπλάσιον γάρ ἐστιν αὐτοῦ: ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΕΗ, ΗΘ τῷ δὶς ὑπὸ τῶν ΕΗ, ΗΘ: καὶ συναμφότερα ἄρα τά τε ἀπὸ τῶν ΕΗ, ΗΘ καὶ τὸ δὶς ὑπὸ τῶν ΕΗ, ΗΘ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΕΘ, ἀσύμμετρόν ἐστι τοῖς ἀπὸ τῶν ΕΗ, ΗΘ. ῥητὰ δὲ τὰ ἀπὸ τῶν ΕΗ, ΗΘ: ἄλογον ἄρα τὸ ἀπὸ τῆς ΕΘ. ἄλογος ἄρα ἐστὶν ἡ ΕΘ. ἀλλὰ καὶ ῥητή: ὅπερ ἐστὶν ἀδύνατον. μέσον ἄρα μέσου οὐχ ὑπερέχει ῥητῷ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",301]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",27]],["Association",["Rule","'VertexLabel'","'10.27'"],["Rule","'Text'","'To find medial straight lines commensurable in square only which contain a rational rectangle.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'μέσας εὑρεῖν δυνάμει μόνον συμμέτρους ῥητὸν περιεχούσας.'"],["Rule","'GreekTextWordCount'",7],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",12]],["List",["Rule","'Book'",6],["Rule","'Theorem'",13]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'Let two rational straight lines A, B commensurable in square only be set out; let C be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as A is to B, so is C to D. [VI. 12]  Then, since A, B are rational and commensurable in square only, the rectangle A, B, that is, the square on C [VI.17], is medial. [X. 21] Therefore C is medial. [X. 21] And since, as A is to B, so is C to D, and A, B are commensurable in square only, therefore C, D are also commensurable in square only. [X. 11] And C is medial; therefore D is also medial. [X. 23, addition] Therefore C, D are medial and commensurable in square only. I say that they also contain a rational rectangle. For since, as A is to B, so is C to D, therefore, alternately, as A is to C, so is B to D. [V. 16] But, as A is to C, so is C to B; therefore also, as C is to B, so is B to D; therefore the rectangle C, D is equal to the square on B. But the square on B is rational; therefore the rectangle C, D is also rational. Therefore medial straight lines commensurable in square only have been found which contain a rational rectangle.'"],["Rule","'ProofWordCount'",231],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Α, Β, καὶ εἰλήφθω τῶν Α, Β μέση ἀνάλογον ἡ Γ, καὶ γεγονέτω ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ. καὶ ἐπεὶ αἱ Α, Β ῥηταί εἰσι δυνάμει μόνον σύμμετροι, τὸ ἄρα ὑπὸ τῶν Α, Β, τουτέστι τὸ ἀπὸ τῆς Γ, μέσον ἐστίν. μέση ἄρα ἡ Γ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, αἱ δὲ Α, Β δυνάμει μόνον εἰσὶ σύμμετροι, καὶ αἱ Γ, Δ ἄρα δυνάμει μόνον εἰσὶ σύμμετροι. καί ἐστι μέση ἡ Γ: μέση ἄρα καὶ ἡ Δ. αἱ Γ, Δ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι. λέγω, ὅτι καὶ ῥητὸν περιέχουσιν. ἐπεὶ γάρ ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ Α πρὸς τὴν Γ, ἡ Β πρὸς τὴν Δ. ἀλλ᾽ ὡς ἡ Α πρὸς τὴν Γ, ἡ Γ πρὸς τὴν Β: καὶ ὡς ἄρα ἡ Γ πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Δ: τὸ ἄρα ὑπὸ τῶν Γ, Δ ἴσον ἐστὶ τῷ ἀπὸ τῆς Β. ῥητὸν δὲ τὸ ἀπὸ τῆς Β: ῥητὸν ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν Γ, Δ. Εὕρηνται ἄρα μέσαι δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",209]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",28]],["Association",["Rule","'VertexLabel'","'10.28'"],["Rule","'Text'","'To find medial straight lines commensurable in square only which contain a medial rectangle.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'μέσας εὑρεῖν δυνάμει μόνον συμμέτρους μέσον περιεχούσας.'"],["Rule","'GreekTextWordCount'",7],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",12]],["List",["Rule","'Book'",6],["Rule","'Theorem'",13]],["List",["Rule","'Book'",6],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]]]],["Rule","'Proof'","'Let the rational straight lines A, B, C commensurable in square only be set out; let D be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as B is to C, so is D to E. [VI. 12]  Since A, B are rational straight lines commensurable in square only, therefore the rectangle A, B, that is, the square on D [VI. 17], is medial. [X. 21] Therefore D is medial. [X. 21] And since B, C are commensurable in square only, and, as B is to C, so is D to E, therefore D, E are also commensurable in square only. [X. 11] But D is medial; therefore E is also medial. [X. 23, addition] Therefore D, E are medial straight lines commensurable in square only. I say next that they also contain a medial rectangle. For since, as B is to C, so is D to E, therefore, alternately, as B is to D, so is C to E. [V. 16] But, as B is to D, so is D to A; therefore also, as D is to A, so is C to E; therefore the rectangle A, C is equal to the rectangle D, E. [VI. 16] But the rectangle A, C is medial; [X. 21] therefore the rectangle D, E is also medial. Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle.'"],["Rule","'ProofWordCount'",239],["Rule","'GreekProof'","'Ἐκκείσθωσαν τρεῖς ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Α, Β, Γ, καὶ εἰλήφθω τῶν Α, Β μέση ἀνάλογον ἡ Δ, καὶ γεγονέτω ὡς ἡ Β πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε. ἐπεὶ αἱ Α, Β ῥηταί εἰσι δυνάμει μόνον σύμμετροι, τὸ ἄρα ὑπὸ τῶν Α, Β, τουτέστι τὸ ἀπὸ τῆς Δ, μέσον ἐστίν. μέση ἄρα ἡ Δ. καὶ ἐπεὶ αἱ Β, Γ δυνάμει μόνον εἰσὶ σύμμετροι, καί ἐστιν ὡς ἡ Β πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε, καὶ αἱ Δ, Ε ἄρα δυνάμει μόνον εἰσὶ σύμμετροι. μέση δὲ ἡ Δ: μέση ἄρα καὶ ἡ Ε: αἱ Δ, Ε ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι. λέγω δή, ὅτι καὶ μέσον περιέχουσιν. ἐπεὶ γάρ ἐστιν ὡς ἡ Β πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε, ἐναλλὰξ ἄρα ὡς ἡ Β πρὸς τὴν Δ, ἡ Γ πρὸς τὴν Ε. ὡς δὲ ἡ Β πρὸς τὴν Δ, ἡ Δ πρὸς τὴν Α: καὶ ὡς ἄρα ἡ Δ πρὸς τὴν Α, ἡ Γ πρὸς τὴν Ε: τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Δ, Ε. μέσον δὲ τὸ ὑπὸ τῶν Α, Γ: μέσον ἄρα καὶ τὸ ὑπὸ τῶν Δ, Ε. Εὕρηνται ἄρα μέσαι δυνάμει μόνον σύμμετροι μέσον περιέχουσαι: ὅπερ ἔδει δεῖξαι. λῆμμα εὑρεῖν δύο τετραγώνους ἀριθμούς, ὥστε καὶ τὸν συγκείμενον ἐξ αὐτῶν εἶναι τετράγωνον. Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΒ, ΒΓ, ἔστωσαν δὲ ἤτοι ἄρτιοι á¼¢ περιττοί. καὶ ἐπεί, ἐάν τε ἀπὸ ἀρτίου ἄρτιος ἀφαιρεθῇ, ἐάν τε ἀπὸ περισσοῦ περισσός, ὁ λοιπὸς ἄρτιός ἐστιν, ὁ λοιπὸς ἄρα ὁ ΑΓ ἄρτιός ἐστιν. τετμήσθω ὁ ΑΓ δίχα κατὰ τὸ Δ. ἔστωσαν δὲ καὶ οἱ ΑΒ, ΒΓ ἤτοι ὅμοιοι ἐπίπεδοι á¼¢ τετράγωνοι, οἳ καὶ αὐτοὶ ὅμοιοί εἰσιν ἐπίπεδοι: ὁ ἄρα ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ τοῦ ΓΔ τετραγώνου ἴσος ἐστὶ τῷ ἀπὸ τοῦ ΒΔ τετραγώνῳ. καί ἐστι τετράγωνος ὁ ἐκ τῶν ΑΒ, ΒΓ, ἐπειδήπερ ἐδείχθη, ὅτι, ἐὰν δύο ὅμοιοι ἐπίπεδοι πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος τετράγωνός ἐστιν. εὕρηνται ἄρα δύο τετράγωνοι ἀριθμοὶ ὅ τε ἐκ τῶν ΑΒ, ΒΓ καὶ ὁ ἀπὸ τοῦ ΓΔ, οἳ συντεθέντες ποιοῦσι τὸν ἀπὸ τοῦ ΒΔ τετράγωνον. καὶ φανερόν, ὅτι εὕρηνται πάλιν δύο τετράγωνοι ὅ τε ἀπὸ τοῦ ΒΔ καὶ ὁ ἀπὸ τοῦ ΓΔ, ὥστε τὴν ὑπεροχὴν αὐτῶν τὸν ὑπὸ ΑΒ, ΒΓ εἶναι τετράγωνον, ὅταν οἱ ΑΒ, ΒΓ ὅμοιοι ὦσιν ἐπίπεδοι. ὅταν δὲ μὴ ὦσιν ὅμοιοι ἐπίπεδοι, εὕρηνται δύο τετράγωνοι ὅ τε ἀπὸ τοῦ ΒΔ καὶ ὁ ἀπὸ τοῦ ΔΓ, ὧν ἡ ὑπεροχὴ ὁ ὑπὸ τῶν ΑΒ, ΒΓ οὐκ ἔστι τετράγωνος: ὅπερ ἔδει δεῖξαι. λῆμμα εὑρεῖν δύο τετραγώνους ἀριθμούς, ὥστε τὸν ἐξ αὐτῶν συγκείμενον μὴ εἶναι τετράγωνον. ἔστω γὰρ ὁ ἐκ τῶν ΑΒ, ΒΓ, ὡς ἔφαμεν, τετράγωνος, καὶ ἄρτιος ὁ ΓΑ, καὶ τετμήσθω ὁ ΓΑ δίχα τῷ Δ. φανερὸν δή, ὅτι ὁ ἐκ τῶν ΑΒ, ΒΓ τετράγωνος μετὰ τοῦ ἀπὸ τοῦ ΓΔ τετραγώνου ἴσος ἐστὶ τῷ ἀπὸ τοῦ ΒΔ τετραγώνῳ. ἀφῃρήσθω μονὰς ἡ ΔΕ: ὁ ἄρα ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ τοῦ ΓΕ ἐλάσσων ἐστὶ τοῦ ἀπὸ τοῦ ΒΔ τετραγώνου. λέγω οὖν, ὅτι ὁ ἐκ τῶν ΑΒ, ΒΓ τετράγωνος μετὰ τοῦ ἀπὸ τοῦ ΓΕ οὐκ ἔσται τετράγωνος. εἰ γὰρ ἔσται τετράγωνος, ἤτοι ἴσος ἐστὶ τῷ ἀπὸ τοῦ ΒΕ á¼¢ ἐλάσσων τοῦ ἀπὸ τοῦ ΒΕ, οὐκέτι δὲ καὶ μείζων, ἵνα μὴ τμηθῇ ἡ μονάς. ἔστω, εἰ δυνατόν, πρότερον ὁ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ ἴσος τῷ ἀπὸ ΒΕ, καὶ ἔστω τῆς ΔΕ μονάδος διπλασίων ὁ ΗΑ. ἐπεὶ οὖν ὅλος ὁ ΑΓ ὅλου τοῦ ΓΔ ἐστι διπλασίων, ὧν ὁ ΑΗ τοῦ ΔΕ ἐστι διπλασίων, καὶ λοιπὸς ἄρα ὁ ΗΓ λοιποῦ τοῦ ΕΓ ἐστι διπλασίων: δίχα ἄρα τέτμηται ὁ ΗΓ τῷ Ε. ὁ ἄρα ἐκ τῶν ΗΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ ἴσος ἐστὶ τῷ ἀπὸ ΒΕ τετραγώνῳ. ἀλλὰ καὶ ὁ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ ἴσος ὑπόκειται τῷ ἀπὸ τοῦ ΒΕ τετραγώνῳ: ὁ ἄρα ἐκ τῶν ΗΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ ἴσος ἐστὶ τῷ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ. καὶ κοινοῦ ἀφαιρεθέντος τοῦ ἀπὸ ΓΕ συνάγεται ὁ ΑΒ ἴσος τῷ ΗΒ: ὅπερ ἄτοπον. οὐκ ἄρα ὁ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ τοῦ ΓΕ ἴσος ἐστὶ τῷ ἀπὸ ΒΕ. λέγω δή, ὅτι οὐδὲ ἐλάσσων τοῦ ἀπὸ ΒΕ. εἰ γὰρ δυνατόν, ἔστω τῷ ἀπὸ ΒΖ ἴσος, καὶ τοῦ ΔΖ διπλασίων ὁ ΘΑ. καὶ συναχθήσεται πάλιν διπλασίων ὁ ΘΓ τοῦ ΓΖ: ὥστε καὶ τὸν ΓΘ δίχα τετμῆσθαι κατὰ τὸ Ζ, καὶ διὰ τοῦτο τὸν ἐκ τῶν ΘΒ, ΒΓ μετὰ τοῦ ἀπὸ ΖΓ ἴσον γίνεσθαι τῷ ἀπὸ ΒΖ. ὑπόκειται δὲ καὶ ὁ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ ἴσος τῷ ἀπὸ ΒΖ. ὥστε καὶ ὁ ἐκ τῶν ΘΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΖ ἴσος ἔσται τῷ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ: ὅπερ ἄτοπον. οὐκ ἄρα ὁ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ ἴσος ἐστὶ τῷ ἐλάσσονι τοῦ ἀπὸ ΒΕ. ἐδείχθη δέ, ὅτι οὐδὲ αὐτῷ τῷ ἀπὸ ΒΕ. οὐκ ἄρα ὁ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ τετράγωνός ἐστιν. δυνατοῦ δὲ ὄντος καὶ κατὰ πλείονας τρόπους τοὺς εἰρημένους ἀριθμοὺς ἐπιδεικνύειν, ἀρκείσθωσαν ἡμῖν οἱ εἰρημένοι, ἵνα μὴ μακροτέρας οὔσης τῆς πραγματείας ἐπὶ πλέον αὐτὴν μηκύνωμεν. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",839]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",29]],["Association",["Rule","'VertexLabel'","'10.29'"],["Rule","'Text'","'To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.'"],["Rule","'TextWordCount'",39],["Rule","'GreekText'","'εὑρεῖν δύο ῥητὰς δυνάμει μόνον συμμέτρους, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]]]],["Rule","'Proof'","'For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1] let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6]  Let FB be joined. Since, as the square on BA is to the square on AF, so is DC to CE, therefore the square on BA has to the square on AF the ratio which the number DC has to the number CE; therefore the square on BA is commensurable with the square on AF. [X. 6] But the square on AB is rational; [X. Def. 4] therefore the square on AF is also rational; [id.] therefore AF is also rational. And, since DC has not to CE the ratio which a square number has to a square number, neither has the square on BA to the square on AF the ratio which a square number has to a square number; therefore AB is incommensurable in length with AF. [X. 9] Therefore BA, AF are rational straight lines commensurable in square only. And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, III. 31, I. 47] But CD has to DE the ratio which a square number has to a square number: therefore also the square on AB has to the square on BF the ratio which a square number has to a square number; therefore AB is commensurable in length with BF. [X. 9] And the square on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB. Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB.'"],["Rule","'ProofWordCount'",369],["Rule","'GreekProof'","'Ἐκκείσθω γάρ τις ῥητὴ ἡ ΑΒ καὶ δύο τετράγωνοι ἀριθμοὶ οἱ ΓΔ, ΔΕ, ὥστε τὴν ὑπεροχὴν αὐτῶν τὸν ΓΕ μὴ εἶναι τετράγωνον, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΖΒ, καὶ πεποιήσθω ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ τῆς ΒΑ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΑΖ τετράγωνον καὶ ἐπεζεύχθω ἡ ΖΒ. ἐπεὶ οὖν ἐστιν ὡς τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, οὕτως ὁ ΔΓ πρὸς τὸν ΓΕ, τὸ ἀπὸ τῆς ΒΑ ἄρα πρὸς τὸ ἀπὸ τῆς ΑΖ λόγον ἔχει, ὃν ἀριθμὸς ὁ ΔΓ πρὸς ἀριθμὸν τὸν ΓΕ: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΑ τῷ ἀπὸ τῆς ΑΖ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΑΒ: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΑΖ: ῥητὴ ἄρα καὶ ἡ ΑΖ. καὶ ἐπεὶ ὁ ΔΓ πρὸς τὸν ΓΕ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΒΑ ἄρα πρὸς τὸ ἀπὸ τῆς ΑΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ ΑΖ μήκει: αἱ ΒΑ, ΑΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ ἐπεί ἐστιν ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, ἀναστρέψαντι ἄρα ὡς ὁ ΓΔ πρὸς τὸν ΔΕ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΖ. ὁ δὲ ΓΔ πρὸς τὸν ΔΕ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ τὸ ἀπὸ τῆς ΑΒ ἄρα πρὸς τὸ ἀπὸ τῆς ΒΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: σύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ ΒΖ μήκει. καί ἐστι τὸ ἀπὸ τῆς ΑΒ ἴσον τοῖς ἀπὸ τῶν ΑΖ, ΖΒ: ἡ ΑΒ ἄρα τῆς ΑΖ μεῖζον δύναται τῇ ΒΖ συμμέτρῳ ἑαυτῇ. Εὕρηνται ἄρα δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ ΒΑ, ΑΖ, ὥστε τὴν μείζονα τὴν ΑΒ τῆς ἐλάσσονος τῆς ΑΖ μεῖζον δύνασθαι τῷ ἀπὸ τῆς ΒΖ συμμέτρου ἑαυτῇ μήκει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",306]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",30]],["Association",["Rule","'VertexLabel'","'10.30'"],["Rule","'Text'","'To find two rational straight lines commensurable in square only and such that the square on the greater is greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater.'"],["Rule","'TextWordCount'",41],["Rule","'GreekText'","'εὑρεῖν δύο ῥητὰς δυνάμει μόνον συμμέτρους, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]]]],["Rule","'Proof'","'Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not square; [Lemma 2] let there be described on AB the semicircle AFB, let it be contrived that, as DC is to CE, so is the square on BA to the square on AF, [X. 6] and let FB be joined. Then, in a similar manner to the preceding, we can prove that BA, AF are rational straight lines commensurable in square only. And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, III. 31, I. 47] But CD has not to DE the ratio which a square number has to a square number; therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number; therefore AB is incommensurable in length with BF. [X. 9] And the square on AB is greater than the square on AF by the square on FB incommensurable with AB. Therefore AB, AF are rational straight lines commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB.'"],["Rule","'ProofWordCount'",230],["Rule","'GreekProof'","'Ἐκκείσθω ῥητὴ ἡ ΑΒ καὶ δύο τετράγωνοι ἀριθμοὶ οἱ ΓΕ, ΕΔ, ὥστε τὸν συγκείμενον ἐξ αὐτῶν τὸν ΓΔ μὴ εἶναι τετράγωνον, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΖΒ, καὶ πεποιήσθω ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, καὶ ἐπεζεύχθω ἡ ΖΒ. ὁμοίως δὴ δείξομεν τῷ πρὸ τούτου, ὅτι αἱ ΒΑ, ΑΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ ἐπεί ἐστιν ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, ἀναστρέψαντι ἄρα ὡς ὁ ΓΔ πρὸς τὸν ΔΕ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΖ. ὁ δὲ ΓΔ πρὸς τὸν ΔΕ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ ΒΖ μήκει. καὶ δύναται ἡ ΑΒ τῆς ΑΖ μεῖζον τῷ ἀπὸ τῆς ΖΒ ἀσυμμέτρου ἑαυτῇ. αἱ ΑΒ, ΑΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΒ τῆς ΑΖ μεῖζον δύναται τῷ ἀπὸ τῆς ΖΒ ἀσυμμέτρου ἑαυτῇ μήκει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",185]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",31]],["Association",["Rule","'VertexLabel'","'10.31'"],["Rule","'Text'","'To find two medial straight lines commensurable in square only, containing a rational rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.'"],["Rule","'TextWordCount'",43],["Rule","'GreekText'","'εὑρεῖν δύο μέσας δυνάμει μόνον συμμέτρους ῥητὸν περιεχούσας, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",14]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",29]],["List",["Rule","'Book'",10],["Rule","'Theorem'",30]]]],["Rule","'Proof'","'Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [X. 29] And let the square on C be equal to the rectangle A, B. Now the rectangle A, B is medial; [X. 21] therefore the square on C is also medial; therefore C is also medial. [X. 21] Let the rectangle C, D be equal to the square on B. Now the square on B is rational; therefore the rectangle C, D is also rational. And since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle C, D. But, as the square on C is to the rectangle C, D, so is C to D; therefore also, as A is to B, so is C to D. But A is commensurable with B in square only; therefore C is also commensurable with D in square only. [X. 11] And C is medial; therefore D is also medial. [X. 23, addition] And since, as A is to B, so is C to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A, therefore also the square on C is greater than the square on D by the square on a straight line commensurable with C. [X. 14] Therefore two medial straight lines C, D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C. Similarly also it can be proved that the square on C exceeds the square on D by the square on a straight line incommensurable with C, when the square on A is greater than the square on B by the square on a straight line incommensurable with A. [X. 30]'"],["Rule","'ProofWordCount'",385],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ α, Β, ὥστε τὴν Α μείζονα οὖσαν τῆς ἐλάσσονος τῆς Β μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καὶ τῷ ὑπὸ τῶν Α, Β ἴσον ἔστω τὸ ἀπὸ τῆς Γ. μέσον δὲ τὸ ὑπὸ τῶν Α, Β: μέσον ἄρα καὶ τὸ ἀπὸ τῆς Γ: μέση ἄρα καὶ ἡ Γ. τῷ δὲ ἀπὸ τῆς Β ἴσον ἔστω τὸ ὑπὸ τῶν Γ, Δ. ῥητὸν δὲ τὸ ἀπὸ τῆς Β: ῥητὸν ἄρα καὶ τὸ ὑπὸ τῶν Γ, Δ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως τὸ ὑπὸ τῶν Α, Β πρὸς τὸ ἀπὸ τῆς Β, ἀλλὰ τῷ μὲν ὑπὸ τῶν Α, Β ἴσον ἐστὶ τὸ ἀπὸ τῆς Γ, τῷ δὲ ἀπὸ τῆς Β ἴσον τὸ ὑπὸ τῶν Γ, δ, ὡς ἄρα ἡ Α πρὸς τὴν Β, οὕτως τὸ ἀπὸ τῆς Γ πρὸς τὸ ὑπὸ τῶν Γ, Δ. ὡς δὲ τὸ ἀπὸ τῆς Γ πρὸς τὸ ὑπὸ τῶν Γ, Δ, οὕτως ἡ Γ πρὸς τὴν Δ: καὶ ὡς ἄρα ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ. σύμμετρος δὲ ἡ Α τῇ Β δυνάμει μόνον: σύμμετρος ἄρα καὶ ἡ Γ τῇ Δ δυνάμει μόνον. καί ἐστι μέση ἡ Γ: μέση ἄρα καὶ ἡ Δ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, ἡ Γ πρὸς τὴν Δ, ἡ δὲ Α τῆς Β μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ Γ ἄρα τῆς Δ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. Εὕρηνται ἄρα δύο μέσαι δυνάμει μόνον σύμμετροι αἱ Γ, Δ ῥητὸν περιέχουσαι, καὶ ἡ Γ τῆς Δ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. ὁμοίως δὴ δειχθήσεται καὶ τῷ ἀπὸ ἀσυμμέτρου, ὅταν ἡ Α τῆς Β μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.'"],["Rule","'GreekProofWordCount'",282]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",32]],["Association",["Rule","'VertexLabel'","'10.32'"],["Rule","'Text'","'To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.'"],["Rule","'TextWordCount'",41],["Rule","'GreekText'","'εὑρεῖν δύο μέσας δυνάμει μόνον συμμέτρους μέσον περιεχούσας, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",8]],["List",["Rule","'Book'",6],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",14]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",29]],["List",["Rule","'Book'",10],["Rule","'Theorem'",30]]]],["Rule","'Proof'","'Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square on A is greater than the square on C by the square on a straight line commensurable with A, [X. 29] and let the square on D be equal to the rectangle A, B. Therefore the square on D is medial; therefore D is also medial. [X. 21] Let the rectangle D, E be equal to the rectangle B, C. Then since, as the rectangle A, B is to the rectangle B, C, so is A to C; while the square on D is equal to the rectangle A, B, and the rectangle D, E is equal to the rectangle B, C, therefore, as A is to C, so is the square on D to the rectangle D, E. But, as the square on D is to the rectangle D, E, so is D to E; therefore also, as A is to C, so is D to E. But A is commensurable with C in square only; therefore D is also commensurable with E in square only. [X. 11] But D is medial; therefore E is also medial. [X. 23, addition] And, since, as A is to C, so is D to E, while the square on A is greater than the square on C by the square on a straight line commensurable with A, therefore also the square on D will be greater than the square on E by the square on a straight line commensurable with D.[X. 14] I say next that the rectangle D, E is also medial. For, since the rectangle B, C is equal to the rectangle D, E, while the rectangle B, C is medial, [X. 21] therefore the rectangle D, E is also medial. Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [X. 30]   LEMMA. Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn; I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC. And first that the rectangle CB, BD is equal to the square on BA. For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD, ADC are similar both to the whole ABC and to one another. [VI. 8] And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4] therefore the rectangle CB, BD is equal to the square on AB. [VI. 17] For the same reason the rectangle BC, CD is also equal to the square on AC. And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8] therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DC is equal to the square on AD. [VI. 17] I say that the rectangle BC, AD is also equal to the rectangle BA, AC. For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD. [VI. 4] Therefore the rectangle BC, AD is equal to the rectangle BA, AC. [VI. 16]'"],["Rule","'ProofWordCount'",680],["Rule","'GreekProof'","'Ἐκκείσθωσαν τρεῖς ῥηταὶ δυνάμει μόνον σύμμετροι αἱ α, Β, Γ, ὥστε τὴν Α τῆς Γ μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ τῷ μὲν ὑπὸ τῶν Α, Β ἴσον ἔστω τὸ ἀπὸ τῆς Δ. μέσον ἄρα τὸ ἀπὸ τῆς Δ: καὶ ἡ Δ ἄρα μέση ἐστίν. τῷ δὲ ὑπὸ τῶν Β, Γ ἴσον ἔστω τὸ ὑπὸ τῶν Δ, Ε. καὶ ἐπεί ἐστιν ὡς τὸ ὑπὸ τῶν Α, Β πρὸς τὸ ὑπὸ τῶν Β, Γ, οὕτως ἡ Α πρὸς τὴν Γ, ἀλλὰ τῷ μὲν ὑπὸ τῶν Α, Β ἴσον ἐστὶ τὸ ἀπὸ τῆς Δ, τῷ δὲ ὑπὸ τῶν Β, Γ ἴσον τὸ ὑπὸ τῶν Δ, Ε, ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Γ, οὕτως τὸ ἀπὸ τῆς Δ πρὸς τὸ ὑπὸ τῶν Δ, Ε. ὡς δὲ τὸ ἀπὸ τῆς Δ πρὸς τὸ ὑπὸ τῶν Δ, Ε, οὕτως ἡ Δ πρὸς τὴν Ε: καὶ ὡς ἄρα ἡ Α πρὸς τὴν Γ, οὕτως ἡ Δ πρὸς τὴν Ε: σύμμετρος δὲ ἡ Α τῇ Γ δυνάμει μόνον. σύμμετρος ἄρα καὶ ἡ Δ τῇ Ε δυνάμει μόνον. μέση δὲ ἡ Δ: μέση ἄρα καὶ ἡ Ε. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε, ἡ δὲ Α τῆς Γ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ Δ ἄρα τῆς Ε μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ. λέγω δή, ὅτι καὶ μέσον ἐστὶ τὸ ὑπὸ τῶν Δ, Ε. ἐπεὶ γὰρ ἴσον ἐστὶ τὸ ὑπὸ τῶν Β, Γ τῷ ὑπὸ τῶν Δ, Ε, μέσον δὲ τὸ ὑπὸ τῶν Β, Γ αἱ γὰρ Β, Γ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, μέσον ἄρα καὶ τὸ ὑπὸ τῶν Δ, Ε. Εὕρηνται ἄρα δύο μέσαι δυνάμει μόνον σύμμετροι αἱ Δ, Ε μέσον περιέχουσαι, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ. ὁμοίως δὴ πάλιν δειχθήσεται καὶ τῷ ἀπὸ ἀσυμμέτρου, ὅταν ἡ Α τῆς Γ μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. λῆμμα ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν α, καὶ ἤχθω κάθετος ἡ ΑΔ: λέγω, ὅτι τὸ μὲν ὑπὸ τῶν ΓΒΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΑ, τὸ δὲ ὑπὸ τῶν ΒΓΔ ἴσον τῷ ἀπὸ τῆς ΓΑ, καὶ τὸ ὑπὸ τῶν ΒΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΑΔ, καὶ ἔτι τὸ ὑπὸ τῶν ΒΓ, ΑΔ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΒΑ, ΑΓ. καὶ πρῶτον, ὅτι τὸ ὑπὸ τῶν ΓΒΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΑ. ἐπεὶ γὰρ ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἦκται ἡ ΑΔ, τὰ ΑΒΔ, ΑΔΓ ἄρα τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ τῷ ΑΒΓ καὶ ἀλλήλοις. καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ, ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΒΔ: τὸ ἄρα ὑπὸ τῶν ΓΒΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ὑπὸ τῶν ΒΓΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ. καὶ ἐπεί, ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, ἡ ἀχθεῖσα τῶν τῆς βάσεως τμημάτων μέση ἀνάλογόν ἐστιν, ἔστιν ἄρα ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΑΔ πρὸς τὴν ΔΓ: τὸ ἄρα ὑπὸ τῶν ΒΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΑ. λέγω, ὅτι καὶ τὸ ὑπὸ τῶν ΒΓ, ΑΔ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΒΑ, ΑΓ. ἐπεὶ γάρ, ὡς ἔφαμεν, ὅμοιόν ἐστι τὸ ΑΒΓ τῷ ΑΒΔ, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΔ. ἐὰν δὲ τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων. τὸ ἄρα ὑπὸ τῶν ΒΓ, ΑΔ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΒΑ, ΑΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",587]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",33]],["Association",["Rule","'VertexLabel'","'10.33'"],["Rule","'Text'","'To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δ᾽ ὑπ᾽ αὐτῶν μέσον.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",6],["Rule","'Theorem'",28]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",30]]]],["Rule","'Proof'","'Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB, [X. 30] let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28] let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined. Then, since AB, BC are unequal straight lines, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB, while there has been applied to AB a parallelogram equal to the fourth part of the square on BC, that is, to the square on half of it, and deficient by a square figure, making the rectangle AE, EB, therefore AE is incommensurable with EB. [X. 18] And, as AE is to EB, so is the rectangle BA, AE to the rectangle AB, BE, while the rectangle BA, AE is equal to the square on AF, and the rectangle AB, BE to the square on BF; therefore the square on AF is incommensurable with the square on FB; therefore AF, FB are incommensurable in square. And, since AB is rational, therefore the square on AB is also rational; so that the sum of the squares on AF, FB is also rational. [I. 47] And since, again, the rectangle AE, EB is equal to the square on EF, and, by hypothesis, the rectangle AE, EB is also equal to the square on BD, therefore FE is equal to BD; therefore BC is double of FE, so that the rectangle AB, BC is also commensurable with the rectangle AB, EF. But the rectangle AB, BC is medial; [X. 21] therefore the rectangle AB, EF is also medial. [X. 23] But the rectangle AB, EF is equal to the rectangle AF, FB; [Lemma]therefore the rectangle AF, FB is also medial. But it was also proved that the sum of the squares on these straight lines is rational. Therefore two straight lines AF, FB incommensurable in square have been found which make the sum of the squares on them rational, but the rectangle contained by them medial.'"],["Rule","'ProofWordCount'",414],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ, ὥστε τὴν μείζονα τὴν ΑΒ τῆς ἐλάσσονος τῆς ΒΓ μεῖζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Δ, καὶ τῷ ἀφ᾽ ὁποτέρας τῶν ΒΔ, ΔΓ ἴσον παρὰ τὴν ΑΒ παραβεβλήσθω παραλληλόγραμμον ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΕΒ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΖΒ, καὶ ἤχθω τῇ ΑΒ πρὸς ὀρθὰς ἡ ΕΖ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΖΒ. καὶ ἐπεὶ δύο εὐθεῖαι ἄνισοί εἰσιν αἱ ΑΒ, ΒΓ, καὶ ἡ ΑΒ τῆς ΒΓ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ΒΓ, τουτέστι τῷ ἀπὸ τῆς ἡμισείας αὐτῆς, ἴσον παρὰ τὴν ΑΒ παραβέβληται παραλληλόγραμμον ἐλλεῖπον εἴδει τετραγώνῳ καὶ ποιεῖ τὸ ὑπὸ τῶν ΑΕΒ, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΕ τῇ ΕΒ. καί ἐστιν ὡς ἡ ΑΕ πρὸς ΕΒ, οὕτως τὸ ὑπὸ τῶν ΒΑ, ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΕ, ἴσον δὲ τὸ μὲν ὑπὸ τῶν ΒΑ, ΑΕ τῷ ἀπὸ τῆς ΑΖ, τὸ δὲ ὑπὸ τῶν ΑΒ, ΒΕ τῷ ἀπὸ τῆς ΒΖ: ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΖ τῷ ἀπὸ τῆς ΖΒ: αἱ ΑΖ, ΖΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι. καὶ ἐπεὶ ἡ ΑΒ ῥητή ἐστιν, ῥητὸν ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΒ: ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΖ, ΖΒ ῥητόν ἐστιν. καὶ ἐπεὶ πάλιν τὸ ὑπὸ τῶν ΑΕ, ΕΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΖ, ὑπόκειται δὲ τὸ ὑπὸ τῶν ΑΕ, ΕΒ καὶ τῷ ἀπὸ τῆς ΒΔ ἴσον, ἴση ἄρα ἐστὶν ἡ ΖΕ τῇ ΒΔ: διπλῆ ἄρα ἡ ΒΓ τῆς ΖΕ: ὥστε καὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τῷ ὑπὸ τῶν ΑΒ, ΕΖ. μέσον δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ: μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΕΖ. ἴσον δὲ τὸ ὑπὸ τῶν ΑΒ, ΕΖ τῷ ὑπὸ τῶν ΑΖ, ΖΒ: μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΖ, ΖΒ. ἐδείχθη δὲ καὶ ῥητὸν τὸ συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων. Εὕρηνται ἄρα δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΖ, ΖΒ ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δὲ ὑπ᾽ αὐτῶν μέσον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",346]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",34]],["Association",["Rule","'VertexLabel'","'10.34'"],["Rule","'Text'","'To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δ᾽ ὑπ᾽ αὐτῶν ῥητόν.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",6],["Rule","'Theorem'",28]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",31]]]],["Rule","'Proof'","'Let there be set out two medial straight lines AB, BC, commensurable in square only, such that the rectangle which they contain is rational, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 31, ad fin.]let the semicircle ADB be described on AB, let BC be bisected at E, let there be applied to AB a parallelogram equal to the square on BE and deficient by a square figure, namely the rectangle AF, FB; [VI. 28] therefore AF is incommensurable in length with FB. [X. 18] Let FD be drawn from F at right angles to AB, and let AD, DB be joined. Since AF is incommensurable in length with FB, therefore the rectangle BA, AF is also incommensurable with the rectangle AB, BF. [X. 11] But the rectangle BA, AF is equal to the square on AD, and the rectangle AB, BF to the square on DB; therefore the square on AD is also incommensurable with the square on DB. And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31, I. 47] And, since BC is double of DF, therefore the rectangle AB, BC is also double of the rectangle AB, FD. But the rectangle AB, BC is rational; therefore the rectangle AB, FD is also rational. [X. 6] But the rectangle AB, FD is equal to the rectangle AD, DB; [Lemma]so that the rectangle AD, DB is also rational. Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial, but the rectangle contained by them rational.'"],["Rule","'ProofWordCount'",288],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο μέσαι δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ ῥητὸν περιέχουσαι τὸ ὑπ᾽ αὐτῶν, ὥστε τὴν ΑΒ τῆς ΒΓ μεῖζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ τὸ ΑΔΒ ἡμικύκλιον, καὶ τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε, καὶ παραβεβλήσθω παρὰ τὴν ΑΒ τῷ ἀπὸ τῆς ΒΕ ἴσον παραλληλόγραμμον ἐλλεῖπον εἴδει τετραγώνῳ τὸ ὑπὸ τῶν ΑΖΒ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΒ μήκει. καὶ ἤχθω ἀπὸ τοῦ Ζ τῇ ΑΒ πρὸς ὀρθὰς ἡ ΖΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΒ. ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΒ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΒΑ, ΑΖ τῷ ὑπὸ τῶν ΑΒ, ΒΖ. ἴσον δὲ τὸ μὲν ὑπὸ τῶν ΒΑ, ΑΖ τῷ ἀπὸ τῆς ΑΔ, τὸ δὲ ὑπὸ τῶν ΑΒ, ΒΖ τῷ ἀπὸ τῆς ΔΒ: ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΔ τῷ ἀπὸ τῆς ΔΒ. καὶ ἐπεὶ μέσον ἐστὶ τὸ ἀπὸ τῆς ΑΒ, μέσον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΒΓ τῆς ΔΖ, διπλάσιον ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ τοῦ ὑπὸ τῶν ΑΒ, ΖΔ. ῥητὸν δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ: ῥητὸν ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ. τὸ δὲ ὑπὸ τῶν ΑΒ, ΖΔ ἴσον τῷ ὑπὸ τῶν ΑΔ, ΔΒ: ὥστε καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΒ ῥητόν ἐστιν. Εὕρηνται ἄρα δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΔ, ΔΒ ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δ᾽ ὑπ᾽ αὐτῶν ῥητόν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",245]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",35]],["Association",["Rule","'VertexLabel'","'10.35'"],["Rule","'Text'","'To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾽ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνῳ.'"],["Rule","'GreekTextWordCount'",33],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",32]],["List",["Rule","'Book'",10],["Rule","'Theorem'",32]]]],["Rule","'Proof'","'Let there be set out two medial straight lines AB, BC commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 32, ad fin.]let the semicircle ADB be described on AB, and let the rest of the construction be as above. Then, since AF is incommensurable in length with FB, [X. 18] AD is also incommensurable in square with DB. [X. 11] And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31, I. 47] And, since the rectangle AF, FB is equal to the square on each of the straight lines BE, DF, therefore BE is equal to DF; therefore BC is double of FD, so that the rectangle AB, BC is also double of the rectangle AB, FD. But the rectangle AB, BC is medial; therefore the rectangle AB, FD is also medial. [X. 32] And it is equal to the rectangle AD, DB; [Lemma after X. 32] therefore the rectangle AD, DB is also medial. And, since AB is incommensurable in length with BC, while CB is commensurable with BE, therefore AB is also incommensurable in length with BE, [X. 13] so that the square on AB is also incommensurable with the rectangle AB, BE. [X. 11] But the squares on AD, DB are equal to the square on AB, [I. 47] and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB. Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.'"],["Rule","'ProofWordCount'",324],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο μέσαι δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ μέσον περιέχουσαι, ὥστε τὴν ΑΒ τῆς ΒΓ μεῖζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ τὰ λοιπὰ γεγονέτω τοῖς ἐπάνω ὁμοίως. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΒ μήκει, ἀσύμμετρός ἐστι καὶ ἡ ΑΔ τῇ ΔΒ δυνάμει. καὶ ἐπεὶ μέσον ἐστὶ τὸ ἀπὸ τῆς ΑΒ, μέσον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΖ, ΖΒ ἴσον ἐστὶ τῷ ἀφ᾽ ἑκατέρας τῶν ΒΕ, ΔΖ, ἴση ἄρα ἐστὶν ἡ ΒΕ τῇ ΔΖ: διπλῆ ἄρα ἡ ΒΓ τῆς ΖΔ: ὥστε καὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ διπλάσιόν ἐστι τοῦ ὑπὸ τῶν ΑΒ, ΖΔ. μέσον δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ: μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ. καί ἐστιν ἴσον τῷ ὑπὸ τῶν ΑΔ, ΔΒ: μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, σύμμετρος δὲ ἡ ΓΒ τῇ ΒΕ, ἀσύμμετρος ἄρα καὶ ἡ ΑΒ τῇ ΒΕ μήκει: ὥστε καὶ τὸ ἀπὸ τῆς ΑΒ τῷ ὑπὸ τῶν ΑΒ, ΒΕ ἀσύμμετρόν ἐστιν. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΒ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ, τῷ δὲ ὑπὸ τῶν ΑΒ, ΒΕ ἴσον ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ, τουτέστι τὸ ὑπὸ τῶν ΑΔ, ΔΒ: ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ τῷ ὑπὸ τῶν ΑΔ, ΔΒ. Εὕρηνται ἄρα δύο εὐθεῖαι αἱ ΑΔ, ΔΒ δυνάμει ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν μέσον καὶ τὸ ὑπ᾽ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",270]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",36]],["Association",["Rule","'VertexLabel'","'10.36'"],["Rule","'Text'","'If two rational straight lines commensurable in square only be added together, the whole is irrational; and let it be called binomial.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐὰν δύο ῥηταὶ δυνάμει μόνον σύμμετροι συντεθῶσιν, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο ὀνομάτων.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'For let two rational straight lines AB, BC commensurablein square only be added together; I say that the whole AC is irrational. For, since AB is incommensurable in length with BC —for they are commensurable in square only— and, as AB is to BC, so is the rectangle AB, BC to the square on BC, therefore the rectangle AB, BC is incommensurable with the square on BC. [X. 11] But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [X. 6], and the squares on AB, BC are commensurable with the square on BC —for AB, BC are rational straight lines commensurable in square only— [X. 15] therefore twice the rectangle AB, BC is incommensurablewith the squares on AB, BC. [X. 13] And, componendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on AC [II. 4], is incommensurable with the sum of the squares on AB, BC. [X. 16] But the sum of the squares on AB, BC is rational;therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4] And let it be called binomial.'"],["Rule","'ProofWordCount'",193],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ: λέγω, ὅτι ὅλη ἡ ΑΓ ἄλογός ἐστιν. ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει: δυνάμει γὰρ μόνον εἰσὶ σύμμετροι: ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ὑπὸ τῶν ΑΒΓ πρὸς τὸ ἀπὸ τῆς ΒΓ, ἀσύμμετρον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ τῷ ἀπὸ τῆς ΒΓ. ἀλλὰ τῷ μὲν ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, τῷ δὲ ἀπὸ τῆς ΒΓ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ: αἱ γὰρ ΑΒ, ΒΓ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀσύμμετρον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τοῖς ἀπὸ τῶν ΑΒ, ΒΓ. καὶ συνθέντι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μετὰ τῶν ἀπὸ τῶν ΑΒ, ΒΓ, τουτέστι τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ: ἄλογον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΓ: ὥστε καὶ ἡ ΑΓ ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο ὀνομάτων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",170]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",37]],["Association",["Rule","'VertexLabel'","'10.37'"],["Rule","'Text'","'If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational; and let it be called a first bimedial straight line.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι ῥητὸν περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο μέσων πρώτη.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'For let two medial straight lines AB, BC commensurable in square only and containing a rational rectangle be added together; I say that the whole AC is irrational. For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC; [cf. X. 36, ll. 9-20] and, componendo, the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on AC [II. 4], is incommensurable with the rectangle AB, BC. [X. 16] But the rectangle AB, BC is rational, for, by hypothesis, AB, BC are straight lines containing a rational rectangle; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4] And let it be called a first bimedial straight line.'"],["Rule","'ProofWordCount'",131],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ δύο μέσαι δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ ῥητὸν περιέχουσαι: λέγω, ὅτι ὅλη ἡ ΑΓ ἄλογός ἐστιν. ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ ἄρα ἀσύμμετρά ἐστι τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: καὶ συνθέντι τὰ ἀπὸ τῶν ΑΒ, ΒΓ μετὰ τοῦ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ ὑπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ: ὑπόκεινται γὰρ αἱ ΑΒ, ΒΓ ῥητὸν περιέχουσαι: ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ: ἄλογος ἄρα ἡ ΑΓ, καλείσθω δὲ ἐκ δύο μέσων πρώτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",103]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",38]],["Association",["Rule","'VertexLabel'","'10.38'"],["Rule","'Text'","'If two medial straight lines commensurable in square only and containing a medial rectangle be added together, the whole is irrational; and let it be called a second bimedial straight line.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι μέσον περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο μέσων δευτέρα.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",44]],["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'For let two medial straight lines AB, BC commensurable in square only and containing a medial rectangle be added together; I say that AC is irrational. For let a rational straight line DE be set out, and let the parallelogram DF equal to the square on AC be applied to DE, producing DG as breadth. [I. 44] Then, since the square on AC is equal to the squares on AB, BC and twice the rectangle AB, BC, [II. 4] let EH, equal to the squares on AB, BC, be applied to DE; therefore the remainder HF is equal to twice the rectangle AB, BC. And, since each of the straight lines AB, BC is medial, therefore the squares on AB, BC are also medial. But, by hypothesis, twice the rectangle AB, BC is also medial. And EH is equal to the squares on AB, BC,while FH is equal to twice the rectangle AB, BC; therefore each of the rectangle EH, HF is medial. And they are applied to the rational straight line DE; therefore each of the straight lines DH, HG is rational and incommensurable in length with DE. [X. 22] Since then AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11] But the sum of the squares on AB, BC is commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6] Therefore the sum of the squares on AB, BC is incommensurablewith twice the rectangle AB, BC. [X. 13] But EH is equal to the squares on AB, BC, and HF is equal to twice the rectangle AB, BC. Therefore EH is incommensurable with HF, so that DH is also incommensurable in length with HG. [VI. 1, X. 11] Therefore DH, HG are rational straight lines commensurable in square only; so that DG is irrational. [X. 36] But DE is rational; and the rectangle contained by an irrational and a rationalstraight line is irrational; [cf. X. 20] therefore the area DF is irrational, and the side of the square equal to it is irrational. [X. Def. 4] But AC is the side of the square equal to DF; therefore AC is irrational. And let it be called a second bimedial straight line.'"],["Rule","'ProofWordCount'",406],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ δύο μέσαι δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ μέσον περιέχουσαι: λέγω, ὅτι ἄλογός ἐστιν ἡ ΑΓ. Ἐκκείσθω γὰρ ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΓ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ. καὶ ἐπεὶ τὸ ἀπὸ τῆς ΑΓ ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΒ, ΒΓ καὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, παραβεβλήσθω δὴ τοῖς ἀπὸ τῶν ΑΒ, ΒΓ παρὰ τὴν ΔΕ ἴσον τὸ ΕΘ: λοιπὸν ἄρα τὸ ΘΖ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καὶ ἐπεὶ μέση ἐστὶν ἑκατέρα τῶν ΑΒ, ΒΓ, μέσα ἄρα ἐστὶ καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ. μέσον δὲ ὑπόκειται καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καί ἐστι τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΕΘ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΖΘ: μέσον ἄρα ἑκάτερον τῶν ΕΘ, ΘΖ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται: ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΘ, ΘΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. ἐπεὶ οὖν ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, καί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τῷ ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΒ σύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τετραγώνων, τῷ δὲ ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τὸ ΕΘ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τὸ ΘΖ. ἀσύμμετρον ἄρα ἐστὶ τὸ ΕΘ τῷ ΘΖ: ὥστε καὶ ἡ ΔΘ τῇ ΘΗ ἐστιν ἀσύμμετρος μήκει. αἱ ΔΘ, ΘΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. ὥστε ἡ ΔΗ ἄλογός ἐστιν. ῥητὴ δὲ ἡ ΔΕ: τὸ δὲ ὑπὸ ἀλόγου καὶ ῥητῆς περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν: ἄλογον ἄρα ἐστὶ τὸ ΔΖ χωρίον, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν. δύναται δὲ τὸ ΔΖ ἡ ΑΓ: ἄλογος ἄρα ἐστὶν ἡ ΑΓ, καλείσθω δὲ ἐκ δύο μέσων δευτέρα. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",343]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",39]],["Association",["Rule","'VertexLabel'","'10.39'"],["Rule","'Text'","'If two straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole straight line is irrational: and let it be called major.'"],["Rule","'TextWordCount'",39],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δ᾽ ὑπ᾽ αὐτῶν μέσον, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, καλείσθω δὲ μείζων.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",33]]]],["Rule","'Proof'","'For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 33], be added together; I say that AC is irrational. For, since the rectangle AB, BC is medial, twice the rectangle AB, BC is also medial. [X. 6 and 23] But the sum of the squares on AB, BC is rational; therefore twice the rectangle AB, BC is incommensurable with the sum of the squares on AB, BC, so that the squares on AB, BC together with twice the rectangle AB, BC that is, the square on AC, is also incommensurable with the sum of the squares on AB, BC; [X. 16] therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4] And let it be called major.'"],["Rule","'ProofWordCount'",130],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα: λέγω, ὅτι ἄλογός ἐστιν ἡ ΑΓ. ἐπεὶ γὰρ τὸ ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν, καὶ τὸ δὶς ἄρα ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν. τὸ δὲ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ ῥητόν: ἀσύμμετρον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ: ὥστε καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ μετὰ τοῦ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ ῥητὸν δὲ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ: ἄλογον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΓ. ὥστε καὶ ἡ ΑΓ ἄλογός ἐστιν, καλείσθω δὲ μείζων. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",123]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",40]],["Association",["Rule","'VertexLabel'","'10.40'"],["Rule","'Text'","'If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational, be added together, the whole straight line is irrational; and let it be called the side of a rational plus a medial area.'"],["Rule","'TextWordCount'",47],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δ᾽ ὑπ᾽ αὐτῶν ῥητόν, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, καλείσθω δὲ ῥητὸν καὶ μέσον δυναμένη.'"],["Rule","'GreekTextWordCount'",35],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",34]]]],["Rule","'Proof'","'For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 34], be added together; I say that AC is irrational. For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC; so that the square on AC is also incommensurable with twice the rectangle AB, BC. [X. 16] But twice the rectangle AB, BC is rational; therefore the square on AC is irrational. Therefore AC is irrational. [X. Def. 4] And let it be called the side of a rational plus a medial area.'"],["Rule","'ProofWordCount'",116],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα: λέγω, ὅτι ἄλογός ἐστιν ἡ ΑΓ. ἐπεὶ γὰρ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν, τὸ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ῥητόν, ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ὥστε καὶ τὸ ἀπὸ τῆς ΑΓ ἀσύμμετρόν ἐστι τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ. ἄλογος ἄρα ἡ ΑΓ, καλείσθω δὲ ῥητὸν καὶ μέσον δυναμένη. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",96]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",41]],["Association",["Rule","'VertexLabel'","'10.41'"],["Rule","'Text'","'If two straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them, be added together, the whole straight line is irrational; and let it be called the   side of the sum of two medial areas.'"],["Rule","'TextWordCount'",58],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾽ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, καλείσθω δὲ δύο μέσα δυναμένη.'"],["Rule","'GreekTextWordCount'",44],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",35]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [X. 35] be added together; I say that AC is irrational. Let a rational straight line DE be set out, and let there be applied to DE the rectangle DF equal to the squares on AB, BC, and the rectangle GH equal to twice the rectangle AB, BC; therefore the whole DH is equal to the square on AC. [II. 4] Now, since the sum of the squares on AB, BC is medial, and is equal to DF, therefore DF is also medial. And it is applied to the rational straight line DE; therefore DG is rational and incommensurable in length with DE. [X. 22] For the same reason GK is also rational and incommensurable in length with GF, that is, DE. And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, DF is incommensurable with GH; so that DG is also incommensurable with GK. [VI. 1, X. 11] And they are rational; therefore DG, GK are rational straight lines commensurable in square only; therefore DK is irrational and what is called binomial. [X. 36] But DE is rational; therefore DH is irrational, and the side of the square which is equal to it is irrational. [X. Def. 4] But AC is the side of the square equal to HD; therefore AC is irrational. And let it be called the side of the sum of two medial areas.'"],["Rule","'ProofWordCount'",248],["Rule","'GreekProof'","'Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα: λέγω, ὅτι ἡ ΑΓ ἄλογός ἐστιν. Ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ παραβεβλήσθω παρὰ τὴν ΔΕ τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΔΖ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΗΘ: ὅλον ἄρα τὸ ΔΘ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ. καὶ ἐπεὶ μέσον ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ, καί ἐστιν ἴσον τῷ ΔΖ, μέσον ἄρα ἐστὶ καὶ τὸ ΔΖ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται: ῥητὴ ἄρα ἐστὶν ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΗΚ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΗΖ, τουτέστι τῇ ΔΕ, μήκει. καὶ ἐπεὶ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρόν ἐστι τὸ ΔΖ τῷ ΗΘ: ὥστε καὶ ἡ ΔΗ τῇ ΗΚ ἀσύμμετρός ἐστιν. καί εἰσι ῥηταί: αἱ ΔΗ, ΗΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἄλογος ἄρα ἐστὶν ἡ ΔΚ ἡ καλουμένη ἐκ δύο ὀνομάτων. ῥητὴ δὲ ἡ ΔΕ: ἄλογον ἄρα ἐστὶ τὸ ΔΘ καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν. δύναται δὲ τὸ ΘΔ ἡ ΑΓ: ἄλογος ἄρα ἐστὶν ἡ ΑΓ, καλείσθω δὲ δύο μέσα δυναμένη. ὅπερ ἔδει δεῖξαι. λῆμμα ὅτι δὲ αἱ εἰρημέναι ἄλογοι μοναχῶς διαιροῦνται εἰς τὰς εὐθείας, ἐξ ὧν σύγκεινται ποιουσῶν τὰ προκείμενα εἴδη, δείξομεν ἤδη προεκθέμενοι λημμάτιον τοιοῦτον: Ἐκκείσθω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω ἡ ὅλη εἰς ἄνισα καθ᾽ ἑκάτερον τῶν Γ, Δ, ὑποκείσθω δὲ μείζων ἡ ΑΓ τῆς ΔΒ: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τῶν ἀπὸ τῶν ΑΔ, ΔΒ. τετμήσθω γὰρ ἡ ΑΒ δίχα κατὰ τὸ Ε. καὶ ἐπεὶ μείζων ἐστὶν ἡ ΑΓ τῆς ΔΒ, κοινὴ ἀφῃρήσθω ἡ ΔΓ: λοιπὴ ἄρα ἡ ΑΔ λοιπῆς τῆς ΓΒ μείζων ἐστίν. ἴση δὲ ἡ ΑΕ τῇ ΕΒ: ἐλάττων ἄρα ἡ ΔΕ τῆς ΕΓ: τὰ Γ, Δ ἄρα σημεῖα οὐκ ἴσον ἀπέχουσι τῆς διχοτομίας. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΒ, ἀλλὰ μὴν καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΒ μετὰ τοῦ ἀπὸ ΔΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΒ, τὸ ἄρα ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΔ, ΔΒ μετὰ τοῦ ἀπὸ τῆς ΔΕ: ὧν τὸ ἀπὸ τῆς ΔΕ ἔλασσόν ἐστι τοῦ ἀπὸ τῆς ΕΓ: καὶ λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΓ, ΓΒ ἔλασσόν ἐστι τοῦ ὑπὸ τῶν ΑΔ, ΔΒ. ὥστε καὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἔλασσόν ἐστι τοῦ δὶς ὑπὸ ΑΔ, ΔΒ. καὶ λοιπὸν ἄρα τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μεῖζόν ἐστι τοῦ συγκειμένου ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",433]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",42]],["Association",["Rule","'VertexLabel'","'10.42'"],["Rule","'Text'","'A binomial straight line is divided into its terms at one point only.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἡ ἐκ δύο ὀνομάτων κατὰ ἓν μόνον σημεῖον διαιρεῖται εἰς τὰ ὀνόματα.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let AB be a binomial straight line divided into its terms at C; therefore AC, CB are rational straight lines commensurable in square only. [X. 36] I say that AB is not divided at another point into two rational straight lines commensurable in square only. For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only. It is then manifest that AC is not the same with DB. For, if possible, let it be so. Then AD will also be the same as CB, and, as AC is to CB, so will BD be to DA; thus AB will be divided at D also in the same way as by the division at C: which is contrary to the hypothesis. Therefore AC is not the same with DB. For this reason also the points C, D are not equidistant from the point of bisection. Therefore that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, because both the squares on AC, CB together with twice the rectangle AC, CB, and the squares on AD, DB together with twice the rectangle AD, DB, are equal to the square on AB. [II. 4] But the squares on AC, CB differ from the squares on AD, DB by a rational area, for both are rational; therefore twice the rectangle AD, DB also differs from twice the rectangle AC, CB by a rational area, though they are medial [X. 21]: which is absurd, for a medial area does not exceed a medial by a rational area. [x. 26] Therefore a binomial straight line is not divided at different points; therefore it is divided at one point only.'"],["Rule","'ProofWordCount'",307],["Rule","'GreekProof'","'ἔστω ἐκ δύο ὀνομάτων ἡ ΑΒ διῃρημένη εἰς τὰ ὀνόματα κατὰ τὸ Γ: αἱ ΑΓ, ΓΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. λέγω, ὅτι ἡ ΑΒ κατ᾽ ἄλλο σημεῖον οὐ διαιρεῖται εἰς δύο ῥητὰς δυνάμει μόνον συμμέτρους. εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ τὰς ΑΔ, ΔΒ ῥητὰς εἶναι δυνάμει μόνον συμμέτρους. φανερὸν δή, ὅτι ἡ ΑΓ τῇ ΔΒ οὐκ ἔστιν ἡ αὐτή. εἰ γὰρ δυνατόν, ἔστω. ἔσται δὴ καὶ ἡ ΑΔ τῇ ΓΒ ἡ αὐτή: καὶ ἔσται ὡς ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΒΔ πρὸς τὴν ΔΑ, καὶ ἔσται ἡ ΑΒ κατὰ τὸ αὐτὸ τῇ κατὰ τὸ Γ διαιρέσει διαιρεθεῖσα καὶ κατὰ τὸ Δ: ὅπερ οὐχ ὑπόκειται. οὐκ ἄρα ἡ ΑΓ τῇ ΔΒ ἐστιν ἡ αὐτή. διὰ δὴ τοῦτο καὶ τὰ Γ, Δ σημεῖα οὐκ ἴσον ἀπέχουσι τῆς διχοτομίας. ᾧ ἄρα διαφέρει τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ, τούτῳ διαφέρει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ διὰ τὸ καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ μετὰ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ἴσα εἶναι τῷ ἀπὸ τῆς ΑΒ. ἀλλὰ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ διαφέρει ῥητῷ: ῥητὰ γὰρ ἀμφότερα: καὶ τὸ δὶς ἄρα ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ διαφέρει ῥητῷ μέσα ὄντα: ὅπερ ἄτοπον: μέσον γὰρ μέσου οὐχ ὑπερέχει ῥητῷ. οὐκ ἄρα ἡ ἐκ δύο ὀνομάτων κατ᾽ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται: καθ᾽ ἓν ἄρα μόνον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",261]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",43]],["Association",["Rule","'VertexLabel'","'10.43'"],["Rule","'Text'","'A first bimedial straight line is divided at one point only.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἡ ἐκ δύο μέσων πρώτη καθ᾽ ἓν μόνον σημεῖον διαιρεῖται.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List"]],["Rule","'Proof'","'Let AB be a first bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a rational rectangle; I say that AB is not so divided at another point. For, if possible, let it be divided at D also, so that AD, DB are also medial straight lines commensurable in square only and containing a rational rectangle. Since, then, that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB is that by which the squares on AC, CB differ from the squares on AD, DB, while twice the rectangle AD, DB differs from twice the rectangle AC, CB by a rational area—for both are rational— therefore the squares on AC, CB also differ from the squares on AD, DB by a rational area, though they are medial: which is absurd. [x. 26] Therefore a first bimedial straight line is not divided into its terms at different points; therefore it is so divided at one point only.'"],["Rule","'ProofWordCount'",172],["Rule","'GreekProof'","'ἔστω ἐκ δύο μέσων πρώτη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ μέσας εἶναι δυνάμει μόνον συμμέτρους ῥητὸν περιεχούσας: λέγω, ὅτι ἡ ΑΒ κατ᾽ ἄλλο σημεῖον οὐ διαιρεῖται. εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ τὰς ΑΔ, ΔΒ μέσας εἶναι δυνάμει μόνον συμμέτρους ῥητὸν περιεχούσας. ἐπεὶ οὖν, ᾧ διαφέρει τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τούτῳ διαφέρει τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ, ῥητῷ δὲ διαφέρει τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ: ῥητὰ γὰρ ἀμφότερα: ῥητῷ ἄρα διαφέρει καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ μέσα ὄντα: ὅπερ ἄτοπον. οὐκ ἄρα ἡ ἐκ δύο μέσων πρώτη κατ᾽ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται εἰς τὰ ὀνόματα: καθ᾽ ἓν ἄρα μόνον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",141]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",44]],["Association",["Rule","'VertexLabel'","'10.44'"],["Rule","'Text'","'A second bimedial straight line is divided at one point only.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἡ ἐκ μέσων δευτέρα καθ᾽ ἓν μόνον σημεῖον διαιρεῖται.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",38]]]],["Rule","'Proof'","'Let AB be a second bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a medial rectangle; [X. 38] it is then manifest that C is not at the point of bisection, because the segments are not commensurable in length. I say that AB is not so divided at another point. For, if possible, let it be divided at D also, so that AC is not the same with DB, but AC is supposed greater; it is then clear that the squares on AD, DB are also, as we proved above [Lemma], less than the squares on AC, CB; and suppose that AD, DB are medial straight lines commensurable in square only and containing a medial rectangle. Now let a rational straight line EF be set out, let there be applied to EF the rectangular parallelogram EK equal to the square on AB, and let EG equal to the squares on AC, CB be subtracted; therefore the remainder HK is equal to twice the rectangle AC, CB. [II. 4] Again, let there be subtracted EL, equal to the squares on AD, DB, which were proved less than the squares on AC, CB [Lemma]; therefore the remainder MK is also equal to twice the rectangle AD, DB. Now, since the squares on AC, CB are medial, therefore EG is medial. And it is applied to the rational straight line EF; therefore EH is rational and incommensurable in length with EF. [X. 22] For the same reason HN is also rational and incommensurable in length with EF. And, since AC, CB are medial straight lines commensurable in square only, therefore AC is incommensurable in length with CB. But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11] But the squares on AC, CB are commensurable with the square on AC; for AC, CB are commensurable in square. [X. 15] And twice the rectangle AC, CB is commensurable with the rectangle AC, CB. [X. 6] Therefore the squares on AC, CB are also incommensurable with twice the rectangle AC, CB. [X. 13] But EG is equal to the squares on AC, CB, and HK is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HK, so that EH is also incommensurable in length with HN. [VI. 1, X. 11] And they are rational; therefore EH, HN are rational straight lines commensurable in square only. But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [X. 36] Therefore EN is a binomial straight line divided at H. In the same way EM, MN will also be proved to be rational straight lines commensurable in square only; and EN will be a binomial straight line divided at different points, H and M. And EH is not the same with MN. For the squares on AC, CB are greater than the squares on AD, DB. But the squares on AD, DB are greater than twice the rectangle AD, DB; therefore also the squares on AC, CB, that is, EG, are much greater than twice the rectangle AD, DB, that is, MK, so that EH is also greater than MN. Therefore EH is not the same with MN.'"],["Rule","'ProofWordCount'",568],["Rule","'GreekProof'","'ἔστω ἐκ δύο μέσων δευτέρα ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ μέσας εἶναι δυνάμει μόνον συμμέτρους μέσον περιεχούσας: φανερὸν δή, ὅτι τὸ Γ οὐκ ἔστι κατὰ τῆς διχοτομίας, ὅτι οὐκ εἰσὶ μήκει σύμμετροι. λέγω, ὅτι ἡ ΑΒ κατ᾽ ἄλλο σημεῖον οὐ διαιρεῖται. εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε τὴν ΑΓ τῇ ΔΒ μὴ εἶναι τὴν αὐτήν, ἀλλὰ μείζονα καθ᾽ ὑπόθεσιν τὴν ΑΓ: δῆλον δή, ὅτι καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ, ὡς ἐπάνω ἐδείξαμεν, ἐλάσσονα τῶν ἀπὸ τῶν ΑΓ, ΓΒ: καὶ τὰς ΑΔ, ΔΒ μέσας εἶναι δυνάμει μόνον συμμέτρους μέσον περιεχούσας. καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τῷ μὲν ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΕΖ παραλληλόγραμμον ὀρθογώνιον παραβεβλήσθω τὸ ΕΚ, τοῖς δὲ ἀπὸ τῶν ΑΓ, ΓΒ ἴσον ἀφῃρήσθω τὸ ΕΗ: λοιπὸν ἄρα τὸ ΘΚ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. πάλιν δὴ τοῖς ἀπὸ τῶν ΑΔ, ΔΒ, ἅπερ ἐλάσσονα ἐδείχθη τῶν ἀπὸ τῶν ΑΓ, ΓΒ, ἴσον ἀφῃρήσθω τὸ ΕΛ: καὶ λοιπὸν ἄρα τὸ ΜΚ ἴσον τῷ δὶς ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ μέσα ἐστὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ, μέσον ἄρα καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται: ῥητὴ ἄρα ἐστὶν ἡ ΕΘ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΝ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ αἱ ΑΓ, ΓΒ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΓ τῇ ΓΒ μήκει. ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ὑπὸ τῶν ΑΓ, ΓΒ: ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΓ τῷ ὑπὸ τῶν ΑΓ, ΓΒ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΓ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΓ, ΓΒ: δυνάμει γάρ εἰσι σύμμετροι αἱ ΑΓ, ΓΒ. τῷ δὲ ὑπὸ τῶν ΑΓ, ΓΒ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ ἄρα ἀσύμμετρά ἐστι τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον ἐστὶ τὸ ΕΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΘΚ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΕΗ τῷ ΘΚ: ὥστε καὶ ἡ ΕΘ τῇ ΘΝ ἀσύμμετρός ἐστι μήκει. καί εἰσι ῥηταί: αἱ ΕΘ, ΘΝ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. ἐὰν δὲ δύο ῥηταὶ δυνάμει μόνον σύμμετροι συντεθῶσιν, ἡ ὅλη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο ὀνομάτων: ἡ ΕΝ ἄρα ἐκ δύο ὀνομάτων ἐστὶ διῃρημένη κατὰ τὸ Θ. κατὰ τὰ αὐτὰ δὴ δειχθήσονται καὶ αἱ ΕΜ, ΜΝ ῥηταὶ δυνάμει μόνον σύμμετροι: καὶ ἔσται ἡ ΕΝ ἐκ δύο ὀνομάτων κατ᾽ ἄλλο καὶ ἄλλο διῃρημένη τό τε Θ καὶ τὸ Μ, καὶ οὐκ ἔστιν ἡ ΕΘ τῇ ΜΝ ἡ αὐτή, ὅτι τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τῶν ἀπὸ τῶν ΑΔ, ΔΒ. ἀλλὰ τὰ ἀπὸ τῶν ΑΔ, ΔΒ μείζονά ἐστι τοῦ δὶς ὑπὸ ΑΔ, ΔΒ: πολλῷ ἄρα καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ, τουτέστι τὸ ΕΗ, μεῖζόν ἐστι τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, τουτέστι τοῦ ΜΚ: ὥστε καὶ ἡ ΕΘ τῆς ΜΝ μείζων ἐστίν. ἡ ἄρα ΕΘ τῇ ΜΝ οὐκ ἔστιν ἡ αὐτή: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",500]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",45]],["Association",["Rule","'VertexLabel'","'10.45'"],["Rule","'Text'","'A major straight line is divided at one and the same point only.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἡ μείζων κατὰ τὸ αὐτὸ μόνον σημεῖον διαιρεῖται.'"],["Rule","'GreekTextWordCount'",8],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Theorem'",39]]]],["Rule","'Proof'","'Let AB be a major straight line divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB rational, but the rectangle AC, CB medial; [X. 39] I say that AB is not so divided at another point. For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB rational, but the rectangle contained by them medial. Then, since that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, while the squares on AC, CB exceed the squares on AD, DB by a rational area—for both are rational— therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area, though they are medial: which is impossible. [X. 26] Therefore a major straight line is not divided at different points; therefore it is only divided at one and the same point.'"],["Rule","'ProofWordCount'",187],["Rule","'GreekProof'","'ἔστω μείζων ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνων ῥητόν, τὸ δ᾽ ὑπὸ τῶν ΑΓ, ΓΒ μέσον: λέγω, ὅτι ἡ ΑΒ κατ᾽ ἄλλο σημεῖον οὐ διαιρεῖται. εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ τὰς ΑΔ, ΔΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ ῥητόν, τὸ δ᾽ ὑπ᾽ αὐτῶν μέσον. καὶ ἐπεί, ᾧ διαφέρει τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ, τούτῳ διαφέρει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ἀλλὰ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ ὑπερέχει ῥητῷ: ῥητὰ γὰρ ἀμφότερα: καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ἄρα τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ μέσα ὄντα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ μείζων κατ᾽ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται: κατὰ τὸ αὐτὸ ἄρα μόνον διαιρεῖται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",165]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",46]],["Association",["Rule","'VertexLabel'","'10.46'"],["Rule","'Text'","'The side of a rational plus a medial area is divided at one point only.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἡ ῥητὸν καὶ μέσον δυναμένη καθ᾽ ἓν μόνον σημεῖον διαιρεῖται.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Theorem'",40]]]],["Rule","'Proof'","'Let AB be the side of a rational plus a medial area divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, but twice the rectangle AC, CB rational; [X. 40] I say that AB is not so divided at another point. For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB medial, but twice the rectangle AD, DB rational. Since then that by which twice the rectangle AC, CB differs from twice the rectangle AD, DB is also that by which the squares on AD, DB differ from the squares on AC, CB, while twice the rectangle AC, CB exceeds twice the rectangle AD, DB by a rational area, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area, though they are medial: which is impossible. [X. 26] Therefore the side of a rational plus a medial area is not divided at different points; therefore it is divided at one point only.'"],["Rule","'ProofWordCount'",191],["Rule","'GreekProof'","'ἔστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον, τὸ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ῥητόν: λέγω, ὅτι ἡ ΑΒ κατ᾽ ἄλλο σημεῖον οὐ διαιρεῖται. εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ τὰς ΑΔ, ΔΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ μέσον, τὸ δὲ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ῥητόν. ἐπεὶ οὖν, ᾧ διαφέρει τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, τούτῳ διαφέρει καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τὸ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ὑπερέχει ῥητῷ, καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ ἄρα τῶν ἀπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ μέσα ὄντα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ῥητὸν καὶ μέσον δυναμένη κατ᾽ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται. κατὰ ἓν ἄρα σημεῖον διαιρεῖται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",167]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",47]],["Association",["Rule","'VertexLabel'","'10.47'"],["Rule","'Text'","'The side of the sum of two medial areas is divided at one point only.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἡ δύο μέσα δυναμένη καθ᾽ ἓν μόνον σημεῖον διαιρεῖται.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",42]]]],["Rule","'Proof'","'Let AB be divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, and the rectangle AC, CB medial and also incommensurable with the sum of the squares on them; I say that AB is not divided at another point so as to fulfil the given conditions. For, if possible, let it be divided at D, so that again AC is of course not the same as BD, but AC is supposed greater; let a rational straight line EF be set out, and let there be applied to EF the rectangle EG equal to the squares on AC, CB, and the rectangle HK equal to twice the rectangle AC, CB; therefore the whole EK is equal to the square on AB. [II. 4] Again, let EL, equal to the squares on AD, DB, be applied to EF; therefore the remainder, twice the rectangle AD, DB, is equal to the remainder MK. And since, by hypothesis, the sum of the squares on AC, CB is medial, therefore EG is also medial. And it is applied to the rational straight line EF; therefore HE is rational and incommensurable in length with EF. [X. 22] For the same reason HN is also rational and incommensurable in length with EF. And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore EG is also incommensurable with GN, so that EH is also incommensurable with HN. [VI. 1, X. 11] And they are rational; therefore EH, HN are rational straight lines commensurable in square only; therefore EN is a binomial straight line divided at H. [X. 36] Similarly we can prove that it is also divided at M.  And EH is not the same with MN; therefore a binomial has been divided at different points: which is absurd. [X. 42] Therefore a side of the sum of two medial areas is not divided at different points; therefore it is divided at one point only.'"],["Rule","'ProofWordCount'",339],["Rule","'GreekProof'","'ἔστω δύο μέσα δυναμένη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τό τε συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον καὶ τὸ ὑπὸ τῶν ΑΓ, ΓΒ μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπ᾽ αὐτῶν. λέγω, ὅτι ἡ ΑΒ κατ᾽ ἄλλο σημεῖον οὐ διαιρεῖται ποιοῦσα τὰ προκείμενα. εἰ γὰρ δυνατόν, διῃρήσθω κατὰ τὸ Δ, ὥστε πάλιν δηλονότι τὴν ΑΓ τῇ ΔΒ μὴ εἶναι τὴν αὐτήν, ἀλλὰ μείζονα καθ᾽ ὑπόθεσιν τὴν ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ παραβεβλήσθω παρὰ τὴν ΕΖ τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΕΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΘΚ: ὅλον ἄρα τὸ ΕΚ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. πάλιν δὴ παραβεβλήσθω παρὰ τὴν ΕΖ τοῖς ἀπὸ τῶν ΑΔ, ΔΒ ἴσον τὸ ΕΛ: λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ λοιπῷ τῷ ΜΚ ἴσον ἐστίν. καὶ ἐπεὶ μέσον ὑπόκειται τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, μέσον ἄρα ἐστὶ καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται: ῥητὴ ἄρα ἐστὶν ἡ ΘΕ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΝ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, καὶ τὸ ΕΗ ἄρα τῷ ΗΝ ἀσύμμετρόν ἐστιν: ὥστε καὶ ἡ ΕΘ τῇ ΘΝ ἀσύμμετρός ἐστιν. καί εἰσι ῥηταί: αἱ ΕΘ, ΘΝ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ΕΝ ἄρα ἐκ δύο ὀνομάτων ἐστὶ διῃρημένη κατὰ τὸ Θ. ὁμοίως δὴ δείξομεν, ὅτι καὶ κατὰ τὸ Μ διῄρηται. καὶ οὐκ ἔστιν ἡ ΕΘ τῇ ΜΝ ἡ αὐτή: ἡ ἄρα ἐκ δύο ὀνομάτων κατ᾽ ἄλλο καὶ ἄλλο σημεῖον διῄρηται: ὅπερ ἐστὶν ἄτοπον. οὐκ ἄρα ἡ δύο μέσα δυναμένη κατ᾽ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται: καθ᾽ ἓν ἄρα μόνον σημεῖον διαιρεῖται.'"],["Rule","'GreekProofWordCount'",308]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",48]],["Association",["Rule","'VertexLabel'","'10.48'"],["Rule","'Text'","'To find the first binomial straight line.'"],["Rule","'TextWordCount'",7],["Rule","'GreekText'","'εὑρεῖν τὴν ἐκ δύο ὀνομάτων πρώτην.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",28]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to CA the ratio which a square number has to a square number; [Lemma I after X. 28] let any rational straight line D be set out, and let EF be commensurable in length with D. Therefore EF is also rational. Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG. [X. 6] But AB has to AC the ratio which a number has to a number; therefore the square on EF also has to the square on FG the ratio which a number has to a number, so that the square on EF is commensurable with the square on FG. [X. 6] And EF is rational; therefore FG is also rational. And, since BA has not to AC the ratio which a square number has to a square number. neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9] Therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36] I say that it is also a first binomial straight line. For since, as the number BA is to AC, so is the square on EF to the square on FG, while BA is greater than AC, therefore the square on EF is also greater than the square on FG. Let then the squares on FG, H be equal to the square on EF. Now since, as BA is to AC, so is the square on EF to the square on FG, therefore, convertendo, as AB is to BC, so is the square on EF to the square on H. [V. 19] But AB has to BC the ratio which a square number has to a square number; therefore the square on EF also has to the square on H the ratio which a square number has to a square number. Therefore EF is commensurable in length with H; [X. 9] therefore the square on EF is greater than the square on FG by the square on a straight line commensurable with EF. And EF, FG are rational, and EF is commensurable in length with D. Therefore EF is a first binomial straight line.'"],["Rule","'ProofWordCount'",419],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, πρὸς δὲ τὸν ΓΑ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω τις ῥητὴ ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω μήκει ἡ ΕΖ. ῥητὴ ἄρα ἐστὶ καὶ ἡ ΕΖ. καὶ γεγονέτω ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ. ὁ δὲ ΑΒ πρὸς τὸν ΑΓ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν: καὶ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν: ὥστε σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΖΗ. καί ἐστι ῥητὴ ἡ ΕΖ: ῥητὴ ἄρα καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ μήκει. αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ. λέγω, ὅτι καὶ πρώτη. ἐπεὶ γάρ ἐστιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, μείζων δὲ ὁ ΒΑ τοῦ ΑΓ, μεῖζον ἄρα καὶ τὸ ἀπὸ τῆς ΕΖ τοῦ ἀπὸ τῆς ΖΗ. ἔστω οὖν τῷ ἀπὸ τῆς ΕΖ ἴσα τὰ ἀπὸ τῶν ΖΗ, Θ. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. σύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ Θ μήκει: ἡ ΕΖ ἄρα τῆς ΖΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί εἰσι ῥηταὶ αἱ ΕΖ, ΖΗ, καὶ σύμμετρος ἡ ΕΖ τῇ Δ μήκει. ἡ ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πρώτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",353]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",49]],["Association",["Rule","'VertexLabel'","'10.49'"],["Rule","'Text'","'To find the second binomial straight line.'"],["Rule","'TextWordCount'",7],["Rule","'GreekText'","'εὑρεῖν τὴν ἐκ δύο ὀνομάτων δευτέραν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number; let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is rational. Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG; [X. 6] therefore the square on EF is commensurable with the square on FG. [X. 6] Therefore FG is also rational. Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number. Therefore EF is incommensurable in length with FG; [X. 9] therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36] It is next to be proved that it is also a second binomial straight line. For since, inversely, as the number BA is to AC, so is the square on GF to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE. Let the squares on EF, H be equal to the square on GF; therefore, convertendo, as AB is to BC, so is the square on FG to the square on H. [V. 19] But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on H the ratio which a square number has to a square number. Therefore FG is commensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG. And FG, FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out. Therefore EG is a second binomial straight line.'"],["Rule","'ProofWordCount'",377],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, πρὸς δὲ τὸν ΑΓ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω ῥητὴ ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω ἡ ΕΖ μήκει: ῥητὴ ἄρα ἐστὶν ἡ ΕΖ. γεγονέτω δὴ καὶ ὡς ὁ ΓΑ ἀριθμὸς πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΖΗ. ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΓΑ ἀριθμὸς πρὸς τὸν ΑΒ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ μήκει: αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ. δεικτέον δή, ὅτι καὶ δευτέρα. ἐπεὶ γὰρ ἀνάπαλίν ἐστιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΗΖ πρὸς τὸ ἀπὸ τῆς ΖΕ, μείζων δὲ ὁ ΒΑ τοῦ ΑΓ, μεῖζον ἄρα καὶ τὸ ἀπὸ τῆς ΗΖ τοῦ ἀπὸ τῆς ΖΕ. ἔστω τῷ ἀπὸ τῆς ΗΖ ἴσα τὰ ἀπὸ τῶν ΕΖ, Θ: ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Θ. ἀλλ᾽ ὁ ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ τὸ ἀπὸ τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. σύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Θ μήκει: ὥστε ἡ ΖΗ τῆς ΖΕ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί εἰσι ῥηταὶ αἱ ΖΗ, ΖΕ δυνάμει μόνον σύμμετροι, καὶ τὸ ΕΖ ἔλασσον ὄνομα τῇ ἐκκειμένῃ ῥητῇ σύμμετρόν ἐστι τῇ Δ μήκει. ἡ ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ δευτέρα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",311]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",50]],["Association",["Rule","'VertexLabel'","'10.50'"],["Rule","'Text'","'To find the third binomial straight line.'"],["Rule","'TextWordCount'",7],["Rule","'GreekText'","'εὑρεῖν τὴν ἐκ δύο ὀνομάτων τρίτην.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number. Let any other number D, not square, be set out also, and let it not have to either of the numbers BA. AC the ratio which a square number has to a square number. Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6] therefore the square on E is commensurable with the square on FG. [X. 6] And E is rational; therefore FG is also rational. And, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9] Next let it be contrived that, as the number BA is to AC, so is the square on FG to the square on GH; [X. 6] therefore the square on FG is commensurable with the square on GH. [X. 6] But FG is rational; therefore GH is also rational. And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on HG the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9] Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36] I say next that it is also a third binomial straight line. For since, as D is to AB, so is the square on E to the square on FG, and, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22] But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9] And since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH. Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19] But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number; therefore FG is commensurable in length with K. [X. 9] Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG. And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E. Therefore FH is a third binomial straight line.'"],["Rule","'ProofWordCount'",583],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, πρὸς δὲ τὸν ΑΓ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἐκκείσθω δέ τις καὶ ἄλλος μὴ τετράγωνος ἀριθμὸς ὁ Δ, καὶ πρὸς ἑκάτερον τῶν ΒΑ, ΑΓ λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ ἐκκείσθω τις ῥητὴ εὐθεῖα ἡ Ε, καὶ γεγονέτω ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς Ε τῷ ἀπὸ τῆς ΖΗ. καί ἐστι ῥητὴ ἡ Ε: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ Δ πρὸς τὸν ΑΒ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ ΖΗ μήκει. γεγονέτω δὴ πάλιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΗΘ. ῥητὴ δὲ ἡ ΖΗ: ῥητὴ ἄρα καὶ ἡ ΗΘ. καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὅν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΘΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΘ μήκει. αἱ ΖΗ, ΗΘ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστίν. λέγω δή, ὅτι καὶ τρίτη. ἐπεὶ γάρ ἐστιν ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Δ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΗΘ. ὁ δὲ Δ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ ΗΘ μήκει. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, μεῖζον ἄρα τὸ ἀπὸ τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ. ἔστω οὖν τῷ ἀπὸ τῆς ΖΗ ἴσα τὰ ἀπὸ τῶν ΗΘ, Κ: ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ τὸ ἀπὸ τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: σύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Κ μήκει. ἡ ΖΗ ἄρα τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΖΗ, ΗΘ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα αὐτῶν σύμμετρός ἐστι τῇ Ε μήκει. ἡ ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",494]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",51]],["Association",["Rule","'VertexLabel'","'10.51'"],["Rule","'Text'","'To find the fourth binomial straight line.'"],["Rule","'TextWordCount'",7],["Rule","'GreekText'","'εὑρεῖν τὴν ἐκ δύο ὀνομάτων τετάρτην.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]]]],["Rule","'Proof'","'Let two numbers AC, CB be set out such that AB neither has to BC, nor yet to AC, the ratio which a square number has to a square number. Let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is also rational. Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG; [X. 6] therefore the square on EF is commensurable with the square on FG; [X. 6] therefore FG is also rational. Now, since BA has not to AC the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9] Therefore EF, FG are rational straight lines commensurable in square only; so that EG is binomial. I say next that it is also a fourth binomial straight line. For since, as BA is to AC, so is the square on EF to the square on FG, therefore the square on EF is greater than the square on FG. Let then the squares on FG, H be equal to the square on EF; therefore, convertendo, as the number AB is to BC, so is the square on EF to the square on H. [V. 19] But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on EF to the square on H the ratio which a square number has to a square number. Therefore EF is incommensurable in length with H; [X. 9] therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF. And EF, FG are rational straight lines commensurable in square only, and EF is commensurable in length with D. Therefore EG is a fourth binomial straight line.'"],["Rule","'ProofWordCount'",340],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν ΑΒ πρὸς τὸν ΒΓ λόγον μὴ ἔχειν μήτε μὴν πρὸς τὸν ΑΓ, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ ἐκκείσθω ῥητὴ ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω μήκει ἡ ΕΖ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΕΖ. καὶ γεγονέτω ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΖΗ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ μήκει. αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ὥστε ἡ ΕΗ ἐκ δύο ὀνομάτων ἐστίν. λέγω δή, ὅτι καὶ τετάρτη. ἐπεὶ γάρ ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ μείζων δὲ ὁ ΒΑ τοῦ ΑΓ, μεῖζον ἄρα τὸ ἀπὸ τῆς ΕΖ τοῦ ἀπὸ τῆς ΖΗ. ἔστω οὖν τῷ ἀπὸ τῆς ΕΖ ἴσα τὰ ἀπὸ τῶν ΖΗ, Θ: ἀναστρέψαντι ἄρα ὡς ὁ ΑΒ ἀριθμὸς πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ Θ μήκει: ἡ ΕΖ ἄρα τῆς ΗΖ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΕΖ, ΖΗ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ ἡ ΕΖ τῇ Δ σύμμετρός ἐστι μήκει. ἡ ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τετάρτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",292]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",52]],["Association",["Rule","'VertexLabel'","'10.52'"],["Rule","'Text'","'To find the fifth binomial straight line.'"],["Rule","'TextWordCount'",7],["Rule","'GreekText'","'εὑρεῖν τὴν ἐκ δύο ὀνομάτων πέμπτην.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; let any rational straight line D be set out, and let EF be commensurable with D; therefore EF is rational. Let it be contrived that, as CA is to AB, so is the square on EF to the square on FG. [X. 6] But CA has not to AB the ratio which a square number has to a square number; therefore neither has the square on EF to the square on FG the ratio which a square number has to a square number. Therefore EF, FG are rational straight lines commensurable in square only; [X. 9] therefore EG is binomial. [X. 36] I say next that it is also a fifth binomial straight line. For since, as CA is to AB, so is the square on EF to the square on FG, inversely, as BA is to AC, so is the square on FG to the square on FE; therefore the square on GF is greater than the square on FE. Let then the squares on EF, H be equal to the square on GF; therefore, convertendo, as the number AB is to BC, so is the square on GF to the square on H. [V. 19] But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on FG to the square on H the ratio which a square number has to a square number. Therefore FG is incommensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line incommensurable with FG. And GF, FE are rational straight lines commensurable in square only, and the lesser term EF is commensurable in length with the rational straight line D set out. Therefore EG is a fifth binomial straight line.'"],["Rule","'ProofWordCount'",334],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν ΑΒ πρὸς ἑκάτερον αὐτῶν λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω ῥητή τις εὐθεῖα ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω μήκει ἡ ΕΖ: ῥητὴ ἄρα ἡ ΕΖ. καὶ γεγονέτω ὡς ὁ ΓΑ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ. ὁ δὲ ΓΑ πρὸς τὸν ΑΒ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδὲ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ. λέγω δή, ὅτι καὶ πέμπτη. ἐπεὶ γάρ ἐστιν ὡς ὁ ΓΑ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, ἀνάπαλιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΖΕ: μεῖζον ἄρα τὸ ἀπὸ τῆς ΗΖ τοῦ ἀπὸ τῆς ΖΕ. ἔστω οὖν τῷ ἀπὸ τῆς ΗΖ ἴσα τὰ ἀπὸ τῶν ΕΖ, Θ: ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΑΒ ἀριθμὸς πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΗΖ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Θ μήκει: ὥστε ἡ ΖΗ τῆς ΖΕ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΗΖ, ΖΕ ῥηταὶ δυνάμει μόνον σύμμετροι καὶ τὸ ΕΖ ἔλαττον ὄνομα σύμμετρόν ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ Δ μήκει. ἡ ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πέμπτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",278]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",53]],["Association",["Rule","'VertexLabel'","'10.53'"],["Rule","'Text'","'To find the sixth binomial straight line.'"],["Rule","'TextWordCount'",7],["Rule","'GreekText'","'εὑρεῖν τὴν ἐκ δύο ὀνομάτων ἕκτην.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; and let there also be another number D which is not square and which has not to either of the numbers BA, AC the ratio which a square number has to a square number. Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6] therefore the square on E is commensurable with the square on FG. [X. 6] And E is rational; therefore FG is also rational. Now, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9] Again, let it be contrived that, as BA is to AC, so is the square on FG to the square on GH. [X. 6] Therefore the square on FG is commensurable with the square on HG. [X. 6] Therefore the square on HG is rational; therefore HG is rational. And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9] Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36] It is next to be proved that it is also a sixth binomial straight line. For since, as D is to AB, so is the square on E to the square on FG, and also, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22] But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9] But it was also proved incommensurable with FG; therefore each of the straight lines FG, GH is incommensurable in length with E. And, since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH. Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19] But AB has not to BC the ratio which a square number has to a square number; so that neither has the square on FG to the square on K the ratio which a square number has to a square number. Therefore FG is incommensurable in length with K; [X. 9] therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable with FG. And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out. Therefore FH is a sixth binomial straight line.'"],["Rule","'ProofWordCount'",600],["Rule","'GreekProof'","'Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν ΑΒ πρὸς ἑκάτερον αὐτῶν λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἔστω δὲ καὶ ἕτερος ἀριθμὸς ὁ Δ μὴ τετράγωνος ὢν μηδὲ πρὸς ἑκάτερον τῶν ΒΑ, ΑΓ λόγον ἔχων, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ ἐκκείσθω τις ῥητὴ εὐθεῖα ἡ Ε, καὶ γεγονέτω ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ: σύμμετρον ἄρα τὸ ἀπὸ τῆς Ε τῷ ἀπὸ τῆς ΖΗ. καί ἐστι ῥητὴ ἡ Ε: ῥητὴ ἄρα καὶ ἡ ΖΗ. καὶ ἐπεὶ οὐκ ἔχει ὁ Δ πρὸς τὸν ΑΒ λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἡ Ε τῇ ΖΗ μήκει. γεγονέτω δὴ πάλιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ. σύμμετρον ἄρα τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΘΗ. ῥητὸν ἄρα τὸ ἀπὸ τῆς ΘΗ: ῥητὴ ἄρα ἡ ΘΗ. καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΘ μήκει. αἱ ΖΗ, ΗΘ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΖΘ. δεικτέον δή, ὅτι καὶ ἕκτη. ἐπεὶ γάρ ἐστιν ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ, ἔστι δὲ καὶ ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Δ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΗΘ. ὁ δὲ Δ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ ΗΘ μήκει. ἐδείχθη δὲ καὶ τῇ ΖΗ ἀσύμμετρος: ἑκατέρα ἄρα τῶν ΖΗ, ΗΘ ἀσύμμετρός ἐστι τῇ Ε μήκει. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, μεῖζον ἄρα τὸ ἀπὸ τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ. ἔστω οὖν τῷ ἀπὸ τῆς ΖΗ ἴσα τὰ ἀπὸ τῶν ΗΘ, Κ: ἀναστρέψαντι ἄρα ὡς ὁ ΑΒ πρὸς ΒΓ, οὕτως τὸ ἀπὸ ΖΗ πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ὥστε οὐδὲ τὸ ἀπὸ ΖΗ πρὸς τὸ ἀπὸ τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Κ μήκει: ἡ ΖΗ ἄρα τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΖΗ, ΗΘ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα αὐτῶν σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ Ε. ἡ ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστὶν ἕκτη: ὅπερ ἔδει δεῖξαι. λῆμμα ἔστω δύο τετράγωνα τὰ ΑΒ, ΒΓ καὶ κείσθωσαν ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΔΒ τῇ ΒΕ: ἐπ᾽ εὐθείας ἄρα ἐστὶ καὶ ἡ ΖΒ τῇ ΒΗ. καὶ συμπεπληρώσθω τὸ ΑΓ παραλληλόγραμμον: λέγω, ὅτι τετράγωνόν ἐστι τὸ ΑΓ, καὶ ὅτι τῶν ΑΒ, ΒΓ μέσον ἀνάλογόν ἐστι τὸ ΔΗ, καὶ ἔτι τῶν ΑΓ, ΓΒ μέσον ἀνάλογόν ἐστι τὸ ΔΓ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ μὲν ΔΒ τῇ ΒΖ, ἡ δὲ ΒΕ τῇ ΒΗ, ὅλη ἄρα ἡ ΔΕ ὅλῃ τῇ ΖΗ ἐστιν ἴση. ἀλλ᾽ ἡ μὲν ΔΕ ἑκατέρᾳ τῶν ΑΘ, ΚΓ ἐστιν ἴση, ἡ δὲ ΖΗ ἑκατέρᾳ τῶν ΑΚ, ΘΓ ἐστιν ἴση: καὶ ἑκατέρα ἄρα τῶν ΑΘ, ΚΓ ἑκατέρᾳ τῶν ΑΚ, ΘΓ ἐστιν ἴση. ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΓ παραλληλόγραμμον: ἔστι δὲ καὶ ὀρθογώνιον: τετράγωνον ἄρα ἐστὶ τὸ ΑΓ. καὶ ἐπεί ἐστιν ὡς ἡ ΖΒ πρὸς τὴν ΒΗ, οὕτως ἡ ΔΒ πρὸς τὴν ΒΕ, ἀλλ᾽ ὡς μὲν ἡ ΖΒ πρὸς τὴν ΒΗ, οὕτως τὸ ΑΒ πρὸς τὸ ΔΗ, ὡς δὲ ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως τὸ ΔΗ πρὸς τὸ ΒΓ, καὶ ὡς ἄρα τὸ ΑΒ πρὸς τὸ ΔΗ, οὕτως τὸ ΔΗ πρὸς τὸ ΒΓ. τῶν ΑΒ, ΒΓ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΔΗ. λέγω δή, ὅτι καὶ τῶν ΑΓ, ΓΒ μέσον ἀνάλογόν ἐστι τὸ ΔΓ. ἐπεὶ γάρ ἐστιν ὡς ἡ ΑΔ πρὸς τὴν ΔΚ, οὕτως ἡ ΚΗ πρὸς τὴν ΗΓ: ἴση γάρ ἐστιν ἑκατέρα ἑκατέρᾳ: καὶ συνθέντι ὡς ἡ ΑΚ πρὸς ΚΔ, οὕτως ἡ ΚΓ πρὸς ΓΗ, ἀλλ᾽ ὡς μὲν ἡ ΑΚ πρὸς ΚΔ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΔ, ὡς δὲ ἡ ΚΓ πρὸς ΓΗ, οὕτως τὸ ΔΓ πρὸς ΓΒ, καὶ ὡς ἄρα τὸ ΑΓ πρὸς ΔΓ, οὕτως τὸ ΔΓ πρὸς τὸ ΒΓ. τῶν ΑΓ, ΓΒ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΔΓ: ἃ προέκειτο δεῖξαι.'"],["Rule","'GreekProofWordCount'",780]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",54]],["Association",["Rule","'VertexLabel'","'10.54'"],["Rule","'Text'","'If an area be contained by a rational straight line and the first binomial, the “side” of the area is the irrational straight line which is called binomial.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων πρώτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο ὀνομάτων.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",14]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the “side” of the area AC is the irrational straight line which is called binomial. For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term. It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1] Let ED be bisected at the point F. Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, therefore, if there be applied to the greater AE a parallelogram equal to the fourth part of the square on the less, that is, to the square on EF, and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let then the rectangle AG, GE equal to the square on EF be applied to AE; therefore AG is commensurable in length with EG. Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD; let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14] and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP. And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma] Now, since the rectangle AG, GE is equal to the square on EF, therefore, as AG is to EF, so is FE to EG; [VI. 17] therefore also, as AH is to EL, so is EL to KG; [VI. 1] therefore EL is a mean proportional between AH, GK. But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ. But MR is also a mean proportional between the same SN, NQ; [Lemma]therefore EL is equal to MR, so that it is also equal to PO. But AH, GK are also equal to SN, NQ; therefore the whole AC is equal to the whole SQ, that is, to the square on MO; therefore MO is the “side” of AC. I say next that MO is binomial. For, since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines AG, GE. [X. 15] But AE is also, by hypothesis, commensurable with AB; therefore AG, GE are also commensurable with AB. [X. 12] And AB is rational; therefore each of the straight lines AG, GE is also rational; therefore each of the rectangles AH, GK is rational, [X. 19] and AH is commensurable with GK. But AH is equal to SN, and GK to NQ; therefore SN, NQ, that is, the squares on MN, NO, are rational and commensurable. And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and DE is commensurable with EF, therefore AG is also incommensurable with EF, [X. 13] so that AH is also incommensurable with EL. [VI. 1, X. 11] But AH is equal to SN, and EL to MR; therefore SN is also incommensurable with MR. But, as SN is to MR, so is PN to NR; [VI. 1] therefore PN is incommensurable with NR. [X. 11] But PN is equal to MN, and NR to NO; therefore MN is incommensurable with NO. And the square on MN is commensurable with the square on NO, and each is rational; therefore MN, NO are rational straight lines commensurable in square only. Therefore MO is binomial [X. 36] and the “side” of AC.'"],["Rule","'ProofWordCount'",656],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων πρώτης τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο ὀνομάτων. ἐπεὶ γὰρ ἐκ δύο ὀνομάτων ἐστὶ πρώτη ἡ ΑΔ, διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, καὶ ἔστω τὸ μεῖζον ὄνομα τὸ ΑΕ. φανερὸν δή, ὅτι αἱ ΑΕ, ΕΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΑΕ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΒ μήκει. τετμήσθω δὴ ἡ ΕΔ δίχα κατὰ τὸ Ζ σημεῖον. καὶ ἐπεὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος, τουτέστι τῷ ἀπὸ τῆς ΕΖ, ἴσον παρὰ τὴν μείζονα τὴν ΑΕ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διαιρεῖ. παραβεβλήσθω οὖν παρὰ τὴν ΑΕ τῷ ἀπὸ τῆς ΕΖ ἴσον τὸ ὑπὸ ΑΗ, ΗΕ: σύμμετρος ἄρα ἐστὶν ἡ ΑΗ τῇ ΕΗ μήκει. καὶ ἤχθωσαν ἀπὸ τῶν Η, Ε, Ζ ὁποτέρᾳ τῶν ΑΒ, ΓΔ παράλληλοι αἱ ΗΘ, ΕΚ, ΖΛ: καὶ τῷ μὲν ΑΘ παραλληλογράμμῳ ἴσον τετράγωνον συνεστάτω τὸ ΣΝ, τῷ δὲ ΗΚ ἴσον τὸ ΝΠ, καὶ κείσθω ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΜΝ τῇ ΝΞ: ἐπ᾽ εὐθείας ἄρα ἐστὶ καὶ ἡ ΡΝ τῇ ΝΟ. καὶ συμπεπληρώσθω τὸ ΣΠ παραλληλόγραμμον: τετράγωνον ἄρα ἐστὶ τὸ ΣΠ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΗ, ΗΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΖ, ἔστιν ἄρα ὡς ἡ ΑΗ πρὸς ΕΖ, οὕτως ἡ ΖΕ πρὸς ΕΗ: καὶ ὡς ἄρα τὸ ΑΘ πρὸς ΕΛ, τὸ ΕΛ πρὸς ΚΗ: τῶν ΑΘ, ΗΚ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΕΛ. ἀλλὰ τὸ μὲν ΑΘ ἴσον ἐστὶ τῷ ΣΝ, τὸ δὲ ΗΚ ἴσον τῷ ΝΠ: τῶν ΣΝ, ΝΠ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΕΛ. ἔστι δὲ τῶν αὐτῶν τῶν ΣΝ, ΝΠ μέσον ἀνάλογον καὶ τὸ ΜΡ: ἴσον ἄρα ἐστὶ τὸ ΕΛ τῷ ΜΡ: ὥστε καὶ τῷ ΟΞ ἴσον ἐστίν. ἔστι δὲ καὶ τὰ ΑΘ, ΗΚ τοῖς ΣΝ, ΝΠ ἴσα: ὅλον ἄρα τὸ ΑΓ ἴσον ἐστὶν ὅλῳ τῷ ΣΠ, τουτέστι τῷ ἀπὸ τῆς ΜΞ τετραγώνῳ: τὸ ΑΓ ἄρα δύναται ἡ ΜΞ. λέγω, ὅτι ἡ ΜΞ ἐκ δύο ὀνομάτων ἐστίν. ἐπεὶ γὰρ σύμμετρός ἐστιν ἡ ΑΗ τῇ ΗΕ, σύμμετρός ἐστι καὶ ἡ ΑΕ ἑκατέρᾳ τῶν ΑΗ, ΗΕ. ὑπόκειται δὲ καὶ ἡ ΑΕ τῇ ΑΒ σύμμετρος: καὶ αἱ ΑΗ, ΗΕ ἄρα τῇ ΑΒ σύμμετροί εἰσιν. καί ἐστι ῥητὴ ἡ ΑΒ: ῥητὴ ἄρα ἐστὶ καὶ ἑκατέρα τῶν ΑΗ, ΗΕ: ῥητὸν ἄρα ἐστὶν ἑκάτερον τῶν ΑΘ, ΗΚ, καί ἐστι σύμμετρον τὸ ΑΘ τῷ ΗΚ. ἀλλὰ τὸ μὲν ΑΘ τῷ ΣΝ ἴσον ἐστίν, τὸ δὲ ΗΚ τῷ ΝΠ: καὶ τὰ ΣΝ, ΝΠ ἄρα, τουτέστι τὰ ἀπὸ τῶν ΜΝ, ΝΞ, ῥητά ἐστι καὶ σύμμετρα. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΕ τῇ ΕΔ μήκει, ἀλλ᾽ ἡ μὲν ΑΕ τῇ ΑΗ ἐστι σύμμετρος, ἡ δὲ ΔΕ τῇ ΕΖ σύμμετρος, ἀσύμμετρος ἄρα καὶ ἡ ΑΗ τῇ ΕΖ: ὥστε καὶ τὸ ΑΘ τῷ ΕΛ ἀσύμμετρόν ἐστιν. ἀλλὰ τὸ μὲν ΑΘ τῷ ΣΝ ἐστιν ἴσον, τὸ δὲ ΕΛ τῷ ΜΡ: καὶ τὸ ΣΝ ἄρα τῷ ΜΡ ἀσύμμετρόν ἐστιν. ἀλλ᾽ ὡς τὸ ΣΝ πρὸς ΜΡ, ἡ ΟΝ πρὸς τὴν ΝΡ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΟΝ τῇ ΝΡ. ἴση δὲ ἡ μὲν ΟΝ τῇ ΜΝ, ἡ δὲ ΝΡ τῇ ΝΞ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΜΝ τῇ ΝΞ. καί ἐστι τὸ ἀπὸ τῆς ΜΝ σύμμετρον τῷ ἀπὸ τῆς ΝΞ, καὶ ῥητὸν ἑκάτερον: αἱ ΜΝ, ΝΞ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. ἡ ΜΞ ἄρα ἐκ δύο ὀνομάτων ἐστὶ καὶ δύναται τὸ ΑΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",577]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",55]],["Association",["Rule","'VertexLabel'","'10.55'"],["Rule","'Text'","'If an area be contained by a rational straight line and the second binomial, the “side” of the area is the irrational straight line which is called a first bimedial.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων δευτέρας, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο μέσων πρώτη.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.2]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",37]]]],["Rule","'Proof'","'For let the area ABCD be contained by the rationalstraight line AB and the second binomial AD; I say that the “side” of the area AC is a first bimedial straight line. For, since AD is a second binomial straight line, let it be divided into its terms at E, so that AE is the greater term;therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and the lesser term ED is commensurable in length with AB. [X. Deff. II. 2] Let ED be bisected at F, and let there be applied to AE the rectangle AG, GE equal to the square on EF and deficient by a square figure; therefore AG is commensurable in length with GE. [X. 17] Through G, E, F let GH, EK, FL be drawn parallel toAB, CD, let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, and let them be placed so that MN is in a straight line with NO;therefore RN is also in a straight line with NP. Let the square SQ be completed. It is then manifest from what was proved before that MR is a mean proportional between SN, NQ and is equal to EL, and that MO is the “side” of the area AC. It is now to be proved that MO is a first bimedial straight line. Since AE is incommensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. [X. 13] And, since AG is commensurable with EG,AE is also commensurable with each of the straight lines AG, GE. [X. 15] But AE is incommensurable in length with AB; therefore AG, GE are also incommensurable with AB. [X. 13] Therefore BA, AG and BA, GE are pairs of rationalstraight lines commensurable in square only; so that each of the rectangles AH, GK is medial. [X. 21] Hence each of the squares SN, NQ is medial. Therefore MN, NO are also medial. And, since AG is commensurable in length with GE,AH is also commensurable with GK, [VI. 1. X. 11] that is, SN is commensurable with NQ, that is, the square on MN with the square on NO. And, since AE is incommensurable in length with ED, while AE is commensurable with AG,and ED is commensurable with EF, therefore AG is incommensurable with EF; [X. 13] so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, [VI. 1, X. 11]that is, MN is incommensurable in length with NO. But MN, NO were proved to be both medial and commensurable in square; therefore MN, NO are medial straight lines commensurable in square only. I say next that they also contain a rational rectangle. For, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF, therefore EF is also commensurable with EK. [X. 12] And each of them is rational;therefore EL, that is, MR is rational, [X. 19] and MR is the rectangle MN, NO. But, if two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational and is called a first bimedial straight line. [X. 37] Therefore MO is a first bimedial straight line.'"],["Rule","'ProofWordCount'",574],["Rule","'GreekProof'","'περιεχέσθω γὰρ χωρίον τὸ ΑΒΓΔ ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων δευτέρας τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐκ δύο μέσων πρώτη ἐστίν. ἐπεὶ γὰρ ἐκ δύο ὀνομάτων δευτέρα ἐστὶν ἡ ΑΔ, διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ: αἱ ΑΕ, ΕΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ τὸ ἔλαττον ὄνομα ἡ ΕΔ σύμμετρόν ἐστι τῇ ΑΒ μήκει. τετμήσθω ἡ ΕΔ δίχα κατὰ τὸ Ζ, καὶ τῷ ἀπὸ τῆς ΕΖ ἴσον παρὰ τὴν ΑΕ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ τὸ ὑπὸ τῶν ΑΗΕ: σύμμετρος ἄρα ἡ ΑΗ τῇ ΗΕ μήκει. καὶ διὰ τῶν Η, Ε, Ζ παράλληλοι ἤχθωσαν ταῖς ΑΒ, ΓΔ αἱ ΗΘ, ΕΚ, ΖΛ, καὶ τῷ μὲν ΑΘ παραλληλογράμμῳ ἴσον τετράγωνον συνεστάτω τὸ ΣΝ, τῷ δὲ ΗΚ ἴσον τετράγωνον τὸ ΝΠ, καὶ κείσθω ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΜΝ τῇ ΝΞ: ἐπ᾽ εὐθείας ἄρα ἐστὶ καὶ ἡ ΡΝ τῇ ΝΟ. καὶ συμπεπληρώσθω τὸ ΣΠ τετράγωνον: φανερὸν δὴ ἐκ τοῦ προδεδειγμένου, ὅτι τὸ ΜΡ μέσον ἀνάλογόν ἐστι τῶν ΣΝ, ΝΠ, καὶ ἴσον τῷ ΕΛ, καὶ ὅτι τὸ ΑΓ χωρίον δύναται ἡ ΜΞ. δεικτέον δή, ὅτι ἡ ΜΞ ἐκ δύο μέσων ἐστὶ πρώτη. ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΕ τῇ ΕΔ μήκει, σύμμετρος δὲ ἡ ΕΔ τῇ ΑΒ, ἀσύμμετρος ἄρα ἡ ΑΕ τῇ ΑΒ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΗ τῇ ΕΗ, σύμμετρός ἐστι καὶ ἡ ΑΕ ἑκατέρᾳ τῶν ΑΗ, ΗΕ. ἀλλὰ ἡ ΑΕ ἀσύμμετρος τῇ ΑΒ μήκει: καὶ αἱ ΑΗ, ΗΕ ἄρα ἀσύμμετροί εἰσι τῇ ΑΒ. αἱ ΒΑ, ΑΗ, ΗΕ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ὥστε μέσον ἐστὶν ἑκάτερον τῶν ΑΘ, ΗΚ. ὥστε καὶ ἑκάτερον τῶν ΣΝ, ΝΠ μέσον ἐστίν. καὶ αἱ ΜΝ, ΝΞ ἄρα μέσαι εἰσίν. καὶ ἐπεὶ σύμμετρος ἡ ΑΗ τῇ ΗΕ μήκει, σύμμετρόν ἐστι καὶ τὸ ΑΘ τῷ ΗΚ, τουτέστι τὸ ΣΝ τῷ ΝΠ, τουτέστι τὸ ἀπὸ τῆς ΜΝ τῷ ἀπὸ τῆς ΝΞ ὥστε δυνάμει εἰσὶ σύμμετροι αἱ ΜΝ, ΝΞ. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΕ τῇ ΕΔ μήκει, ἀλλ᾽ ἡ μὲν ΑΕ σύμμετρός ἐστι τῇ ΑΗ, ἡ δὲ ΕΔ τῇ ΕΖ σύμμετρος, ἀσύμμετρος ἄρα ἡ ΑΗ τῇ ΕΖ: ὥστε καὶ τὸ ΑΘ τῷ ΕΛ ἀσύμμετρόν ἐστιν, τουτέστι τὸ ΣΝ τῷ ΜΡ, τουτέστιν ἡ ΟΝ τῇ ΝΡ, τουτέστιν ἡ ΜΝ τῇ ΝΞ ἀσύμμετρός ἐστι μήκει. ἐδείχθησαν δὲ αἱ ΜΝ, ΝΞ καὶ μέσαι οὖσαι καὶ δυνάμει σύμμετροι: αἱ ΜΝ, ΝΞ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι. λέγω δή, ὅτι καὶ ῥητὸν περιέχουσιν. ἐπεὶ γὰρ ἡ ΔΕ ὑπόκειται ἑκατέρᾳ τῶν ΑΒ, ΕΖ σύμμετρος, σύμμετρος ἄρα καὶ ἡ ΕΖ τῇ ΕΚ. καὶ ῥητὴ ἑκατέρα αὐτῶν: ῥητὸν ἄρα τὸ ΕΛ, τουτέστι τὸ ΜΡ: τὸ δὲ ΜΡ ἐστι τὸ ὑπὸ τῶν ΜΝΞ. ἐὰν δὲ δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι ῥητὸν περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλεῖται δὲ ἐκ δύο μέσων πρώτη. ἡ ἄρα ΜΞ ἐκ δύο μέσων ἐστὶ πρώτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",478]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",56]],["Association",["Rule","'VertexLabel'","'10.56'"],["Rule","'Text'","'If an area be contained by a rational straight line and the third binomial, the “side” of the area is the irrational straight line called a second bimedial.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων τρίτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο μέσων δευτέρα.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",2.3]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",38]]]],["Rule","'Proof'","'For let the area ABCD be contained by the rational straight line AB and the third binomial AD divided into its terms at E, of which terms AE is the greater; I say that the “side” of the area AC is the irrational straight line called a second bimedial. For let the same construction be made as before. Now, since AD is a third binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and neither of the terms AE, ED is commensurable in length with AB. [X. Deff. II. 3] Then, in manner similar to the foregoing, we shall prove that MO is the “side” of the area AC, and MN, NO are medial straight lines commensurable in square only; so that MO is bimedial. It is next to be proved that it is also a second bimedial straight line. Since DE is incommensurable in length with AB, that is, with EK, and DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13] And they are rational; therefore FE, EK are rational straight lines commensurable in square only. Therefore EL, that is, MR, is medial. [X. 21] And it is contained by MN, NO; therefore the rectangle MN, NO is medial. Therefore MO is a second bimedial straight line. [X. 38]'"],["Rule","'ProofWordCount'",242],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΒΓΔ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων τρίτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὧν μεῖζόν ἐστι τὸ ΑΕ: λέγω, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο μέσων δευτέρα. Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον. καὶ ἐπεὶ ἐκ δύο ὀνομάτων ἐστὶ τρίτη ἡ ΑΔ, αἱ ΑΕ, ΕΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ οὐδετέρα τῶν ΑΕ, ΕΔ σύμμετρός ἐστι τῇ ΑΒ μήκει. ὁμοίως δὴ τοῖς προδεδειγμένοις δείξομεν, ὅτι ἡ ΜΞ ἐστιν ἡ τὸ ΑΓ χωρίον δυναμένη, καὶ αἱ ΜΝ, ΝΞ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι: ὥστε ἡ ΜΞ ἐκ δύο μέσων ἐστίν. δεικτέον δή, ὅτι καὶ δευτέρα. Καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, σύμμετρος δὲ ἡ ΔΕ τῇ ΕΖ, ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΕΚ μήκει. καί εἰσι ῥηταί: αἱ ΖΕ, ΕΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. μέσον ἄρα ἐστὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ: καὶ περιέχεται ὑπὸ τῶν ΜΝΞ: μέσον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΜΝΞ. ἡ ΜΞ ἄρα ἐκ δύο μέσων ἐστὶ δευτέρα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",193]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",57]],["Association",["Rule","'VertexLabel'","'10.57'"],["Rule","'Text'","'If an area be contained by a rational straight line and the fourth binomial, the “side” of the area is the irrational straight line called major.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων τετάρτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη μείζων.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",39]]]],["Rule","'Proof'","'For let the area AC be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let AE be the greater; I say that the “side” of the area AC is the irrational straight line called major. For, since AD is a fourth binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line incommensurable with AE, and AE is commensurable in length with AB. [X. Deff. II. 4] Let DE be bisected at F, and let there be applied to AE a parallelogram, the rectangle AG, GE, equal to the square on EF; therefore AG is incommensurable in length with GE. [X. 18] Let GH, EK, FL be drawn parallel to AB, and let the rest of the construction be as before; it is then manifest that MO is the “side” of the area AC. It is next to be proved that MO is the irrational straight line called major. Since AG is incommensurable with EG, AH is also incommensurable with GK, that is, SN with NQ; [VI. 1, X. 11] therefore MN, NO are incommensurable in square. And, since AE is commensurable with AB, AK is rational; [X. 19] and it is equal to the squares on MN, NO; therefore the sum of the squares on MN, NO is also rational. And, since DE is incommensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13] Therefore EK, EF are rational straight lines commensurable in square only; therefore LE, that is, MR, is medial. [X. 21] And it is contained by MN, NO; therefore the rectangle MN, NO is medial. And the [sum]of the squares on MN, NO is rational, and MN, NO are incommensurable in square. But, if two straight lines incommensurable in square and making the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole is irrational and is called major. [X. 39] Therefore MO is the irrational straight line called major and is the “side” of the area AC.'"],["Rule","'ProofWordCount'",381],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων τετάρτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὧν μεῖζον ἔστω τὸ ΑΕ: λέγω, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη μείζων. ἐπεὶ γὰρ ἡ ΑΔ ἐκ δύο ὀνομάτων ἐστὶ τετάρτη, αἱ ΑΕ, ΕΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΑΕ τῇ ΑΒ σύμμετρός ἐστι μήκει. τετμήσθω ἡ ΔΕ δίχα κατὰ τὸ Ζ, καὶ τῷ ἀπὸ τῆς ΕΖ ἴσον παρὰ τὴν ΑΕ παραβεβλήσθω παραλληλόγραμμον τὸ ὑπὸ ΑΗ, ΗΕ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΗ τῇ ΗΕ μήκει. ἤχθωσαν παράλληλοι τῇ ΑΒ αἱ ΗΘ, ΕΚ, ΖΛ, καὶ τὰ λοιπὰ τὰ αὐτὰ τοῖς πρὸ τούτου γεγονέτω: φανερὸν δή, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐστὶν ἡ ΜΞ. δεικτέον δή, ὅτι ἡ ΜΞ ἄλογός ἐστιν ἡ καλουμένη μείζων. ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΕΗ μήκει, ἀσύμμετρόν ἐστι καὶ τὸ ΑΘ τῷ ΗΚ, τουτέστι τὸ ΣΝ τῷ ΝΠ: αἱ ΜΝ, ΝΞ ἄρα δυνάμει εἰσὶν ἀσύμμετροι. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΕ τῇ ΑΒ μήκει, ῥητόν ἐστι τὸ ΑΚ: καί ἐστιν ἴσον τοῖς ἀπὸ τῶν ΜΝ, ΝΞ: ῥητὸν ἄρα ἐστὶ καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, ἀλλὰ ἡ ΔΕ σύμμετρός ἐστι τῇ ΕΖ, ἀσύμμετρος ἄρα ἡ ΕΖ τῇ ΕΚ μήκει. αἱ ΕΚ, ΕΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: μέσον ἄρα τὸ ΛΕ, τουτέστι τὸ ΜΡ. καὶ περιέχεται ὑπὸ τῶν ΜΝ, ΝΞ: μέσον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΜΝ, ΝΞ. καὶ ῥητὸν τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ, καί εἰσιν ἀσύμμετροι αἱ ΜΝ, ΝΞ δυνάμει. ἐὰν δὲ δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δ᾽ ὑπ᾽ αὐτῶν μέσον, ἡ ὅλη ἄλογός ἐστιν, καλεῖται δὲ μείζων. ἡ ΜΞ ἄρα ἄλογός ἐστιν ἡ καλουμένη μείζων, καὶ δύναται τὸ ΑΓ χωρίον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",327]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",58]],["Association",["Rule","'VertexLabel'","'10.58'"],["Rule","'Text'","'If an area be contained by a rational straight line and the fifth binomial, the “side” of the area is the irrational straight line called the side of a rational plus a medial area.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων πέμπτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ῥητὸν καὶ μέσον δυναμένη.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",40]]]],["Rule","'Proof'","'For let the area AC be contained by the rational straight line AB and the fifth binomial AD divided into its terms at E, so that AE is the greater term; I say that the “side” of the area AC is the irrational straight line called the side of a rational plus a medial area. For let the same construction be made as before shown; it is then manifest that MO is the “side” of the area AC. It is then to be proved that MO is the side of a rational plus a medial area. For, since AG is incommensurable with GE, [X. 18] therefore AH is also commensurable with HE, [VI. 1, X. 11] that is, the square on MN with the square on NO; therefore MN, NO are incommensurable in square. And, since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. [X. Deff. II. 5] But AE is incommensurable with ED; therefore AB is also incommensurable in length with AE. [X. 13] Therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21] And, since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. [X. 12] And EK is rational; therefore EL, that is, MR, that is, the rectangle MN, NO, is also rational. [X. 19] Therefore MN, NO are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. Therefore MO is the side of a rational plus a medial area [X. 40] and is the “side” of the area AC.'"],["Rule","'ProofWordCount'",289],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων πέμπτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ: λέγω δή, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ῥητὸν καὶ μέσον δυναμένη. Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον δεδειγμένοις: φανερὸν δή, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐστὶν ἡ ΜΞ. δεικτέον δή, ὅτι ἡ ΜΞ ἐστιν ἡ ῥητὸν καὶ μέσον δυναμένη. ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΗΕ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΑΘ τῷ ΘΕ, τουτέστι τὸ ἀπὸ τῆς ΜΝ τῷ ἀπὸ τῆς ΝΞ: αἱ ΜΝ, ΝΞ ἄρα δυνάμει εἰσὶν ἀσύμμετροι. καὶ ἐπεὶ ἡ ΑΔ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη, καί ἐστιν ἔλασσον αὐτῆς τμῆμα τὸ ΕΔ, σύμμετρος ἄρα ἡ ΕΔ τῇ ΑΒ μήκει. ἀλλὰ ἡ ΑΕ τῇ ΕΔ ἐστιν ἀσύμμετρος: καὶ ἡ ΑΒ ἄρα τῇ ΑΕ ἐστιν ἀσύμμετρος μήκει. αἱ ΒΑ, ΑΕ ῥηταί εἰσι δυνάμει μόνον σύμμετροι. μέσον ἄρα ἐστὶ τὸ ΑΚ, τουτέστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, ἀλλὰ ἡ ΔΕ τῇ ΕΖ σύμμετρός ἐστιν, καὶ ἡ ΕΖ ἄρα τῇ ΕΚ σύμμετρός ἐστιν. καὶ ῥητὴ ἡ ΕΚ: ῥητὸν ἄρα καὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ, τουτέστι τὸ ὑπὸ ΜΝΞ: αἱ ΜΝ, ΝΞ ἄρα δυνάμει ἀσύμμετροί εἰσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δ᾽ ὑπ᾽ αὐτῶν ῥητόν. ἡ ΜΞ ἄρα ῥητὸν καὶ μέσον δυναμένη ἐστὶ καὶ δύναται τὸ ΑΓ χωρίον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",253]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",59]],["Association",["Rule","'VertexLabel'","'10.59'"],["Rule","'Text'","'If an area be contained by a rational straight line and the sixth binomial, the “side” of the area is the irrational straight line called the side of the sum of two medial areas.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων ἕκτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη δύο μέσα δυναμένη.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",41]]]],["Rule","'Proof'","'For let the area ABCD be contained by the rational straight line AB and the sixth binomial AD, divided into its terms at E, so that AE is the greater term; I say that the “side” of AC is the side of the sum of two medial areas. Let the same construction be made as before shown. It is then manifest that MO is the “side” of AC, and that MN is incommensurable in square with NO. Now, since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only; therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21] Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK; [X. 13] therefore FE, EK are rational straight lines commensurable in square only; therefore EL, that is, MR, that is, the rectangle MN, NO, is medial. [X. 21] And, since AE is incommensurable with EF, AK is also incommensurable with EL. [VI. 1, X. 11] But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO; therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO. And each of them is medial, and MN, NO are incommensurable in square. Therefore MO is the side of the sum of two medial areas [X. 41], and is the “side” of AC.'"],["Rule","'ProofWordCount'",241],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΒΓΔ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων ἕκτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ: λέγω, ὅτι ἡ τὸ ΑΓ δυναμένη ἡ δύο μέσα δυναμένη ἐστίν. Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς προδεδειγμένοις. φανερὸν δή, ὅτι ἡ τὸ ΑΓ δυναμένη ἐστὶν ἡ ΜΞ, καὶ ὅτι ἀσύμμετρός ἐστι ἡ ΜΝ τῇ ΝΞ δυνάμει. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΕΑ τῇ ΑΒ μήκει, αἱ ΕΑ, ΑΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: μέσον ἄρα ἐστὶ τὸ ΑΚ, τουτέστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ. πάλιν, ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΕΔ τῇ ΑΒ μήκει, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΖΕ τῇ ΕΚ: αἱ ΖΕ, ΕΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: μέσον ἄρα ἐστὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ, τουτέστι τὸ ὑπὸ τῶν ΜΝΞ. καὶ ἐπεὶ ἀσύμμετρος ἡ ΑΕ τῇ ΕΖ, καὶ τὸ ΑΚ τῷ ΕΛ ἀσύμμετρόν ἐστιν. ἀλλὰ τὸ μὲν ΑΚ ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ, τὸ δὲ ΕΛ ἐστι τὸ ὑπὸ τῶν ΜΝΞ: ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝΞ τῷ ὑπὸ τῶν ΜΝΞ. καί ἐστι μέσον ἑκάτερον αὐτῶν, καὶ αἱ ΜΝ, ΝΞ δυνάμει εἰσὶν ἀσύμμετροι. ἡ ΜΞ ἄρα δύο μέσα δυναμένη ἐστὶ καὶ δύναται τὸ ΑΓ: ὅπερ ἔδει δεῖξαι. λῆμμα ἐὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἄνισα, τὰ ἀπὸ τῶν ἀνίσων τετράγωνα μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ἀνίσων περιεχομένου ὀρθογωνίου. ἔστω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω εἰς ἄνισα κατὰ τὸ Γ, καὶ ἔστω μείζων ἡ ΑΓ: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. τετμήσθω γὰρ ἡ ΑΒ δίχα κατὰ τὸ Δ. ἐπεὶ οὖν εὐθεῖα γραμμὴ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Δ, εἰς δὲ ἄνισα κατὰ τὸ Γ, τὸ ἄρα ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ ΓΔ ἴσον ἐστὶ τῷ ἀπὸ ΑΔ: ὥστε τὸ ὑπὸ τῶν ΑΓ, ΓΒ ἔλαττόν ἐστι τοῦ ἀπὸ ΑΔ: τὸ ἄρα δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἔλαττον á¼¢ διπλάσιόν ἐστι τοῦ ἀπὸ ΑΔ. ἀλλὰ τὰ ἀπὸ τῶν ΑΓ, ΓΒ διπλάσιά [ἐστι] τῶν ἀπὸ τῶν ΑΔ, ΔΓ: τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",360]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",60]],["Association",["Rule","'VertexLabel'","'10.60'"],["Rule","'Text'","'The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'τὸ ἀπὸ τῆς ἐκ δύο ὀνομάτων παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πρώτην.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",53]]]],["Rule","'Proof'","'Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term; let a rational straight line DE be set out, and let DEFG equal to the square on AB be applied to DE producing DG as its breadth; I say that DG is a first binomial straight line. For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC; therefore the remainder, twice the rectangle AC, CB, is equal to MF. Let MG be bisected at N, and let NO be drawn parallel [to ML or GF]. Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB. Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [X. 36] therefore the squares on AC, CB are rational and commensurable with one another, so that the sum of the squares on AC, CB is also rational. [X. 15] And it is equal to DL; therefore DL is rational. And it is applied to the rational straight line DE; therefore DM is rational and commensurable in length with DE. [X. 20] Again, since AC, CB are rational straight lines commensurable in square only, therefore twice the rectangle AC, CB, that is MF, is medial. [X. 21] And it is applied to the rational straight line ML; therefore MG is also rational and incommensurable in length with ML, that is, DE. [X. 22] But MD is also rational and is commensurable in length with DE; therefore DM is incommensurable in length with MG. [X. 13] And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] It is next to be proved that it is also a first binomial straight line. Since the rectangle AC, CB is a mean proportional between the squares on AC, CB, [cf. Lemma after X. 53] therefore MO is also a mean proportional between DH, KL. Therefore, as DH is to MO, so is MO to KL, that is, as DK is to MN, so is MN to MK; [VI. 1] therefore the rectangle DK, KM is equal to the square on MN. [VI. 17] And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11] And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [Lemma] therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1] And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM. But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [X. 17] therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. And DM, MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out. Therefore DG is a first binomial straight line. [X. Deff. II. 1]'"],["Rule","'ProofWordCount'",606],["Rule","'GreekProof'","'ἔστω ἐκ δύο ὀνομάτων ἡ ΑΒ διῃρημένη εἰς τὰ ὀνόματα κατὰ τὸ Γ, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΕΖΗ πλάτος ποιοῦν τὴν ΔΗ: λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ πρώτη. παραβεβλήσθω γὰρ παρὰ τὴν ΔΕ τῷ μὲν ἀπὸ τῆς ΑΓ ἴσον τὸ ΔΘ, τῷ δὲ ἀπὸ τῆς ΒΓ ἴσον τὸ ΚΛ: λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον ἐστὶ τῷ ΜΖ. τετμήσθω ἡ ΜΗ δίχα κατὰ τὸ Ν, καὶ παράλληλος ἤχθω ἡ ΝΞ ἑκατέρᾳ τῶν ΜΛ, ΗΖ. ἑκάτερον ἄρα τῶν ΜΞ, ΝΖ ἴσον ἐστὶ τῷ ἅπαξ ὑπὸ τῶν ΑΓΒ. καὶ ἐπεὶ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ διῃρημένη εἰς τὰ ὀνόματα κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΒ ῥητά ἐστι καὶ σύμμετρα ἀλλήλοις: ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ σύμμετρόν ἐστι τοῖς ἀπὸ τῶν ΑΓ, ΓΒ: ῥητὸν ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ. καί ἐστιν ἴσον τῷ ΔΛ: ῥητὸν ἄρα ἐστὶ τὸ ΔΛ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται: ῥητὴ ἄρα ἐστὶν ἡ ΔΜ καὶ σύμμετρος τῇ ΔΕ μήκει. πάλιν, ἐπεὶ αἱ ΑΓ, ΓΒ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, μέσον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τουτέστι τὸ ΜΖ. καὶ παρὰ ῥητὴν τὴν ΜΛ παράκειται: ῥητὴ ἄρα καὶ ἡ ΜΗ ἐστι καὶ ἀσύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ, μήκει. ἔστι δὲ καὶ ἡ ΜΔ ῥητὴ καὶ τῇ ΔΕ μήκει σύμμετρος: ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΜ τῇ ΜΗ μήκει. καί εἰσι ῥηταί: αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον δή, ὅτι καὶ πρώτη. ἐπεὶ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΓΒ, καὶ τῶν ΔΘ, ΚΛ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΜΞ. ἔστιν ἄρα ὡς τὸ ΔΘ πρὸς τὸ ΜΞ, οὕτως τὸ ΜΞ πρὸς τὸ ΚΛ, τουτέστιν ὡς ἡ ΔΚ πρὸς τὴν ΜΝ, ἡ ΜΝ πρὸς τὴν ΜΚ: τὸ ἄρα ὑπὸ τῶν ΔΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς ΓΒ, σύμμετρόν ἐστι καὶ τὸ ΔΘ τῷ ΚΛ: ὥστε καὶ ἡ ΔΚ τῇ ΚΜ σύμμετρός ἐστιν. καὶ ἐπεὶ μείζονά ἐστι τὰ ἀπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, μεῖζον ἄρα καὶ τὸ ΔΛ τοῦ ΜΖ: ὥστε καὶ ἡ ΔΜ τῆς ΜΗ μείζων ἐστίν. καί ἐστιν ἴσον τὸ ὑπὸ τῶν ΔΚ, ΚΜ τῷ ἀπὸ τῆς ΜΝ, τουτέστι τῷ τετάρτῳ τοῦ ἀπὸ τῆς ΜΗ, καὶ σύμμετρος ἡ ΔΚ τῇ ΚΜ. ἐὰν δὲ ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς σύμμετρα αὐτὴν διαιρῇ, ἡ μείζων τῆς ἐλάσσονος μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ: ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί εἰσι ῥηταὶ αἱ ΔΜ, ΜΗ, καὶ ἡ ΔΜ μεῖζον ὄνομα σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πρώτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",506]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",61]],["Association",["Rule","'VertexLabel'","'10.61'"],["Rule","'Text'","'The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'τὸ ἀπὸ τῆς ἐκ δύο μέσων πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων δευτέραν.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.2]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",37]]]],["Rule","'Proof'","'Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a second binominal straight line. For let the same construction as before be made. Then, since AB is a first bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [X. 37] so that the squares on AC, CB are also medial. [X. 21] Therefore DL is medial. [X. 15 and 23] And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE. [X. 22] Again, since twice the rectangle AC, CB is rational, MF is also rational. And it is applied to the rational straight line ML; therefore MG is also rational and commensurable in length with ML, that is, DE; [X. 20] therefore DM is incommensurable in length with MG. [X. 13] And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] It is next to be proved that it is also a second binomial straight line. For, since the squares on AC, CB are greater than twice the rectangle AC, CB, therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1] And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11] And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. [X. 17] And MG is commensurable is length with DE. Therefore DG is a second binomial straight line. [X. Deff. II. 2]'"],["Rule","'ProofWordCount'",345],["Rule","'GreekProof'","'ἔστω ἐκ δύο μέσων πρώτη ἡ ΑΒ διῃρημένη εἰς τὰς μέσας κατὰ τὸ Γ, ὧν μείζων ἡ ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ παραβεβλήσθω παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραλληλόγραμμον τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ: λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ δευτέρα. Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρὸ τούτου. καὶ ἐπεὶ ἡ ΑΒ ἐκ δύο μέσων ἐστὶ πρώτη διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι: ὥστε καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μέσα ἐστίν. μέσον ἄρα ἐστὶ τὸ ΔΛ. καὶ παρὰ ῥητὴν τὴν ΔΕ παραβέβληται: ῥητὴ ἄρα ἐστίν ἡ ΜΔ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. πάλιν, ἐπεὶ ῥητόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ῥητόν ἐστι καὶ τὸ ΜΖ. καὶ παρὰ ῥητὴν τὴν ΜΛ παράκειται: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΜΗ καὶ μήκει σύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΜ τῇ ΜΗ μήκει. καί εἰσι ῥηταί: αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον δή, ὅτι καὶ δευτέρα. ἐπεὶ γὰρ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, μεῖζον ἄρα καὶ τὸ ΔΛ τοῦ ΜΖ: ὥστε καὶ ἡ ΔΜ τῆς ΜΗ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς ΓΒ, σύμμετρόν ἐστι καὶ τὸ ΔΘ τῷ ΚΛ: ὥστε καὶ ἡ ΔΚ τῇ ΚΜ σύμμετρός ἐστιν. καί ἐστι τὸ ὑπὸ τῶν ΔΚΜ ἴσον τῷ ἀπὸ τῆς ΜΝ: ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί ἐστιν ἡ ΜΗ σύμμετρος τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ δευτέρα.'"],["Rule","'GreekProofWordCount'",270]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",62]],["Association",["Rule","'VertexLabel'","'10.62'"],["Rule","'Text'","'The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'τὸ ἀπὸ τῆς ἐκ δύο μέσων δευτέρας παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τρίτην.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.3]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",38]]]],["Rule","'Proof'","'Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment; let DE be any rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a third binomial straight line. Let the same construction be made as before shown. Then, since AB is a second bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only and containing a medial rectangle, [X. 38] so that the sum of the squares on AC, CB is also medial. [X. 15 and 23 Por.] And it is equal to DL; therefore DL is also medial. And it is applied to the rational straight line DE; therefore MD is also rational and incommensurable in length with DE. [X. 22] For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. And, since AC is incommensurable in length with CB, and, as AC is to CB, so is the square on AC to the rectangle AC, CB, therefore the square on AC is also incommensurable with the rectangle AC, CB. [X. 11] Hence the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, [X. 12, 13] that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [VI. 1, X. 11] And they are rational; therefore DG is binomial. [X. 36] It is to be proved that it is also a third binomial straight line. In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM. And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. And neither of the straight lines DM, MG is commensurable in length with DE. Therefore DG is a third binomial straight line. [X. Deff. II. 3]'"],["Rule","'ProofWordCount'",369],["Rule","'GreekProof'","'ἔστω ἐκ δύο μέσων δευτέρα ἡ ΑΒ διῃρημένη εἰς τὰς μέσας κατὰ τὸ Γ, ὥστε τὸ μεῖζον τμῆμα εἶναι τὸ ΑΓ, ῥητὴ δέ τις ἔστω ἡ ΔΕ, καὶ παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραλληλόγραμμον παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ: λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ τρίτη. Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις. καὶ ἐπεὶ ἐκ δύο μέσων δευτέρα ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι: ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον ἐστίν. καί ἐστιν ἴσον τῷ ΔΛ: μέσον ἄρα καὶ τὸ ΔΛ. καὶ παράκειται παρὰ ῥητὴν τὴν ΔΕ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΜΔ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΜΗ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ, μήκει: ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΜ, ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΓ τῇ ΓΒ μήκει, ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ὑπὸ τῶν ΑΓΒ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΑΓ τῷ ὑπὸ τῶν ΑΓΒ. ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓΒ ἀσύμμετρόν ἐστιν, τουτέστι τὸ ΔΛ τῷ ΜΖ: ὥστε καὶ ἡ ΔΜ τῇ ΜΗ ἀσύμμετρός ἐστιν. καί εἰσι ῥηταί: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον δή, ὅτι καὶ τρίτη. ὁμοίως δὴ τοῖς προτέροις ἐπιλογιούμεθα, ὅτι μείζων ἐστὶν ἡ ΔΜ τῆς ΜΗ, καὶ σύμμετρος ἡ ΔΚ τῇ ΚΜ. καί ἐστι τὸ ὑπὸ τῶν ΔΚΜ ἴσον τῷ ἀπὸ τῆς ΜΝ: ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ οὐδετέρα τῶν ΔΜ, ΜΗ σύμμετρός ἐστι τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",293]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",63]],["Association",["Rule","'VertexLabel'","'10.63'"],["Rule","'Text'","'The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'τὸ ἀπὸ τῆς μείζονος παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τετάρτην.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",39]]]],["Rule","'Proof'","'Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a fourth binomial straight line. Let the same construction be made as before shown. Then, since AB is a major straight line divided at C, AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39] Since then the sum of the squares on AC, CB is rational, therefore DL is rational; therefore DM is also rational and commensurable in length with DE. [X. 20] Again, since twice the rectangle AC, CB, that is, MF, is medial, and it is applied to the rational straight line ML, therefore MG is also rational and incommensurable in length with DE; [X. 22] therefore DM is also incommensurable in length with MG. [X. 13] Therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] It is to be proved that it is also a fourth binomial straight line. In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK, KM is equal to the square on MN. Since then the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM. [VI. 1, X. 11] But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [X. 18] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. And DM, MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out. Therefore DG is a fourth binomial straight line. [X. Deff. II. 4]'"],["Rule","'ProofWordCount'",398],["Rule","'GreekProof'","'ἔστω μείζων ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε μείζονα εἶναι τὴν ΑΓ τῆς ΓΒ, ῥητὴ δὲ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ παραλληλόγραμμον πλάτος ποιοῦν τὴν ΔΗ: λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ τετάρτη. Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις. καὶ ἐπεὶ μείζων ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δὲ ὑπ᾽ αὐτῶν μέσον. ἐπεὶ οὖν ῥητόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, ῥητὸν ἄρα ἐστὶ τὸ ΔΛ: ῥητὴ ἄρα καὶ ἡ ΔΜ καὶ σύμμετρος τῇ ΔΕ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τουτέστι τὸ ΜΖ, καὶ παρὰ ῥητήν ἐστι τὴν ΜΛ, ῥητὴ ἄρα ἐστὶ καὶ ἡ ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει: ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΜ τῇ ΜΗ μήκει. αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον δή, ὅτι καὶ τετάρτη. ὁμοίως δὴ δείξομεν τοῖς πρότερον, ὅτι μείζων ἐστὶν ἡ ΔΜ τῆς ΜΗ, καὶ ὅτι τὸ ὑπὸ ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ. ἐπεὶ οὖν ἀσύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς ΓΒ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΔΘ τῷ ΚΛ: ὥστε ἀσύμμετρος καὶ ἡ ΔΚ τῇ ΚΜ ἐστιν. ἐὰν δὲ ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παραλληλόγραμμον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς ἀσύμμετρα αὐτὴν διαιρῇ, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει: ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΔΜ, ΜΗ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ ἡ ΔΜ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τετάρτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",301]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",64]],["Association",["Rule","'VertexLabel'","'10.64'"],["Rule","'Text'","'The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'τὸ ἀπὸ τῆς ῥητὸν καὶ μέσον δυναμένης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πέμπτην.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",40]]]],["Rule","'Proof'","'Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a fifth binomial straight line. Let the same construction as before be made. Since then AB is the side of a rational plus a medial area, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40] Since then the sum of the squares on AC, CB is medial, therefore DL is medial, so that DM is rational and incommensurable in length with DE. [X. 22] Again, since twice the rectangle AC, CB, that is MF, is rational, therefore MG is rational and commensurable with DE. [X. 20] Therefore DM is incommensurable with MG; [X. 13] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] I say next that it is also a fifth binomial straight line. For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. [X. 18] And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE. Therefore DG is a fifth binomial.'"],["Rule","'ProofWordCount'",280],["Rule","'GreekProof'","'ἔστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ διῃρημένη εἰς τὰς εὐθείας κατὰ τὸ Γ, ὥστε μείζονα εἶναι τὴν ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ: λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη. Κατεσκευάσθω τὰ αὐτὰ τοῖς πρὸ τούτου. ἐπεὶ οὖν ῥητὸν καὶ μέσον δυναμένη ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δ᾽ ὑπ᾽ αὐτῶν ῥητόν. ἐπεὶ οὖν μέσον ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, μέσον ἄρα ἐστὶ τὸ ΔΛ: ὥστε ῥητή ἐστιν ἡ ΔΜ καὶ μήκει ἀσύμμετρος τῇ ΔΕ. πάλιν, ἐπεὶ ῥητόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓΒ, τουτέστι τὸ ΜΖ, ῥητὴ ἄρα ἡ ΜΗ καὶ σύμμετρος τῇ ΔΕ. ἀσύμμετρος ἄρα ἡ ΔΜ τῇ ΜΗ: αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. λέγω δή, ὅτι καὶ πέμπτη. ὁμοίως γὰρ δειχθήσεται, ὅτι τὸ ὑπὸ τῶν ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, καὶ ἀσύμμετρος ἡ ΔΚ τῇ ΚΜ μήκει: ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΔΜ, ΜΗ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ ἡ ἐλάσσων ἡ ΜΗ σύμμετρος τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πέμπτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",228]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",65]],["Association",["Rule","'VertexLabel'","'10.65'"],["Rule","'Text'","'The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'τὸ ἀπὸ τῆς δύο μέσα δυναμένης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων ἕκτην.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",41]]]],["Rule","'Proof'","'Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a sixth binomial straight line. For let the same construction be made as before. Then, since AB is the side of the sum of two medial areas, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them, [X. 41] so that, in accordance with what was before proved, each of the rectangles DL, MF is medial. And they are applied to the rational straight line DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. [X. 22] And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore DL is incommensurable with MF. Therefore DM is also incommensurable with MG; [VI. 1, X. 11] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36] I say next that it is also a sixth binomial straight line. Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM. And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out. Therefore DG is a sixth binomial straight line.'"],["Rule","'ProofWordCount'",312],["Rule","'GreekProof'","'ἔστω δύο μέσα δυναμένη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ῥητὴ δὲ ἔστω ἡ ΔΕ. καὶ παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ: λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶν ἕκτη. Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον. καὶ ἐπεὶ ἡ ΑΒ δύο μέσα δυναμένη ἐστὶ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾽ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων συγκείμενον τῷ ὑπ᾽ αὐτῶν: ὥστε κατὰ τὰ προδεδειγμένα μέσον ἐστὶν ἑκάτερον τῶν ΔΛ, ΜΖ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται: ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΜ, ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ἀσύμμετρον ἄρα ἐστὶ τὸ ΔΛ τῷ ΜΖ. ἀσύμμετρος ἄρα καὶ ἡ ΔΜ τῇ ΜΗ: αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. λέγω δή, ὅτι καὶ ἕκτη. ὁμοίως δὴ πάλιν δείξομεν, ὅτι τὸ ὑπὸ τῶν ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, καὶ ὅτι ἡ ΔΚ τῇ ΚΜ μήκει ἐστὶν ἀσύμμετρος: καὶ διὰ τὰ αὐτὰ δὴ ἡ ΔΜ τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. καὶ οὐδετέρα τῶν ΔΜ, ΜΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶν ἕκτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",242]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",66]],["Association",["Rule","'VertexLabel'","'10.66'"],["Rule","'Text'","'A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ἡ τῇ ἐκ δύο ὀνομάτων μήκει σύμμετρος καὶ αὐτὴ ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτή.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",6],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.2]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.3]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.4]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.5]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",14]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let AB be binomial, and let CD be commensurable in length with AB; I say that CD is binomial and the same in order with AB. For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only. [X. 36] Let it be contrived that, as AB is to CD, so is AE to CF; [VI. 12] therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19] But AB is commensurable in length with CD; therefore AE is also commensurable with CF, and EB with FD. [X. 11] And AE, EB are rational; therefore CF, FD are also rational. And, as AE is to CF, so is EB to FD. [V. 11] Therefore, alternately, as AE is to EB, so is CF to FD. [V. 16] But AE, EB are commensurable in square only; therefore CF, FD are also commensurable in square only. [X. 11] And they are rational; therefore CD is binomial. [X. 36] I say next that it is the same in order with AB. For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it. If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14] And, if AE is commensurable with the rational straight line set out, CF will also be commensurable with it, [X. 12] and for this reason each of the straight lines AB, CD is a first binomial, that is, the same in order. [X. Deff. II. 1] But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, [X. 12] and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. [X. Deff. II. 2] But, if neither of the straight lines AE, EB is commensurable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [X. 13] and each of the straight lines AB, CD is a third binomial. [X. Deff. II. 3] But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14] And, if AE is commensurable with the rational straight line set out, CF is also commensurable with it, and each of the straight lines AB, CD is a fourth binomial. [X. Deff. II. 4] But, if EB is so commensurable, so is FD also, and each of the straight lines AB, CD will be a fifth binomial. [X. Deff. II. 5] But, if neither of the straight lines AE, EB is so commensurable, neither of the straight lines CF, FD is commensurable with the rational straight line set out, and each of the straight lines AB, CD will be a sixth binomial. [X. Deff. II. 6] Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order.'"],["Rule","'ProofWordCount'",589],["Rule","'GreekProof'","'ἔστω ἐκ δύο ὀνομάτων ἡ ΑΒ, καὶ τῇ ΑΒ μήκει σύμμετρος ἔστω ἡ ΓΔ: λέγω, ὅτι ἡ ΓΔ ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ. ἐπεὶ γὰρ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, καὶ ἔστω μεῖζον ὄνομα τὸ ΑΕ: αἱ ΑΕ, ΕΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. γεγονέτω ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΑΕ πρὸς τὴν ΓΖ: καὶ λοιπὴ ἄρα ἡ ΕΒ πρὸς λοιπὴν τὴν ΖΔ ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ. σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ μήκει. σύμμετρος ἄρα ἐστὶ καὶ ἡ μὲν ΑΕ τῇ ΓΖ, ἡ δὲ ΕΒ τῇ ΖΔ. καί εἰσι ῥηταὶ αἱ ΑΕ, ΕΒ: ῥηταὶ ἄρα εἰσὶ καὶ αἱ ΓΖ, ΖΔ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΕ πρὸς ΓΖ, ἡ ΕΒ πρὸς ΖΔ. ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ. αἱ δὲ ΑΕ, ΕΒ δυνάμει μόνον εἰσὶ σύμμετροι: καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει μόνον εἰσὶ σύμμετροι. καί εἰσι ῥηταί: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΓΔ. λέγω δή, ὅτι τῇ τάξει ἐστὶν ἡ αὐτὴ τῇ ΑΒ. ἡ γὰρ ΑΕ τῆς ΕΒ μεῖζον δύναται ἤτοι τῷ ἀπὸ συμμέτρου ἑαυτῇ á¼¢ τῷ ἀπὸ ἀσυμμέτρου. εἰ μὲν οὖν ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΑΕ τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΓΖ σύμμετρος αὐτῇ ἔσται, καὶ διὰ τοῦτο ἑκατέρα τῶν ΑΒ, ΓΔ ἐκ δύο ὀνομάτων ἐστὶ πρώτη, τουτέστι τῇ τάξει ἡ αὐτή. εἰ δὲ ἡ ΕΒ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΖΔ σύμμετρός ἐστιν αὐτῇ, καὶ διὰ τοῦτο πάλιν τῇ τάξει ἡ αὐτὴ ἔσται τῇ ΑΒ: ἑκατέρα γὰρ αὐτῶν ἔσται ἐκ δύο ὀνομάτων δευτέρα. εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, οὐδετέρα τῶν ΓΖ, ΖΔ σύμμετρος αὐτῇ ἔσται, καί ἐστιν ἑκατέρα τρίτη. εἰ δὲ ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ εἰ μὲν ἡ ΑΕ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΓΖ σύμμετρός ἐστιν αὐτῇ, καί ἐστιν ἑκατέρα τετάρτη. εἰ δὲ ἡ ΕΒ, καὶ ἡ ΖΔ, καὶ ἔσται ἑκατέρα πέμπτη. εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ, καὶ τῶν ΓΖ, ΖΔ οὐδετέρα σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἔσται ἑκατέρα ἕκτη. ὥστε ἡ τῇ ἐκ δύο ὀνομάτων μήκει σύμμετρος ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτή: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",406]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",67]],["Association",["Rule","'VertexLabel'","'10.67'"],["Rule","'Text'","'A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'ἡ τῇ ἐκ δύο μέσων μήκει σύμμετρος καὶ αὐτὴ ἐκ δύο μέσων ἐστὶ καὶ τῇ τάξει ἡ αὐτή.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",37]],["List",["Rule","'Book'",10],["Rule","'Theorem'",38]]]],["Rule","'Proof'","'Let AB be bimedial, and let CD be commensurable in length with AB; I say that CD is bimedial and the same in order with AB. For, since AB is bimedial, let it be divided into its medials at E; therefore AE, EB are medial straight lines commensurable in square only. [X. 37, 38] And let it be contrived that, as AB is to CD, so is AE to CF; therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19] But AB is commensurable in length with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11] But AE, EB are medial; therefore CF, FD are also medial. [X. 23] And since, as AE is to EB, so is CF to FD, [V. 11] and AE, EB are commensurable in square only, CF, FD are also commensurable in square only. [X. 11] But they were also proved medial; therefore CD is bimedial. I say next that it is also the same in order with AB. For since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD; therefore, alternately, as the square on AE is to the square on CF, so is the rectangle AE, EB to the rectangle CF, FD. [V. 16] But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedial]; [X. 37] but if medial, medial, [X. 23] and each of the straight lines AB, CD is a second bimedial. [X. 38] And for this reason CD will be the same in order with AB.'"],["Rule","'ProofWordCount'",321],["Rule","'GreekProof'","'ἔστω ἐκ δύο μέσων ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἔστω μήκει ἡ ΓΔ: λέγω, ὅτι ἡ ΓΔ ἐκ δύο μέσων ἐστὶ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ. ἐπεὶ γὰρ ἐκ δύο μέσων ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰς μέσας κατὰ τὸ Ε: αἱ ΑΕ, ΕΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι. καὶ γεγονέτω ὡς ἡ ΑΒ πρὸς ΓΔ, ἡ ΑΕ πρὸς ΓΖ: καὶ λοιπὴ ἄρα ἡ ΕΒ πρὸς λοιπὴν τὴν ΖΔ ἐστιν, ὡς ἡ ΑΒ πρὸς ΓΔ. σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ μήκει: σύμμετρος ἄρα καὶ ἑκατέρα τῶν ΑΕ, ΕΒ ἑκατέρᾳ τῶν ΓΖ, ΖΔ. μέσαι δὲ αἱ ΑΕ, ΕΒ: μέσαι ἄρα καὶ αἱ ΓΖ, ΖΔ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ, αἱ δὲ ΑΕ, ΕΒ δυνάμει μόνον σύμμετροί εἰσιν, καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει μόνον σύμμετροί εἰσιν. ἐδείχθησαν δὲ καὶ μέσαι: ἡ ΓΔ ἄρα ἐκ δύο μέσων ἐστίν. λέγω δή, ὅτι καὶ τῇ τάξει ἡ αὐτή ἐστι τῇ ΑΒ. ἐπεὶ γάρ ἐστιν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ, καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖΔ: ἐναλλὰξ ὡς τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ἀπὸ τῆς ΓΖ, οὕτως τὸ ὑπὸ τῶν ΑΕΒ πρὸς τὸ ὑπὸ τῶν ΓΖΔ. σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς ΓΖ: σύμμετρον ἄρα καὶ τὸ ὑπὸ τῶν ΑΕΒ τῷ ὑπὸ τῶν ΓΖΔ. εἴτε οὖν ῥητόν ἐστι τὸ ὑπὸ τῶν ΑΕΒ, καὶ τὸ ὑπὸ τῶν ΓΖΔ ῥητόν ἐστιν καὶ διὰ τοῦτό ἐστιν ἐκ δύο μέσων πρώτη. εἴτε μέσον, μέσον, καί ἐστιν ἑκατέρα δευτέρα. καὶ διὰ τοῦτο ἔσται ἡ ΓΔ τῇ ΑΒ τῇ τάξει ἡ αὐτή: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",280]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",68]],["Association",["Rule","'VertexLabel'","'10.68'"],["Rule","'Text'","'A straight line commensurable with a major straight line is itself also major.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἡ τῇ μείζονι σύμμετρος καὶ αὐτὴ μείζων ἐστίν.'"],["Rule","'GreekTextWordCount'",8],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",18]],["List",["Rule","'Book'",6],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",39]]]],["Rule","'Proof'","'Let AB be major, and let CD be commensurable with AB; I say that CD is major. Let AB be divided at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39] Let the same construction be made as before. Then since, as AB is to CD, so is AE to CF, and EB to FD, therefore also, as AE is to CF, so is EB to FD. [V. 11] But AB is commensurable with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11] And since, as AE is to CF, so is EB to FD, alternately also, as AE is to EB, so is CF to FD; [V. 16] therefore also, componendo, as AB is to BE, so is CD to DF; [V. 18] therefore also, as the square on AB is to the square on BE, so is the square on CD to the square on DF. [VI. 20] Similarly we can prove that, as the square on AB is to the square on AE, so also is the square on CD to the square on CF. Therefore also, as the square on AB is to the squares on AE, EB, so is the square on CD to the squares on CF, FD; therefore also, alternately, as the square on AB is to the square on CD, so are the squares on AE, EB to the squares on CF, FD. [V. 16] But the square on AB is commensurable with the square on CD; therefore the squares on AE, EB are also commensurable with the squares on CF, FD. And the squares on AE, EB together are rational; therefore the squares on CF, FD together are rational. Similarly also twice the rectangle AE, EB is commensurable with twice the rectangle CF, FD. And twice the rectangle AE, EB is medial; therefore twice the rectangle CF, FD is also medial. [X. 23] Therefore CF, FD are straight lines incommensurable in square which make, at the same time, the sum of the squares on them rational, but the rectangle contained by them medial; therefore the whole CD is the irrational straight line called major. [X. 39] Therefore a straight line commensurable with the major straight line is major.'"],["Rule","'ProofWordCount'",391],["Rule","'GreekProof'","'ἔστω μείζων ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἔστω ἡ ΓΔ: λέγω, ὅτι ἡ ΓΔ μείζων ἐστίν. διῃρήσθω ἡ ΑΒ κατὰ τὸ Ε: αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δ᾽ ὑπ᾽ αὐτῶν μέσον: καὶ γεγονέτω τὰ αὐτὰ τοῖς πρότερον. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως á¼¥ τε ΑΕ πρὸς τὴν ΓΖ καὶ ἡ ΕΒ πρὸς τὴν ΖΔ, καὶ ὡς ἄρα ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΕΒ πρὸς τὴν ΖΔ. σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ. σύμμετρος ἄρα καὶ ἑκατέρα τῶν ΑΕ, ΕΒ ἑκατέρᾳ τῶν ΓΖ, ΖΔ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΕΒ πρὸς τὴν ΖΔ, καὶ ἐναλλὰξ ὡς ἡ ΑΕ πρὸς ΕΒ, οὕτως ἡ ΓΖ πρὸς ΖΔ, καὶ συνθέντι ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΓΔ πρὸς τὴν ΔΖ: καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΕ, οὕτως τὸ ἀπὸ τῆς ΓΔ πρὸς τὸ ἀπὸ τῆς ΔΖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ὡς τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΑΕ, οὕτως τὸ ἀπὸ τῆς ΓΔ πρὸς τὸ ἀπὸ τῆς ΓΖ. καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΑΒ πρὸς τὰ ἀπὸ τῶν ΑΕ, ΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΔ πρὸς τὰ ἀπὸ τῶν ΓΖ, ΖΔ: καὶ ἐναλλὰξ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΓΔ, οὕτως τὰ ἀπὸ τῶν ΑΕ, ΕΒ πρὸς τὰ ἀπὸ τῶν ΓΖ, ΖΔ. σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΒ τῷ ἀπὸ τῆς ΓΔ: σύμμετρα ἄρα καὶ τὰ ἀπὸ τῶν ΑΕ, ΕΒ τοῖς ἀπὸ τῶν ΓΖ, ΖΔ. καί ἐστι τὰ ἀπὸ τῶν ΑΕ, ΕΒ ἅμα ῥητόν, καὶ τὰ ἀπὸ τῶν ΓΖ, ΖΔ ἅμα ῥητόν ἐστιν. ὁμοίως δὲ καὶ τὸ δὶς ὑπὸ τῶν ΑΕ, ΕΒ σύμμετρόν ἐστι τῷ δὶς ὑπὸ τῶν ΓΖ, ΖΔ. καί ἐστι μέσον τὸ δὶς ὑπὸ τῶν ΑΕ, ΕΒ: μέσον ἄρα καὶ τὸ δὶς ὑπὸ τῶν ΓΖ, ΖΔ. αἱ ΓΖ, ΖΔ ἄρα δυνάμει ἀσύμμετροί εἰσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ἅμα ῥητόν, τὸ δὲ δὶς ὑπ᾽ αὐτῶν μέσον: ὅλη ἄρα ἡ ΓΔ ἄλογός ἐστιν ἡ καλουμένη μείζων. ἡ ἄρα τῇ μείζονι σύμμετρος μείζων ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",368]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",69]],["Association",["Rule","'VertexLabel'","'10.69'"],["Rule","'Text'","'A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἡ τῇ ῥητὸν καὶ μέσον δυναμένῃ σύμμετρος καὶ αὐτὴ ῥητὸν καὶ μέσον δυναμένη ἐστίν.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",40]]]],["Rule","'Proof'","'Let AB be the side of a rational plus a medial area, and let CD be commensurable with AB; it is to be proved that CD is also the side of a rational plus a medial area. Let AB be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40] Let the same construction be made as before. We can then prove similarly that CF, FD are incommensurable in square, and the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, and the rectangle CF, FD rational. Therefore CD is the side of a rational plus a medial area.'"],["Rule","'ProofWordCount'",156],["Rule","'GreekProof'","'ἔστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἔστω ἡ ΓΔ: δεικτέον, ὅτι καὶ ἡ ΓΔ ῥητὸν καὶ μέσον δυναμένη ἐστίν. διῃρήσθω ἡ ΑΒ εἰς τὰς εὐθείας κατὰ τὸ Ε: αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δ᾽ ὑπ᾽ αὐτῶν ῥητόν: καὶ τὰ αὐτὰ κατεσκευάσθω τοῖς πρότερον. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ΓΖ, ΖΔ δυνάμει εἰσὶν ἀσύμμετροι, καὶ σύμμετρον τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ, τὸ δὲ ὑπὸ ΑΕ, ΕΒ τῷ ὑπὸ ΓΖ, ΖΔ: ὥστε καὶ τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων ἐστὶ μέσον, τὸ δ᾽ ὑπὸ τῶν ΓΖ, ΖΔ ῥητόν. ῥητὸν ἄρα καὶ μέσον δυναμένη ἐστὶν ἡ ΓΔ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",136]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",70]],["Association",["Rule","'VertexLabel'","'10.70'"],["Rule","'Text'","'A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'ἡ τῇ δύο μέσα δυναμένῃ σύμμετρος δύο μέσα δυναμένη ἐστίν.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",41]]]],["Rule","'Proof'","'Let AB be the side of the sum of two medial areas, and CD commensurable with AB; it is to be proved that CD is also the side of the sum of two medial areas. For, since AB is the side of the sum of two medial areas, let it be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and furthermore the sum of the squares on AE, EB incommensurable with the rectangle AE, EB. [X. 41] Let the same construction be made as before. We can then prove similarly that CF, FD are also incommensurable in square, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, the rectangle CF, FD is medial, and moreover the sum of the squares on CF, FD is incommensurable with the rectangle CF, FD. Therefore CD is the side of the sum of two medial areas.'"],["Rule","'ProofWordCount'",199],["Rule","'GreekProof'","'ἔστω δύο μέσα δυναμένη ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἡ ΓΔ: δεικτέον, ὅτι καὶ ἡ ΓΔ δύο μέσα δυναμένη ἐστίν. ᾿επεὶ γὰρ δύο μέσα δυναμένη ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰς εὐθείας κατὰ τὸ Ε: αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾽ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων τῷ ὑπὸ τῶν ΑΕ, ΕΒ: καὶ κατεσκευάσθω τὰ αὐτὰ τοῖς πρότερον. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ΓΖ, ΖΔ δυνάμει εἰσὶν ἀσύμμετροι καὶ σύμμετρον τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ, τὸ δὲ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ: ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων μέσον ἐστὶ καὶ τὸ ὑπὸ τῶν ΓΖ, ΖΔ μέσον καὶ ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων τῷ ὑπὸ τῶν ΓΖ, ΖΔ. ἡ ἄρα ΓΔ δύο μέσα δυναμένη ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",172]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",71]],["Association",["Rule","'VertexLabel'","'10.71'"],["Rule","'Text'","'If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'ῥητοῦ καὶ μέσου συντιθεμένου τέσσαρες ἄλογοι γίγνονται ἤτοι ἐκ δύο ὀνομάτων á¼¢ ἐκ δύο μέσων πρώτη á¼¢ μείζων á¼¢ ῥητὸν καὶ μέσον δυναμένη.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.2]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.4]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",54]],["List",["Rule","'Book'",10],["Rule","'Theorem'",55]],["List",["Rule","'Book'",10],["Rule","'Theorem'",57]],["List",["Rule","'Book'",10],["Rule","'Theorem'",58]]]],["Rule","'Proof'","'Let AB be rational, and CD medial; I say that the “side” of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area. For AB is either greater or less than CD. First, let it be greater; let a rational straight line EF be set out, let there be applied to EF the rectangle EG equal to AB, producing EH as breadth, and let HI, equal to DC, be applied to EF, producing HK as breadth. Then, since AB is rational and is equal to EG, therefore EG is also rational. And it has been applied to EF, producing EH as breadth; therefore EH is rational and commensurable in length with EF. [X. 20] Again, since CD is medial and is equal to HI, therefore HI is also medial. And it is applied to the rational straight line EF, producing HK as breadth; therefore HK is rational and incommensurable in length with EF [X. 22] And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that EG is also incommensurable with HI. But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is also incommensurable in length with HK. [X. 11] And both are rational; therefore EH, HK are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [X. 36] And, since AB is greater than CD, while AB is equal to EG and CD to HI, therefore EG is also greater than HI; therefore EH is also greater than HK. The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable with itself. Now the greater straight line HE is commensurable in length with the rational straight line EF set out; therefore EK is a first binomial. [X. Deff. II. 1] But EF is rational; and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial. [X. 54] Therefore the “side” of EI is binomial; so that the “side” of AD is also binomial. Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH. Now the greater straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a fourth binomial. [X. Deff. II. 4] But EF is rational; and, if an area be contained by a rational straight line and the fourth binomial, the “side” of the area is the irrational straight line called major. [X. 57] Therefore the “side” of the area EI is major; so that the “side” of the area AD is also major. Next, let AB be less than CD; therefore EG is also less than HI, so that EH is also less than HK. Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable in length with itself. Now the lesser straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a second binomial. [X. Deff. II. 2] But EF is rational, and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [X. 55] therefore the “side” of the area EI is a first bimedial, so that the “side” of AD is also a first bimedial. Next, let the square on HK be greater than the square on HE by the square on a straight line incommensurable with HK. Now the lesser straight line EH is commensurable with the rational straight line EF set out; therefore EK is a fifth binomial. [X. Deff. II. 5] But EF is rational; and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a side of a rational plus a medial area. [X. 58] Therefore the “side” of the area EI is a side of a rational plus a medial area, so that the “side” of the area AD is also a side of a rational plus a medial area.'"],["Rule","'ProofWordCount'",791],["Rule","'GreekProof'","'ἔστω ῥητὸν μὲν τὸ ΑΒ, μέσον δὲ τὸ ΓΔ: λέγω, ὅτι ἡ τὸ ΑΔ χωρίον δυναμένη ἤτοι ἐκ δύο ὀνομάτων ἐστὶν á¼¢ ἐκ δύο μέσων πρώτη á¼¢ μείζων á¼¢ ῥητὸν καὶ μέσον δυναμένη. τὸ γὰρ ΑΒ τοῦ ΓΔ ἤτοι μεῖζόν ἐστιν á¼¢ ἔλασσον. ἔστω πρότερον μεῖζον: καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ παραβεβλήσθω παρὰ τὴν ΕΖ τῷ ΑΒ ἴσον τὸ ΕΗ πλάτος ποιοῦν τὴν ΕΘ: τῷ δὲ ΔΓ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΘΙ πλάτος ποιοῦν τὴν ΘΚ. καὶ ἐπεὶ ῥητόν ἐστι τὸ ΑΒ καί ἐστιν ῥητόν ἐστι τὸ ΑΒ καί ἐστιν ἴσον τῷ ΕΗ, ῥητὸν ἄρα καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παραβέβληται πλάτος ποιοῦν τὴν ΕΘ: ἡ ΕΘ ἄρα ῥητή ἐστι καὶ σύμμετρος τῇ ΕΖ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ ΓΔ καί ἐστιν ἴσον τῷ ΘΙ, μέσον ἄρα ἐστὶ καὶ τὸ ΘΙ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν ΘΚ: ῥητὴ ἄρα ἐστὶν ἡ ΘΚ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ μέσον ἐστὶ τὸ ΓΔ, ῥητὸν δὲ τὸ ΑΒ, ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΒ τῷ ΓΔ: ὥστε καὶ τὸ ΕΗ ἀσύμμετρόν ἐστι τῷ ΘΙ. ὡς δὲ τὸ ΕΗ πρὸς τὸ ΘΙ, οὕτως ἐστὶν ἡ ΕΘ πρὸς τὴν ΘΚ: ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΕΘ τῇ ΘΚ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΕΘ, ΘΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΚ διῃρημένη κατὰ τὸ Θ. καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΑΒ τοῦ ΓΔ, ἴσον δὲ τὸ μὲν ΑΒ τῷ ΕΗ, τὸ δὲ ΓΔ τῷ ΘΙ, μεῖζον ἄρα καὶ τὸ ΕΗ τοῦ ΘΙ: καὶ ἡ ΕΘ ἄρα μείζων ἐστὶ τῆς ΘΚ. ἤτοι οὖν ἡ ΕΘ τῆς ΘΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει á¼¢ τῷ ἀπὸ ἀσυμμέτρου. δυνάσθω πρότερον τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί ἐστιν ἡ μείζων ἡ ΘΕ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΕΖ: ἡ ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶ πρώτη. ῥητὴ δὲ ἡ ΕΖ: ἐὰν δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων πρώτης, ἡ τὸ χωρίον δυναμένη ἐκ δύο ὀνομάτων ἐστίν. ἡ ἄρα τὸ ΕΙ δυναμένη ἐκ δύο ὀνομάτων ἐστίν: ὥστε καὶ ἡ τὸ ΑΔ δυναμένη ἐκ δύο ὀνομάτων ἐστίν. ἀλλὰ δὴ δυνάσθω ἡ ΕΘ τῆς ΘΚ μεῖζον τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ: καί ἐστιν ἡ μείζων ἡ ΕΘ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΕΖ μήκει: ἡ ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶ τετάρτη. ῥητὴ δὲ ἡ ΕΖ: ἐὰν δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων τετάρτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη μείζων. ἡ ἄρα τὸ ΕΙ χωρίον δυναμένη μείζων ἐστίν: ὥστε καὶ ἡ τὸ ΑΔ δυναμένη μείζων ἐστίν. ἀλλὰ δὴ ἔστω ἔλασσον τὸ ΑΒ τοῦ ΓΔ: καὶ τὸ ΕΗ ἄρα ἔλασσόν ἐστι τοῦ ΘΙ: ὥστε καὶ ἡ ΕΘ ἐλάσσων ἐστὶ τῆς ΘΚ. ἤτοι δὲ ἡ ΘΚ τῆς ΕΘ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ á¼¢ τῷ ἀπὸ ἀσυμμέτρου. δυνάσθω πρότερον τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει: καί ἐστιν ἡ ἐλάσσων ἡ ΕΘ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΕΖ μήκει: ἡ ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶ δευτέρα. ῥητὴ δὲ ἡ ΕΖ: ἐὰν δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων δευτέρας, ἡ τὸ χωρίον δυναμένη ἐκ δύο μέσων ἐστὶ πρώτη. ἡ ἄρα τὸ ΕΙ χωρίον δυναμένη ἐκ δύο μέσων ἐστὶ πρώτη: ὥστε καὶ ἡ τὸ ΑΔ δυναμένη ἐκ δύο μέσων ἐστὶ πρώτη. ἀλλὰ δὴ ἡ ΘΚ τῆς ΘΕ μεῖζον δυνάσθω τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί ἐστιν ἡ ἐλάσσων ἡ ΕΘ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΕΖ: ἡ ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη. ῥητὴ δὲ ἡ ΕΖ: ἐὰν δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων πέμπτης, ἡ τὸ χωρίον δυναμένη ῥητὸν καὶ μέσον δυναμένη ἐστίν. ἡ ἄρα τὸ ΕΙ χωρίον δυναμένη ῥητὸν καὶ μέσον δυναμένη ἐστίν: ὥστε καὶ ἡ τὸ ΑΔ χωρίον δυναμένη ῥητὸν καὶ μέσον δυναμένη ἐστίν. ῥητοῦ ἄρα καὶ μέσου συντιθεμένου τέσσαρες ἄλογοι γίγνονται ἤτοι ἐκ δύο ὀνομάτων á¼¢ ἐκ δύο μέσων πρώτη á¼¢ μείζων á¼¢ ῥητὸν καὶ μέσον δυναμένη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",647]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",72]],["Association",["Rule","'VertexLabel'","'10.72'"],["Rule","'Text'","'If two medial areas incommensurable with one another be added together, the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas.'"],["Rule","'TextWordCount'",33],["Rule","'GreekText'","'δύο μέσων ἀσυμμέτρων ἀλλήλοις συντιθεμένων αἱ λοιπαὶ δύο ἄλογοι γίγνονται ἤτοι ἐκ δύο μέσων δευτέρα á¼¢ ἡ δύο μέσα δυναμένη.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.3]],["List",["Rule","'Book'",10],["Rule","'Definition'",2.6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",56]],["List",["Rule","'Book'",10],["Rule","'Theorem'",59]]]],["Rule","'Proof'","'For let two medial areas AB, CD incommensurable with one another be added together; I say that the “side” of the area AD is either a second bimedial or a side of the sum of two medial areas. For AB is either greater or less than CD. First, if it so chance, let AB be greater than CD. Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth. Now, since each of the areas AB, CD is medial, therefore each of the areas EG, HI is also medial. And they are applied to the rational straight line FE, producing EH, HK as breadth; therefore each of the straight lines EH, HK is rational and incommensurable in length with EF. [X. 22] And, since AB is incommensurable with CD, and AB is equal to EG, and CD to HI, therefore EG is also incommensurable with HI. But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is incommensurable in length with HK. [X. 11] Therefore EH, HK are rational straight lines commensurable in square only; therefore EK is binomial. [X. 36] But the square on EH is greater than the square on HK either by the square on a straight line commensurable with EH or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable in length with itself. Now neither of the straight lines EH, HK is commensurable in length with the rational straight line EF set out; therefore EK is a third binomial. [X. Deff. II. 3] But EF is rational; and, if an area be contained by a rational straight line and the third binomial, the “side” of the area is a second bimedial; [X. 56] therefore the “side” of EI, that is, of AD, is a second bimedial. Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable in length with EH. Now each of the straight lines EH, HK is incommensurable in length with EF; therefore EK is a sixth binomial. [X. Deff. II. 6] But, if an area be contained by a rational straight line and the sixth binomial, the “side” of the area is the side of the sum of two medial areas; [X. 59] so that the “side” of the area AD is also the side of the sum of two medial areas.'"],["Rule","'ProofWordCount'",441],["Rule","'GreekProof'","'Συγκείσθω γὰρ δύο μέσα ἀσύμμετρα ἀλλήλοις τὰ ΑΒ, ΓΔ: λέγω, ὅτι ἡ τὸ ΑΔ χωρίον δυναμένη ἤτοι ἐκ δύο μέσων ἐστὶ δευτέρα á¼¢ δύο μέσα δυναμένη. τὸ γὰρ ΑΒ τοῦ ΓΔ ἤτοι μεῖζόν ἐστιν á¼¢ ἔλασσον. ἔστω, εἰ τύχοι, πρότερον μεῖζον τὸ ΑΒ τοῦ ΓΔ: καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τῷ μὲν ΑΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΕΗ πλάτος ποιοῦν τὴν ΕΘ, τῷ δὲ ΓΔ ἴσον τὸ ΘΙ πλάτος ποιοῦν τὴν ΘΚ. καὶ ἐπεὶ μέσον ἐστὶν ἑκάτερον τῶν ΑΒ, ΓΔ, μέσον ἄρα καὶ ἑκάτερον τῶν ΕΗ, ΘΙ. καὶ παρὰ ῥητὴν τὴν ΖΕ παράκειται πλάτος ποιοῦν τὰς ΕΘ, ΘΚ: ἑκατέρα ἄρα τῶν ΕΘ, ΘΚ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ΑΒ τῷ ΓΔ, καί ἐστιν ἴσον τὸ μὲν ΑΒ τῷ ΕΗ, τὸ δὲ ΓΔ τῷ ΘΙ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΕΗ τῷ ΘΙ. ὡς δὲ τὸ ΕΗ πρὸς τὸ ΘΙ, οὕτως ἐστὶν ἡ ΕΘ πρὸς ΘΚ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΘ τῇ ΘΚ μήκει. αἱ ΕΘ, ΘΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΚ. ἤτοι δὲ ἡ ΕΘ τῆς ΘΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ á¼¢ τῷ ἀπὸ ἀσυμμέτρου. δυνάσθω πρότερον τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει: καὶ οὐδετέρα τῶν ΕΘ, ΘΚ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΕΖ μήκει: ἡ ΕΚ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη. ῥητὴ δὲ ἡ ΕΖ: ἐὰν δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων τρίτης, ἡ τὸ χωρίον δυναμένη ἐκ δύο μέσων ἐστὶ δευτέρα: ἡ ἄρα τὸ ΕΙ, τουτέστι τὸ ΑΔ, δυναμένη ἐκ δύο μέσων ἐστὶ δευτέρα. ἀλλὰ δὴ ἡ ΕΘ τῆς ΘΚ μεῖζον δυνάσθω τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει: καὶ ἀσύμμετρός ἐστιν ἑκατέρα τῶν ΕΘ, ΘΚ τῇ ΕΖ μήκει: ἡ ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶν ἕκτη. ἐὰν δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων ἕκτης, ἡ τὸ χωρίον δυναμένη ἡ δύο μέσα δυναμένη ἐστίν: ὥστε καὶ ἡ τὸ ΑΔ χωρίον δυναμένη ἡ δύο μέσα δυναμένη ἐστίν. Ὁμοίως δὴ δείξομεν, ὅτι κἂν ἔλαττον ᾖ τὸ ΑΒ τοῦ ΓΔ, ἡ τὸ ΑΔ χωρίον δυναμένη á¼¢ ἐκ δύο μέσων δευτέρα ἐστὶν ἤτοι δύο μέσα δυναμένη. δύο ἄρα μέσων ἀσυμμέτρων ἀλλήλοις συντιθεμένων αἱ λοιπαὶ δύο ἄλογοι γίγνονται ἤτοι ἐκ δύο μέσων δευτέρα á¼¢ δύο μέσα δυναμένη. ἡ ἐκ δύο ὀνομάτων καὶ αἱ μετ᾽ αὐτὴν ἄλογοι οὔτε τῇ μέσῃ οὔτε ἀλλήλαις εἰσὶν αἱ αὐταί. τὸ μὲν γὰρ ἀπὸ μέσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ῥητὴν καὶ ἀσύμμετρον τῇ παρ᾽ ἣν παράκειται μήκει. τὸ δὲ ἀπὸ τῆς ἐκ δύο ὀνομάτων παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πρώτην. τὸ δὲ ἀπὸ τῆς ἐκ δύο μέσων πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων δευτέραν. τὸ δὲ ἀπὸ τῆς ἐκ δύο μέσων δευτέρας παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τρίτην. τὸ δὲ ἀπὸ τῆς μείζονος παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τετάρτην. τὸ δὲ ἀπὸ τῆς ῥητὸν καὶ μέσον δυναμένης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πέμπτην. τὸ δὲ ἀπὸ τῆς δύο μέσα δυναμένης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων ἕκτην. τὰ δ᾽ εἰρημένα πλάτη διαφέρει τοῦ τε πρώτου καὶ ἀλλήλων, τοῦ μὲν πρώτου, ὅτι ῥητή ἐστιν, ἀλλήλων δέ, ὅτι τῇ τάξει οὐκ εἰσὶν αἱ αὐταί: ὥστε καὶ αὐταὶ αἱ ἄλογοι διαφέρουσιν ἀλλήλων.'"],["Rule","'GreekProofWordCount'",544]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",73]],["Association",["Rule","'VertexLabel'","'10.73'"],["Rule","'Text'","'If from a rational straight line there be subtracted a rational straight line commensurable with the whole in square only, the remainder is irrational; and let it be called   an apotome.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν ἀπὸ ῥητῆς ῥητὴ ἀφαιρεθῇ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, ἡ λοιπὴ ἄλογός ἐστιν: καλείσθω δὲ ἀποτομή.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",10],["Rule","'Definition'",1.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'For from the rational straight line AB let the rational straight line BC, commensurable with the whole in square only, be subtracted; I say that the remainder AC is the irrational straight line called apotome. For, since AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11] But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6] And, inasmuch as the squares on AB, BC are equal to twice the rectangle AB, BC together with the square on CA, [II. 7] therefore the squares on AB, BC are also incommensurable with the remainder, the square on AC. [X. 13, 16] But the squares on AB, BC are rational; therefore AC is irrational. [X. Def. 4] And let it be called an apotome.'"],["Rule","'ProofWordCount'",170],["Rule","'GreekProof'","'ἀπὸ γὰρ ῥητῆς τῆς ΑΒ ῥητὴ ἀφῃρήσθω ἡ ΒΓ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ: λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν ἡ καλουμένη ἀποτομή. ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, καί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τῷ ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΒ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα, τῷ δὲ ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καὶ ἐπειδήπερ τὰ ἀπὸ τῶν ΑΒ, ΒΓ ἴσα ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΑ, καὶ λοιπῷ ἄρα τῷ ἀπὸ τῆς ΑΓ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ. ῥητὰ δὲ τὰ ἀπὸ τῶν ΑΒ, ΒΓ: ἄλογος ἄρα ἐστὶν ἡ ΑΓ: καλείσθω δὲ ἀποτομή. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",145]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",74]],["Association",["Rule","'VertexLabel'","'10.74'"],["Rule","'Text'","'If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a rational rectangle, the remainder is irrational. And let it be called a first apotome of a medial straight line.'"],["Rule","'TextWordCount'",48],["Rule","'GreekText'","'ἐὰν ἀπὸ μέσης μέση ἀφαιρεθῇ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ῥητὸν περιέχουσα, ἡ λοιπὴ ἄλογός ἐστιν: καλείσθω δὲ μέσης ἀποτομὴ πρώτη.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",10],["Rule","'Definition'",1.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'For from the medial straight line AB let there be subtracted the medial straight line BC which is commensurable with AB in square only and with AB makes the rectangle AB, BC rational; I say that the remainder AC is irrational; and let it be called a first apotome of a medial straight line. For, since AB, BC are medial, the squares on AB, BC are also medial. But twice the rectangle AB, BC is rational; therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore twice the rectangle AB, BC is also incommensurable with the remainder, the square on AC, [Cf. II. 7] since, if the whole is incommensurable with one of the magnitudes, the original magnitudes will also be incommensurable. [X. 16] But twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4] And let it be called a first apotome of a medial straight line.'"],["Rule","'ProofWordCount'",164],["Rule","'GreekProof'","'ἀπὸ γὰρ μέσης τῆς ΑΒ μέση ἀφῃρήσθω ἡ ΒΓ δυνάμει μόνον σύμμετρος οὖσα τῇ ΑΒ, μετὰ δὲ τῆς ΑΒ ῥητὸν ποιοῦσα τὸ ὑπὸ τῶν ΑΒ, ΒΓ: λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν: καλείσθω δὲ μέσης ἀποτομὴ πρώτη. ἐπεὶ γὰρ αἱ ΑΒ, ΒΓ μέσαι εἰσίν, μέσα ἐστὶ καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἀσύμμετρα ἄρα τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: καὶ λοιπῷ ἄρα τῷ ἀπὸ τῆς ΑΓ ἀσύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ἐπεὶ κἂν τὸ ὅλον ἑνὶ αὐτῶν ἀσύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη ἀσύμμετρα ἔσται. ῥητὸν δὲ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ: ἄλογος ἄρα ἐστὶν ἡ ΑΓ: καλείσθω δὲ μέσης ἀποτομὴ πρώτη.'"],["Rule","'GreekProofWordCount'",129]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",75]],["Association",["Rule","'VertexLabel'","'10.75'"],["Rule","'Text'","'If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, the remainder is irrational; and let it be called a second apotome of a medial straight line.'"],["Rule","'TextWordCount'",48],["Rule","'GreekText'","'ἐὰν ἀπὸ μέσης μέση ἀφαιρεθῇ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλη, μετὰ δὲ τῆς ὅλης μέσον περιέχουσα, ἡ λοιπὴ ἄλογός ἐστιν: καλείσθω δὲ μέσης ἀποτομὴ δευτέρα.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",28]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'For from the medial straight line AB let there be subtracted the medial straight line CB which is commensurable with the whole AB in square only and such that the rectangle AB, BC, which it contains with the whole AB, is medial; [X. 28] I say that the remainder AC is irrational; and let it be called a second apotome of a medial straight line. For let a rational straight line DI be set out, let DE equal to the squares on AB, BC be applied to DI, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be applied to DI, producing DF as breadth; therefore the remainder FE is equal to the square on AC. [II. 7] Now, since the squares on AB, BC are medial and commensurable, therefore DE is also medial. [X. 15 and 23] And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22] Again, since the rectangle AB, BC is medial, therefore twice the rectangle AB, BC is also medial. [X. 23] And it is equal to DH; therefore DH is also medial. And it has been applied to the rational straight line DI, producing DF as breadth; therefore DF is rational and incommensurable in length with DI. [X. 22] And, since AB, BC are commensurable in square only, therefore AB is incommensurable in length with BC; therefore the square on AB is also incommensurable with the rectangle AB, BC. [X. 11] But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC; [X. 6] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13] But DE is equal to the squares on AB, BC, and DH to twice the rectangle AB, BC; therefore DE is incommensurable with DH. But, as DE is to DH, so is GD to DF; [VI. 1] therefore GD is incommensurable with DF. [X. 11] And both are rational; therefore GD, DF are rational straight lines commensurable in square only; therefore FG is an apotome. [X. 73] But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, [deduction from X. 20] and its ’side’ is irrational. And AC is the ’side’ of FE; therefore AC is irrational. And let it be called a second apotome of a medial straight line.'"],["Rule","'ProofWordCount'",424],["Rule","'GreekProof'","'ἀπὸ γὰρ μέσης τῆς ΑΒ μέση ἀφῃρήσθω ἡ ΓΒ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ τῇ ΑΒ, μετὰ δὲ τῆς ὅλης τῆς ΑΒ μέσον περιέχουσα τὸ ὑπὸ τῶν ΑΒ, ΒΓ: λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν: καλείσθω δὲ μέσης ἀποτομὴ δευτέρα. Ἐκκείσθω γὰρ ῥητὴ ἡ ΔΙ, καὶ τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον παρὰ τὴν ΔΙ παραβεβλήσθω τὸ ΔΕ πλάτος ποιοῦν τὴν ΔΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον παρὰ τὴν ΔΙ παραβεβλήσθω τὸ ΔΘ πλάτος ποιοῦν τὴν ΔΖ: λοιπὸν ἄρα τὸ ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ. καὶ ἐπεὶ μέσα καὶ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ, μέσον ἄρα καὶ τὸ ΔΕ. καὶ παρὰ ῥητὴν τὴν ΔΙ παράκειται πλάτος ποιοῦν τὴν ΔΗ: ῥητὴ ἄρα ἐστὶν ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΔΙ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ, καὶ τὸ δὶς ἄρα ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν. καί ἐστιν ἴσον τῷ ΔΘ: καὶ τὸ ΔΘ ἄρα μέσον ἐστίν. καὶ παρὰ ῥητὴν τὴν ΔΙ παραβέβληται πλάτος ποιοῦν τὴν ΔΖ: ῥητὴ ἄρα ἐστὶν ἡ ΔΖ καὶ ἀσύμμετρος τῇ ΔΙ μήκει. καὶ ἐπεὶ αἱ ΑΒ, ΒΓ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ ΒΓ μήκει: ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΑΒ τετράγωνον τῷ ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΒ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ, τῷ δὲ ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἀσύμμετρον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τοῖς ἀπὸ τῶν ΑΒ, ΒΓ. ἴσον δὲ τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ τὸ ΔΕ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τὸ ΔΘ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΔΕ τῷ ΔΘ. ὡς δὲ τὸ ΔΕ πρὸς τὸ ΔΘ, οὕτως ἡ ΗΔ πρὸς τὴν ΔΖ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΗΔ τῇ ΔΖ. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ἄρα ΗΔ, ΔΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ΖΗ ἄρα ἀποτομή ἐστιν. ῥητὴ δὲ ἡ ΔΙ: τὸ δὲ ὑπὸ ῥητῆς καὶ ἀλόγου περιεχόμενον ἄλογόν ἐστιν, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν. καὶ δύναται τὸ ΖΕ ἡ ΑΓ: ἡ ΑΓ ἄρα ἄλογός ἐστιν: καλείσθω δὲ μέσης ἀποτομὴ δευτέρα. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",353]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",76]],["Association",["Rule","'VertexLabel'","'10.76'"],["Rule","'Text'","'If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the squares on them added together rational, but the rectangle contained by them medial, the remainder is irrational; and let it be called minor.'"],["Rule","'TextWordCount'",49],["Rule","'GreekText'","'ἐὰν ἀπὸ εὐθείας εὐθεῖα ἀφαιρεθῇ δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τὰ μὲν ἀπ᾽ αὐτῶν ἅμα ῥητόν, τὸ δ᾽ ὑπ᾽ αὐτῶν μέσον, ἡ λοιπὴ ἄλογός ἐστιν: καλείσθω δὲ ἐλάσσων.'"],["Rule","'GreekTextWordCount'",36],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",33]]]],["Rule","'Proof'","'For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with the whole and fulfils the given conditions. [X. 33] I say that the remainder AC is the irrational straight line called minor. For, since the sum of the squares on AB, BC is rational, while twice the rectangle AB, BC is medial, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; and, convertendo, the squares on AB, BC are incommensurable with the remainder, the square on AC. [II. 7, X. 16] But the squares on AB, BC are rational; therefore the square on AC is irrational; therefore AC is irrational. And let it be called minor.'"],["Rule","'ProofWordCount'",121],["Rule","'GreekProof'","'ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ ποιοῦσα τὰ προκείμενα. λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων. ἐπεὶ γὰρ τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τετραγώνων ῥητόν ἐστιν, τὸ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μέσον, ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: καὶ ἀναστρέψαντι λοιπῷ τῷ ἀπὸ τῆς ΑΓ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ. ῥητὰ δὲ τὰ ἀπὸ τῶν ΑΒ, ΒΓ. ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ: ἄλογος ἄρα ἡ ΑΓ: καλείσθω δὲ ἐλάσσων. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",101]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",77]],["Association",["Rule","'VertexLabel'","'10.77'"],["Rule","'Text'","'If from a straight line there be subtracted a straight line which is incommensurable in square with the whole, and which with the whole makes the sum of the squares on them medial, but twice the rectangle contained by them rational, the remainder is irrational: and let it be called that which produces with a rational area a medial whole. '"],["Rule","'TextWordCount'",60],["Rule","'GreekText'","'ἐὰν ἀπὸ εὐθείας εὐθεῖα ἀφαιρεθῇ δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δὲ δὶς ὑπ᾽ αὐτῶν ῥητόν, ἡ λοιπὴ ἄλογός ἐστιν: καλείσθω δὲ ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα.'"],["Rule","'GreekTextWordCount'",45],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",10],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",34]]]],["Rule","'Proof'","'For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfils the given conditions; [X. 34] I say that the remainder AC is the irrational straight line aforesaid. For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore the remainder also, the square on AC, is incommensurable with twice the rectangle AB, BC. [II. 7, X. 16] And twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational. And let it be called that which produces with a rational area a medial whole.'"],["Rule","'ProofWordCount'",128],["Rule","'GreekProof'","'ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ δυνάμει ἀσύμμετρος οὖσα τῇ ΑΒ ποιοῦσα τὰ προκείμενα: λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν ἡ προειρημένη. ἐπεὶ γὰρ τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τετραγώνων μέσον ἐστίν, τὸ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ῥητόν, ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: καὶ λοιπὸν ἄρα τὸ ἀπὸ τῆς ΑΓ ἀσύμμετρόν ἐστι τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καί ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ῥητόν: τὸ ἄρα ἀπὸ τῆς ΑΓ ἄλογόν ἐστιν: ἄλογος ἄρα ἐστὶν ἡ ΑΓ: καλείσθω δὲ ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",111]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",78]],["Association",["Rule","'VertexLabel'","'10.78'"],["Rule","'Text'","'If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them, the remainder is irrational; and let it be called that which produces with a medial area a medial whole.'"],["Rule","'TextWordCount'",73],["Rule","'GreekText'","'ἐὰν ἀπὸ εὐθείας εὐθεῖα ἀφαιρεθῇ δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον τό τε δὶς ὑπ᾽ αὐτῶν μέσον καὶ ἔτι τὰ ἀπ᾽ αὐτῶν τετράγωνα ἀσύμμετρα τῷ δὶς ὑπ᾽ αὐτῶν, ἡ λοιπὴ ἄλογός ἐστιν: καλείσθω δὲ ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα.'"],["Rule","'GreekTextWordCount'",58],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",35]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'For from the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions; [X. 35] I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole. For let a rational straight line DI be set out, to DI let there be applied DE equal to the squares on AB, BC, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be subtracted. Therefore the remainder FE is equal to the square on AC, [II. 7] so that AC is the “side” of FE. Now, since the sum of the squares on AB, BC is medial and is equal to DE, therefore DE is medial. And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22] Again, since twice the rectangle AB, BC is medial and is equal to DH, therefore DH is medial. And it is applied to the rational straight line DI, producing DF as breadth; therefore DF is also rational and incommensurable in length with DI. [X. 22] And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, therefore DE is also incommensurable with DH. But, as DE is to DH, so also is DG to DF; [VI. 1] therefore DG is incommensurable with DF. [X. 11] And both are rational; therefore GD, DF are rational straight lines commensurable in square only. Therefore FG is an apotome. [X. 73] And FH is rational; but the rectangle contained by a rational straight line and an apotome is irrational, [deduction from X. 20] and its “side” is irrational. And AC is the “side” of FE; therefore AC is irrational. And let it be called that which produces with a medial area a medial whole.'"],["Rule","'ProofWordCount'",320],["Rule","'GreekProof'","'ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ δυνάμει ἀσύμμετρος οὖσα τῇ ΑΒ ποιοῦσα τὰ προκείμενα: λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν ἡ καλουμένη ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα. Ἐκκείσθω γὰρ ῥητὴ ἡ ΔΙ, καὶ τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον παρὰ τὴν ΔΙ παραβεβλήσθω τὸ ΔΕ πλάτος ποιοῦν τὴν ΔΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἀφῃρήσθω τὸ ΔΘ πλάτος ποιοῦν τὴν ΔΖ. λοιπὸν ἄρα τὸ ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ: ὥστε ἡ ΑΓ δύναται τὸ ΖΕ. καὶ ἐπεὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τετραγώνων μέσον ἐστὶ καί ἐστιν ἴσον τῷ ΔΕ, μέσον ἄρα ἐστὶ τὸ ΔΕ. καὶ παρὰ ῥητὴν τὴν ΔΙ παράκειται πλάτος ποιοῦν τὴν ΔΗ: ῥητὴ ἄρα ἐστὶν ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΔΙ μήκει. πάλιν, ἐπεὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστὶ καί ἐστιν ἴσον τῷ ΔΘ, τὸ ἄρα ΔΘ μέσον ἐστίν. καὶ παρὰ ῥητὴν τὴν ΔΙ παράκειται πλάτος ποιοῦν τὴν ΔΖ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΔΖ καὶ ἀσύμμετρος τῇ ΔΙ μήκει. καὶ ἐπεὶ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρον ἄρα καὶ τὸ ΔΕ τῷ ΔΘ. ὡς δὲ τὸ ΔΕ πρὸς τὸ ΔΘ, οὕτως ἐστὶ καὶ ἡ ΔΗ πρὸς τὴν ΔΖ: ἀσύμμετρος ἄρα ἡ ΔΗ τῇ ΔΖ. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΗΔ, ΔΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. ἀποτομὴ ἄρα ἐστὶν ἡ ΖΗ: ῥητὴ δὲ ἡ ΖΘ. τὸ δὲ ὑπὸ ῥητῆς καὶ ἀποτομῆς περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν. καὶ δύναται τὸ ΖΕ ἡ ΑΓ: ἡ ΑΓ ἄρα ἄλογός ἐστιν: καλείσθω δὲ ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",276]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",79]],["Association",["Rule","'VertexLabel'","'10.79'"],["Rule","'Text'","'To an apotome only one rational straight line can be annexed which is commensurable with the whole in square only.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'τῇ ἀποτομῇ μία μόνον προσαρμόζει εὐθεῖα ῥητὴ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let AB be an apotome, and BC an annex to it; therefore AC, CB are rational straight lines commensurable in square only. [X. 73] I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only. For, if possible, let BD be so annexed; therefore AD, DB are also rational straight lines commensurable in square only. [X. 73] Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for both exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is the excess of twice the rectangle AD, DB over twice the rectangle AC, CB. But the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational; therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial [X. 21], and a medial area does not exceed a medial by a rational area. [X. 26] Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only. Therefore only one rational straight line can be annexed to an apotome which is commensurable with the whole in square only.'"],["Rule","'ProofWordCount'",245],["Rule","'GreekProof'","'ἔστω ἀποτομὴ ἡ ΑΒ, προσαρμόζουσα δὲ αὐτῇ ἡ ΒΓ: αἱ ΑΓ, ΓΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: λέγω, ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόζει ῥητὴ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ. εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ: καὶ αἱ ΑΔ, ΔΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ ἐπεί, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, τούτῳ ὑπερέχει καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ: τῷ γὰρ αὐτῷ τῷ ἀπὸ τῆς ΑΒ ἀμφότερα ὑπερέχει: ἐναλλὰξ ἄρα, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τούτῳ ὑπερέχει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. τὰ δὲ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ: ῥητὰ γὰρ ἀμφότερα. καὶ τὸ δὶς ἄρα ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ: ὅπερ ἐστὶν ἀδύνατον: μέσα γὰρ ἀμφότερα, μέσον δὲ μέσου οὐχ ὑπερέχει ῥητῷ. τῇ ἄρα ΑΒ ἑτέρα οὐ προσαρμόζει ῥητὴ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ. μία ἄρα μόνη τῇ ἀποτομῇ προσαρμόζει ῥητὴ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",188]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",80]],["Association",["Rule","'VertexLabel'","'10.80'"],["Rule","'Text'","'To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'τῇ μέσης ἀποτομῇ πρώτῃ μία μόνον προσαρμόζει εὐθεῖα μέση δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ῥητὸν περιέχουσα.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Theorem'",74]]]],["Rule","'Proof'","'For let AB be a first apotome of a medial straight line, and let BC be an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is rational; [X. 74] I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a rational area. For, if possible, let DB also be so annexed; therefore AD, DB are medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is rational. [X. 74] Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for they exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB. But twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational. Therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area. which is impossible, for both are medial [X. 15 and 23], and a medial area does not exceed a medial by a rational area. [X. 26]'"],["Rule","'ProofWordCount'",249],["Rule","'GreekProof'","'ἔστω γὰρ μέσης ἀποτομὴ πρώτη ἡ ΑΒ, καὶ τῇ ΑΒ προσαρμοζέτω ἡ ΒΓ: αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι τὸ ὑπὸ τῶν ΑΓ, ΓΒ: λέγω, ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόζει μέση δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ῥητὸν περιέχουσα. εἰ γὰρ δυνατόν, προσαρμοζέτω καὶ ἡ ΔΒ. αἱ ἄρα ΑΔ, ΔΒ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι τὸ ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεί, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, τούτῳ ὑπερέχει καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ: τῷ γὰρ αὐτῷ πάλιν ὑπερέχουσι τῷ ἀπὸ τῆς ΑΒ: ἐναλλὰξ ἄρα, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τούτῳ ὑπερέχει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. τὸ δὲ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ: ῥητὰ γὰρ ἀμφότερα. καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ ἄρα τῶν ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνων ὑπερέχει ῥητῷ: ὅπερ ἐστὶν ἀδύνατον: μέσα γάρ ἐστιν ἀμφότερα, μέσον δὲ μέσου οὐχ ὑπερέχει ῥητῷ. τῇ ἄρα μέσης ἀποτομῇ πρώτῃ μία μόνον προσαρμόζει εὐθεῖα μέση δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ῥητὸν περιέχουσα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",210]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",81]],["Association",["Rule","'VertexLabel'","'10.81'"],["Rule","'Text'","'To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'τῇ μέσης ἀποτομῇ δευτέρᾳ μία μόνον προσαρμόζει εὐθεῖα μέση δυνάμει μόνον σύμμετρος τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης μέσον περιέχουσα.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",75]],["List",["Rule","'Book'",10],["Rule","'Theorem'",79]]]],["Rule","'Proof'","'Let AB be a second apotome of a medial straight line and BC an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is medial. [X. 75] I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a medial rectangle. For, if possible, let BD also be so annexed; therefore AD, DB are also medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is medial. [X. 75] Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be subtracted, producing HM as breadth; therefore the remainder EL is equal to the square on AB, [II. 7] so that AB is the “side” of EL. Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth. But EL is also equal to the square on AB; therefore the remainder HI is equal to twice the rectangle AD, DB. [II. 7] Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial. And they are equal to EG; therefore EG is also medial. [X. 15 and 23] And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22] Again, since the rectangle AC, CB is medial, twice the rectangle AC, CB is also medial. [X. 23] And it is equal to HG; therefore HG is also medial. And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is also rational and incommensurable in length with EF. [X. 22] And, since AC, CB are commensurable in square only, therefore AC is incommensurable in length with CB. But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11] But the squares on AC, CB are commensurable with the square on AC, while twice the rectangle AC, CB is commensurable with the rectangle AC, CB; [X. 6] therefore the squares on AC, CB are incommensurable with twice the rectangle AC, CB. [X. 13] And EG is equal to the squares on AC, CB, while GH is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HG. But, as EG is to HG, so is EM to HM; [VI. 1] therefore EM is incommensurable in length with MH. [X. 11] And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73] Similarly we can prove that HN is also an annex to it; therefore to an apotome different straight lines are annexed which are commensurable with the wholes in square only: which is impossible. [X. 79]'"],["Rule","'ProofWordCount'",531],["Rule","'GreekProof'","'ἔστω μέσης ἀποτομὴ δευτέρα ἡ ΑΒ καὶ τῇ ΑΒ προσαρμόζουσα ἡ ΒΓ: αἱ ἄρα ΑΓ, ΓΒ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι τὸ ὑπὸ τῶν ΑΓ, ΓΒ: λέγω, ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόσει εὐθεῖα μέση δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης μέσον περιέχουσα. εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ: καὶ αἱ ΑΔ, ΔΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι τὸ ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΕΗ πλάτος ποιοῦν τὴν ΕΜ: τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον ἀφῃρήσθω τὸ ΘΗ πλάτος ποιοῦν τὴν ΘΜ: λοιπὸν ἄρα τὸ ΕΛ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ: ὥστε ἡ ΑΒ δύναται τὸ ΕΛ. πάλιν δὴ τοῖς ἀπὸ τῶν ΑΔ, ΔΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΕΙ πλάτος ποιοῦν τὴν ΕΝ: ἔστι δὲ καὶ τὸ ΕΛ ἴσον τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ: λοιπὸν ἄρα τὸ ΘΙ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ μέσαι εἰσὶν αἱ ΑΓ, ΓΒ, μέσα ἄρα ἐστὶ καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ. καί ἐστιν ἴσα τῷ ΕΗ: μέσον ἄρα καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν ΕΜ: ῥητὴ ἄρα ἐστὶν ἡ ΕΜ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ ὑπὸ τῶν ΑΓ, ΓΒ, καὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ μέσον ἐστίν. καί ἐστιν ἴσον τῷ ΘΗ: καὶ τὸ ΘΗ ἄρα μέσον ἐστίν. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν ΘΜ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΘΜ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ αἱ ΑΓ, ΓΒ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΓ τῇ ΓΒ μήκει. ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως ἐστὶ τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ὑπὸ τῶν ΑΓ, ΓΒ: ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΓ τῷ ὑπὸ τῶν ΑΓ, ΓΒ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΓ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΓ, ΓΒ, τῷ δὲ ὑπὸ τῶν ΑΓ, ΓΒ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ: ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. καί ἐστι τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΕΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΗΘ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΕΗ τῷ ΘΗ. ὡς δὲ τὸ ΕΗ πρὸς τὸ ΘΗ, οὕτως ἐστὶν ἡ ΕΜ πρὸς τὴν ΘΜ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΜ τῇ ΜΘ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΕΜ, ΜΘ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΕΘ, προσαρμόζουσα δὲ αὐτῇ ἡ ΘΜ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΘΝ αὐτῇ προσαρμόζει: τῇ ἄρα ἀποτομῇ ἄλλη καὶ ἄλλη προσαρμόζει εὐθεῖα δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ: ὅπερ ἐστὶν ἀδύνατον. τῇ ἄρα μέσης ἀποτομῇ δευτέρᾳ μία μόνον προσαρμόζει εὐθεῖα μέση δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης μέσον περιέχουσα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",476]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",82]],["Association",["Rule","'VertexLabel'","'10.82'"],["Rule","'Text'","'To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of the squares on them rational but twice the rectangle contained by them medial.'"],["Rule","'TextWordCount'",42],["Rule","'GreekText'","'τῇ ἐλάσσονι μία μόνον προσαρμόζει εὐθεῖα δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ ποιοῦσα μετὰ τῆς ὅλης τὸ μὲν ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δὲ δὶς ὑπ᾽ αὐτῶν μέσον.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Theorem'",76]]]],["Rule","'Proof'","'Let AB be the minor straight line, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. [X. 76] I say that no other straight line can be annexed to AB fulfilling the same conditions. For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 76] Now, since the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational, therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial. [X. 26] Therefore to a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes the squares on them added together rational, but twice the rectangle contained by them medial.'"],["Rule","'ProofWordCount'",202],["Rule","'GreekProof'","'ἔστω ἡ ἐλάσσων ἡ ΑΒ, καὶ τῇ ΑΒ προσαρμόζουσα ἔστω ἡ ΒΓ: αἱ ἄρα ΑΓ, ΓΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δὲ δὶς ὑπ᾽ αὐτῶν μέσον: λέγω, ὅτι τῇ ΑΒ ἑτέρα εὐθεῖα οὐ προσαρμόσει τὰ αὐτὰ ποιοῦσα. εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ: καὶ αἱ ΑΔ, ΔΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὰ προειρημένα. καὶ ἐπεί, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τούτῳ ὑπερέχει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τὰ δὲ ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα τῶν ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνων ὑπερέχει ῥητῷ: ῥητὰ γάρ ἐστιν ἀμφότερα: καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ἄρα τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ: ὅπερ ἐστὶν ἀδύνατον: μέσα γάρ ἐστιν ἀμφότερα. τῇ ἄρα ἐλάσσονι μία μόνον προσαρμόζει εὐθεῖα δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ καὶ ποιοῦσα τὰ μὲν ἀπ᾽ αὐτῶν τετράγωνα ἅμα ῥητόν, τὸ δὲ δὶς ὑπ᾽ αὐτῶν μέσον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",168]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",83]],["Association",["Rule","'VertexLabel'","'10.83'"],["Rule","'Text'","'To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.'"],["Rule","'TextWordCount'",54],["Rule","'GreekText'","'τῇ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιούσῃ μία μόνον προσαρμόζει εὐθεῖα δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δὲ δὶς ὑπ᾽ αὐτῶν ῥητόν.'"],["Rule","'GreekTextWordCount'",38],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Theorem'",77]]]],["Rule","'Proof'","'Let AB be the straight line which produces with a rational area a medial whole, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which fulfil the given conditions. [X. 77] I say that no other straight line can be annexed to AB which fulfils the same conditions. For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the given conditions. [X. 77] Since then, as in the preceding cases, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area: which is impossible, for both are medial. [X. 26] Therefore no other straight line can be annexed to AB which is incommensurable in square with the whole and which with the whole fulfils the aforesaid conditions; therefore only one straight line can be so annexed.'"],["Rule","'ProofWordCount'",200],["Rule","'GreekProof'","'ἔστω ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, καὶ τῇ ΑΒ προσαρμοζέτω ἡ ΒΓ: αἱ ἄρα ΑΓ, ΓΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὰ προκείμενα: λέγω, ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόσει τὰ αὐτὰ ποιοῦσα. εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ: καὶ αἱ ΑΔ, ΔΒ ἄρα εὐθεῖαι δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὰ προκείμενα. ἐπεὶ οὖν, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τούτῳ ὑπερέχει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἀκολούθως τοῖς πρὸ αὐτοῦ, τὸ δὲ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ: ῥητὰ γάρ ἐστιν ἀμφότερα: καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ ἄρα τῶν ἀπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ: ὅπερ ἐστὶν ἀδύνατον: μέσα γάρ ἐστιν ἀμφότερα. οὐκ ἄρα τῇ ΑΒ ἑτέρα προσαρμόσει εὐθεῖα δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τὰ προειρημένα: μία ἄρα μόνον προσαρμόσει: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",153]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",84]],["Association",["Rule","'VertexLabel'","'10.84'"],["Rule","'Text'","'To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.'"],["Rule","'TextWordCount'",66],["Rule","'GreekText'","'τῇ μετὰ μέσου μέσον τὸ ὅλον ποιούσῃ μία μόνη προσαρμόζει εὐθεῖα δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον τό τε δὶς ὑπ᾽ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπ᾽ αὐτῶν.'"],["Rule","'GreekTextWordCount'",48],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",78]],["List",["Rule","'Book'",10],["Rule","'Theorem'",79]]]],["Rule","'Proof'","'Let AB be the straight line which produces with a medial area a medial whole, and BC an annex to it; therefore AC, CB are straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 78] I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions. For, if possible, let BD be so annexed, so that AD, DB are also straight lines incommensurable in square which make the squares on AD, DB added together medial, twice the rectangle AD, DB medial, and also the squares on AD, DB incommensurable with twice the rectangle AD, DB. [X. 78] Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be applied to EF, producing HM as breadth; therefore the remainder, the square on AB [II. 7], is equal to EL; therefore AB is the “side” of EL. Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth. But the square on AB is also equal to EL; therefore the remainder, twice the rectangle AD, DB [II. 7], is equal to HI. Now, since the sum of the squares on AC, CB is medial and is equal to EG, therefore EG is also medial. And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22] Again, since twice the rectangle AC, CB is medial and is equal to HG, therefore HG is also medial. And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is rational and incommensurable in length with EF. [X. 22] And, since the squares on AC, CB are incommensurable with twice the rectangle AC, CB, EG is also incommensurable with HG; therefore EM is also incommensurable in length with MH. [VI. 1, X. 11] And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73] Similarly we can prove that EH is again an apotome and HN an annex to it. Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only: which was proved impossible. [X. 79] Therefore no other straight line can be so annexed to AB. Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and also the squares on them incommensurable with twice the rectangle contained by them.'"],["Rule","'ProofWordCount'",470],["Rule","'GreekProof'","'ἔστω ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, προσαρμόζουσα δὲ αὐτῇ ἡ ΒΓ: αἱ ἄρα ΑΓ, ΓΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὰ προειρημένα. λέγω, ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόσει ποιοῦσα τὰ προειρημένα. εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ, ὥστε καὶ τὰς ΑΔ, ΔΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τά τε ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα ἅμα μέσον καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ μέσον καὶ ἔτι τὰ ἀπὸ τῶν ΑΔ, ΔΒ ἀσύμμετρα τῷ δὶς ὑπὸ τῶν ΑΔ, ΔΒ: καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΕΗ πλάτος ποιοῦν τὴν ΕΜ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΘΗ πλάτος ποιοῦν τὴν ΘΜ: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τῷ ΕΛ: ἡ ἄρα ΑΒ δύναται τὸ ΕΛ. πάλιν τοῖς ἀπὸ τῶν ΑΔ, ΔΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΕΙ πλάτος ποιοῦν τὴν ΕΝ. ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΑΒ ἴσον τῷ ΕΛ: λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ἴσον ἐστὶ τῷ ΘΙ. καὶ ἐπεὶ μέσον ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ καί ἐστιν ἴσον τῷ ΕΗ, μέσον ἄρα ἐστὶ καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν ΕΜ: ῥητὴ ἄρα ἐστὶν ἡ ΕΜ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ καί ἐστιν ἴσον τῷ ΘΗ, μέσον ἄρα καὶ τὸ ΘΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν ΘΜ: ῥητὴ ἄρα ἐστὶν ἡ ΘΜ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ἀσύμμετρόν ἐστι καὶ τὸ ΕΗ τῷ ΘΗ: ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΕΜ τῇ ΜΘ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ἄρα ΕΜ, ΜΘ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΕΘ, προσαρμόζουσα δὲ αὐτῇ ἡ ΘΜ. ὁμοίως δὴ δείξομεν, ὅτι ἡ ΕΘ πάλιν ἀποτομή ἐστιν, προσαρμόζουσα δὲ αὐτῇ ἡ ΘΝ. τῇ ἄρα ἀποτομῇ ἄλλη καὶ ἄλλη προσαρμόζει ῥητὴ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ: ὅπερ ἐδείχθη ἀδύνατον. οὐκ ἄρα τῇ ΑΒ ἑτέρα προσαρμόσει εὐθεῖα. τῇ ἄρα ΑΒ μία μόνον προσαρμόζει εὐθεῖα δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τά τε ἀπ᾽ αὐτῶν τετράγωνα ἅμα μέσον καὶ τὸ δὶς ὑπ᾽ αὐτῶν μέσον καὶ ἔτι τὰ ἀπ᾽ αὐτῶν τετράγωνα ἀσύμμετρα τῷ δὶς ὑπ᾽ αὐτῶν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",400]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",85]],["Association",["Rule","'VertexLabel'","'10.85'"],["Rule","'Text'","'To find the first apotome.'"],["Rule","'TextWordCount'",5],["Rule","'GreekText'","'εὑρεῖν τὴν πρώτην ἀποτομήν.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let a rational straight line A be set out, and let BG be commensurable in length with A; therefore BG is also rational. Let two square numbers DE, EF be set out, and let their difference FD not be square; therefore neither has ED to DF the ratio which a square number has to a square number. Let it be contrived that, as ED is to DF, so is the square on BG to the square on GC; [X. 6] therefore the square on BG is commensurable with the square on GC. [X. 6] But the square on BG is rational; therefore the square on GC is also rational; therefore GC is also rational. And, since ED has not to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] I say next that it is also a first apotome. For let the square on H be that by which the square on BG is greater than the square on GC. Now since. as ED is to FD, so is the square on BG to the square on GC, therefore also, convertendo, [v. 19] as DE is to EF, so is the square on GB to the square on H. But DE has to EF the ratio which a square number has to a square number, for each is square; therefore the square on GB also has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a first apotome. [X. Deff. III. 1] Therefore the first apotome BC has been found. (Being) that which it was required to find.'"],["Rule","'ProofWordCount'",376],["Rule","'GreekProof'","'Ἐκκείσθω ῥητὴ ἡ Α, καὶ τῇ Α μήκει σύμμετρος ἔστω ἡ ΒΗ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΗ. καὶ ἐκκείσθωσαν δύο τετράγωνοι ἀριθμοὶ οἱ ΔΕ, ΕΖ, ὧν ἡ ὑπεροχὴ ὁ ΖΔ μὴ ἔστω τετράγωνος: οὐδ᾽ ἄρα ὁ ΕΔ πρὸς τὸν ΔΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ πεποιήσθω ὡς ὁ ΕΔ πρὸς τὸν ΔΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΓ τετράγωνον: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΗ τῷ ἀπὸ τῆς ΗΓ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΒΗ: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΓ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΗΓ. καὶ ἐπεὶ ὁ ΕΔ πρὸς τὸν ΔΖ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ ΗΓ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΒΗ, ΗΓ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ἄρα ΒΓ ἀποτομή ἐστιν. λέγω δή, ὅτι καὶ πρώτη. ὧι γὰρ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, ἔστω τὸ ἀπὸ τῆς Θ. καὶ ἐπεί ἐστιν ὡς ὁ ΕΔ πρὸς τὸν ΖΔ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, καὶ ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΔΕ πρὸς τὸν ΕΖ, οὕτως τὸ ἀπὸ τῆς ΗΒ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΔΕ πρὸς τὸν ΕΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἑκάτερος γὰρ τετράγωνός ἐστιν: καὶ τὸ ἀπὸ τῆς ΗΒ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: σύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ Θ μήκει. καὶ δύναται ἡ ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ τῆς Θ: ἡ ΒΗ ἄρα τῆς ΗΓ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καί ἐστιν ἡ ὅλη ἡ ΒΗ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ Α. ἡ ΒΓ ἄρα ἀποτομή ἐστι πρώτη. εὕρηται ἄρα ἡ πρώτη ἀποτομὴ ἡ ΒΓ: ὅπερ ἔδει εὑρεῖν.'"],["Rule","'GreekProofWordCount'",320]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",86]],["Association",["Rule","'VertexLabel'","'10.86'"],["Rule","'Text'","'To find the second apotome.'"],["Rule","'TextWordCount'",5],["Rule","'GreekText'","'εὑρεῖν τὴν δευτέραν ἀποτομήν.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.2]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let a rational straight line A be set out, and GC commensurable in length with A; therefore GC is rational. Let two square numbers DE, EF be set out, and let their difference DF not be square. Now let it be contrived that, as FD is to DE, so is the square on CG to the square on GB. [X. 6] Therefore the square on CG is commensurable with the square on GB. [X. 6] But the square on CG is rational; therefore the square on GB is also rational; therefore BG is rational. And, since the square on GC has not to the square on GB the ratio which a square number has to a square number, CG is incommensurable in length with GB. [X. 9] And both are rational; therefore CG, GB are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] I say next that it is also a second apotome. For let the square on H be that by which the square on BG is greater than the square on GC. Since then, as the square on BG is to the square on GC, so is the number ED to the number DF, therefore, convertendo, as the square on BG is to the square on H, so is DE to EF. [V. 19] And each of the numbers DE, EF is square; therefore the square on BG has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC  by the square on a straight line commensurable in length with BG. And CG, the annex, is commensurable with the rational straight line A set out. Therefore BC is a second apotome. [X. Deff. III. 2] Therefore the second apotome BC has been found.'"],["Rule","'ProofWordCount'",336],["Rule","'GreekProof'","'Ἐκκείσθω ῥητὴ ἡ Α καὶ τῇ Α σύμμετρος μήκει ἡ ΗΓ. ῥητὴ ἄρα ἐστὶν ἡ ΗΓ. καὶ ἐκκείσθωσαν δύο τετράγωνοι ἀριθμοὶ οἱ ΔΕ, ΕΖ, ὧν ἡ ὑπεροχὴ ὁ ΔΖ μὴ ἔστω τετράγωνος. καὶ πεποιήσθω ὡς ὁ ΖΔ πρὸς τὸν ΔΕ, οὕτως τὸ ἀπὸ τῆς ΓΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΒ τετράγωνον. σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΓΗ τετράγωνον τῷ ἀπὸ τῆς ΗΒ τετραγώνῳ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΗΓ. ῥητὸν ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΗΒ: ῥητὴ ἄρα ἐστὶν ἡ ΒΗ. καὶ ἐπεὶ τὸ ἀπὸ τῆς ΗΓ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΒ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ἀσύμμετρός ἐστιν ἡ ΓΗ τῇ ΗΒ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΓΗ, ΗΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ΒΓ ἄρα ἀποτομή ἐστιν. λέγω δή, ὅτι καὶ δευτέρα. ὧι γὰρ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, ἔστω τὸ ἀπὸ τῆς Θ. ἐπεὶ οὖν ἐστιν ὡς τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, οὕτως ὁ ΕΔ ἀριθμὸς πρὸς τὸν ΔΖ ἀριθμόν, ἀναστρέψαντι ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς Θ, οὕτως ὁ ΔΕ πρὸς τὸν ΕΖ. καί ἐστιν ἑκάτερος τῶν ΔΕ, ΕΖ τετράγωνος: τὸ ἄρα ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: σύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ Θ μήκει. καὶ δύναται ἡ ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ τῆς Θ: ἡ ΒΗ ἄρα τῆς ΗΓ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καί ἐστιν ἡ προσαρμόζουσα ἡ ΓΗ τῇ ἐκκειμένῃ ῥητῇ σύμμετρος τῇ Α. ἡ ΒΓ ἄρα ἀποτομή ἐστι δευτέρα. εὕρηται ἄρα δευτέρα ἀποτομὴ ἡ ΒΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",275]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",87]],["Association",["Rule","'VertexLabel'","'10.87'"],["Rule","'Text'","'To find the third apotome.'"],["Rule","'TextWordCount'",5],["Rule","'GreekText'","'εὑρεῖν τὴν τρίτην ἀποτομήν.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.3]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let a rational straight line A be set out, let three numbers E, BC, CD be set out which have not to one another the ratio which a square number has to a square number, but let CB have to BD the ratio which a square number has to a square number. Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6] Since then, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6] But the square on A is rational; therefore the square on FG is also rational; therefore FG is rational. And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9] Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6] But the square on FG is rational; therefore the square on GH is also rational; therefore GH is rational. And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9] And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73] I say next that it is also a third apotome. For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on HG, therefore, ex aequali, as E is to CD, so is the square on A to the square on HG. [V. 22] But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH. [X. 9] Therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out. Now let the square on K be that by which the square on FG is greater than the square on GH. Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as BC is to BD, so is the square on FG to the square on K. [V. 19] But BC has to BD the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number. Therefore FG is commensurable in length with K, [X. 9] and the square on FG is greater than the square on GH by the square on a straight line commensurable with FG. And neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out; therefore FH is a third apotome. [X. Deff. III. 3] Therefore the third apotome FH has been found.'"],["Rule","'ProofWordCount'",622],["Rule","'GreekProof'","'Ἐκκείσθω ῥητὴ ἡ Α, καὶ ἐκκείσθωσαν τρεῖς ἀριθμοὶ οἱ Ε, ΒΓ, ΓΔ λόγον μὴ ἔχοντες πρὸς ἀλλήλους, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ὁ δὲ ΓΒ πρὸς τὸν ΒΔ λόγον ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ πεποιήσθω ὡς μὲν ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΗ τετράγωνον, ὡς δὲ ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΘ. ἐπεὶ οὖν ἐστιν ὡς ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΗ τετράγωνον, σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς Α τετράγωνον τῷ ἀπὸ τῆς ΖΗ τετραγώνῳ. ῥητὸν δὲ τὸ ἀπὸ τῆς Α τετράγωνον. ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΖΗ: ῥητὴ ἄρα ἐστὶν ἡ ΖΗ. καὶ ἐπεὶ ὁ Ε πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾽ ἄρα τὸ ἀπὸ τῆς α τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΗ τετράγωνον λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ ΖΗ μήκει. πάλιν, ἐπεί ἐστιν ὡς ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΘ, σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΗΘ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΖΗ: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΘ: ῥητὴ ἄρα ἐστὶν ἡ ΗΘ. καὶ ἐπεὶ ὁ ΒΓ πρὸς τὸν ΓΔ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΘ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΖΗ, ΗΘ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΖΘ. λέγω δή, ὅτι καὶ τρίτη. ἐπεὶ γάρ ἐστιν ὡς μὲν ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΘΗ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Ε πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΘΗ. ὁ δὲ Ε πρὸς τὸν ΓΔ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδ᾽ ἄρα τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἡ Α τῇ ΗΘ μήκει. οὐδετέρα ἄρα τῶν ΖΗ, ΗΘ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ Α μήκει. ᾧ οὖν μεῖζόν ἐστι τὸ ἀπὸ τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ, ἔστω τὸ ἀπὸ τῆς Κ. ἐπεὶ οὖν ἐστιν ὡς ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΒΓ πρὸς τὸν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ ΒΓ πρὸς τὸν ΒΔ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ τὸ ἀπὸ τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. σύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Κ μήκει, καὶ δύναται ἡ ΖΗ τῆς ΗΘ μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ οὐδετέρα τῶν ΖΗ, ΗΘ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ Α μήκει: ἡ ΖΘ ἄρα ἀποτομή ἐστι τρίτη. εὕρηται ἄρα ἡ τρίτη ἀποτομὴ ἡ ΖΘ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",542]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",88]],["Association",["Rule","'VertexLabel'","'10.88'"],["Rule","'Text'","'To find the fourth apotome.'"],["Rule","'TextWordCount'",5],["Rule","'GreekText'","'εὑρεῖν τὴν τετάρτην ἀποτομήν.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let a rational straight line A be set out, and BG commensurable in length with it; therefore BG is also rational. Let two numbers DF, FE be set out such that the whole DE has not to either of the numbers DF, EF the ratio which a square number has to a square number. Let it be contrived that, as DE is to EF, so is the square on BG to the square on GC; [X. 6] therefore the square on BG is commensurable with the square on GC. [X. 6] But the square on BG is rational; therefore the square on GC is also rational; therefore GC is rational. Now, since DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] Now let the square on H be that by which the square on BG is greater than the square on GC. Since then, as DE is to EF, so is the square on BG to the square on GC, therefore also, convertendo, as ED is to DF, so is the square on GB to the square on H. [v. 19] But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on GB to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC by the square on a straight line incommensurable with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a fourth apotome. [X. Deff. III. 4] Therefore the fourth apotome has been found.'"],["Rule","'ProofWordCount'",365],["Rule","'GreekProof'","'Ἐκκείσθω ῥητὴ ἡ Α καὶ τῇ Α μήκει σύμμετρος ἡ ΒΗ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΗ. καὶ ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΔΖ, ΖΕ, ὥστε τὸν ΔΕ ὅλον πρὸς ἑκάτερον τῶν ΔΖ, ΕΖ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ πεποιήσθω ὡς ὁ ΔΕ πρὸς τὸν ΕΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΓ. σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΗ τῷ ἀπὸ τῆς ΗΓ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΒΗ: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΓ: ῥητὴ ἄρα ἐστὶν ἡ ΗΓ. καὶ ἐπεὶ ὁ ΔΕ πρὸς τὸν ΕΖ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ ΗΓ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΒΗ, ΗΓ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΒΓ. Λέγω δή, ὅτι καὶ τετάρτη. ὧι οὖν μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, ἔστω τὸ ἀπὸ τῆς Θ. ἐπεὶ οὖν ἐστιν ὡς ὁ ΔΕ πρὸς τὸν ΕΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, καὶ ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΕΔ πρὸς τὸν ΔΖ, οὕτως τὸ ἀπὸ τῆς ΗΒ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΕΔ πρὸς τὸν ΔΖ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΗΒ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ Θ μήκει. καὶ δύναται ἡ ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ τῆς Θ: ἡ ἄρα ΒΗ τῆς ΗΓ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί ἐστιν ὅλη ἡ ΒΗ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ α. ἡ ἄρα ΒΓ ἀποτομή ἐστι τετάρτη. εὕρηται ἄρα ἡ τετάρτη ἀποτομή: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",304]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",89]],["Association",["Rule","'VertexLabel'","'10.89'"],["Rule","'Text'","'To find the fifth apotome.'"],["Rule","'TextWordCount'",5],["Rule","'GreekText'","'εὑρεῖν τὴν πέμπτην ἀποτομήν.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let a rational straight line A be set out, and let CG be commensurable in length with A; therefore CG is rational. Let two numbers DF, FE be set out such that DE again has not to either of the numbers DF, FE the ratio which a square number has to a square number; and let it be contrived that, as FE is to ED, so is the square on CG to the square on GB. Therefore the square on GB is also rational; [X. 6] therefore BG is also rational. Now since, as DE is to EF, so is the square on BG to the square on GC, while DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] I say next that it is also a fifth apotome. For let the square on H be that by which the square on BG is greater than the square on GC. Since then, as the square on BG is to the square on GC, so is DE to EF, therefore, convertendo, as ED is to DF, so is the square on BG to the square on H. [V. 19] But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on BG to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on H; therefore the square on GB is greater than the square on GC by the square on a straight line incommensurable in length with GB. And the annex CG is commensurable in length with the rational straight line A set out; therefore BC is a fifth apotome. [X. Deff. III. 5] Therefore the fifth apotome BC has been found.'"],["Rule","'ProofWordCount'",375],["Rule","'GreekProof'","'Ἐκκείσθω ῥητὴ ἡ Α, καὶ τῇ Α μήκει σύμμετρος ἔστω ἡ ΓΗ: ῥητὴ ἄρα ἐστὶν ἡ ΓΗ. καὶ ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΔΖ, ΖΕ, ὥστε τὸν ΔΕ πρὸς ἑκάτερον τῶν ΔΖ, ΖΕ λόγον πάλιν μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ πεποιήσθω ὡς ὁ ΖΕ πρὸς τὸν ΕΔ, οὕτως τὸ ἀπὸ τῆς ΓΗ πρὸς τὸ ἀπὸ τῆς ΗΒ. ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΒ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΗ. καὶ ἐπεί ἐστιν ὡς ὁ ΔΕ πρὸς τὸν ΕΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, ὁ δὲ ΔΕ πρὸς τὸν ΕΖ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ ΗΓ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΒΗ, ΗΓ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ΒΓ ἄρα ἀποτομή ἐστιν. λέγω δή, ὅτι καὶ πέμπτη. ὧι γὰρ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, ἔστω τὸ ἀπὸ τῆς Θ. ἐπεὶ οὖν ἐστιν ὡς τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, οὕτως ὁ ΔΕ πρὸς τὸν ΕΖ, ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΕΔ πρὸς τὸν ΔΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΕΔ πρὸς τὸν ΔΖ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ Θ μήκει. καὶ δύναται ἡ ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ τῆς Θ: ἡ ΗΒ ἄρα τῆς ΗΓ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. καί ἐστιν ἡ προσαρμόζουσα ἡ ΓΗ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ Α μήκει: ἡ ἄρα ΒΓ ἀποτομή ἐστι πέμπτη. εὕρηται ἄρα ἡ πέμπτη ἀποτομὴ ἡ ΒΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",308]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",90]],["Association",["Rule","'VertexLabel'","'10.90'"],["Rule","'Text'","'To find the sixth apotome.'"],["Rule","'TextWordCount'",5],["Rule","'GreekText'","'εὑρεῖν τὴν ἕκτην ἀποτομήν.'"],["Rule","'GreekTextWordCount'",4],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let a rational straight line A be set out, and three numbers E, BC, CD not having to one another the ratio which a square number has to a square number; and further let CB also not have to BD the ratio which a square number has to a square number. Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6] Now since, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6] But the square on A is rational; therefore the square on FG is also rational; therefore FG is also rational. And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9] Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6] But the square on FG is rational; therefore the square on GH is also rational; therefore GH is also rational. And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9] And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73] I say next that it is also a sixth apotome. For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH, therefore, ex aequali, as E is to CD, so is the square on A to the square on GH. [v. 22] But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH; [X. 9] therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A. Now let the square on K be that by which the square on FG is greater than the square on GH. Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as CB is to BD, so is the square on FG to the square on K. [v. 19] But CB has not to BD the ratio which a square number has to a square number; therefore neither has the square on FG to the square on K the ratio which a square number has to a square number; therefore FG is incommensurable in length with K. [X. 9] And the square on FG is greater than the square on GH by the square on K; therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG. And neither of the straight lines FG, GH is commensurable with the rational straight line A set out. Therefore FH is a sixth apotome. [X. Deff. III. 6] Therefore the sixth apotome FH has been found.'"],["Rule","'ProofWordCount'",639],["Rule","'GreekProof'","'Ἐκκείσθω ῥητὴ ἡ Α καὶ τρεῖς ἀριθμοὶ οἱ Ε, ΒΓ, ΓΔ λόγον μὴ ἔχοντες πρὸς ἀλλήλους, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἔτι δὲ καὶ ὁ ΓΒ πρὸς τὸν ΒΔ λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ πεποιήσθω ὡς μὲν ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ. ἐπεὶ οὖν ἐστιν ὡς ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΖΗ, σύμμετρον ἄρα τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς ΖΗ. ῥητὸν δὲ τὸ ἀπὸ τῆς Α: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΖΗ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ Ε πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾽ ἄρα τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ ΖΗ μήκει. πάλιν, ἐπεί ἐστιν ὡς ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, σύμμετρον ἄρα τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΗΘ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΖΗ: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΘ: ῥητὴ ἄρα καὶ ἡ ΗΘ. καὶ ἐπεὶ ὁ ΒΓ πρὸς τὸν ΓΔ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΘ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΖΗ, ΗΘ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ἄρα ΖΘ ἀποτομή ἐστιν. λέγω δή, ὅτι καὶ ἕκτη. ἐπεὶ γάρ ἐστιν ὡς μὲν ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ὁ Ε πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΗΘ. ὁ δὲ Ε πρὸς τὸν ΓΔ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδ᾽ ἄρα τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ ΗΘ μήκει: οὐδετέρα ἄρα τῶν ΖΗ, ΗΘ σύμμετρός ἐστι τῇ Α ῥητῇ μήκει. ᾧ οὖν μεῖζόν ἐστι τὸ ἀπὸ τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ, ἔστω τὸ ἀπὸ τῆς Κ. ἐπεὶ οὖν ἐστιν ὡς ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΓΒ πρὸς τὸν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ ΓΒ πρὸς τὸν ΒΔ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: οὐδ᾽ ἄρα τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Κ μήκει. καὶ δύναται ἡ ΖΗ τῆς ΗΘ μεῖζον τῷ ἀπὸ τῆς Κ: ἡ ΖΗ ἄρα τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. καὶ οὐδετέρα τῶν ΖΗ, ΗΘ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ Α. ἡ ἄρα ΖΘ ἀποτομή ἐστιν ἕκτη. εὕρηται ἄρα ἡ ἕκτη ἀποτομὴ ἡ ΖΘ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",543]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",91]],["Association",["Rule","'VertexLabel'","'10.91'"],["Rule","'Text'","'If an area be contained by a rational straight line and a first apotome, the “side” of the area is an apotome.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς πρώτης, ἡ τὸ χωρίον δυναμένη ἀποτομή ἐστιν.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",6],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",53]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'For let the area AB be contained by the rational straight line AC and the first apotome AD; I say that the “side” of the area AB is an apotome. For, since AD is a first apotome, let DG be its annex; therefore AG, GD are rational straight lines commensurable in square only. [X. 73] And the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG; [X. Deff. III. 1] if therefore there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable with FG. And through the points E, F, G let EH, FI, GK be drawn parallel to AC. Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15] But AG is commensurable with AC; therefore each of the straight lines AF, FG is commensurable in length with AC. [X. 12] And AC is rational; therefore each of the straight lines AF, FG is also rational, so that each of the rectangles AI, FK is also rational. [X. 19] Now, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15] But DG is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21] Now let the square LM be made equal to AI, and let there be subtracted the square NO having a common angle with it, the angle LPM, and equal to FK; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Since then the rectangle contained by AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17] But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to KF; [VI. 1] therefore EK is a mean proportional between AI, KF. [V. 11] But MN is also a mean proportional between LM, NO, as was before proved, [Lemma after X. 53] and AI is equal to the square LM, and KF to NO; therefore MN is also equal to EK. But EK is equal to DH, and MN to LO; therefore DK is equal to the gnomon UVW and NO. But AK is also equal to the squares LM, NO; therefore the remainder AB is equal to ST. But ST is the square on LN; therefore the square on LN is equal to AB; therefore LN is the “side” of AB. I say next that LN is an apotome. For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO, therefore each of the squares LM, NO, that is, the squares on LP, PN respectively, is also rational; therefore each of the straight lines LP, PN is also rational. Again, since DH is medial and is equal to LO, therefore LO is also medial. Since then LO is medial, while NO is rational, therefore LO is incommensurable with NO. But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP is incommensurable in length with PN. [X. 11] And both are rational; therefore LP, PN are rational straight lines commensurable in square only; therefore LN is an apotome. [X. 73] And it is the “side” of the area AB; therefore the “side” of the area AB is an apotome.'"],["Rule","'ProofWordCount'",691],["Rule","'GreekProof'","'περιεχέσθω γὰρ χωρίον τὸ ΑΒ ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς πρώτης τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη ἀποτομή ἐστιν. ἐπεὶ γὰρ ἀποτομή ἐστι πρώτη ἡ ΑΔ, ἔστω αὐτῇ προσαρμόζουσα ἡ ΔΗ: αἱ ΑΗ, ΗΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ ὅλη ἡ ΑΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ, καὶ ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει: ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διαιρεῖ. τετμήσθω ἡ ΔΗ δίχα κατὰ τὸ Ε, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ: σύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ. καὶ διὰ τῶν Ε, Ζ, Η σημείων τῇ ΑΓ παράλληλοι ἤχθωσαν αἱ ΕΘ, ΖΙ, ΗΚ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΗ μήκει, καὶ ἡ ΑΗ ἄρα ἑκατέρᾳ τῶν ΑΖ, ΖΗ σύμμετρός ἐστι μήκει. ἀλλὰ ἡ ΑΗ σύμμετρός ἐστι τῇ ΑΓ: καὶ ἑκατέρα ἄρα τῶν ΑΖ, ΖΗ σύμμετρός ἐστι τῇ ΑΓ μήκει. καί ἐστι ῥητὴ ἡ ΑΓ: ῥητὴ ἄρα καὶ ἑκατέρα τῶν ΑΖ, ΖΗ: ὥστε καὶ ἑκάτερον τῶν ΑΙ, ΖΚ ῥητόν ἐστιν. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ μήκει, καὶ ἡ ΔΗ ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστι μήκει. ῥητὴ δὲ ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει: ῥητὴ ἄρα καὶ ἑκατέρα τῶν ΔΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει: ἑκάτερον ἄρα τῶν ΔΘ, ΕΚ μέσον ἐστίν. κείσθω δὴ τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον τετράγωνον ἀφῃρήσθω κοινὴν γωνίαν ἔχον αὐτῷ τὴν ὑπὸ ΛΟΜ τὸ ΝΞ: περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ὑπὸ τῶν ΑΖ, ΖΗ περιεχόμενον ὀρθογώνιον τῷ ἀπὸ τῆς ΕΗ τετραγώνῳ, ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ. ἀλλ᾽ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως τὸ ΑΙ πρὸς τὸ ΕΚ, ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΚΖ: τῶν ἄρα ΑΙ, ΚΖ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν ΛΜ, ΝΞ μέσον ἀνάλογον τὸ ΜΝ, ὡς ἐν τοῖς ἔμπροσθεν ἐδείχθη, καί ἐστι τὸ μὲν ΑΙ τῷ ΛΜ τετραγώνῳ ἴσον, τὸ δὲ ΚΖ τῷ ΝΞ: καὶ τὸ ΜΝ ἄρα τῷ ΕΚ ἴσον ἐστίν. ἀλλὰ τὸ μὲν ΕΚ τῷ ΔΘ ἐστιν ἴσον, τὸ δὲ ΜΝ τῷ ΛΞ: τὸ ἄρα ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἔστι δὲ καὶ τὸ ΑΚ ἴσον τοῖς ΛΜ, ΝΞ τετραγώνοις: λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΣΤ. τὸ δὲ ΣΤ τὸ ἀπὸ τῆς ΛΝ ἐστι τετράγωνον: τὸ ἄρα ἀπὸ τῆς ΛΝ τετράγωνον ἴσον ἐστὶ τῷ ΑΒ: ἡ ΛΝ ἄρα δύναται τὸ ΑΒ. λέγω δή, ὅτι ἡ ΛΝ ἀποτομή ἐστιν. ἐπεὶ γὰρ ῥητόν ἐστιν ἑκάτερον τῶν ΑΙ, ΖΚ, καί ἐστιν ἴσον τοῖς ΛΜ, ΝΞ, καὶ ἑκάτερον ἄρα τῶν ΛΜ, ΝΞ ῥητόν ἐστιν, τουτέστι τὸ ἀπὸ ἑκατέρας τῶν ΛΟ, ΟΝ: καὶ ἑκατέρα ἄρα τῶν ΛΟ, ΟΝ ῥητή ἐστιν. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ ΔΘ καί ἐστιν ἴσον τῷ ΛΞ, μέσον ἄρα ἐστὶ καὶ τὸ ΛΞ. ἐπεὶ οὖν τὸ μὲν ΛΞ μέσον ἐστίν, τὸ δὲ ΝΞ ῥητόν, ἀσύμμετρον ἄρα ἐστὶ τὸ ΛΞ τῷ ΝΞ: ὡς δὲ τὸ ΛΞ πρὸς τὸ ΝΞ, οὕτως ἐστὶν ἡ ΛΟ πρὸς τὴν ΟΝ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΛΟ τῇ ΟΝ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΛΟ, ΟΝ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΛΝ. καὶ δύναται τὸ ΑΒ χωρίον: ἡ ἄρα τὸ ΑΒ χωρίον δυναμένη ἀποτομή ἐστιν. ἐὰν ἄρα χωρίον περιέχηται ὑπὸ ῥητῆς, καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",599]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",92]],["Association",["Rule","'VertexLabel'","'10.92'"],["Rule","'Text'","'If an area be contained by a rational straight line and a second apotome, the “side” of the area is a first apotome of a medial straight line.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς δευτέρας, ἡ τὸ χωρίον δυναμένη μέσης ἀποτομή ἐστι πρώτη.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",6],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.2]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",74]]]],["Rule","'Proof'","'For let the area AB be contained by the rational straight line AC and the second apotome AD; I say that the “side” of the area AB is a first apotome of a medial straight line. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, [X. 73] and the annex DG is commensurable with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. [X. Deff. III. 2] Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on GD and deficient by a square figure, it divides it into commensurable parts. [X. 17] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable in length with FG. Therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15] But AG is rational and incommensurable in length with AC; therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles AI, FK is medial. [X. 21] Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE, EG. [X. 15] But DG is commensurable in length with AC. Therefore each of the rectangles DH, EK is rational. [X. 19] Let then the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM, namely the angle LPM; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Since then AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial; therefore LP, PN are also medial straight lines commensurable in square only. And, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG, [VI. 17] while, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11] But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore MN is also equal to EK. But DH is equal to EK, and LO equal to MN; therefore the whole DK is equal to the gnomon UVW and NO. Since then the whole AK is equal to LM, NO, and, in these, DK is equal to the gnomon UVW and NO, therefore the remainder AB is equal to TS. But TS is the square on LN; therefore the square on LN is equal to the area AB; therefore LN is the “side” of the area AB. I say that LN is a first apotome of a medial straight line. For, since EK is rational and is equal to LO, therefore LO, that is, the rectangle LP, PN, is rational. But NO was proved medial; therefore LO is incommensurable with NO. But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP, PN are incommensurable in length. [X. 11] Therefore LP, PN are medial straight lines commensurable in square only which contain a rational rectangle; therefore LN is a first apotome of a medial straight line. [X. 74] And it is the “side” of the area AB. Therefore the “side” of the area AB is a first apotome of a medial straight line.'"],["Rule","'ProofWordCount'",679],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς δευτέρας τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη μέσης ἀποτομή ἐστι πρώτη. ἔστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ: αἱ ἄρα ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ προσαρμόζουσα ἡ ΔΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΗΔ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διαιρεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε: καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ: σύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει. καὶ ἡ ΑΗ ἄρα ἑκατέρᾳ τῶν ΑΖ, ΖΗ σύμμετρός ἐστι μήκει. ῥητὴ δὲ ἡ ΑΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει: καὶ ἑκατέρα ἄρα τῶν ΑΖ, ΖΗ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΑΓ μήκει: ἑκάτερον ἄρα τῶν ΑΙ, ΖΚ μέσον ἐστίν. πάλιν, ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ, καὶ ἡ ΔΗ ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστιν. ἀλλ᾽ ἡ ΔΗ σύμμετρός ἐστι τῇ ΑΓ μήκει. ῥητὴ ἄρα καὶ ἑκατέρα τῶν ΔΕ, ΕΗ καὶ σύμμετρος τῇ ΑΓ μήκει. ἑκάτερον ἄρα τῶν ΔΘ, ΕΚ ῥητόν ἐστιν. συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον ἀφῃρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν γωνίαν ὂν τῷ ΛΜ τὴν ὑπὸ τῶν ΛΟΜ: περὶ τὴν αὐτὴν ἄρα ἐστὶ διάμετρον τὰ ΛΜ, ΝΞ τετράγωνα. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν τὰ ΑΙ, ΖΚ μέσα ἐστὶ καί ἐστιν ἴσα τοῖς ἀπὸ τῶν ΛΟ, ΟΝ, καὶ τὰ ἀπὸ τῶν ΛΟ, ΟΝ ἄρα μέσα ἐστίν: καὶ αἱ ΛΟ, ΟΝ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΖ, ΖΗ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΗ, ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ: ἀλλ᾽ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως τὸ ΑΙ πρὸς τὸ ΕΚ: ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΖΚ: τῶν ἄρα ΑΙ, ΖΚ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν ΛΜ, ΝΞ τετραγώνων μέσον ἀνάλογον τὸ ΜΝ: καί ἐστιν ἴσον τὸ μὲν ΑΙ τῷ ΛΜ, τὸ δὲ ΖΚ τῷ ΝΞ: καὶ τὸ ΜΝ ἄρα ἴσον ἐστὶ τῷ ΕΚ. ἀλλὰ τῷ μὲν ΕΚ ἴσον ἐστὶ τὸ ΔΘ, τῷ δὲ ΜΝ ἴσον τὸ ΛΞ: ὅλον ἄρα τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἐπεὶ οὖν ὅλον τὸ ΑΚ ἴσον ἐστὶ τοῖς ΛΜ, ΝΞ, ὧν τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ, λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΤΣ. τὸ δὲ ΤΣ ἐστι τὸ ἀπὸ τῆς ΛΝ: τὸ ἀπὸ τῆς ΛΝ ἄρα ἴσον ἐστὶ τῷ ΑΒ χωρίῳ: ἡ ΛΝ ἄρα δύναται τὸ ΑΒ χωρίον. λέγω δή, ὅτι ἡ ΛΝ μέσης ἀποτομή ἐστι πρώτη. ἐπεὶ γὰρ ῥητόν ἐστι τὸ ΕΚ καί ἐστιν ἴσον τῷ ΛΞ, ῥητὸν ἄρα ἐστὶ τὸ ΛΞ, τουτέστι τὸ ὑπὸ τῶν ΛΟ, ΟΝ. μέσον δὲ ἐδείχθη τὸ ΝΞ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΛΞ τῷ ΝΞ: ὡς δὲ τὸ ΛΞ πρὸς τὸ ΝΞ, οὕτως ἐστὶν ἡ ΛΟ πρὸς ΟΝ: αἱ ΛΟ, ΟΝ ἄρα ἀσύμμετροί εἰσι μήκει. αἱ ἄρα ΛΟ, ΟΝ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι: ἡ ΛΝ ἄρα μέσης ἀποτομή ἐστι πρώτη: καὶ δύναται τὸ ΑΒ χωρίον. ἡ ἄρα τὸ ΑΒ χωρίον δυναμένη μέσης ἀποτομή ἐστι πρώτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",580]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",93]],["Association",["Rule","'VertexLabel'","'10.93'"],["Rule","'Text'","'If an area be contained by a rational straight line and a third apotome, the “side” of the area is a second apotome of a medial straight line.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς τρίτης, ἡ τὸ χωρίον δυναμένη μέσης ἀποτομή ἐστι δευτέρα.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",6],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.3]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",75]]]],["Rule","'Proof'","'For let the area AB be contained by the rational straight line AC and the third apotome AD; I say that the “side” of the area AB is a second apotome of a medial straight line. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, and neither of the straight lines AG, GD is commensurable in length with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex DG by the square on a straight line commensurable with AG. [X. Deff. III. 3] Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into commensurable parts. [X. 17] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG. Let EH, FI, GK be drawn through the points E, F, G parallel to AC. Therefore AF, FG are commensurable; therefore AI is also commensurable with FK. [VI. 1, X. 11] And, since AF, FG are commensurable in length, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15] But AG is rational and incommensurable in length with AC; so that AF, FG are so also. [X. 13] Therefore each of the rectangles AI, FK is medial. [X. 21] Again, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15] But GD is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21] And, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD. But AG is commensurable in length with AF, and DG with EG; therefore AF is incommensurable in length with EG. [X. 13] But, as AF is to EG, so is AI to EK; [VI. 1] therefore AI is incommensurable with EK. [X. 11] Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM; therefore LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Now, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17] But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore also, as AI is to EK, so is EK to FK; [V. 11] therefore EK is a mean proportional between AI, FK. But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN. But MN is equal to LO, and EK equal to DH; therefore the whole DK is also equal to the gnomon UVW and NO. But AK is also equal to LM, NO; therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the “side” of the area AB. I say that LN is a second apotome of a medial straight line. For, since AI, FK were proved medial, and are equal to the squares on LP, PN, therefore each of the squares on LP, PN is also medial; therefore each of the straight lines LP, PN is medial. And, since AI is commensurable with FK, [VI. 1, X. 11] therefore the square on LP is also commensurable with the square on PN. Again, since AI was proved incommensurable with EK, therefore LM is also incommensurable with MN, that is, the square on LP with the rectangle LP, PN; so that LP is also incommensurable in length with PN; [VI. 1, X. 11] therefore LP, PN are medial straight lines commensurable in square only. I say next that they also contain a medial rectangle. For, since EK was proved medial, and is equal to the rectangle LP, PN, therefore the rectangle LP, PN is also medial, so that LP, PN are medial straight lines commensurable in square only which contain a medial rectangle. Therefore LN is a second apotome of a medial straight line; [X. 75] and it is the “side” of the area AB. Therefore the “side” of the area AB is a second apotome of a medial straight line.'"],["Rule","'ProofWordCount'",827],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς τρίτης τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη μέσης ἀποτομή ἐστι δευτέρα. ἔστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ: αἱ ΑΗ, ΗΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα τῶν ΑΗ, ΗΔ σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διελεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ. καὶ ἤχθωσαν διὰ τῶν Ε, Ζ, Η σημείων τῇ ΑΓ παράλληλοι αἱ ΕΘ, ΖΙ, ΗΚ: σύμμετροι ἄρα εἰσὶν αἱ ΑΖ, ΖΗ: σύμμετρον ἄρα καὶ τὸ ΑΙ τῷ ΖΚ. καὶ ἐπεὶ αἱ ΑΖ, ΖΗ σύμμετροί εἰσι μήκει, καὶ ἡ ΑΗ ἄρα ἑκατέρᾳ τῶν ΑΖ, ΖΗ σύμμετρός ἐστι μήκει. ῥητὴ δὲ ἡ ΑΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει: ὥστε καὶ αἱ ΑΖ, ΖΗ. ἑκάτερον ἄρα τῶν ΑΙ, ΖΚ μέσον ἐστίν. πάλιν, ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ μήκει, καὶ ἡ ΔΗ ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστι μήκει. ῥητὴ δὲ ἡ ΗΔ καὶ ἀσύμμετρος τῇ ΑΓ μήκει: ῥητὴ ἄρα καὶ ἑκατέρα τῶν ΔΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει: ἑκάτερον ἄρα τῶν ΔΘ, ΕΚ μέσον ἐστίν. καὶ ἐπεὶ αἱ ΑΗ, ΗΔ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα ἐστὶ μήκει ἡ ΑΗ τῇ ΗΔ. ἀλλ᾽ ἡ μὲν ΑΗ τῇ ΑΖ σύμμετρός ἐστι μήκει, ἡ δὲ ΔΗ τῇ ΕΗ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΕΗ μήκει. ὡς δὲ ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΕΚ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΙ τῷ ΕΚ. συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον ἀφῃρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν γωνίαν ὂν τῷ ΛΜ: περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν τὸ ὑπὸ τῶν ΑΖ, ΖΗ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΗ, ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ. ἀλλ᾽ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΕΚ: ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΖΚ: καὶ ὡς ἄρα τὸ ΑΙ πρὸς τὸ ΕΚ, οὕτως τὸ ΕΚ πρὸς τὸ ΖΚ: τῶν ἄρα ΑΙ, ΖΚ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν ΛΜ, ΝΞ τετραγώνων μέσον ἀνάλογον τὸ ΜΝ: καί ἐστιν ἴσον τὸ μὲν ΑΙ τῷ ΛΜ, τὸ δὲ ΖΚ τῷ ΝΞ: καὶ τὸ ΕΚ ἄρα ἴσον ἐστὶ τῷ ΜΝ. ἀλλὰ τὸ μὲν ΜΝ ἴσον ἐστὶ τῷ ΛΞ, τὸ δὲ ΕΚ ἴσον ἐστὶ τῷ ΔΘ: καὶ ὅλον ἄρα τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἔστι δὲ καὶ τὸ ΑΚ ἴσον τοῖς ΛΜ, ΝΞ: λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΣΤ, τουτέστι τῷ ἀπὸ τῆς ΛΝ τετραγώνῳ: ἡ ΛΝ ἄρα δύναται τὸ ΑΒ χωρίον. λέγω, ὅτι ἡ ΛΝ μέσης ἀποτομή ἐστι δευτέρα. ἐπεὶ γὰρ μέσα ἐδείχθη τὰ ΑΙ, ΖΚ καί ἐστιν ἴσα τοῖς ἀπὸ τῶν ΛΟ, ΟΝ, μέσον ἄρα καὶ ἑκάτερον τῶν ἀπὸ τῶν ΛΟ, ΟΝ: μέση ἄρα ἑκατέρα τῶν ΛΟ, ΟΝ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ΑΙ τῷ ΖΚ, σύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ. πάλιν, ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΙ τῷ ΕΚ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΛΜ τῷ ΜΝ, τουτέστι τὸ ἀπὸ τῆς ΛΟ τῷ ὑπὸ τῶν ΛΟ, ΟΝ: ὥστε καὶ ἡ ΛΟ ἀσύμμετρός ἐστι τῇ ΟΝ: αἱ ΛΟ, ΟΝ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι. λέγω δή, ὅτι καὶ μέσον περιέχουσιν. ἐπεὶ γὰρ μέσον ἐδείχθη τὸ ΕΚ καί ἐστιν ἴσον τῷ ὑπὸ τῶν ΛΟ, ΟΝ, μέσον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΛΟ, ΟΝ: ὥστε αἱ ΛΟ, ΟΝ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι. ἡ ΛΝ ἄρα μέσης ἀποτομή ἐστι δευτέρα: καὶ δύναται τὸ ΑΒ χωρίον. ἡ ἄρα τὸ ΑΒ χωρίον δυναμένη μέσης ἀποτομή ἐστι δευτέρα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",683]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",94]],["Association",["Rule","'VertexLabel'","'10.94'"],["Rule","'Text'","'If an area be contained by a rational straight line and a fourth apotome, the “side” of the area is minor.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς τετάρτης, ἡ τὸ χωρίον δυναμένη ἐλάσσων ἐστίν.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",6],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",76]]]],["Rule","'Proof'","'For let the area AB be contained by the rational straight line AC and the fourth apotome AD; I say that the “side” of the area AB is minor. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, AG is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG, [X. Deff. III. 4] Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG. Let EH, FI, GK be drawn through E, F, G parallel to AC, BD. Since then AG is rational and commensurable in length with AC, therefore the whole AK is rational. [X. 19] Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial. [X. 21] Again, since AF is incommensurable in length with FG, therefore AI is also incommensurable with FK. [VI. 1, X. 11] Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and about the same angle, the angle LPM. Therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Since then the rectangle AF, FG is equal to the square on EG, therefore, proportionally, as AF is to EG, so is EG to FG. [VI. 17] But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11] But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN. But DH is equal to EK, and LO is equal to MN; therefore the whole DK is equal to the gnomon UVW and NO. Since, then, the whole AK is equal to the squares LM, NO, and, in these, DK is equal to the gnomon UVW and the square NO, therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the “side” of the area AB. I say that LN is the irrational straight line called minor. For, since AK is rational and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is rational. Again, since DK is medial, and DK is equal to twice the rectangle LP, PN, therefore twice the rectangle LP, PN is medial. And, since AI was proved incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN. Therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. Therefore LN is the irrational straight line called minor; [X. 76] and it is the “side” of the area AB. Therefore the “side” of the area AB is minor.'"],["Rule","'ProofWordCount'",618],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς τετάρτης τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη ἐλάσσων ἐστίν. ἔστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ: αἱ ἄρα ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ μήκει, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διελεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ: ἀσύμμετρος ἄρα ἐστὶ μήκει ἡ ΑΖ τῇ ΖΗ. ἤχθωσαν οὖν διὰ τῶν Ε, Ζ, Η παράλληλοι ταῖς ΑΓ, ΒΔ αἱ ΕΘ, ΖΙ, ΗΚ. ἐπεὶ οὖν ῥητή ἐστιν ἡ ΑΗ καὶ σύμμετρος τῇ ΑΓ μήκει, ῥητὸν ἄρα ἐστὶν ὅλον τὸ ΑΚ. πάλιν, ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΔΗ τῇ ΑΓ μήκει, καί εἰσιν ἀμφότεραι ῥηταί, μέσον ἄρα ἐστὶ τὸ ΔΚ. πάλιν, ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΗ μήκει, ἀσύμμετρον ἄρα καὶ τὸ ΑΙ τῷ ΖΚ. συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον ἀφῃρήσθω περὶ τὴν αὐτὴν γωνίαν τὴν ὑπὸ τῶν ΛΟΜ τὸ ΝΞ. περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν τὸ ὑπὸ τῶν ΑΖ, ΖΗ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΗ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ. ἀλλ᾽ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΕΚ, ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΖΚ: τῶν ἄρα ΑΙ, ΖΚ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν ΛΜ, ΝΞ τετραγώνων μέσον ἀνάλογον τὸ ΜΝ, καί ἐστιν ἴσον τὸ μὲν ΑΙ τῷ ΛΜ, τὸ δὲ ΖΚ τῷ ΝΞ: καὶ τὸ ΕΚ ἄρα ἴσον ἐστὶ τῷ ΜΝ. ἀλλὰ τῷ μὲν ΕΚ ἴσον ἐστὶ τὸ ΔΘ, τῷ δὲ ΜΝ ἴσον ἐστὶ τὸ ΛΞ: ὅλον ἄρα τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἐπεὶ οὖν ὅλον τὸ ΑΚ ἴσον ἐστὶ τοῖς ΛΜ, ΝΞ τετραγώνοις, ὧν τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ τετραγώνῳ, λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΣΤ, τουτέστι τῷ ἀπὸ τῆς ΛΝ τετραγώνῳ: ἡ ΛΝ ἄρα δύναται τὸ ΑΒ χωρίον. λέγω, ὅτι ἡ ΛΝ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων. ἐπεὶ γὰρ ῥητόν ἐστι τὸ ΑΚ καί ἐστιν ἴσον τοῖς ἀπὸ τῶν ΛΟ, ΟΝ τετραγώνοις, τὸ ἄρα συγκείμενον ἐκ τῶν ἀπὸ τῶν ΛΟ, ΟΝ ῥητόν ἐστιν. πάλιν, ἐπεὶ τὸ ΔΚ μέσον ἐστίν, καί ἐστιν ἴσον τὸ ΔΚ τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, τὸ ἄρα δὶς ὑπὸ τῶν ΛΟ, ΟΝ μέσον ἐστίν. καὶ ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΙ τῷ ΖΚ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τετράγωνον τῷ ἀπὸ τῆς ΟΝ τετραγώνῳ. αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δὲ δὶς ὑπ᾽ αὐτῶν μέσον. ἡ ΛΝ ἄρα ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων: καὶ δύναται τὸ ΑΒ χωρίον. ἡ ἄρα τὸ ΑΒ χωρίον δυναμένη ἐλάσσων ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",546]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",95]],["Association",["Rule","'VertexLabel'","'10.95'"],["Rule","'Text'","'If an area be contained by a rational straight line and a fifth apotome, the “side” of the area is a straight line which produces with a rational area a medial whole.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς πέμπτης, ἡ τὸ χωρίον δυναμένη ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",77]]]],["Rule","'Proof'","'For let the area AB be contained by the rational straight line AC and the fifth apotome AD; I say that the “side” of the area AB is a straight line which produces with a rational area a medial whole. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. [X. Deff. III. 5] Therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] Let then DG be bisected at the point E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG. Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. [X. 21] Again, since DG is rational and commensurable in length with AC, DK is rational. [X. 19] Now let the square LM be constructed equal to AI, and let the square NO equal to FK and about the same angle, the angle LPM, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Similarly then we can prove that LN is the “side” of the area AB. I say that LN is the straight line which produces with a rational area a medial whole. For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial. Again, since DK is rational and is equal to twice the rectangle LP, PN, the latter is itself also rational. And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole; [X. 77] and it is the “side” of the area AB. Therefore the “side” of the area AB is a straight line which produces with a rational area a medial whole.'"],["Rule","'ProofWordCount'",444],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς πέμπτης τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν. ἔστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ: αἱ ἄρα ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ προσαρμόζουσα ἡ ΗΔ σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διελεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε σημεῖον, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΓΑ μήκει, καί εἰσιν ἀμφότεραι ῥηταί, μέσον ἄρα ἐστὶ τὸ ΑΚ. πάλιν, ἐπεὶ ῥητή ἐστιν ἡ ΔΗ καὶ σύμμετρος τῇ ΑΓ μήκει, ῥητόν ἐστι τὸ ΔΚ. συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον τετράγωνον ἀφῃρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν γωνίαν τὴν ὑπὸ ΛΟΜ: περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ὁμοίως δὴ δείξομεν, ὅτι ἡ ΛΝ δύναται τὸ ΑΒ χωρίον. λέγω, ὅτι ἡ ΛΝ ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν. ἐπεὶ γὰρ μέσον ἐδείχθη τὸ ΑΚ καί ἐστιν ἴσον τοῖς ἀπὸ τῶν ΛΟ, ΟΝ, τὸ ἄρα συγκείμενον ἐκ τῶν ἀπὸ τῶν ΛΟ, ΟΝ μέσον ἐστίν. πάλιν, ἐπεὶ ῥητόν ἐστι τὸ ΔΚ καί ἐστιν ἴσον τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, καὶ αὐτὸ ῥητόν ἐστιν. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ΑΙ τῷ ΖΚ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ: αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δὲ δὶς ὑπ᾽ αὐτῶν ῥητόν. ἡ λοιπὴ ἄρα ἡ ΛΝ ἄλογός ἐστιν ἡ καλουμένη μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα: καὶ δύναται τὸ ΑΒ χωρίον. ἡ τὸ ΑΒ ἄρα χωρίον δυναμένη μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",354]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",96]],["Association",["Rule","'VertexLabel'","'10.96'"],["Rule","'Text'","'If an area be contained by a rational straight line and a sixth apotome, the “side” of the area is a straight line which produces with a medial area a medial whole.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς ἕκτης, ἡ τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",26]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",21]],["List",["Rule","'Book'",10],["Rule","'Theorem'",78]]]],["Rule","'Proof'","'For let the area AB be contained by the rational straight line AC and the sixth apotome AD; I say that the “side” of the area AB is a straight line which produces with a medial area a medial whole. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, neither of them is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. [X. Deff. III. 6] Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG. But, as AF is to FG, so is AI to FK. [VI. 1] therefore AI is incommensurable with FK. [X. 11] And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21] Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21] Now, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD. But, as AG is to GD, so is AK to KD; [VI. 1] therefore AK is incommensurable with KD. [X. 11] Now let the square LM be constructed equal to AI, and let NO equal to FK, and about the same angle, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Then in manner similar to the above we can prove that LN is the “side” of the area AB. I say that LN is a straight line which produces with a medial area a medial whole. For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial. Again, since DK was proved medial and is equal to twice the rectangle LP, PN, twice the rectangle LP, PN is also medial. And, since AK was proved incommensurable with DK, the squares on LP, PN are also incommensurable with twice the rectangle LP, PN. And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them. Therefore LN is the irrational straight line called that which produces with a medial area a medial whole; [X. 78] and it is the “side” of the area AB. Therefore the “side” of the area is a straight line which produces with a medial area a medial whole.'"],["Rule","'ProofWordCount'",563],["Rule","'GreekProof'","'χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς ἕκτης τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν. ἔστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ: αἱ ἄρα ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα αὐτῶν σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ μήκει, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διελεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε σημεῖον, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει. ὡς δὲ ἡ ΑΖ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΖΚ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΙ τῷ ΖΚ. καὶ ἐπεὶ αἱ ΑΗ, ΑΓ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, μέσον ἐστὶ τὸ ΑΚ. πάλιν, ἐπεὶ αἱ ΑΓ, ΔΗ ῥηταί εἰσι καὶ ἀσύμμετροι μήκει, μέσον ἐστὶ καὶ τὸ ΔΚ. ἐπεὶ οὖν αἱ ΑΗ, ΗΔ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΗ τῇ ΗΔ μήκει. ὡς δὲ ἡ ΑΗ πρὸς τὴν ΗΔ, οὕτως ἐστὶ τὸ ΑΚ πρὸς τὸ ΚΔ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΚ τῷ ΚΔ. συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον ἀφῃρήσθω περὶ τὴν αὐτὴν γωνίαν τὸ ΝΞ: περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ὁμοίως δὴ τοῖς ἐπάνω δείξομεν, ὅτι ἡ ΛΝ δύναται τὸ ΑΒ χωρίον. λέγω, ὅτι ἡ ΛΝ ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν. ἐπεὶ γὰρ μέσον ἐδείχθη τὸ ΑΚ καί ἐστιν ἴσον τοῖς ἀπὸ τῶν ΛΟ, ΟΝ, τὸ ἄρα συγκείμενον ἐκ τῶν ἀπὸ τῶν ΛΟ, ΟΝ μέσον ἐστίν. πάλιν, ἐπεὶ μέσον ἐδείχθη τὸ ΔΚ καί ἐστιν ἴσον τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, καὶ τὸ δὶς ὑπὸ τῶν ΛΟ, ΟΝ μέσον ἐστίν. καὶ ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΚ τῷ ΔΚ, ἀσύμμετρα ἄρα ἐστὶ καὶ τὰ ἀπὸ τῶν ΛΟ, ΟΝ τετράγωνα τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ΑΙ τῷ ΖΚ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ: αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον καὶ τὸ δὶς ὑπ᾽ αὐτῶν μέσον ἔτι τε τὰ ἀπ᾽ αὐτῶν τετράγωνα ἀσύμμετρα τῷ δὶς ὑπ᾽ αὐτῶν. ἡ ἄρα ΛΝ ἄλογός ἐστιν ἡ καλουμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα: καὶ δύναται τὸ ΑΒ χωρίον. ἡ ἄρα τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",457]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",97]],["Association",["Rule","'VertexLabel'","'10.97'"],["Rule","'Text'","'The square on an apotome applied to a rational straight line produces as breadth a first apotome.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'τὸ ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν πρώτην.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let AB be an apotome, and CD rational, and to CD let there be applied CE equal to the square on AB and producing CF as breadth; I say that CF is a first apotome. For let BG be the annex to AB; therefore AG, GB are rational straight lines commensurable in square only. [X. 73] To CD let there be applied CH equal to the square on AG, and KL equal to the square on BG. Therefore the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB; therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7] Let FM be bisected at the point N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, LN is equal to the rectangle AG, GB. Now, since the squares on AG, GB are rational, and DM is equal to the squares on AG, GB,. therefore DM is rational. And it has been applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and commensurable in length with CD. [X. 20] Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial. And it is applied to the rational straight line CD, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22] And, since the squares on AG, GB are rational, while twice the rectangle AG, GB is medial, therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB. And CL is equal to the squares on AG, GB, and FL to twice the rectangle AG, GB; therefore DM is incommensurable with FL. But, as DM is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a first apotome. For, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on BG, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM. And, since the square on AG is commensurable with the square on GB, CH is also commensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is commensurable with KM. [X. 11] Since then CM, MF are two unequal straight lines, and to CM there has been applied the rectangle CK, KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17] And CM is commensurable in length with the rational straight line CD set out; therefore CF is a first apotome. [X. Deff. III. 1]'"],["Rule","'ProofWordCount'",597],["Rule","'GreekProof'","'ἔστω ἀποτομὴ ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος ποιοῦν τὴν ΓΖ: λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι πρώτη. ἔστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ: αἱ ἄρα ΑΗ, ΗΒ ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΒΗ τὸ ΚΛ. ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ: ὧν τὸ ΓΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ: λοιπὸν ἄρα τὸ ΖΛ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. τετμήσθω ἡ ΖΜ δίχα κατὰ τὸ Ν σημεῖον, καὶ ἤχθω διὰ τοῦ Ν τῇ ΓΔ παράλληλος ἡ ΝΞ: ἑκάτερον ἄρα τῶν ΖΞ, ΛΝ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ ῥητά ἐστιν, καί ἐστι τοῖς ἀπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΔΜ, ῥητὸν ἄρα ἐστὶ τὸ ΔΜ. καὶ παρὰ ῥητὴν τὴν ΓΔ παραβέβληται πλάτος ποιοῦν τὴν ΓΜ: ῥητὴ ἄρα ἐστὶν ἡ ΓΜ καὶ σύμμετρος τῇ ΓΔ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ, καὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΖΛ, μέσον ἄρα τὸ ΖΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ παράκειται πλάτος ποιοῦν τὴν ΖΜ: ῥητὴ ἄρα ἐστὶν ἡ ΖΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὰ μὲν ἀπὸ τῶν ΑΗ, ΗΒ ῥητά ἐστιν, τὸ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον, ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. καὶ τοῖς μὲν ἀπὸ τῶν ΑΗ, ΗΒ ἴσον ἐστὶ τὸ ΓΛ, τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ τὸ ΖΛ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΔΜ τῷ ΖΛ. ὡς δὲ τὸ ΔΜ πρὸς τὸ ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΖΜ. ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΜ τῇ ΖΜ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ἄρα ΓΜ, ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ΓΖ ἄρα ἀποτομή ἐστιν. λέγω δή, ὅτι καὶ πρώτη. ἐπεὶ γὰρ τῶν ἀπὸ τῶν ΑΗ, ΗΒ μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΒΗ ἴσον τὸ ΚΛ, τῷ δὲ ὑπὸ τῶν ΑΗ, ΗΒ τὸ ΝΛ, καὶ τῶν ΓΘ, ΚΛ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΝΛ: ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. ἀλλ᾽ ὡς μὲν τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΝΜ: ὡς δὲ τὸ ΝΛ πρὸς τὸ ΚΛ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΚΜ: τὸ ἄρα ὑπὸ τῶν ΓΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΝΜ, τουτέστι τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΗΒ, σύμμετρόν ἐστι καὶ τὸ ΓΘ τῷ ΚΛ. ὡς δὲ τὸ ΓΘ πρὸς τὸ ΚΛ, οὕτως ἡ ΓΚ πρὸς τὴν ΚΜ: σύμμετρος ἄρα ἐστὶν ἡ ΓΚ τῇ ΚΜ. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί εἰσιν αἱ ΓΜ, ΜΖ, καὶ τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ ἴσον παρὰ τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ τὸ ὑπὸ τῶν ΓΚ, ΚΜ, καί ἐστι σύμμετρος ἡ ΓΚ τῇ ΚΜ, ἡ ἄρα ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καί ἐστιν ἡ ΓΜ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ μήκει: ἡ ἄρα ΓΖ ἀποτομή ἐστι πρώτη. τὸ ἄρα ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν πρώτην: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",555]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",98]],["Association",["Rule","'VertexLabel'","'10.98'"],["Rule","'Text'","'The square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'τὸ ἀπὸ μέσης ἀποτομῆς πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν δευτέραν.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.2]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",74]]]],["Rule","'Proof'","'Let AB be a first apotome of a medial straight line and CD a rational straight line, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a second apotome. For let BG be the annex to AB;. therefore AG, GB are medial straight lines commensurable in square only which contain a rational rectangle. [X. 74] To CD let there be applied CH equal to the square on AG, producing CK as breadth, and KL equal to the square on GB, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23] And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22] Now, since CL is equal to the squares on AG, GB, and, in these, the square on AB is equal to CE, therefore the remainder, twice the rectangle AG, GB, is equal to FL. [II. 7] But twice the rectangle AG, GB is rational; therefore FL is rational. And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is also rational and commensurable in length with CD. [X. 20] Now, since the sum of the squares on AG, GB, that is, CL, is medial, while twice the rectangle AG, GB, that is, FL, is rational, therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a second apotome. For let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. Now, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the rectangle AG, GB to NL, and the square on BG to KL, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to MK; [VI. 1] therefore, as CK is to NM, so is NM, so is KM; [V. 11] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM. Since the CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to the greater, CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17] And the annex FM is commensurable in length with the rational straight line CD set out; therefore CF is a second apotome. [X. Deff. III. 2]'"],["Rule","'ProofWordCount'",555],["Rule","'GreekProof'","'ἔστω μέσης ἀποτομὴ πρώτη ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος ποιοῦν τὴν ΓΖ: λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι δευτέρα. ἔστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ: αἱ ἄρα ΑΗ, ΗΒ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι. καὶ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ πλάτος ποιοῦν τὴν ΓΚ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ ΚΛ πλάτος ποιοῦν τὴν ΚΜ: ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ: μέσον ἄρα καὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ παράκειται πλάτος ποιοῦν τὴν ΓΜ: ῥητὴ ἄρα ἐστὶν ἡ ΓΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, ὧν τὸ ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τῷ ΓΕ, λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον ἐστὶ τῷ ΖΛ. ῥητὸν δὲ ἐστι τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ: ῥητὸν ἄρα τὸ ΖΛ. καὶ παρὰ ῥητὴν τὴν ΖΕ παράκειται πλάτος ποιοῦν τὴν ΖΜ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΜ καὶ σύμμετρος τῇ ΓΔ μήκει. ἐπεὶ οὖν τὰ μὲν ἀπὸ τῶν ΑΗ, ΗΒ, τουτέστι τὸ ΓΛ, μέσον ἐστίν, τὸ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ, τουτέστι τὸ ΖΛ, ῥητόν, ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΖΜ: ἀσύμμετρος ἄρα ἡ ΓΜ τῇ ΖΜ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ἄρα ΓΜ, ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ΓΖ ἄρα ἀποτομή ἐστιν. λέγω δή, ὅτι καὶ δευτέρα. τετμήσθω γὰρ ἡ ΖΜ δίχα κατὰ τὸ Ν, καὶ ἤχθω διὰ τοῦ Ν τῇ ΓΔ παράλληλος ἡ ΝΞ: ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ τῶν ἀπὸ τῶν ΑΗ, ΗΒ τετραγώνων μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστιν ἴσον τὸ μὲν ἀπὸ τῆς ΑΗ τῷ ΓΘ, τὸ δὲ ὑπὸ τῶν ΑΗ, ΗΒ τῷ ΝΛ, τὸ δὲ ἀπὸ τῆς ΒΗ τῷ ΚΛ, καὶ τῶν ΓΘ, ΚΛ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΝΛ: ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. ἀλλ᾽ ὡς μὲν τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΝΜ, ὡς δὲ τὸ ΝΛ πρὸς τὸ ΚΛ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΜΚ: ὡς ἄρα ἡ ΓΚ πρὸς τὴν ΝΜ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΚΜ: τὸ ἄρα ὑπὸ τῶν ΓΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΝΜ, τουτέστι τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΒΗ, σύμμετρόν ἐστι καὶ τὸ ΓΘ τῷ ΚΛ, τουτέστιν ἡ ΓΚ τῇ ΚΜ. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί εἰσιν αἱ ΓΜ, ΜΖ, καὶ τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΜΖ ἴσον παρὰ τὴν μείζονα τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ τὸ ὑπὸ τῶν ΓΚ, ΚΜ καὶ εἰς σύμμετρα αὐτὴν διαιρεῖ, ἡ ἄρα ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καί ἐστιν ἡ προσαρμόζουσα ἡ ΖΜ σύμμετρος μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ: ἡ ἄρα ΓΖ ἀποτομή ἐστι δευτέρα. τὸ ἄρα ἀπὸ μέσης ἀποτομῆς πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν δευτέραν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",524]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",99]],["Association",["Rule","'VertexLabel'","'10.99'"],["Rule","'Text'","'The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'τὸ ἀπὸ μέσης ἀποτομῆς δευτέρας παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν τρίτην.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.3]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",75]]]],["Rule","'Proof'","'Let AB be a second apotome of a medial straight line, and CD rational, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a third apotome. For let BG be the annex to AB; therefore AG, GB are medial straight lines commensurable in square only which contain a medial rectangle. [X. 75] Let CH equal to the square on AG be applied to CD, producing CK as breadth, and let KL equal to the square on BG be applied to KH, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23] And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22] Now, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder LF is equal to twice the rectangle AG, GB. [II. 7] Let then FM be bisected at the point N, and let NO be drawn parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. But the rectangle AG, GB is medial; therefore FL is also medial. And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is also rational and incommensurable in length with CD. [X. 22] And, since AG, GB are commensurable in square only, therefore AG is incommensurable in length with GB; therefore the square on AG is also incommensurable with the rectangle AG, GB. [VI. 1, X. 11] But the squares on AG, GB are commensurable with the square on AG, and twice the rectangle AG, GB with the rectangle AG, GB; therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB. [X. 13] But CL is equal to the squares on AG, GB, and FL is equal to twice the rectangle AG, GB; therefore CL is also incommensurable with FL. But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a third apotome. For, since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM. [VI. 1, X. 11] And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to [the square on MN, that is, to]the fourth part of the square on FM. Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable with CM. [X. 17] And neither of the straight lines CM, MF is commensurable in length with the rational straight line CD set out; therefore CF is a third apotome. [X. Deff. III. 3]'"],["Rule","'ProofWordCount'",656],["Rule","'GreekProof'","'ἔστω μέσης ἀποτομὴ δευτέρα ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος ποιοῦν τὴν ΓΖ: λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι τρίτη. ἔστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ: αἱ ἄρα ΑΗ, ΗΒ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι. καὶ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ πλάτος ποιοῦν τὴν ΓΚ, τῷ δὲ ἀπὸ τῆς ΒΗ ἴσον παρὰ τὴν ΚΘ παραβεβλήσθω τὸ ΚΛ πλάτος ποιοῦν τὴν ΚΜ: ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ καί ἐστι μέσα τὰ ἀπὸ τῶν ΑΗ, ΗΒ: μέσον ἄρα καὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ παραβέβληται πλάτος ποιοῦν τὴν ΓΜ: ῥητὴ ἄρα ἐστὶν ἡ ΓΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ ὅλον τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, ὧν τὸ ΓΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ, λοιπὸν ἄρα τὸ ΛΖ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. τετμήσθω οὖν ἡ ΖΜ δίχα κατὰ τὸ Ν σημεῖον, καὶ τῇ ΓΔ παράλληλος ἤχθω ἡ ΝΞ: ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. μέσον δὲ τὸ ὑπὸ τῶν ΑΗ, ΗΒ: μέσον ἄρα ἐστὶ καὶ τὸ ΖΛ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν ΖΜ: ῥητὴ ἄρα καὶ ἡ ΖΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ αἱ ΑΗ, ΗΒ δυνάμει μόνον εἰσὶ σύμμετροι, ἀσύμμετρος ἄρα ἐστὶ μήκει ἡ ΑΗ τῇ ΗΒ: ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΗ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΗ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΗ, ΗΒ, τῷ δὲ ὑπὸ τῶν ΑΗ, ΗΒ τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ: ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΑΗ, ΗΒ ἴσον ἐστὶ τὸ ΓΛ, τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον ἐστὶ τὸ ΖΛ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΖΜ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΜ τῇ ΖΜ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ἄρα ΓΜ, ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΓΖ. λέγω δή, ὅτι καὶ τρίτη. ἐπεὶ γὰρ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΗΒ, σύμμετρον ἄρα καὶ τὸ ΓΘ τῷ ΚΛ: ὥστε καὶ ἡ ΓΚ τῇ ΚΜ. καὶ ἐπεὶ τῶν ἀπὸ τῶν ΑΗ, ΗΒ μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ ΚΛ, τῷ δὲ ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΝΛ, καὶ τῶν ΓΘ, ΚΛ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΝΛ: ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. ἀλλ᾽ ὡς μὲν τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΝΜ, ὡς δὲ τὸ ΝΛ πρὸς τὸ ΚΛ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΚΜ: ὡς ἄρα ἡ ΓΚ πρὸς τὴν ΜΝ, οὕτως ἐστὶν ἡ ΜΝ πρὸς τὴν ΚΜ: τὸ ἄρα ὑπὸ τῶν ΓΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, τουτέστι τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί εἰσιν αἱ ΓΜ, ΜΖ, καὶ τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ ἴσον παρὰ τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς σύμμετρα αὐτὴν διαιρεῖ, ἡ ΓΜ ἄρα τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ οὐδετέρα τῶν ΓΜ, ΜΖ σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ: ἡ ἄρα ΓΖ ἀποτομή ἐστι τρίτη. τὸ ἄρα ἀπὸ μέσης ἀποτομῆς δευτέρας παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν τρίτην: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",596]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",100]],["Association",["Rule","'VertexLabel'","'10.100'"],["Rule","'Text'","'The square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'τὸ ἀπὸ ἐλάσσονος παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν τετάρτην.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",76]]]],["Rule","'Proof'","'Let AB be a minor and CD a rational straight line, and to the rational straight line CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fourth apotome. For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on AG, GB rational, but twice the rectangle AG, GB medial. [X. 76] To CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB. And the sum of the squares on AG, GB is rational; therefore CL is also rational. And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is also rational and commensurable in length with CD. [X. 20] And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7] Let then FM be bisected at the point N, and let NO be drawn through N parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. And, since twice the rectangle AG, GB is medial and is equal to FL, therefore FL is also medial. And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22] And, since the sum of the squares on AG, GB is rational, while twice the rectangle AG, GB is medial, the squares on AG, GB are incommensurable with twice the rectangle AG, GB. But CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB; therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say that it is also a fourth apotome. For, since AG, GB are incommensurable in square, therefore the square on AG is also incommensurable with the square on GB. And CH is equal to the square on AG, and KL equal to the square on GB; therefore CH is incommensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11] And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the square on GB to KL, and the rectangle AG, GB to NL, therefore NL is a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to the square on MN [VI. 17], that is, to the fourth part of the square on FM. Since then CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to CM and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18] And the whole CM is commensurable in length with the rational straight line CD set out; therefore CF is a fourth apotome. [X. Deff. III. 4]'"],["Rule","'ProofWordCount'",675],["Rule","'GreekProof'","'ἔστω ἐλάσσων ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ ῥητὴν τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος ποιοῦν τὴν ΓΖ: λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι τετάρτη. ἔστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ: αἱ ἄρα ΑΗ, ΗΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ τετραγώνων ῥητόν, τὸ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον. καὶ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ πλάτος ποιοῦν τὴν ΓΚ, τῷ δὲ ἀπὸ τῆς ΒΗ ἴσον τὸ ΚΛ πλάτος ποιοῦν τὴν ΚΜ: ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ. καί ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ ῥητόν: ῥητὸν ἄρα ἐστὶ καὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ παράκειται πλάτος ποιοῦν τὴν ΓΜ: ῥητὴ ἄρα καὶ ἡ ΓΜ καὶ σύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ ὅλον τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, ὧν τὸ ΓΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ, λοιπὸν ἄρα τὸ ΖΛ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. τετμήσθω οὖν ἡ ΖΜ δίχα κατὰ τὸ Ν σημεῖον, καὶ ἤχθω διὰ τοῦ Ν ὁποτέρᾳ τῶν ΓΔ, ΜΛ παράλληλος ἡ ΝΞ: ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον ἐστὶ καί ἐστιν ἴσον τῷ ΖΛ, καὶ τὸ ΖΛ ἄρα μέσον ἐστίν. καὶ παρὰ ῥητὴν τὴν ΖΕ παράκειται πλάτος ποιοῦν τὴν ΖΜ: ῥητὴ ἄρα ἐστὶν ἡ ΖΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ ῥητόν ἐστιν, τὸ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον, ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. ἴσον δέ ἐστι τὸ ΓΛ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΖΛ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΜΖ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΜ τῇ ΜΖ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ἄρα ΓΜ, ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΓΖ. λέγω δή, ὅτι καὶ τετάρτη. ἐπεὶ γὰρ αἱ ΑΗ, ΗΒ δυνάμει εἰσὶν ἀσύμμετροι, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΗΒ. καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ ΚΛ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΘ τῷ ΚΛ. ὡς δὲ τὸ ΓΘ πρὸς τὸ ΚΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΚΜ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΚ τῇ ΚΜ μήκει. καὶ ἐπεὶ τῶν ἀπὸ τῶν ΑΗ, ΗΒ μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστιν ἴσον τὸ μὲν ἀπὸ τῆς ΑΗ τῷ ΓΘ, τὸ δὲ ἀπὸ τῆς ΗΒ τῷ ΚΛ, τὸ δὲ ὑπὸ τῶν ΑΗ, ΗΒ τῷ ΝΛ, τῶν ἄρα ΓΘ, ΚΛ μέσον ἀνάλογόν ἐστι τὸ ΝΛ: ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. ἀλλ᾽ ὡς μὲν τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΝΜ, ὡς δὲ τὸ ΝΛ πρὸς τὸ ΚΛ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΚΜ: ὡς ἄρα ἡ ΓΚ πρὸς τὴν ΜΝ, οὕτως ἐστὶν ἡ ΜΝ πρὸς τὴν ΚΜ: τὸ ἄρα ὑπὸ τῶν ΓΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, τουτέστι τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί εἰσιν αἱ ΓΜ, ΜΖ, καὶ τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΜΖ ἴσον παρὰ τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ τὸ ὑπὸ τῶν ΓΚ, ΚΜ καὶ εἰς ἀσύμμετρα αὐτὴν διαιρεῖ, ἡ ἄρα ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί ἐστιν ὅλη ἡ ΓΜ σύμμετρος μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ: ἡ ἄρα ΓΖ ἀποτομή ἐστι τετάρτη. τὸ ἄρα ἀπὸ ἐλάσσονος καὶ τὰ ἑξῆς.'"],["Rule","'GreekProofWordCount'",621]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",101]],["Association",["Rule","'VertexLabel'","'10.101'"],["Rule","'Text'","'The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'τὸ ἀπὸ τῆς μετὰ ῥητοῦ μέσον τὸ ὅλον ποιούσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν πέμπτην.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",77]]]],["Rule","'Proof'","'Let AB be the straight line which produces with a rational area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fifth apotome. For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. [X. 77] To CD let there be applied CH equal to the square on AG, and KL equal to the square on GB; therefore the whole CL is equal to the squares on AG, GB. But the sum of the squares on AG, GB together is medial; therefore CL is medial. And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable with CD. [X. 22] And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7] Let then FM be bisected at N, and through N let NO be drawn parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB: And, since twice the rectangle AG, GB is rational and equal to FL, therefore FL is rational. And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is rational and commensurable in length with CD. [X. 20] Now, since CL is medial, and FL rational, therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11] And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a fifth apotome. For we can prove similarly that the rectangle CK, KM is equal to the square on NM, that is, to the fourth part of the square on FM. And, since the square on AG is incommensurable with the square on GB, while the square on AG is equal to CH, and the square on GB to KL, therefore CH is incommensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11] Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18] And the annex FM is commensurable with the rational straight line CD set out; therefore CF is a fifth apotome. [X. Deff. III. 5]'"],["Rule","'ProofWordCount'",517],["Rule","'GreekProof'","'ἔστω ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος ποιοῦν τὴν ΓΖ: λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι πέμπτη. ἔστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ: αἱ ἄρα ΑΗ, ΗΒ εὐθεῖαι δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον, τὸ δὲ δὶς ὑπ᾽ αὐτῶν ῥητόν. καὶ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ ΚΛ: ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ. τὸ δὲ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ ἅμα μέσον ἐστίν: μέσον ἄρα ἐστὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ παράκειται πλάτος ποιοῦν τὴν ΓΜ: ῥητὴ ἄρα ἐστὶν ἡ ΓΜ καὶ ἀσύμμετρος τῇ ΓΔ. καὶ ἐπεὶ ὅλον τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, ὧν τὸ ΓΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ, λοιπὸν ἄρα τὸ ΖΛ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. τετμήσθω οὖν ἡ ΖΜ δίχα κατὰ τὸ Ν, καὶ ἤχθω διὰ τοῦ Ν ὁποτέρᾳ τῶν ΓΔ, ΜΛ παράλληλος ἡ ΝΞ: ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ῥητόν ἐστι καί ἐστιν ἴσον τῷ ΖΛ, ῥητὸν ἄρα ἐστὶ τὸ ΖΛ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν ΖΜ: ῥητὴ ἄρα ἐστὶν ἡ ΖΜ καὶ σύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὸ μὲν ΓΛ μέσον ἐστίν, τὸ δὲ ΖΛ ῥητόν, ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ ΖΛ, οὕτως ἡ ΓΜ πρὸς τὴν ΜΖ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΜ τῇ ΜΖ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ἄρα ΓΜ, ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΓΖ. λέγω δή, ὅτι καὶ πέμπτη. ὁμοίως γὰρ δείξομεν, ὅτι τὸ ὑπὸ τῶν ΓΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΝΜ, τουτέστι τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΗΒ, ἴσον δὲ τὸ μὲν ἀπὸ τῆς ΑΗ τῷ ΓΘ, τὸ δὲ ἀπὸ τῆς ΗΒ τῷ ΚΛ, ἀσύμμετρον ἄρα τὸ ΓΘ τῷ ΚΛ. ὡς δὲ τὸ ΓΘ πρὸς τὸ ΚΛ, οὕτως ἡ ΓΚ πρὸς τὴν ΚΜ: ἀσύμμετρος ἄρα ἡ ΓΚ τῇ ΚΜ μήκει. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί εἰσιν αἱ ΓΜ, ΜΖ, καὶ τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ ἴσον παρὰ τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς ἀσύμμετρα αὐτὴν διαιρεῖ, ἡ ἄρα ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί ἐστιν ἡ προσαρμόζουσα ἡ ΖΜ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ: ἡ ἄρα ΓΖ ἀποτομή ἐστι πέμπτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",443]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",102]],["Association",["Rule","'VertexLabel'","'10.102'"],["Rule","'Text'","'The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'τὸ ἀπὸ τῆς μετὰ μέσου μέσον τὸ ὅλον ποιούσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν ἕκτην.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",2],["Rule","'Theorem'",7]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",18]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",78]]]],["Rule","'Proof'","'Let AB be the straight line which produces with a medial area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a sixth apotome. For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle AG, GB medial, and the squares on AG, GB incommensurable with twice the rectangle AG, GB. [X. 78] Now to CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22] Since now CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7] And twice the rectangle AG, GB is medial; therefore FL is also medial. And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22] And, since the squares on AG, GB are incommensurable with twice the rectangle AG, GB, and CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB, therefore CL is incommensurable with FL. But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11] And both are rational. Therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73] I say next that it is also a sixth apotome. For, since FL is equal to twice the rectangle AG, GB, let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB. And, since AG, GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB. But CH is equal to the square on AG, and KL is equal to the square on GB; therefore CH is incommensurable with KL. But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable with KM. [X. 11] And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL. And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18] And neither of them is commensurable with the rational straight line CD set out; therefore CF is a sixth apotome. [X. Deff. III. 6]'"],["Rule","'ProofWordCount'",556],["Rule","'GreekProof'","'ἔστω ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος ποιοῦν τὴν ΓΖ: λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστιν ἕκτη. ἔστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ: αἱ ἄρα ΑΗ, ΗΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον καὶ τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον καὶ ἀσύμμετρον τὰ ἀπὸ τῶν ΑΗ, ΗΒ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. παραβεβλήσθω οὖν παρὰ τὴν ΓΔ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ πλάτος ποιοῦν τὴν ΓΚ, τῷ δὲ ἀπὸ τῆς ΒΗ τὸ ΚΛ: ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ: μέσον ἄρα ἐστὶ καὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ παράκειται πλάτος ποιοῦν τὴν ΓΜ: ῥητὴ ἄρα ἐστὶν ἡ ΓΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. ἐπεὶ οὖν τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, ὧν τὸ ΓΕ ἴσον τῷ ἀπὸ τῆς ΑΒ, λοιπὸν ἄρα τὸ ΖΛ ἴσον ἐστὶ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. καί ἐστι τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον: καὶ τὸ ΖΛ ἄρα μέσον ἐστίν. καὶ παρὰ ῥητὴν τὴν ΖΕ παράκειται πλάτος ποιοῦν τὴν ΖΜ: ῥητὴ ἄρα ἐστὶν ἡ ΖΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ ἀσύμμετρά ἐστι τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστι τοῖς μὲν ἀπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΓΛ, τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΖΛ, ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΜΖ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΜ τῇ ΜΖ μήκει. καί εἰσιν ἀμφότεραι ῥηταί. αἱ ΓΜ, ΜΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΓΖ. λέγω δή, ὅτι καὶ ἕκτη. ἐπεὶ γὰρ τὸ ΖΛ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ, τετμήσθω δίχα ἡ ΖΜ κατὰ τὸ Ν, καὶ ἤχθω διὰ τοῦ Ν τῇ ΓΔ παράλληλος ἡ ΝΞ: ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ αἱ ΑΗ, ΗΒ δυνάμει εἰσὶν ἀσύμμετροι, ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΗΒ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον ἐστὶ τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον ἐστὶ τὸ ΚΛ: ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΘ τῷ ΚΛ. ὡς δὲ τὸ ΓΘ πρὸς τὸ ΚΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΚΜ: ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΚ τῇ ΚΜ. καὶ ἐπεὶ τῶν ἀπὸ τῶν ΑΗ, ΗΒ μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ ΚΛ, τῷ δὲ ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΝΛ, καὶ τῶν ἄρα ΓΘ, ΚΛ μέσον ἀνάλογόν ἐστι τὸ ΝΛ: ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. καὶ διὰ τὰ αὐτὰ ἡ ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ οὐδετέρα αὐτῶν σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ: ἡ ΓΖ ἄρα ἀποτομή ἐστιν ἕκτη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",503]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",103]],["Association",["Rule","'VertexLabel'","'10.103'"],["Rule","'Text'","'A straight line commensurable in length with an apotome is an apotome and the same in order.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἡ τῇ ἀποτομῇ μήκει σύμμετρος ἀποτομή ἐστι καὶ τῇ τάξει ἡ αὐτή.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",12]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",14]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let AB be an apotome, and let CD be commensurable in length with AB; I say that CD is also an apotome and the same in order with AB. For, since AB is an apotome, let BE be the annex to it; therefore AE, EB are rational straight lines commensurable in square only. [X. 73] Let it be contrived that the ratio of BE to DF is the same as the ratio of AB to CD; [VI. 12] therefore also, as one is to one, so are all to all; [V. 12] therefore also, as the whole AE is to the whole CF, so is AB to CD. But AB is commensurable in length with CD. Therefore AE is also commensurable with CF, and BE with DF. [X. 11] And AE, EB are rational straight lines commensurable in square only; therefore CF, FD are also rational straight lines commensurable in square only. [X. 13] Now since, as AE is to CF, so is BE to DF, alternately therefore, as AE is to EB, so is CF to FD. [V. 16] And the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it. If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14] And, if AE is commensurable in length with the rational straight line set out, CF is so also, [X. 12] if BE, then DF also, [id.]and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13] But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14] And, if AE is commensurable in length with the rational straight line set out, CF is so also, if BE, then DF also, [X. 12] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13] Therefore CD is an apotome and the same in order with AB.'"],["Rule","'ProofWordCount'",411],["Rule","'GreekProof'","'ἔστω ἀποτομὴ ἡ ΑΒ, καὶ τῇ ΑΒ μήκει σύμμετρος ἔστω ἡ ΓΔ: λέγω, ὅτι καὶ ἡ ΓΔ ἀποτομή ἐστι καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ. ἐπεὶ γὰρ ἀποτομή ἐστιν ἡ ΑΒ, ἔστω αὐτῇ προσαρμόζουσα ἡ ΒΕ: αἱ ΑΕ, ΕΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ τῷ τῆς ΑΒ πρὸς τὴν ΓΔ λόγῳ ὁ αὐτὸς γεγονέτω ὁ τῆς ΒΕ πρὸς τὴν ΔΖ: καὶ ὡς ἓν ἄρα πρὸς ἕν, πάντα ἐστὶ πρὸς πάντα: ἔστιν ἄρα καὶ ὡς ὅλη ἡ ΑΕ πρὸς ὅλην τὴν ΓΖ, οὕτως ἡ ΑΒ πρὸς τὴν ΓΔ. σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ μήκει. σύμμετρος ἄρα καὶ ἡ ΑΕ μὲν τῇ ΓΖ, ἡ δὲ ΒΕ τῇ ΔΖ. καὶ αἱ ΑΕ, ΕΒ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: καὶ αἱ ΓΖ, ΖΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. ἀποτομὴ ἄρα ἐστὶν ἡ ΓΔ. λέγω δή, ὅτι καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΒΕ πρὸς τὴν ΔΖ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς τὴν ΕΒ, οὕτως ἡ ΓΖ πρὸς τὴν ΖΔ. ἤτοι δὴ ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ á¼¢ τῷ ἀπὸ ἀσυμμέτρου. εἰ μὲν οὖν ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΑΕ τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΓΖ, εἰ δὲ ἡ ΒΕ, καὶ ἡ ΔΖ, εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ, καὶ οὐδετέρα τῶν ΓΖ, ΖΔ. εἰ δὲ ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΑΕ τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΓΖ, εἰ δὲ ἡ ΒΕ, καὶ ἡ ΔΖ, εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ, οὐδετέρα τῶν ΓΖ, ΖΔ. ἀποτομὴ ἄρα ἐστὶν ἡ ΓΔ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",315]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",104]],["Association",["Rule","'VertexLabel'","'10.104'"],["Rule","'Text'","'A straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ἡ τῇ μέσης ἀποτομῇ σύμμετρος μέσης ἀποτομή ἐστι καὶ τῇ τάξει ἡ αὐτή.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",12]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Definition'",1.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",74]],["List",["Rule","'Book'",10],["Rule","'Theorem'",75]]]],["Rule","'Proof'","'Let AB be an apotome of a medial straight line, and let CD be commensurable in length with AB; I say that CD is also an apotome of a medial straight line and the same in order with AB. For, since AB is an apotome of a medial straight line, let EB be the annex to it. Therefore AE, EB are medial straight lines commensurable in square only. [X. 74, 75] Let it be contrived that, as AB is to CD, so is BE to DF; [VI. 12] therefore AE is also commensurable with CF, and BE with DF. [V. 12, X. 11] But AE, EB are medial straight lines commensurable in square only; therefore CF, FD are also medial straight lines [X. 23] commensurable in square only; [X. 13] therefore CD is an apotome of a medial straight line. [X. 74, 75] I say next that it is also the same in order with AB. Since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD. But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. [V. 16, X. 11] Therefore, if the rectangle AE, EB is rational, the rectangle CF, FD will also be rational, [X. Def. 4] and if the rectangle AE, EB is medial, the rectangle CF, FD is also medial. [X. 23] Therefore CD is an apotome of a medial straight line and the same in order with AB. [X. 74, 75]'"],["Rule","'ProofWordCount'",276],["Rule","'GreekProof'","'ἔστω μέσης ἀποτομὴ ἡ ΑΒ, καὶ τῇ ΑΒ μήκει σύμμετρος ἔστω ἡ ΓΔ: λέγω, ὅτι καὶ ἡ ΓΔ μέσης ἀποτομή ἐστι καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ. ἐπεὶ γὰρ μέσης ἀποτομή ἐστιν ἡ ΑΒ, ἔστω αὐτῇ προσαρμόζουσα ἡ ΕΒ. αἱ ΑΕ, ΕΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι. καὶ γεγονέτω ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΒΕ πρὸς τὴν ΔΖ: σύμμετρος ἄρα ἐστὶ καὶ ἡ ΑΕ τῇ ΓΖ, ἡ δὲ ΒΕ τῇ ΔΖ. αἱ δὲ ΑΕ, ΕΒ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι: καὶ αἱ ΓΖ, ΖΔ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι: μέσης ἄρα ἀποτομή ἐστιν ἡ ΓΔ. λέγω δή, ὅτι καὶ τῇ τάξει ἐστὶν ἡ αὐτὴ τῇ ΑΒ. ἐπεὶ γάρ ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΕΒ, οὕτως ἡ ΓΖ πρὸς τὴν ΖΔ ἀλλ᾽ ὡς μὲν ἡ ΑΕ πρὸς τὴν ΕΒ, οὕτως τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕ, ΕΒ, ὡς δὲ ἡ ΓΖ πρὸς τὴν ΖΔ, οὕτως τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖ, ΖΔ, ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕ, ΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖ, ΖΔ καὶ ἐναλλὰξ ὡς τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ἀπὸ τῆς ΓΖ, οὕτως τὸ ὑπὸ τῶν ΑΕ, ΕΒ πρὸς τὸ ὑπὸ τῶν ΓΖ, ΖΔ. σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς ΓΖ: σύμμετρον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ. εἴτε οὖν ῥητόν ἐστι τὸ ὑπὸ τῶν ΑΕ, ΕΒ, ῥητὸν ἔσται καὶ τὸ ὑπὸ τῶν ΓΖ, ΖΔ, εἴτε μέσον ἐστὶ τὸ ὑπὸ τῶν ΑΕ, ΕΒ, μέσον ἐστὶ καὶ τὸ ὑπὸ τῶν ΓΖ, ΖΔ. μέσης ἄρα ἀποτομή ἐστιν ἡ ΓΔ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",287]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",105]],["Association",["Rule","'VertexLabel'","'10.105'"],["Rule","'Text'","'A straight line commensurable with a minor straight line is minor.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἡ τῇ ἐλάσσονι σύμμετρος ἐλάσσων ἐστίν.'"],["Rule","'GreekTextWordCount'",6],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",12]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",18]],["List",["Rule","'Book'",6],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Definition'",1.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",23]],["List",["Rule","'Book'",10],["Rule","'Theorem'",76]]]],["Rule","'Proof'","'Let AB be a minor straight line, and CD commensurable with AB; I say that CD is also minor. Let the same construction be made as before; then, since AE, EB are incommensurable in square, [X. 76] therefore CF, FD are also incommensurable in square. [X. 13] Now since, as AE is to EB, so is CF to FD, [V. 12, V. 16] therefore also, as the square on AE is to the square on EB, so is the square on CF to the square on FD. [VI. 22] Therefore, componendo, as the squares on AE, EB are to the square on EB, so are the squares on CF, FD to the square on FD. [V. 18] But the square on BE is commensurable with the square on DF; therefore the sum of the squares on AE, EB is also commensurable with the sum of the squares on CF, FD. [V. 16, X. 11] But the sum of the squares on AE, EB is rational; [X. 76] therefore the sum of the squares on CF, FD is also rational. [X. Def. 4] Again, since, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD, while the square on AE is commensurable with the square on CF, therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. But the rectangle AE, EB is medial; [X. 76] therefore the rectangle CF, FD is also medial; [X. 23] therefore CF, FD are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. Therefore CD is minor. [X. 76]'"],["Rule","'ProofWordCount'",282],["Rule","'GreekProof'","'ἔστω γὰρ ἐλάσσων ἡ ΑΒ καὶ τῇ ΑΒ σύμμετρος ἡ ΓΔ: λέγω, ὅτι καὶ ἡ ΓΔ ἐλάσσων ἐστίν. γεγονέτω γὰρ τὰ αὐτά: καὶ ἐπεὶ αἱ ΑΕ, ΕΒ δυνάμει εἰσὶν ἀσύμμετροι, καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει εἰσὶν ἀσύμμετροι. ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΕΒ, οὕτως ἡ ΓΖ πρὸς τὴν ΖΔ, ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ἀπὸ τῆς ΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ἀπὸ τῆς ΖΔ. συνθέντι ἄρα ἐστὶν ὡς τὰ ἀπὸ τῶν ΑΕ, ΕΒ πρὸς τὸ ἀπὸ τῆς ΕΒ, οὕτως τὰ ἀπὸ τῶν ΓΖ, ΖΔ πρὸς τὸ ἀπὸ τῆς ΖΔ καὶ ἐναλλάξ: σύμμετρον δέ ἐστι τὸ ἀπὸ τῆς ΒΕ τῷ ἀπὸ τῆς ΔΖ: σύμμετρον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων. ῥητὸν δέ ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων: ῥητὸν ἄρα ἐστὶ καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων. πάλιν, ἐπεί ἐστιν ὡς τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕ, ΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖ, ΖΔ, σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΕ τετράγωνον τῷ ἀπὸ τῆς ΓΖ τετραγώνῳ, σύμμετρον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ. μέσον δὲ τὸ ὑπὸ τῶν ΑΕ, ΕΒ: μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΓΖ, ΖΔ: αἱ ΓΖ, ΖΔ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων ῥητόν, τὸ δ᾽ ὑπ᾽ αὐτῶν μέσον. ἐλάσσων ἄρα ἐστὶν ἡ ΓΔ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",259]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",106]],["Association",["Rule","'VertexLabel'","'10.106'"],["Rule","'Text'","'A straight line commensurable with that which produces with a rational area a medial whole is a straight line which produces with a rational area a medial whole.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἡ τῇ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιούσῃ σύμμετρος μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",77]]]],["Rule","'Proof'","'Let AB be a straight line which produces with a rational area a medial whole, and CD commensurable with AB; I say that CD is also a straight line which produces with a rational area a medial whole. For let BE be the annex to AB; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on AE, EB medial, but the rectangle contained by them rational. [X. 77] Let the same construction be made. Then we can prove, in manner similar to the foregoing, that CF, FD are in the same ratio as AE, EB, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that CF, FD are also straight lines incommensurable in square which make the sum of the squares on CF, FD medial, but the rectangle contained by them rational. Therefore CD is a straight line which produces with a rational area a medial whole. [X. 77]'"],["Rule","'ProofWordCount'",177],["Rule","'GreekProof'","'ἔστω μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ καὶ τῇ ΑΒ σύμμετρος ἡ ΓΔ: λέγω, ὅτι καὶ ἡ ΓΔ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν. ἔστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΕ: αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων μέσον, τὸ δ᾽ ὑπ᾽ αὐτῶν ῥητόν. καὶ τὰ αὐτὰ κατεσκευάσθω. ὁμοίως δὴ δείξομεν τοῖς πρότερον, ὅτι αἱ ΓΖ, ΖΔ ἐν τῷ αὐτῷ λόγῳ εἰσὶ ταῖς ΑΕ, ΕΒ, καὶ σύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων, τὸ δὲ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ: ὥστε καὶ αἱ ΓΖ, ΖΔ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων μέσον, τὸ δ᾽ ὑπ᾽ αὐτῶν ῥητόν. ἡ ΓΔ ἄρα μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",153]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",107]],["Association",["Rule","'VertexLabel'","'10.107'"],["Rule","'Text'","'A straight line commensurable with that which produces with a medial area a medial whole is itself also a straight line which produces with a medial area a medial whole.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'ἡ τῇ μετὰ μέσου μέσον τὸ ὅλον ποιούσῃ σύμμετρος καὶ αὐτὴ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Theorem'",78]]]],["Rule","'Proof'","'Let AB be a straight line which produces with a medial area a medial whole, and let CD be commensurable with AB; I say that CD is also a straight line which produces with a medial area a medial whole. For let BE be the annex to AB, and let the same construction be made; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them. [X. 78] Now, as was proved, AE, EB are commensurable with CF, FD, the sum of the squares on AE, EB with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; therefore CF, FD are also straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them. Therefore CD is a straight line which produces with a medial area a medial whole. [X. 78]'"],["Rule","'ProofWordCount'",195],["Rule","'GreekProof'","'ἔστω μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, καὶ τῇ ΑΒ ἔστω σύμμετρος ἡ ΓΔ: λέγω, ὅτι καὶ ἡ ΓΔ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν. ἔστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΕ, καὶ τὰ αὐτὰ κατεσκευάσθω: αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾽ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων τῷ ὑπ᾽ αὐτῶν. καί εἰσιν, ὡς ἐδείχθη, αἱ ΑΕ, ΕΒ σύμμετροι ταῖς ΓΖ, ΖΔ, καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ, τὸ δὲ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ: καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾽ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπ᾽ αὐτῶν τετραγώνων τῷ ὑπ᾽ αὐτῶν. ἡ ΓΔ ἄρα μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",171]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",108]],["Association",["Rule","'VertexLabel'","'10.108'"],["Rule","'Text'","'If from a rational area a medial area be subtracted, the “side” of the remaining area becomes one of two irrational straight lines, either an apotome or a minor straight line.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἀπὸ ῥητοῦ μέσου ἀφαιρουμένου ἡ τὸ λοιπὸν χωρίον δυναμένη μία δύο ἀλόγων γίνεται ἤτοι ἀποτομὴ á¼¢ ἐλάσσων.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",3.1]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",91]],["List",["Rule","'Book'",10],["Rule","'Theorem'",94]]]],["Rule","'Proof'","'For from the rational area BC let the medial area BD be subtracted; I say that the “side” of the remainder EC becomes one of two irrational straight lines, either an apotome or a minor straight line. For let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to BC, and let GK equal to DB be subtracted; therefore the remainder EC is equal to LH. Since then BC is rational, and BD medial, while BC is equal to GH, and BD to GK, therefore GH is rational, and GK medial. And they are applied to the rational straight line FG; therefore FH is rational and commensurable in length with FG, [X. 20] while FK is rational and incommensurable in length with FG; [X. 22] therefore FH is incommensurable in length with FK. [X. 13] Therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome [X. 73], and KF the annex to it. Now the square on HF is greater than the square on FK by the square on a straight line either commensurable with HF or not commensurable. First, let the square on it be greater by the square on a straight line commensurable with it. Now the whole HF is commensurable in length with the rational straight line FG set out; therefore KH is a first apotome. [X. Deff. III. 1] But the “side” of the rectangle contained by a rational straight line and a first apotome is an apotome. [X. 91] Therefore the “side” of LH, that is, of EC, is an apotome. But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the whole FH is commensurable in length with the rational straight line FG set out, KH is a fourth apotome. [X. Deff. III. 4] But the “side” of the rectangle contained by a rational straight line and a fourth apotome is minor. [X. 94]'"],["Rule","'ProofWordCount'",340],["Rule","'GreekProof'","'ἀπὸ γὰρ ῥητοῦ τοῦ ΒΓ μέσον ἀφῃρήσθω τὸ ΒΔ: λέγω, ὅτι ἡ τὸ λοιπὸν δυναμένη τὸ ΕΓ μία δύο ἀλόγων γίνεται ἤτοι ἀποτομὴ á¼¢ ἐλάσσων. Ἐκκείσθω γὰρ ῥητὴ ἡ ΖΗ, καὶ τῷ μὲν ΒΓ ἴσον παρὰ τὴν ΖΗ παραβεβλήσθω ὀρθογώνιον παραλληλόγραμμον τὸ ΗΘ, τῷ δὲ ΔΒ ἴσον ἀφῃρήσθω τὸ ΗΚ: λοιπὸν ἄρα τὸ ΕΓ ἴσον ἐστὶ τῷ ΛΘ. ἐπεὶ οὖν ῥητὸν μέν ἐστι τὸ ΒΓ, μέσον δὲ τὸ ΒΔ, ἴσον δὲ τὸ μὲν ΒΓ τῷ ΗΘ, τὸ δὲ ΒΔ τῷ ΗΚ, ῥητὸν μὲν ἄρα ἐστὶ τὸ ΗΘ, μέσον δὲ τὸ ΗΚ. καὶ παρὰ ῥητὴν τὴν ΖΗ παράκειται: ῥητὴ μὲν ἄρα ἡ ΖΘ καὶ σύμμετρος τῇ ΖΗ μήκει, ῥητὴ δὲ ἡ ΖΚ καὶ ἀσύμμετρος τῇ ΖΗ μήκει: ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΘ τῇ ΖΚ μήκει. αἱ ΖΘ, ΖΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΚΘ, προσαρμόζουσα δὲ αὐτῇ ἡ ΚΖ. ἤτοι δὴ ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου á¼¢ οὔ. δυνάσθω πρότερον τῷ ἀπὸ συμμέτρου. καί ἐστιν ὅλη ἡ ΘΖ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΖΗ: ἀποτομὴ ἄρα πρώτη ἐστὶν ἡ ΚΘ. τὸ δ᾽ ὑπὸ ῥητῆς καὶ ἀποτομῆς πρώτης περιεχόμενον ἡ δυναμένη ἀποτομή ἐστιν. ἡ ἄρα τὸ ΛΘ, τουτέστι τὸ ΕΓ, δυναμένη ἀποτομή ἐστιν. εἰ δὲ ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καί ἐστιν ὅλη ἡ ΖΘ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΖΗ, ἀποτομὴ τετάρτη ἐστὶν ἡ ΚΘ. τὸ δ᾽ ὑπὸ ῥητῆς καὶ ἀποτομῆς τετάρτης περιεχόμενον ἡ δυναμένη ἐλάσσων ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",247]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",109]],["Association",["Rule","'VertexLabel'","'10.109'"],["Rule","'Text'","'If from a medial area a rational area be subtracted, there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.'"],["Rule","'TextWordCount'",39],["Rule","'GreekText'","'ἀπὸ μέσου ῥητοῦ ἀφαιρουμένου ἄλλαι δύο ἄλογοι γίνονται ἤτοι μέσης ἀποτομὴ πρώτη á¼¢ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",3.2]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.5]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",92]],["List",["Rule","'Book'",10],["Rule","'Theorem'",95]]]],["Rule","'Proof'","'For from the medial area BC let the rational area BD be subtracted. I say that the “side” of the remainder EC becomes one of two irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole. For let a rational straight line FG be set out, and let the areas be similarly applied. It follows then that FH is rational and incommensurable in length with FG, while KF is rational and commensurable in length with FG; therefore FH, FK are rational straight lines commensurable in square only; [X. 13] therefore KH is an apotome, and FK the annex to it. [X. 73] Now the square on HF is greater than the square on FK either by the square on a straight line commensurable with HF or by the square on a straight line incommensurable with it. If then the square on HF is greater than the square on FK by the square on a straight line commensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a second apotome. [X. Deff. III. 2] But FG is rational; so that the “side” of LH, that is, of EC, is a first apotome of a medial straight line. [X. 92] But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a fifth apotome; [X. Deff. III. 5] so that the “side” of EC is a straight line which produces with a rational area a medial whole. [X. 95]'"],["Rule","'ProofWordCount'",293],["Rule","'GreekProof'","'ἀπὸ γὰρ μέσου τοῦ ΒΓ ῥητὸν ἀφῃρήσθω τὸ ΒΔ. λέγω, ὅτι ἡ τὸ λοιπὸν τὸ ΕΓ δυναμένη μία δύο ἀλόγων γίνεται ἤτοι μέσης ἀποτομὴ πρώτη á¼¢ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα. Ἐκκείσθω γὰρ ῥητὴ ἡ ΖΗ, καὶ παραβεβλήσθω ὁμοίως τὰ χωρία. ἔστι δὴ ἀκολούθως ῥητὴ μὲν ἡ ΖΘ καὶ ἀσύμμετρος τῇ ΖΗ μήκει, ῥητὴ δὲ ἡ ΚΖ καὶ σύμμετρος τῇ ΖΗ μήκει: αἱ ΖΘ, ΖΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΚΘ, προσαρμόζουσα δὲ ταύτῃ ἡ ΖΚ. ἤτοι δὴ ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ á¼¢ τῷ ἀπὸ ἀσυμμέτρου. εἰ μὲν οὖν ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καί ἐστιν ἡ προσαρμόζουσα ἡ ΖΚ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΖΗ, ἀποτομὴ δευτέρα ἐστὶν ἡ ΚΘ. ῥητὴ δὲ ἡ ΖΗ: ὥστε ἡ τὸ ΛΘ, τουτέστι τὸ ΕΓ, δυναμένη μέσης ἀποτομὴ πρώτη ἐστίν. εἰ δὲ ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου, καί ἐστιν ἡ προσαρμόζουσα ἡ ΖΚ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΖΗ, ἀποτομὴ πέμπτη ἐστὶν ἡ ΚΘ: ὥστε ἡ τὸ ΕΓ δυναμένη μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",189]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",110]],["Association",["Rule","'VertexLabel'","'10.110'"],["Rule","'Text'","'If from a medial area there be subtracted a medial area incommensurable with the whole, the two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.'"],["Rule","'TextWordCount'",44],["Rule","'GreekText'","'ἀπὸ μέσου μέσου ἀφαιρουμένου ἀσυμμέτρου τῷ ὅλῳ αἱ λοιπαὶ δύο ἄλογοι γίνονται ἤτοι μέσης ἀποτομὴ δευτέρα á¼¢ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.3]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",93]],["List",["Rule","'Book'",10],["Rule","'Theorem'",96]]]],["Rule","'Proof'","'For, as in the foregoing figures, let there be subtracted from the medial area BC the medial area BD incommensurable with the whole; I say that the “side” of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole. For, since each of the rectangles BC, BD is medial, and BC is incommensurable with BD, it follows that each of the straight lines FH, FK will be rational and incommensurable in length with FG. [X. 22] And, since BC is incommensurable with BD, that is, GH with GK, HF is also incommensurable with FK; [VI. 1, X. 11] therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome. [X. 73] If then the square on FH is greater than the square on FK by the square on a straight line commensurable with FH, while neither of the straight lines FH, FK is commensurable in length with the rational straight line FG set out, KH is a third apotome. [X. Deff. III. 3] But KL is rational, and the rectangle contained by a rational straight line and a third apotome is irrational, and the “side” of it is irrational, and is called a second apotome of a medial straight line; [X. 93] so that the “side” of LH, that is, of EC, is a second apotome of a medial straight line. But, if the square on FH is greater than the square on FK by the square on a straight line incommensurable with FH, while neither of the straight lines HF, FK is commensurable in length with FG, KH is a sixth apotome. [X. Deff. III. 6] But the “side” of the rectangle contained by a rational straight line and a sixth apotome is a straight line which produces with a medial area a medial whole. [X. 96] Therefore the “side” of LH, that is, of EC, is a straight line which produces with a medial area a medial whole.'"],["Rule","'ProofWordCount'",345],["Rule","'GreekProof'","'ἀφῃρήσθω γὰρ ὡς ἐπὶ τῶν προκειμένων καταγραφῶν ἀπὸ μέσου τοῦ ΒΓ μέσον τὸ ΒΔ ἀσύμμετρον τῷ ὅλῳ: λέγω, ὅτι ἡ τὸ ΕΓ δυναμένη μία ἐστὶ δύο ἀλόγων ἤτοι μέσης ἀποτομὴ δευτέρα ἢ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα. ἐπεὶ γὰρ μέσον ἐστὶν ἑκάτερον τῶν ΒΓ, ΒΔ, καὶ ἀσύμμετρον τὸ ΒΓ τῷ ΒΔ, ἔσται ἀκολούθως ῥητὴ ἑκατέρα τῶν ΖΘ, ΖΚ καὶ ἀσύμμετρος τῇ ΖΗ μήκει. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ΒΓ τῷ ΒΔ, τουτέστι τὸ ΗΘ τῷ ΗΚ, ἀσύμμετρος καὶ ἡ ΘΖ τῇ ΖΚ: αἱ ΖΘ, ΖΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΚΘ προσαρμόζουσα δὲ ἡ ΖΚ. ἤτοι δὴ ἡ ΖΘ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἢ τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. εἰ μὲν δὴ ἡ ΖΘ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ οὐθετέρα τῶν ΖΘ, ΖΚ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΖΗ, ἀποτομὴ τρίτη ἐστὶν ἡ ΚΘ. ῥητὴ δὲ ἡ ΚΛ, τὸ δ᾽ ὑπὸ ῥητῆς καὶ ἀποτομῆς τρίτης περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν, καλεῖται δὲ μέσης ἀποτομὴ δευτέρα: ὥστε ἡ τὸ ΛΘ, τουτέστι τὸ ΕΓ, δυναμένη μέσης ἀποτομή ἐστι δευτέρα. εἰ δὲ ἡ ΖΘ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, καὶ οὐθετέρα τῶν ΘΖ, ΖΚ σύμμετρός ἐστι τῇ ΖΗ μήκει, ἀποτομὴ ἕκτη ἐστὶν ἡ ΚΘ. τὸ δ᾽ ὑπὸ ῥητῆς καὶ ἀποτομῆς ἕκτης ἡ δυναμένη ἐστὶ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα. ἡ τὸ ΛΘ ἄρα, τουτέστι τὸ ΕΓ, δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",250]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",111]],["Association",["Rule","'VertexLabel'","'10.111'"],["Rule","'Text'","'The apotome is not the same with the binomial straight line.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'ἡ ἀποτομὴ οὐκ ἔστιν ἡ αὐτὴ τῇ ἐκ δύο ὀνομάτων.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",2.1]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",13]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",60]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",97]]]],["Rule","'Proof'","'Let AB be an apotome; I say that AB is not the same with the binomial straight line. For, if possible, let it be so; let a rational straight line DC be set out, and to CD let there be applied the rectangle CE equal to the square on AB and producing DE as breadth. Then, since AB is an apotome, DE is a first apotome. [X. 97] Let EF be the annex to it; therefore DF, FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF, and DF is commensurable in length with the rational straight line DC set out. [X. Deff. III. 1] Again, since AB is binomial, therefore DE is a first binomial straight line. [X. 60] Let it be divided into its terms at G, and let DG be the greater term; therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1] Therefore DF is also commensurable in length with DG; [X. 12] therefore the remainder GF is also commensurable in length with DF. [X. 15] But DF is incommensurable in length with EF; therefore FG is also incommensurable in length with EF. [X. 13] Therefore GF, FE are rational straight lines commensurable in square only; therefore EG is an apotome. [X. 73] But it is also rational: which is impossible. Therefore the apotome is not the same with the binomial straight line.'"],["Rule","'ProofWordCount'",290],["Rule","'GreekProof'","'ἔστω ἀποτομὴ ἡ ΑΒ: λέγω, ὅτι ἡ ΑΒ οὐκ ἔστιν ἡ αὐτὴ τῇ ἐκ δύο ὀνομάτων. εἰ γὰρ δυνατόν, ἔστω: καὶ ἐκκείσθω ῥητὴ ἡ ΔΓ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω ὀρθογώνιον τὸ ΓΕ πλάτος ποιοῦν τὴν ΔΕ. ἐπεὶ οὖν ἀποτομή ἐστιν ἡ ΑΒ, ἀποτομὴ πρώτη ἐστὶν ἡ ΔΕ. ἔστω αὐτῇ προσαρμόζουσα ἡ ΕΖ: αἱ ΔΖ, ΖΕ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΔΖ τῆς ΖΕ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΔΖ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΔΓ. πάλιν, ἐπεὶ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ, ἐκ δύο ἄρα ὀνομάτων πρώτη ἐστὶν ἡ ΔΕ. διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Η, καὶ ἔστω μεῖζον ὄνομα τὸ ΔΗ: αἱ ΔΗ, ΗΕ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΔΗ τῆς ΗΕ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ τὸ μεῖζον ἡ ΔΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΔΓ. καὶ ἡ ΔΖ ἄρα τῇ ΔΗ σύμμετρός ἐστι μήκει: καὶ λοιπὴ ἄρα ἡ ΗΖ σύμμετρός ἐστι τῇ ΔΖ μήκει. ἐπεὶ οὖν σύμμετρός ἐστιν ἡ ΔΖ τῇ ΗΖ, ῥητὴ δέ ἐστιν ἡ ΔΖ, ῥητὴ ἄρα ἐστὶ καὶ ἡ ΗΖ. ἐπεὶ οὖν σύμμετρός ἐστιν ἡ ΔΖ τῇ ΗΖ μήκει ἀσύμμετρος δὲ ἡ ΔΖ τῇ ΕΖ μήκει: ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΖΗ τῇ ΕΖ μήκει. αἱ ΗΖ, ΖΕ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΕΗ. ἀλλὰ καὶ ῥητή: ὅπερ ἐστὶν ἀδύνατον. ἡ ἄρα ἀποτομὴ οὐκ ἔστιν ἡ αὐτὴ τῇ ἐκ δύο ὀνομάτων: ὅπερ ἔδει δεῖξαι. Πόρισμα ἡ ἀποτομὴ καὶ αἱ μετ᾽ αὐτὴν ἄλογοι οὔτε τῇ μέσῃ οὔτε ἀλλήλαις εἰσὶν αἱ αὐταί. τὸ μὲν γὰρ ἀπὸ μέσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ῥητὴν καὶ ἀσύμμετρον τῇ, παρ᾽ ἣν παράκειται, μήκει, τὸ δὲ ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν πρώτην, τὸ δὲ ἀπὸ μέσης ἀποτομῆς πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν δευτέραν, τὸ δὲ ἀπὸ μέσης ἀποτομῆς δευτέρας παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν τρίτην, τὸ δὲ ἀπὸ ἐλάσσονος παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν τετάρτην, τὸ δὲ ἀπὸ τῆς μετὰ ῥητοῦ μέσον τὸ ὅλον ποιούσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν πέμπτην, τὸ δὲ ἀπὸ τῆς μετὰ μέσου μέσον τὸ ὅλον ποιούσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν ἕκτην. ἐπεὶ οὖν τὰ εἰρημένα πλάτη διαφέρει τοῦ τε πρώτου καὶ ἀλλήλων, τοῦ μὲν πρώτου, ὅτι ῥητή ἐστιν, ἀλλήλων δέ, ἐπεὶ τῇ τάξει οὐκ εἰσὶν αἱ αὐταί, δῆλον, ὡς καὶ αὐταὶ αἱ ἄλογοι διαφέρουσιν ἀλλήλων. καὶ ἐπεὶ δέδεικται ἡ ἀποτομὴ οὐκ οὖσα ἡ αὐτὴ τῇ ἐκ δύο ὀνομάτων, ποιοῦσι δὲ πλάτη παρὰ ῥητὴν παραβαλλόμεναι αἱ μετὰ τὴν ἀποτομὴν ἀποτομὰς ἀκολούθως ἑκάστη τῇ τάξει τῇ καθ᾽ αὑτήν, αἱ δὲ μετὰ τὴν ἐκ δύο ὀνομάτων τὰς ἐκ δύο ὀνομάτων καὶ αὐταὶ τῇ τάξει ἀκολούθως, ἕτεραι ἄρα εἰσὶν αἱ μετὰ τὴν ἀποτομὴν καὶ ἕτεραι αἱ μετὰ τὴν ἐκ δύο ὀνομάτων, ὡς εἶναι τῇ τάξει πάσας ἀλόγους ιγ, μέσην, ἐκ δύο ὀνομάτων, ἐκ δύο μέσων πρώτην, ἐκ δύο μέσων δευτέραν, μείζονα, ῥητὸν καὶ μέσον δυναμένην, δύο μέσα δυναμένην, ἀποτομήν, μέσης ἀποτομὴν πρώτην, μέσης ἀποτομὴν δευτέραν, ἐλάσσονα, μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσαν, μετὰ μέσου μέσον τὸ ὅλον ποιοῦσαν.'"],["Rule","'GreekProofWordCount'",510]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",112]],["Association",["Rule","'VertexLabel'","'10.112'"],["Rule","'Text'","'The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line.'"],["Rule","'TextWordCount'",52],["Rule","'GreekText'","'τὸ ἀπὸ ῥητῆς παρὰ τὴν ἐκ δύο ὀνομάτων παραβαλλόμενον πλάτος ποιεῖ ἀποτομήν, ἧς τὰ ὀνόματα σύμμετρά ἐστι τοῖς τῆς ἐκ δύο ὀνομάτων ὀνόμασι καὶ ἔτι ἐν τῷ αὐτῷ λόγῳ, καὶ ἔτι ἡ γινομένη ἀποτομὴ τὴν αὐτὴν ἕξει τάξιν τῇ ἐκ δύο ὀνομάτων.'"],["Rule","'GreekTextWordCount'",42],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",12]],["List",["Rule","'Book'",5],["Rule","'Theorem'",14]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",17]],["List",["Rule","'Book'",6],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",22]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",14]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A; I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC. For again let the rectangle BD, G be equal to the square on A. Since then the rectangle BC, EF is equal to the rectangle BD, G, therefore, as CB is to BD, so is G to EF. [VI. 16] But CB is greater than BD; therefore G is also greater than EF. [V. 16, V. 14] Let EH be equal to G; therefore, as CB is to BD, so is HE to EF; therefore, separando, as CD is to BD, so is HF to FE. [V. 17] Let it be contrived that, as HF is to FE, so is FK to KE; therefore also the whole HK is to the whole KF as FK is to KE; for, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. [V. 12] But, as FK is to KE, so is CD to DB; [V. 11] therefore also, as HK is to KF, so is CD to DB. [id.] But the square on CD is commensurable with the square on DB; [X. 36] therefore the square on HK is also commensurable with the square on KF. [VI. 22, X. 11] And, as the square on HK is to the square on KF, so is HK to KE, since the three straight lines HK, KF, KE are proportional. [V. Def. 9] Therefore HK is commensurable in length with KE, so that HE is also commensurable in length with EK. [X. 15] Now, since the square on A is equal to the rectangle EH, BD, while the square on A is rational, therefore the rectangle EH, BD is also rational. And it is applied to the rational straight line BD; therefore EH is rational and commensurable in length with BD; [X. 20] so that EK, being commensurable with it, is also rational and commensurable in length with BD. Since, then, as CD is to DB, so is FK to KE, while CD, DB are straight lines commensurable in square only, therefore FK, KE are also commensurable in square only. [X. 11] But KE is rational; therefore FK is also rational. Therefore FK, KE are rational straight lines commensurable in square only; therefore EF is an apotome. [X. 73] Now the square on CD is greater than the square on DB either by the square on a straight line commensurable with CD or by the square on a straight line incommensurable with it. If then the square on CD is greater than the square on DB by the square on a straight line commensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line commensurable with FK. [X. 14] And, if CD is commensurable in length with the rational straight line set out, so also is FK; [X. 11, 12] if BD is so commensurable, so also is KE; [X. 12] but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so. But, if the square on CD is greater than the square on DB by the square on a straight line incommensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line incommensurable with FK. [X. 14] And, if CD is commensurable with the rational straight line set out, so also is FK; if BD is so commensurable, so also is KE; but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so; so that FE is an apotome, the terms of which FK, KE are commensurable with the terms CD, DB of the binomial straight line and in the same ratio, and it has the same order as BC.'"],["Rule","'ProofWordCount'",700],["Rule","'GreekProof'","'ἔστω ῥητὴ μὲν ἡ Α, ἐκ δύο ὀνομάτων δὲ ἡ ΒΓ, ἧς μεῖζον ὄνομα ἔστω ἡ ΔΓ, καὶ τῷ ἀπὸ τῆς Α ἴσον ἔστω τὸ ὑπὸ τῶν ΒΓ, ΕΖ: λέγω, ὅτι ἡ ΕΖ ἀποτομή ἐστιν, ἧς τὰ ὀνόματα σύμμετρά ἐστι τοῖς ΓΔ, ΔΒ, καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ ἔτι ἡ ΕΖ τὴν αὐτὴν ἕξει τάξιν τῇ ΒΓ. ἔστω γὰρ πάλιν τῷ ἀπὸ τῆς Α ἴσον τὸ ὑπὸ τῶν ΒΔ, Η. ἐπεὶ οὖν τὸ ὑπὸ τῶν ΒΓ, ΕΖ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΒΔ, Η, ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΔ, οὕτως ἡ Η πρὸς τὴν ΕΖ. μείζων δὲ ἡ ΓΒ τῆς ΒΔ: μείζων ἄρα ἐστὶ καὶ ἡ Η τῆς ΕΖ. ἔστω τῇ Η ἴση ἡ ΕΘ: ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΔ, οὕτως ἡ ΘΕ πρὸς τὴν ΕΖ: διελόντι ἄρα ἐστὶν ὡς ἡ ΓΔ πρὸς τὴν ΒΔ, οὕτως ἡ ΘΖ πρὸς τὴν ΖΕ. γεγονέτω ὡς ἡ ΘΖ πρὸς τὴν ΖΕ, οὕτως ἡ ΖΚ πρὸς τὴν ΚΕ: καὶ ὅλη ἄρα ἡ ΘΚ πρὸς ὅλην τὴν ΚΖ ἐστιν, ὡς ἡ ΖΚ πρὸς ΚΕ: ὡς γὰρ ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα. ὡς δὲ ἡ ΖΚ πρὸς ΚΕ, οὕτως ἐστὶν ἡ ΓΔ πρὸς τὴν ΔΒ: καὶ ὡς ἄρα ἡ ΘΚ πρὸς ΚΖ, οὕτως ἡ ΓΔ πρὸς τὴν ΔΒ. σύμμετρον δὲ τὸ ἀπὸ τῆς ΓΔ τῷ ἀπὸ τῆς ΔΒ: σύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΘΚ τῷ ἀπὸ τῆς ΚΖ. καί ἐστιν ὡς τὸ ἀπὸ τῆς ΘΚ πρὸς τὸ ἀπὸ τῆς ΚΖ, οὕτως ἡ ΘΚ πρὸς τὴν ΚΕ, ἐπεὶ αἱ τρεῖς αἱ ΘΚ, ΚΖ, ΚΕ ἀνάλογόν εἰσιν. σύμμετρος ἄρα ἡ ΘΚ τῇ ΚΕ μήκει: ὥστε καὶ ἡ ΘΕ τῇ ΕΚ σύμμετρός ἐστι μήκει. καὶ ἐπεὶ τὸ ἀπὸ τῆς Α ἴσον ἐστὶ τῷ ὑπὸ τῶν ΕΘ, ΒΔ, ῥητὸν δέ ἐστι τὸ ἀπὸ τῆς Α, ῥητὸν ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΕΘ, ΒΔ. καὶ παρὰ ῥητὴν τὴν ΒΔ παράκειται: ῥητὴ ἄρα ἐστὶν ἡ ΕΘ καὶ σύμμετρος τῇ ΒΔ μήκει: ὥστε καὶ ἡ σύμμετρος αὐτῇ ἡ ΕΚ ῥητή ἐστι καὶ σύμμετρος τῇ ΒΔ μήκει. ἐπεὶ οὖν ἐστιν ὡς ἡ ΓΔ πρὸς ΔΒ, οὕτως ἡ ΖΚ πρὸς ΚΕ, αἱ δὲ ΓΔ, ΔΒ δυνάμει μόνον εἰσὶ σύμμετροι, καὶ αἱ ΖΚ, ΚΕ δυνάμει μόνον εἰσὶ σύμμετροι. ῥητὴ δέ ἐστιν ἡ ΚΕ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΚ. αἱ ΖΚ, ΚΕ ἄρα ῥηταὶ δυνάμει μόνον εἰσὶ σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΕΖ. ἤτοι δὲ ἡ ΓΔ τῆς ΔΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου. εἰ μὲν οὖν ἡ ΓΔ τῆς ΔΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΖΚ τῆς ΚΕ μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΓΔ τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΖΚ: εἰ δὲ ἡ ΒΔ, καὶ ἡ ΚΕ: εἰ δὲ οὐδετέρα τῶν ΓΔ, ΔΒ, καὶ οὐδετέρα τῶν ΖΚ, ΚΕ. εἰ δὲ ἡ ΓΔ τῆς ΔΒ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΖΚ τῆς ΚΕ μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ εἰ μὲν ἡ ΓΔ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΖΚ: εἰ δὲ ἡ ΒΔ, καὶ ἡ ΚΕ: εἰ δὲ οὐδετέρα τῶν ΓΔ, ΔΒ, καὶ οὐδετέρα τῶν ΖΚ, ΚΕ: ὥστε ἀποτομή ἐστιν ἡ ΖΕ, ἧς τὰ ὀνόματα τὰ ΖΚ, ΚΕ σύμμετρά ἐστι τοῖς τῆς ἐκ δύο ὀνομάτων ὀνόμασι τοῖς ΓΔ, ΔΒ καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ τὴν αὐτὴν τάξιν ἔχει τῇ ΒΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",561]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",113]],["Association",["Rule","'VertexLabel'","'10.113'"],["Rule","'Text'","'The square on a rational straight line, if applied to an apotome, produces as, breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome.'"],["Rule","'TextWordCount'",49],["Rule","'GreekText'","'τὸ ἀπὸ ῥητῆς παρὰ ἀποτομὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων, ἧς τὰ ὀνόματα σύμμετρά ἐστι τοῖς τῆς ἀποτομῆς ὀνόμασι καὶ ἐν τῷ αὐτῷ λόγῳ, ἔτι δὲ ἡ γινομένη ἐκ δύο ὀνομάτων τὴν αὐτὴν τάξιν ἔχει τῇ ἀποτομῇ.'"],["Rule","'GreekTextWordCount'",39],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",14]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",6],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Definition'",1.3]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",14]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]],["List",["Rule","'Book'",10],["Rule","'Theorem'",36]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]]]],["Rule","'Proof'","'Let A be a rational straight line and BD an apotome, and let the rectangle BD, KH be equal to the square on A, so that the square on the rational straight line A when applied to the apotome BD produces KH as breadth; I say that KH is a binomial straight line the terms of which are commensurable with the terms of BD and in the same ratio; and further KH has the same order as BD. For let DC be the annex to BD; therefore BC, CD are rational straight lines commensurable in square only. [X. 73] Let the rectangle BC, G be also equal to the square on A. But the square on A is rational; therefore the rectangle BC, G is also rational. And it has been applied to the rational straight line BC; therefore G is rational and commensurable in length with BC. [X. 20] Since now the rectangle BC, G is equal to the rectangle BD, KH, therefore, proportionally, as CB is to BD, so is KH to G. [VI. 16] But BC is greater than BD; therefore KH is also greater than G. [V. 16, V. 14] Let KE be made equal to G; therefore KE is commensurable in length with BC. And since, as CB is to BD, so is HK to KE, therefore, convertendo, as BC is to CD, so is KH to HE. [V. 19] Let it be contrived that, as KH is to HE, so is HF to FE; therefore also the remainder KF is to FH as KH is to HE, that is, as BC is to CD. [V. 19] But BC, CD are commensurable in square only; therefore KF, FH are also commensurable in square only. [X. 11] And since, as KH is to HE, so is KF to FH, while, as KH is to HE, so is HF to FE, therefore also, as KF is to FH, so is HF to FE, [V. 11] so that also, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9] therefore also, as KF is to FE, so is the square on KF to the square on FH. But the square on KF is commensurable with the square on FH, for KF, FH are commensurable in square; therefore KF is also commensurable in length with FE, [X. 11] so that KF is also commensurable in length with KE. [X. 15] But KE is rational and commensurable in length with BC; therefore KF is also rational and commensurable in length with BC. [X. 12] And, since, as BC is to CD, so is KF to FH, alternately, as BC is to KF, so is DC to FH. [V. 16] But BC is commensurable with KF; therefore FH is also commensurable in length with CD. [X. 11] But BC, CD are rational straight lines commensurable in square only; therefore KF, FH are also rational straight lines [X. Def. 3] commensurable in square only; therefore KH is binomial. [X. 36] If now the square on BC is greater than the square on CD by the square on a straight line commensurable with BC, the square on KF will also be greater than the square on FH by the square on a straight line commensurable with KF. [X 14] And, if BC is commensurable in length with the rational straight line set out, so also is KF; if CD is commensurable in length with the rational straight line set out, so also is FH, but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH. But, if the square on BC is greater than the square on CD by the square on a straight line incommensurable with BC, the square on KF is also greater than the square on FH by the square on a straight line incommensurable with KF. [X. 14] And, if BC is commensurable with the rational straight line set out, so also is KF; if CD is so commensurable, in length with the rational straight line set out, so also is FH; but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH. Therefore KH is a binomial straight line, the terms of which KF, FH are commensurable with the terms BC, CD of the apotome and in the same ratio, and further KH has the same order as BD.'"],["Rule","'ProofWordCount'",746],["Rule","'GreekProof'","'ἔστω ῥητὴ μὲν ἡ Α, ἀποτομὴ δὲ ἡ ΒΔ, καὶ τῷ ἀπὸ τῆς Α ἴσον ἔστω τὸ ὑπὸ τῶν ΒΔ, ΚΘ, ὥστε τὸ ἀπὸ τῆς Α ῥητῆς παρὰ τὴν ΒΔ ἀποτομὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ΚΘ: λέγω, ὅτι ἐκ δύο ὀνομάτων ἐστὶν ἡ ΚΘ, ἧς τὰ ὀνόματα σύμμετρά ἐστι τοῖς τῆς ΒΔ ὀνόμασι καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ ἔτι ἡ ΚΘ τὴν αὐτὴν ἔχει τάξιν τῇ ΒΔ. ἔστω γὰρ τῇ ΒΔ προσαρμόζουσα ἡ ΔΓ: αἱ ΒΓ, ΓΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ τῷ ἀπὸ τῆς Α ἴσον ἔστω καὶ τὸ ὑπὸ τῶν ΒΓ, Η. ῥητὸν δὲ τὸ ἀπὸ τῆς Α: ῥητὸν ἄρα καὶ τὸ ὑπὸ τῶν ΒΓ, Η. καὶ παρὰ ῥητὴν τὴν ΒΓ παραβέβληται: ῥητὴ ἄρα ἐστὶν ἡ Η καὶ σύμμετρος τῇ ΒΓ μήκει. ἐπεὶ οὖν τὸ ὑπὸ τῶν ΒΓ, Η ἴσον ἐστὶ τῷ ὑπὸ τῶν ΒΔ, ΚΘ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΒ πρὸς ΒΔ, οὕτως ἡ ΚΘ πρὸς Η. μείζων δὲ ἡ ΒΓ τῆς ΒΔ: μείζων ἄρα καὶ ἡ ΚΘ τῆς Η. κείσθω τῇ Η ἴση ἡ ΚΕ: σύμμετρος ἄρα ἐστὶν ἡ ΚΕ τῇ ΒΓ μήκει. καὶ ἐπεί ἐστιν ὡς ἡ ΓΒ πρὸς ΒΔ, οὕτως ἡ ΘΚ πρὸς ΚΕ, ἀναστρέψαντι ἄρα ἐστὶν ὡς ἡ ΒΓ πρὸς τὴν ΓΔ, οὕτως ἡ ΚΘ πρὸς ΘΕ. γεγονέτω ὡς ἡ ΚΘ πρὸς ΘΕ, οὕτως ἡ ΘΖ πρὸς ΖΕ: καὶ λοιπὴ ἄρα ἡ ΚΖ πρὸς ΖΘ ἐστιν, ὡς ἡ ΚΘ πρὸς ΘΕ, τουτέστιν ὡς ἡ ΒΓ πρὸς ΓΔ. αἱ δὲ ΒΓ, ΓΔ δυνάμει μόνον εἰσὶ σύμμετροι: καὶ αἱ ΚΖ, ΖΘ ἄρα δυνάμει μόνον εἰσὶ σύμμετροι. καὶ ἐπεί ἐστιν ὡς ἡ ΚΘ πρὸς ΘΕ, ἡ ΚΖ πρὸς ΖΘ, ἀλλ᾽ ὡς ἡ ΚΘ πρὸς ΘΕ, ἡ ΘΖ πρὸς ΖΕ, καὶ ὡς ἄρα ἡ ΚΖ πρὸς ΖΘ, ἡ ΘΖ πρὸς ΖΕ: ὥστε καὶ ὡς ἡ πρώτη πρὸς τὴν τρίτην, τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ἀπὸ τῆς δευτέρας: καὶ ὡς ἄρα ἡ ΚΖ πρὸς ΖΕ, οὕτως τὸ ἀπὸ τῆς ΚΖ πρὸς τὸ ἀπὸ τῆς ΖΘ. σύμμετρον δέ ἐστι τὸ ἀπὸ τῆς ΚΖ τῷ ἀπὸ τῆς ΖΘ: αἱ γὰρ ΚΖ, ΖΘ δυνάμει εἰσὶ σύμμετροι: σύμμετρος ἄρα ἐστὶ καὶ ἡ ΚΖ τῇ ΖΕ μήκει: ὥστε ἡ ΚΖ καὶ τῇ ΚΕ σύμμετρός ἐστι μήκει. ῥητὴ δέ ἐστιν ἡ ΚΕ καὶ σύμμετρος τῇ ΒΓ μήκει: ῥητὴ ἄρα καὶ ἡ ΚΖ καὶ σύμμετρος τῇ ΒΓ μήκει. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς ΓΔ, οὕτως ἡ ΚΖ πρὸς ΖΘ, ἐναλλὰξ ὡς ἡ ΒΓ πρὸς ΚΖ, οὕτως ἡ ΔΓ πρὸς ΖΘ. σύμμετρος δὲ ἡ ΒΓ τῇ ΚΖ: σύμμετρος ἄρα καὶ ἡ ΖΘ τῇ ΓΔ μήκει. αἱ ΒΓ, ΓΔ δὲ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: καὶ αἱ ΚΖ, ΖΘ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἐκ δύο ὀνομάτων ἐστὶν ἄρα ἡ ΚΘ. εἰ μὲν οὖν ἡ ΒΓ τῆς ΓΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΚΖ τῆς ΖΘ μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΒΓ τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΚΖ, εἰ δὲ ἡ ΓΔ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΖΘ, εἰ δὲ οὐδετέρα τῶν ΒΓ, ΓΔ, οὐδετέρα τῶν ΚΖ, ΖΘ. εἰ δὲ ἡ ΒΓ τῆς ΓΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΚΖ τῆς ΖΘ μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΒΓ τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΚΖ, εἰ δὲ ἡ ΓΔ, καὶ ἡ ΖΘ, εἰ δὲ οὐδετέρα τῶν ΒΓ, ΓΔ, οὐδετέρα τῶν ΚΖ, ΖΘ. ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΚΘ, ἧς τὰ ὀνόματα τὰ ΚΖ, ΖΘ σύμμετρά ἐστι τοῖς τῆς ἀποτομῆς ὀνόμασι τοῖς ΒΓ, ΓΔ καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ ἔτι ἡ ΚΘ τῇ ΒΓ τὴν αὐτὴν ἕξει τάξιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",598]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",114]],["Association",["Rule","'VertexLabel'","'10.114'"],["Rule","'Text'","'If an area be contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, the “side” of the area is rational.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν χωρίον περιέχηται ὑπὸ ἀποτομῆς καὶ τῆς ἐκ δύο ὀνομάτων, ἧς τὰ ὀνόματα σύμμετρά τέ ἐστι τοῖς τῆς ἀποτομῆς ὀνόμασι καὶ ἐν τῷ αὐτῷ λόγῳ, ἡ τὸ χωρίον δυναμένη ῥητή ἐστιν.'"],["Rule","'GreekTextWordCount'",31],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",10],["Rule","'Theorem'",11]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",112]]]],["Rule","'Proof'","'For let an area, the rectangle AB, CD, be contained by the apotome AB and the binomial straight line CD, and let CE be the greater term of the latter; let the terms CE, ED of the binomial straight line be commensurable with the terms AF, FB of the apotome and in the same ratio; and let the “side” of the rectangle AB, CD be G; I say that G is rational. For let a rational straight line H be set out, and to CD let there be applied a rectangle equal to the square on H and producing KL as breadth. Therefore KL is an apotome. Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112] But CE, ED are also commensurable with AF, FB and in the same ratio; therefore, as AF is to FB, so is KM to ML. Therefore, alternately, as AF is to KM, so is BF to LM; therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19] But AF is commensurable with KM; [X. 12] therefore AB is also commensurable with KL. [X. 11] And, as AB is to KL, so is the rectangle CD, AB to the rectangle CD, KL; [VI. 1] therefore the rectangle CD, AB is also commensurable with the rectangle CD, KL. [X. 11] But the rectangle CD, KL is equal to the square on H; therefore the rectangle CD, AB is commensurable with the square on H. But the square on G is equal to the rectangle CD, AB; therefore the square on G is commensurable with the square on H. But the square on H is rational; therefore the square on G is also rational; therefore G is rational. And it is the “side” of the rectangle CD, AB.'"],["Rule","'ProofWordCount'",313],["Rule","'GreekProof'","'περιεχέσθω γὰρ χωρίον τὸ ὑπὸ τῶν ΑΒ, ΓΔ ὑπὸ ἀποτομῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων τῆς ΓΔ, ἧς μεῖζον ὄνομα ἔστω τὸ ΓΕ, καὶ ἔστω τὰ ὀνόματα τῆς ἐκ δύο ὀνομάτων τὰ ΓΕ, ΕΔ σύμμετρά τε τοῖς τῆς ἀποτομῆς ὀνόμασι τοῖς ΑΖ, ΖΒ καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ ἔστω ἡ τὸ ὑπὸ τῶν ΑΒ, ΓΔ δυναμένη ἡ Η: λέγω, ὅτι ῥητή ἐστιν ἡ Η. Ἐκκείσθω γὰρ ῥητὴ ἡ Θ, καὶ τῷ ἀπὸ τῆς Θ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω πλάτος ποιοῦν τὴν ΚΛ: ἀποτομὴ ἄρα ἐστὶν ἡ ΚΛ, ἧς τὰ ὀνόματα ἔστω τὰ ΚΜ, ΜΛ σύμμετρα τοῖς τῆς ἐκ δύο ὀνομάτων ὀνόμασι τοῖς ΓΕ, ΕΔ καὶ ἐν τῷ αὐτῷ λόγῳ. ἀλλὰ καὶ αἱ ΓΕ, ΕΔ σύμμετροί τέ εἰσι ταῖς ΑΖ, ΖΒ καὶ ἐν τῷ αὐτῷ λόγῳ: ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΖΒ, οὕτως ἡ ΚΜ πρὸς ΜΛ. ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΖ πρὸς τὴν ΚΜ, οὕτως ἡ ΒΖ πρὸς τὴν ΛΜ: καὶ λοιπὴ ἄρα ἡ ΑΒ πρὸς λοιπὴν τὴν ΚΛ ἐστιν ὡς ἡ ΑΖ πρὸς ΚΜ. σύμμετρος δὲ ἡ ΑΖ τῇ ΚΜ: σύμμετρος ἄρα ἐστὶ καὶ ἡ ΑΒ τῇ ΚΛ. καί ἐστιν ὡς ἡ ΑΒ πρὸς ΚΛ, οὕτως τὸ ὑπὸ τῶν ΓΔ, ΑΒ πρὸς τὸ ὑπὸ τῶν ΓΔ, ΚΛ: σύμμετρον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΓΔ, ΑΒ τῷ ὑπὸ τῶν ΓΔ, ΚΛ. ἴσον δὲ τὸ ὑπὸ τῶν ΓΔ, ΚΛ τῷ ἀπὸ τῆς Θ: σύμμετρον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΓΔ, ΑΒ τῷ ἀπὸ τῆς Θ. τῷ δὲ ὑπὸ τῶν ΓΔ, ΑΒ ἴσον ἐστὶ τὸ ἀπὸ τῆς Η: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς Η τῷ ἀπὸ τῆς Θ. ῥητὸν δὲ τὸ ἀπὸ τῆς Θ: ῥητὸν ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς Η: ῥητὴ ἄρα ἐστὶν ἡ Η. καὶ δύναται τὸ ὑπὸ τῶν ΓΔ, ΑΒ. ἐὰν ἄρα χωρίον περιέχηται ὑπὸ ἀποτομῆς καὶ τῆς ἐκ δύο ὀνομάτων, ἧς τὰ ὀνόματα σύμμετρά ἐστι τοῖς τῆς ἀποτομῆς ὀνόμασι καὶ ἐν τῷ αὐτῷ λόγῳ, ἡ τὸ χωρίον δυναμένη ῥητή ἐστιν. Πόρισμα καὶ γέγονεν ἡμῖν καὶ διὰ τούτου φανερόν, ὅτι δυνατόν ἐστι ῥητὸν χωρίον ὑπὸ ἀλόγων εὐθειῶν περιέχεσθαι. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",342]]],["Rule",["Association",["Rule","'Book'",10],["Rule","'Theorem'",115]],["Association",["Rule","'VertexLabel'","'10.115'"],["Rule","'Text'","'From a medial straight line there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'ἀπὸ μέσης ἄπειροι ἄλογοι γίνονται, καὶ οὐδεμία οὐδεμιᾷ τῶν πρότερον ἡ αὐτή.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",10],["Rule","'Definition'",1.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let A be a medial straight line; I say that from A there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding. Let a rational straight line B be set out, and let the square on C be equal to the rectangle B, A; therefore C is irrational; [X. Def. 4] for that which is contained by an irrational and a rational straight line is irrational. [deduction from X. 20] And it is not the same with any of the preceding; for the square on none of the preceding, if applied to a rational straight line produces as breadth a medial straight line. Again, let the square on D be equal to the rectangle B, C; therefore the square on D is irrational. [deduction from X. 20] Therefore D is irrational; [X. Def. 4] and it is not the same with any of the preceding, for the square on none of the preceding, if applied to a rational straight line, produces C as breadth. Similarly, if this arrangement proceeds ad infinitum, it is manifest that from the medial straight line there arise irrational straight lines infinite in number, and none is the same with any of the preceding.'"],["Rule","'ProofWordCount'",208],["Rule","'GreekProof'","'ἔστω μέση ἡ Α: λέγω, ὅτι ἀπὸ τῆς Α ἄπειροι ἄλογοι γίνονται, καὶ οὐδεμία οὐδεμιᾷ τῶν πρότερον ἡ αὐτή. Ἐκκείσθω ῥητὴ ἡ Β, καὶ τῷ ὑπὸ τῶν Β, Α ἴσον ἔστω τὸ ἀπὸ τῆς Γ: ἄλογος ἄρα ἐστὶν ἡ Γ: τὸ γὰρ ὑπὸ ἀλόγου καὶ ῥητῆς ἄλογόν ἐστιν. καὶ οὐδεμιᾷ τῶν πρότερον ἡ αὐτή: τὸ γὰρ ἀπ᾽ οὐδεμιᾶς τῶν πρότερον παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ μέσην. πάλιν δὴ τῷ ὑπὸ τῶν Β, Γ ἴσον ἔστω τὸ ἀπὸ τῆς Δ: ἄλογον ἄρα ἐστὶ τὸ ἀπὸ τῆς Δ. ἄλογος ἄρα ἐστὶν ἡ Δ: καὶ οὐδεμιᾷ τῶν πρότερον ἡ αὐτή: τὸ γὰρ ἀπ᾽ οὐδεμιᾶς τῶν πρότερον παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν Γ. ὁμοίως δὴ τῆς τοιαύτης τάξεως ἐπ᾽ ἄπειρον προβαινούσης φανερόν, ὅτι ἀπὸ τῆς μέσης ἄπειροι ἄλογοι γίνονται, καὶ οὐδεμία οὐδεμιᾷ τῶν πρότερον ἡ αὐτή: ὅπερ ἔδει δεῖξαι].'"],["Rule","'GreekProofWordCount'",139]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'11.1'"],["Rule","'Text'","'A part of a straight line cannot be in the plane of reference and a part in a plane more elevated.'"],["Rule","'TextWordCount'",21],["Rule","'GreekText'","'εὐθείας γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑποκειμένῳ, ἐπιπέδῳ, μέρος δέ τι ἐν μετεωροτέρῳ.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List"]],["Rule","'Proof'","'For, if possible, let a part AB of the straight line ABC be in the plane of reference, and a part BC in a plane more elevated. There will then be in the plane of reference some straight line continuous with AB in a straight line. Let it be BD; therefore AB is a common segment of the two straight lines ABC, ABD: which is impossible, inasmuch as, if we describe a circle with centre B and distance AB, the diameters will cut off unequal circumferences of the circle. Therefore a part of a straight line cannot be in the plane of reference, and a part in a plane more elevated.'"],["Rule","'ProofWordCount'",111],["Rule","'GreekProof'","'εἰ γὰρ δυνατόν, εὐθείας γραμμῆς τῆς ΑΒΓ μέρος μέν τι τὸ ΑΒ ἔστω ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, μέρος δέ τι τὸ ΒΓ ἐν μετεωροτέρῳ. ἔσται δή τις τῇ ΑΒ συνεχὴς εὐθεῖα ἐπ᾽ εὐθείας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ. ἔστω ἡ ΒΔ: δύο ἄρα εὐθειῶν τῶν ΑΒΓ, ΑΒΔ κοινὸν τμῆμά ἐστιν ἡ ΑΒ: ὅπερ ἐστὶν ἀδύνατον, ἐπειδήπερ ἐὰν κέντρῳ τῷ Β καὶ διαστήματι τῷ ΑΒ κύκλον γράψωμεν, αἱ διάμετροι ἀνίσους ἀπολήψονται τοῦ κύκλου περιφερείας. εὐθείας ἄρα γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν μετεωροτέρῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",92]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'11.2'"],["Rule","'Text'","'If two straight lines cut one another, they are in one plane, and every triangle is in one plane.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, ἐν ἑνί εἰσιν ἐπιπέδῳ, καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'For let the two straight lines AB, CD cut one another at the point E; I say that AB, CD are in one plane, and every triangle is in one plane. For let points F, G be taken at random on EC, EB, let CB, FG be joined, and let FH, GK be drawn across; I say first that the triangle ECB is in one plane. For, if part of the triangle ECB, either FHC or GBK, is in the plane of reference, and the rest in another, a part also of one of the straight lines EC, EB will be in the plane of reference, and a part in another. But, if the part FCBG of the triangle ECB be in the plane of reference, and the rest in another, a part also of both the straight lines EC, EB will be in the plane of reference and a part in another: which was proved absurd. [XI. 1] Therefore the triangle ECB is in one plane. But, in whatever plane the triangle ECB is, in that plane also is each of the straight lines EC, EB, and, in whatever plane each of the straight lines EC, EB is, in that plane are AB, CD also. [XI. 1] Therefore the straight lines AB, CD are in one plane, and every triangle is in one plane.'"],["Rule","'ProofWordCount'",225],["Rule","'GreekProof'","'δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον: λέγω, ὅτι αἱ ΑΒ, ΓΔ ἐν ἑνί εἰσιν ἐπιπέδῳ, καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. εἰλήφθω γὰρ ἐπὶ τῶν ΕΓ, ΕΒ τυχόντα σημεῖα τὰ Ζ, Η, καὶ ἐπεζεύχθωσαν αἱ ΓΒ, ΖΗ, καὶ διήχθωσαν αἱ ΖΘ, ΗΚ: λέγω πρῶτον, ὅτι τὸ ΕΓΒ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. εἰ γάρ ἐστι τοῦ ΕΓΒ τριγώνου μέρος ἤτοι τὸ ΖΘΓ ἢ τὸ ΗΒΚ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ λοιπὸν ἐν ἄλλῳ, ἔσται καὶ μιᾶς τῶν ΕΓ, ΕΒ εὐθειῶν μέρος μέν τι ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν ἄλλῳ. εἰ δὲ τοῦ ΕΓΒ τριγώνου τὸ ΖΓΒΗ μέρος ᾖ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ λοιπὸν ἐν ἄλλῳ, ἔσται καὶ ἀμφοτέρων τῶν ΕΓ, ΕΒ εὐθειῶν μέρος μέν τι ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν ἄλλῳ: ὅπερ ἄτοπον ἐδείχθη. τὸ ἄρα ΕΓΒ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. ἐν ᾧ δέ ἐστι τὸ ΕΓΒ τρίγωνον, ἐν τούτῳ καὶ ἑκατέρα τῶν ΕΓ, ΕΒ, ἐν ᾧ δὲ ἑκατέρα τῶν ΕΓ, ΕΒ, ἐν τούτῳ καὶ αἱ ΑΒ, ΓΔ. αἱ ΑΒ, ΓΔ ἄρα εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ, καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",192]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'11.3'"],["Rule","'Text'","'If two planes cut one another, their common section is a straight line.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἐὰν δύο ἐπίπεδα τέμνῃ ἄλληλα, ἡ κοινὴ αὐτῶν τομὴ εὐθεῖά ἐστιν.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List"]],["Rule","'Proof'","'For let the two planes AB, BC cut one another, and let the line DB be their common section; I say that the line DB is a straight line. For, if not, from D to B let the straight line DEB be joined in the plane AB, and in the plane BC the straight line DFB. Then the two straight lines DEB, DFB will have the same extremities, and will clearly enclose an area: which is absurd. Therefore DEB, DFB are not straight lines. Similarly we can prove that neither will there be any other straight line joined from D to B except DB the common section of the planes AB, BC.'"],["Rule","'ProofWordCount'",112],["Rule","'GreekProof'","'δύο γὰρ ἐπίπεδα τὰ ΑΒ, ΒΓ τεμνέτω ἄλληλα, κοινὴ δὲ αὐτῶν τομὴ ἔστω ἡ ΔΒ γραμμή: λέγω, ὅτι ἡ ΔΒ γραμμὴ εὐθεῖά ἐστιν. εἰ γὰρ μή, ἐπεζεύχθω ἀπὸ τοῦ Δ ἐπὶ τὸ Β ἐν μὲν τῷ ΑΒ ἐπιπέδῳ εὐθεῖα ἡ ΔΕΒ, ἐν δὲ τῷ ΒΓ ἐπιπέδῳ εὐθεῖα ἡ ΔΖΒ. ἔσται δὴ δύο εὐθειῶν τῶν ΔΕΒ, ΔΖΒ τὰ αὐτὰ πέρατα, καὶ περιέξουσι δηλαδὴ χωρίον: ὅπερ ἄτοπον. οὐκ ἄρα αἱ ΔΕΒ, ΔΖΒ εὐθεῖαί εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἄλλη τις ἀπὸ τοῦ Δ ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἔσται πλὴν τῆς ΔΒ κοινῆς τομῆς τῶν ΑΒ, ΒΓ ἐπιπέδων. ἐὰν ἄρα δύο ἐπίπεδα τέμνῃ ἄλληλα, ἡ κοινὴ αὐτῶν τομὴ εὐθεῖά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",112]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'11.4'"],["Rule","'Text'","'If a straight line be set up at right angles to two straight lines which cut one another, at their common point of section, it will also be at right angles to the plane through them.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν εὐθεῖα δύο εὐθείαις τεμνούσαις ἀλλήλας πρὸς ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, καὶ τῷ δι᾽ αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",15]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]]]],["Rule","'Proof'","'For let a straight line EF be set up at right angles to the two straight lines AB, CD, which cut one another at the point E, from E; I say that EF is also at right angles to the plane through AB, CD. For let AE, EB, CE, ED be cut off equal to one another, and let any straight line GEH be drawn across through E, at random; let AD, CB be joined, and further let FA, FG, FD, FC, FH, FB be joined from the point F taken at random <on EF >. Now, since the two straight lines AE, ED are equal to the two straight lines CE, EB, and contain equal angles, [I. 15] therefore the base AD is equal to the base CB, and the triangle AED will be equal to the triangle CEB; [I. 4] so that the angle DAE is also equal to the angle EBC. But the angle AEG is also equal to the angle BEH; [I. 15] therefore AGE, BEH are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say, AE to EB; therefore they will also have the remaining sides equal to the remaining sides. [I. 26] Therefore GE is equal to EH, and AG to BH. And, since AE is equal to EB, while FE is common and at right angles, therefore the base FA is equal to the base FB. [I. 4] For the same reason FC is also equal to FD. And, since AD is equal to CB, and FA is also equal to FB, the two sides FA, AD are equal to the two sides FB, BC respectively; and the base FD was proved equal to the base FC; therefore the angle FAD is also equal to the angle FBC. [I. 8] And since, again, AG was proved equal to BH, and further FA also equal to FB, the two sides FA, AG are equal to the two sides FB, BH. And the angle FAG was proved equal to the angle FBH; therefore the base FG is equal to the base FH. [I. 4] Now since, again, GE was proved equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the base FG is equal to the base FH; therefore the angle GEF is equal to the angle HEF. [I. 8] Therefore each of the angles GEF, HEF is right. Therefore FE is at right angles to GH drawn at random through E. Similarly we can prove that FE will also make right angles with all the straight lines which meet it and are in the plane of reference. But a straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in that same plane; [XI. Def. 3] therefore FE is at right angles to the plane of reference. But the plane of reference is the plane through the straight lines AB, CD. Therefore FE is at right angles to the plane through AB, CD.'"],["Rule","'ProofWordCount'",533],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΕΖ δύο εὐθείαις ταῖς ΑΒ, ΓΔ τεμνούσαις ἀλλήλας κατὰ τὸ Ε σημεῖον ἀπὸ τοῦ Ε πρὸς ὀρθὰς ἐφεστάτω: λέγω, ὅτι ἡ ΕΖ καὶ τῷ διὰ τῶν ΑΒ, ΓΔ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. Ἀπειλήφθωσαν γὰρ αἱ ΑΕ, ΕΒ, ΓΕ, ΕΔ ἴσαι ἀλλήλαις, καὶ διήχθω τις διὰ τοῦ Ε, ὡς ἔτυχεν, ἡ ΗΕΘ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΓΒ, καὶ ἔτι ἀπὸ τυχόντος τοῦ Ζ ἐπεζεύχθωσαν αἱ ΖΑ, ΖΗ, ΖΔ, ΖΓ, ΖΘ, ΖΒ. καὶ ἐπεὶ δύο αἱ ΑΕ, ΕΔ δυσὶ ταῖς ΓΕ, ΕΒ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΑΔ βάσει τῇ ΓΒ ἴση ἐστίν, καὶ τὸ ΑΕΔ τρίγωνον τῷ ΓΕΒ τριγώνῳ ἴσον ἔσται: ὥστε καὶ γωνία ἡ ὑπὸ ΔΑΕ γωνίᾳ τῇ ὑπὸ ΕΒΓ ἴση ἐστίν. ἔστι δὲ καὶ ἡ ὑπὸ ΑΕΗ γωνία τῇ ὑπὸ ΒΕΘ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΑΗΕ, ΒΕΘ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΑΕ τῇ ΕΒ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν. ἴση ἄρα ἡ μὲν ΗΕ τῇ ΕΘ, ἡ δὲ ΑΗ τῇ ΒΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΖΕ, βάσις ἄρα ἡ ΖΑ βάσει τῇ ΖΒ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΖΓ τῇ ΖΔ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΓΒ, ἔστι δὲ καὶ ἡ ΖΑ τῇ ΖΒ ἴση, δύο δὴ αἱ ΖΑ, ΑΔ δυσὶ ταῖς ΖΒ, ΒΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΖΔ βάσει τῇ ΖΓ ἐδείχθη ἴση: καὶ γωνία ἄρα ἡ ὑπὸ ΖΑΔ γωνίᾳ τῇ ὑπὸ ΖΒΓ ἴση ἐστίν. καὶ ἐπεὶ πάλιν ἐδείχθη ἡ ΑΗ τῇ ΒΘ ἴση, ἀλλὰ μὴν καὶ ἡ ΖΑ τῇ ΖΒ ἴση, δύο δὴ αἱ ΖΑ, ΑΗ δυσὶ ταῖς ΖΒ, ΒΘ ἴσαι εἰσίν. καὶ γωνία ἡ ὑπὸ ΖΑΗ ἐδείχθη ἴση τῇ ὑπὸ ΖΒΘ: βάσις ἄρα ἡ ΖΗ βάσει τῇ ΖΘ ἐστιν ἴση. καὶ ἐπεὶ πάλιν ἴση ἐδείχθη ἡ ΗΕ τῇ ΕΘ, κοινὴ δὲ ἡ ΕΖ, δύο δὴ αἱ ΗΕ, ΕΖ δυσὶ ταῖς ΘΕ, ΕΖ ἴσαι εἰσίν: καὶ βάσις ἡ ΖΗ βάσει τῇ ΖΘ ἴση: γωνία ἄρα ἡ ὑπὸ ΗΕΖ γωνίᾳ τῇ ὑπὸ ΘΕΖ ἴση ἐστίν. ὀρθὴ ἄρα ἑκατέρα τῶν ὑπὸ ΗΕΖ, ΘΕΖ γωνιῶν. ἡ ΖΕ ἄρα πρὸς τὴν ΗΘ τυχόντως διὰ τοῦ Ε ἀχθεῖσαν ὀρθή ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι ἡ ΖΕ καὶ πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. εὐθεῖα δὲ πρὸς ἐπίπεδον ὀρθή ἐστιν, ὅταν πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ αὐτῷ ἐπιπέδῳ ὀρθὰς ποιῇ γωνίας: ἡ ΖΕ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ ὑποκείμενον ἐπίπεδόν ἐστι τὸ διὰ τῶν ΑΒ, ΓΔ εὐθειῶν. ἡ ΖΕ ἄρα πρὸς ὀρθάς ἐστι τῷ διὰ τῶν ΑΒ, ΓΔ ἐπιπέδῳ. ἐὰν ἄρα εὐθεῖα δύο εὐθείαις τεμνούσαις ἀλλήλας πρὸς ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, καὶ τῷ δι᾽ αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",482]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'11.5'"],["Rule","'Text'","'If a straight line be set up at right angles to three straight lines which meet one another, at their common point of section, the three straight lines are in one plane.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν εὐθεῖα τρισὶν εὐθείαις ἁπτομέναις ἀλλήλων πρὸς ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, αἱ τρεῖς εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'For let a straight line AB be set up at right angles to the three straight lines BC, BD, BE, at their point of meeting at B; I say that BC, BD, BE are in one plane. For suppose they are not, but, if possible, let BD, BE be in the plane of reference and BC in one more elevated; let the plane through AB, BC be produced; it will thus make, as common section in the plane of reference, a straight line. [XI. 3] Let it make BF. Therefore the three straight lines AB, BC, BF are in one plane, namely that drawn through AB, BC. Now, since AB is at right angles to each of the straight lines BD, BE, therefore AB is also at right angles to the plane through BD, BE. [XI. 4] But the plane through BD, BE is the plane of reference; therefore AB is at right angles to the plane of reference. Thus AB will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3] But BF which is in the plane of reference meets it; therefore the angle ABF is right. But, by hypothesis, the angle ABC is also right; therefore the angle ABF is equal to the angle ABC. And they are in one plane: which is impossible. Therefore the straight line BC is not in a more elevated plane; therefore the three straight lines BC, BD, BE are in one plane. Therefore, if a straight line be set up at right angles to three straight lines, at their point of meeting, the three straight lines are in one plane.'"],["Rule","'ProofWordCount'",281],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ τρισὶν εὐθείαις ταῖς ΒΓ, ΒΔ, ΒΕ πρὸς ὀρθὰς ἐπὶ τῆς κατὰ τὸ Β ἁφῆς ἐφεστάτω: λέγω, ὅτι αἱ ΒΓ, ΒΔ, ΒΕ ἐν ἑνί εἰσιν ἐπιπέδῳ. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἔστωσαν αἱ μὲν ΒΔ, ΒΕ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, ἡ δὲ ΒΓ ἐν μετεωροτέρῳ, καὶ ἐκβεβλήσθω τὸ διὰ τῶν ΑΒ, ΒΓ ἐπίπεδον: κοινὴν δὴ τομὴν ποιήσει ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν. ποιείτω τὴν ΒΖ. ἐν ἑνὶ ἄρα εἰσὶν ἐπιπέδῳ τῷ διηγμένῳ διὰ τῶν ΑΒ, ΒΓ αἱ τρεῖς εὐθεῖαι αἱ ΑΒ, ΒΓ, ΒΖ. καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς ἑκατέραν τῶν ΒΔ, ΒΕ, καὶ τῷ διὰ τῶν ΒΔ, ΒΕ ἄρα ἐπιπέδῳ ὀρθή ἐστιν ἡ ΑΒ. τὸ δὲ διὰ τῶν ΒΔ, ΒΕ ἐπίπεδον τὸ ὑποκείμενόν ἐστιν: ἡ ΑΒ ἄρα ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον. ὥστε καὶ πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας ἡ ΑΒ. ἅπτεται δὲ αὐτῆς ἡ ΒΖ οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ: ἡ ἄρα ὑπὸ ΑΒΖ γωνία ὀρθή ἐστιν. ὑπόκειται δὲ καὶ ἡ ὑπὸ ΑΒΓ ὀρθή: ἴση ἄρα ἡ ὑπὸ ΑΒΖ γωνία τῇ ὑπὸ ΑΒΓ. καί εἰσιν ἐν ἑνὶ ἐπιπέδῳ: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ΒΓ εὐθεῖα ἐν μετεωροτέρῳ ἐστὶν ἐπιπέδῳ: αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΒΓ, ΒΔ, ΒΕ ἐν ἑνί εἰσιν ἐπιπέδῳ. ἐὰν ἄρα εὐθεῖα τρισὶν εὐθείαις ἁπτομέναις ἀλλήλων ἐπὶ τῆς ἁφῆς πρὸς ὀρθὰς ἐπισταθῇ, αἱ τρεῖς εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",234]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'11.6'"],["Rule","'Text'","'If two straight lines be at right angles to the same plane, the straight lines will be parallel.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, παράλληλοι ἔσονται αἱ εὐθεῖαι.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",28]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",2]],["List",["Rule","'Book'",11],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'For let the two straight lines AB, CD be at right angles to the plane of reference; I say that AB is parallel to CD. For let them meet the plane of reference at the points B, D, let the straight line BD be joined, let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined. Now, since AB is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3] But each of the straight lines BD, BE is in the plane of reference and meets AB; therefore each of the angles ABD, ABE is right. For the same reason each of the angles CDB, CDE is also right. And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and they include right angles; therefore the base AD is equal to the base BE. [I. 4] And, since AB is equal to DE, while AD is also equal to BE, the two sides AB, BE are equal to the two sides ED, DA; and AE is their common base; therefore the angle ABE is equal to the angle EDA. [I. 8] But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to DA. But it is also at right angles to each of the straight lines BD, DC; therefore ED is set up at right angles to the three straight lines BD, DA, DC at their point of meeting; therefore the three straight lines BD, DA, DC are in one plane. [XI. 5] But, in whatever plane DB, DA are, in that plane is AB also, for every triangle is in one plane; [XI. 2] therefore the straight lines AB, BD, DC are in one plane. And each of the angles ABD, BDC is right; therefore AB is parallel to CD. [I. 28]'"],["Rule","'ProofWordCount'",355],["Rule","'GreekProof'","'δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστωσαν: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. συμβαλλέτωσαν γὰρ τῷ ὑποκειμένῳ ἐπιπέδῳ κατὰ τὰ Β, Δ σημεῖα, καὶ ἐπεζεύχθω ἡ ΒΔ εὐθεῖα, καὶ ἤχθω τῇ ΒΔ πρὸς ὀρθὰς ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ἡ ΔΕ, καὶ κείσθω τῇ ΑΒ ἴση ἡ ΔΕ, καὶ ἐπεζεύχθωσαν αἱ ΒΕ, ΑΕ, ΑΔ. καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. ἅπτεται δὲ τῆς ΑΒ ἑκατέρα τῶν ΒΔ, ΒΕ οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΑΒΔ, ΑΒΕ γωνιῶν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΔΒ, ΓΔΕ ὀρθή ἐστιν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, κοινὴ δὲ ἡ ΒΔ, δύο δὴ αἱ ΑΒ, ΒΔ δυσὶ ταῖς ΕΔ, ΔΒ ἴσαι εἰσίν: καὶ γωνίας ὀρθὰς περιέχουσιν: βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΕ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, ἀλλὰ καὶ ἡ ΑΔ τῇ ΒΕ, δύο δὴ αἱ ΑΒ, ΒΕ δυσὶ ταῖς ΕΔ, ΔΑ ἴσαι εἰσίν: καὶ βάσις αὐτῶν κοινὴ ἡ ΑΕ: γωνία ἄρα ἡ ὑπὸ ΑΒΕ γωνίᾳ τῇ ὑπὸ ΕΔΑ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΑΒΕ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΕΔΑ: ἡ ΕΔ ἄρα πρὸς τὴν ΔΑ ὀρθή ἐστιν. ἔστι δὲ καὶ πρὸς ἑκατέραν τῶν ΒΔ, ΔΓ ὀρθή. ἡ ΕΔ ἄρα τρισὶν εὐθείαις ταῖς ΒΔ, ΔΑ, ΔΓ πρὸς ὀρθὰς ἐπὶ τῆς ἁφῆς ἐφέστηκεν: αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΒΔ, ΔΑ, ΔΓ ἐν ἑνί εἰσιν ἐπιπέδῳ. ἐν ᾧ δὲ αἱ ΔΒ, ΔΑ, ἐν τούτῳ καὶ ἡ ΑΒ: πᾶν γὰρ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ: αἱ ἄρα ΑΒ, ΒΔ, ΔΓ εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ. καί ἐστιν ὀρθὴ ἑκατέρα τῶν ὑπὸ ΑΒΔ, ΒΔΓ γωνιῶν: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. ἐὰν ἄρα δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, παράλληλοι ἔσονται αἱ εὐθεῖαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",315]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'11.7'"],["Rule","'Text'","'If two straight lines be parallel and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallel straight lines.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν ὦσι δύο εὐθεῖαι παράλληλοι, ληφθῇ δὲ ἐφ᾽ ἑκατέρας αὐτῶν τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'Let AB, CD be two parallel straight lines, and let points E, F be taken at random on them respectively; I say that the straight line joining the points E, F is in the same plane with the parallel straight lines. For suppose it is not, but, if possible, let it be in a more elevated plane as EGF, and let a plane be drawn through EGF; it will then make, as section in the plane of reference, a straight line. [XI. 3] Let it make it, as EF; therefore the two straight lines EGF, EF will enclose an area: which is impossible. Therefore the straight line joined from E to F is not in a plane more elevated; therefore the straight line joined from E to F is in the plane through the parallel straight lines AB, CD.'"],["Rule","'ProofWordCount'",139],["Rule","'GreekProof'","'ἔστωσαν δύο εὐθεῖαι παράλληλοι αἱ ΑΒ, ΓΔ, καὶ εἰλήφθω ἐφ᾽ ἑκατέρας αὐτῶν τυχόντα σημεῖα τὰ Ε, Ζ: λέγω, ὅτι ἡ ἐπὶ τὰ Ε, Ζ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἔστω ἐν μετεωροτέρῳ ὡς ἡ ΕΗΖ, καὶ διήχθω διὰ τῆς ΕΗΖ ἐπίπεδον: τομὴν δὴ ποιήσει ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν. ποιείτω ὡς τὴν ΕΖ: δύο ἄρα εὐθεῖαι αἱ ΕΗΖ, ΕΖ χωρίον περιέξουσιν: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Ε ἐπὶ τὸ Ζ ἐπιζευγνυμένη εὐθεῖα ἐν μετεωροτέρῳ ἐστὶν ἐπιπέδῳ: ἐν τῷ διὰ τῶν ΑΒ, ΓΔ ἄρα παραλλήλων ἐστὶν ἐπιπέδῳ ἡ ἀπὸ τοῦ Ε ἐπὶ τὸ Ζ ἐπιζευγνυμένη εὐθεῖα. ἐὰν ἄρα ὦσι δύο εὐθεῖαι παράλληλοι, ληφθῇ δὲ ἐφ᾽ ἑκατέρας αὐτῶν τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",140]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'11.8'"],["Rule","'Text'","'If two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'ἐὰν ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ ἑτέρα αὐτῶν ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",2]],["List",["Rule","'Book'",11],["Rule","'Theorem'",4]],["List",["Rule","'Book'",11],["Rule","'Theorem'",7]]]],["Rule","'Proof'","'Let AB, CD be two parallel straight lines, and let one of them, AB, be at right angles to the plane of reference; I say that the remaining one, CD, will also be at right angles to the same plane. For let AB, CD meet the plane of reference at the points B, D, and let BD be joined; therefore AB, CD, BD are in one plane. [XI. 7] Let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined. Now, since AB is at right angles to the plane of reference, therefore AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] therefore each of the angles ABD, ABE is right. And, since the straight line BD has fallen on the parallels AB, CD, therefore the angles ABD, CDB are equal to two right angles. [I. 29] But the angle ABD is right; therefore the angle CDB is also right; therefore CD is at right angles to BD. And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and the angle ABD is equal to the angle EDB, for each is right; therefore the base AD is equal to the base BE. And, since AB is equal to DE, and BE to AD, the two sides AB, BE are equal to the two sides ED, DA respectively, and AE is their common base; therefore the angle ABE is equal to the angle EDA. But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to AD. But it is also at right angles to DB; therefore ED is also at right angles to the plane through BD, DA. [XI. 4] Therefore ED will also make right angles with all the straight lines which meet it and are in the plane through BD, DA. But DC is in the plane through BD, DA, inasmuch as AB, BD are in the plane through BD, DA, [XI. 2] and DC is also in the plane in which AB, BD are. Therefore ED is at right angles to DC, so that CD is also at right angles to DE. But CD is also at right angles to BD. Therefore CD is set up at right angles to the two straight lines DE, DB which cut one another, from the point of section at D; so that CD is also at right angles to the plane through DE, DB. [XI. 4] But the plane through DE, DB is the plane of reference; therefore CD is at right angles to the plane of reference.'"],["Rule","'ProofWordCount'",472],["Rule","'GreekProof'","'ἔστωσαν δύο εὐθεῖαι παράλληλοι αἱ ΑΒ, ΓΔ, ἡ δὲ ἑτέρα αὐτῶν ἡ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστω: λέγω, ὅτι καὶ ἡ λοιπὴ ἡ ΓΔ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται. συμβαλλέτωσαν γὰρ αἱ ΑΒ, ΓΔ τῷ ὑποκειμένῳ ἐπιπέδῳ κατὰ τὰ Β, Δ σημεῖα, καὶ ἐπεζεύχθω ἡ ΒΔ: αἱ ΑΒ, ΓΔ, ΒΔ ἄρα ἐν ἑνί εἰσιν ἐπιπέδῳ. ἤχθω τῇ ΒΔ πρὸς ὀρθὰς ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ἡ ΔΕ, καὶ κείσθω τῇ ΑΒ ἴση ἡ ΔΕ, καὶ ἐπεζεύχθωσαν αἱ ΒΕ, ΑΕ, ΑΔ. καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν ἡ ΑΒ: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΑΒΔ, ΑΒΕ γωνιῶν. καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπέπτωκεν ἡ ΒΔ, αἱ ἄρα ὑπὸ ΑΒΔ, ΓΔΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΑΒΔ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΓΔΒ: ἡ ΓΔ ἄρα πρὸς τὴν ΒΔ ὀρθή ἐστιν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, κοινὴ δὲ ἡ ΒΔ, δύο δὴ αἱ ΑΒ, ΒΔ δυσὶ ταῖς ΕΔ, ΔΒ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΒΔ γωνίᾳ τῇ ὑπὸ ΕΔΒ ἴση: ὀρθὴ γὰρ ἑκατέρα: βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΕ ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΒΕ τῇ ΑΔ, δύο δὴ αἱ ΑΒ, ΒΕ δυσὶ ταῖς ΕΔ, ΔΑ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ. καὶ βάσις αὐτῶν κοινὴ ἡ ΑΕ: γωνία ἄρα ἡ ὑπὸ ΑΒΕ γωνίᾳ τῇ ὑπὸ ΕΔΑ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΑΒΕ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΕΔΑ: ἡ ΕΔ ἄρα πρὸς τὴν ΑΔ ὀρθή ἐστιν. ἔστι δὲ καὶ πρὸς τὴν ΔΒ ὀρθή: ἡ ΕΔ ἄρα καὶ τῷ διὰ τῶν ΒΔ, ΔΑ ἐπιπέδῳ ὀρθή ἐστιν. καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας ἡ ΕΔ. ἐν δὲ τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ ἐστὶν ἡ ΔΓ, ἐπειδήπερ ἐν τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ εἰσὶν αἱ ΑΒ, ΒΔ, ἐν ᾧ δὲ αἱ ΑΒ, ΒΔ, ἐν τούτῳ ἐστὶ καὶ ἡ ΔΓ. ἡ ΕΔ ἄρα τῇ ΔΓ πρὸς ὀρθάς ἐστιν: ὥστε καὶ ἡ ΓΔ τῇ ΔΕ πρὸς ὀρθάς ἐστιν. ἔστι δὲ καὶ ἡ ΓΔ τῇ ΒΔ πρὸς ὀρθάς. ἡ ΓΔ ἄρα δύο εὐθείαις τεμνούσαις ἀλλήλας ταῖς ΔΕ, ΔΒ ἀπὸ τῆς κατὰ τὸ Δ τομῆς πρὸς ὀρθὰς ἐφέστηκεν: ὥστε ἡ ΓΔ καὶ τῷ διὰ τῶν ΔΕ, ΔΒ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ τῶν ΔΕ, ΔΒ ἐπίπεδον τὸ ὑποκείμενόν ἐστιν: ἡ ΓΔ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. ἐὰν ἄρα ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ μία αὐτῶν ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",444]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'11.9'"],["Rule","'Text'","'Straight lines which are parallel to the same straight line and are not in the same plane with it are also parallel to one another.'"],["Rule","'TextWordCount'",25],["Rule","'GreekText'","'αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ μὴ οὖσαι αὐτῇ ἐν τῷ αὐτῷ ἐπιπέδῳ καὶ ἀλλήλαις εἰσὶ παράλληλοι.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Theorem'",4]],["List",["Rule","'Book'",11],["Rule","'Theorem'",6]],["List",["Rule","'Book'",11],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'For let each of the straight lines AB, CD be parallel to EF, not being in the same plane with it; I say that AB is parallel to CD. For let a point G be taken at random on EF, and from it let there be drawn GH, in the plane through EF, AB, at right angles to EF, and GK in the plane through FE, CD again at right angles to EF. Now, since EF is at right angles to each of the straight lines GH, GK, therefore EF is also at right angles to the plane through GH, GK. [XI. 4] And EF is parallel to AB; therefore AB is also at right angles to the plane through HG, GK. [XI. 8] For the same reason CD is also at right angles to the plane through HG, GK; therefore each of the straight lines AB, CD is at right angles to the plane through HG, GK. But if two straight lines be at right angles to the same plane, the straight lines are parallel; [XI. 6] therefore AB is parallel to CD.'"],["Rule","'ProofWordCount'",184],["Rule","'GreekProof'","'ἔστω γὰρ ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος μὴ οὖσαι αὐτῇ ἐν τῷ αὐτῷ ἐπιπέδῳ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. εἰλήφθω γὰρ ἐπὶ τῆς ΕΖ τυχὸν σημεῖον τὸ Η, καὶ ἀπ᾽ αὐτοῦ τῇ ΕΖ ἐν μὲν τῷ διὰ τῶν ΕΖ, ΑΒ ἐπιπέδῳ πρὸς ὀρθὰς ἤχθω ἡ ΗΘ, ἐν δὲ τῷ διὰ τῶν ΖΕ, ΓΔ τῇ ΕΖ πάλιν πρὸς ὀρθὰς ἤχθω ἡ ΗΚ. καὶ ἐπεὶ ἡ ΕΖ πρὸς ἑκατέραν τῶν ΗΘ, ΗΚ ὀρθή ἐστιν, ἡ ΕΖ ἄρα καὶ τῷ διὰ τῶν ΗΘ, ΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. καί ἐστιν ἡ ΕΖ τῇ ΑΒ παράλληλος: καὶ ἡ ΑΒ ἄρα τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΔ τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν: ἑκατέρα ἄρα τῶν ΑΒ, ΓΔ τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. ἐὰν δὲ δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, παράλληλοί εἰσιν αἱ εὐθεῖαι: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",162]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'11.10'"],["Rule","'Text'","'If two straight lines meeting one another be parallel to two straight lines meeting one another not in the same plane, they will contain equal angles.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἴσας γωνίας περιέξουσιν.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",33]],["List",["Rule","'Book'",11],["Rule","'Theorem'",9]]]],["Rule","'Proof'","'For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not in the same plane; I say that the angle ABC is equal to the angle DEF. For let BA, BC, ED, EF be cut off equal to one another, and let AD, CF, BE, AC, DF be joined. Now, since BA is equal and parallel to ED, therefore AD is also equal and parallel to BE. [I. 33] For the same reason CF is also equal and parallel to BE. Therefore each of the straight lines AD, CF is equal and parallel to BE. But straight lines which are parallel to the same straight line and are not in the same plane with it are parallel to one another; [XI. 9] therefore AD is parallel and equal to CF. And AC, DF join them; therefore AC is also equal and parallel to DF. [I. 33] Now, since the two sides AB, BC are equal to the two sides DE, EF, and the base AC is equal to the base DF, therefore the angle ABC is equal to the angle DEF. [I. 8]'"],["Rule","'ProofWordCount'",196],["Rule","'GreekProof'","'δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΒΓ ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας τὰς ΔΕ, ΕΖ ἁπτομένας ἀλλήλων ἔστωσαν μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ: λέγω, ὅτι ἴση ἐστὶν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ. Ἀπειλήφθωσαν γὰρ αἱ ΒΑ, ΒΓ, ΕΔ, ΕΖ ἴσαι ἀλλήλαις, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΓΖ, ΒΕ, ΑΓ, ΔΖ. καὶ ἐπεὶ ἡ ΒΑ τῇ ΕΔ ἴση ἐστὶ καὶ παράλληλος, καὶ ἡ ΑΔ ἄρα τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΖ τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος: ἑκατέρα ἄρα τῶν ΑΔ, ΓΖ τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος. αἱ δὲ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ μὴ οὖσαι αὐτῇ ἐν τῷ αὐτῷ ἐπιπέδῳ καὶ ἀλλήλαις εἰσὶ παράλληλοι: παράλληλος ἄρα ἐστὶν ἡ ΑΔ τῇ ΓΖ καὶ ἴση. καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΑΓ, ΔΖ: καὶ ἡ ΑΓ ἄρα τῇ ΔΖ ἴση ἐστὶ καὶ παράλληλος. καὶ ἐπεὶ δύο αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσίν, καὶ βάσις ἡ ΑΓ βάσει τῇ ΔΖ ἴση, γωνία ἄρα ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. ἐὰν ἄρα δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἴσας γωνίας περιέξουσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",191]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'11.11'"],["Rule","'Text'","'From a given elevated point to draw a straight line perpendicular to a given plane.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'ἀπὸ τοῦ δοθέντος σημείου μετεώρου ἐπὶ τὸ δοθὲν ἐπίπεδον κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",11]],["List",["Rule","'Book'",1],["Rule","'Theorem'",12]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",4]],["List",["Rule","'Book'",11],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let A be the given elevated point, and the plane of reference the given plane; thus it is required to draw from the point A a straight line perpendicular to the plane of reference. Let any straight line BC be drawn, at random, in the plane of reference, and let AD be drawn from the point A perpendicular to BC. [I. 12] If then AD is also perpendicular to the plane of reference, that which was enjoined will have been done. But, if not, let DE be drawn from the point D at right angles to BC and in the plane of reference, [I. 11] let AF be drawn from A perpendicular to DE, [I. 12] and let GH be drawn through the point F parallel to BC. [I. 31] Now, since BC is at right angles to each of the straight lines DA, DE, therefore BC is also at right angles to the plane through ED, DA. [XI. 4] And GH is parallel to it; but, if two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane; [XI. 8] therefore GH is also at right angles to the plane through ED, DA. Therefore GH is also at right angles to all the straight lines which meet it and are in the plane through ED, DA. [XI. Def. 3] But AF meets it and is in the plane through ED, DA; therefore GH is at right angles to FA, so that FA is also at right angles to GH. But AF is also at right angles to DE; therefore AF is at right angles to each of the straight lines GH, DE. But, if a straight line be set up at right angles to two straight lines which cut one another, at the point of section, it will also be at right angles to the plane through them; [XI. 4] therefore FA is at right angles to the plane through ED, GH. But the plane through ED, GH is the plane of reference; therefore AF is at right angles to the plane of reference. Therefore from the given elevated point A the straight line AF has been drawn perpendicular to the plane of reference.'"],["Rule","'ProofWordCount'",383],["Rule","'GreekProof'","'ἔστω τὸ μὲν δοθὲν σημεῖον μετέωρον τὸ Α, τὸ δὲ δοθὲν ἐπίπεδον τὸ ὑποκείμενον: δεῖ δὴ ἀπὸ τοῦ Α σημείου ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. διήχθω γάρ τις ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖα, ὡς ἔτυχεν, ἡ ΒΓ, καὶ ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ. εἰ μὲν οὖν ἡ ΑΔ κάθετός ἐστι καὶ ἐπὶ τὸ ὑποκείμενον ἐπίπεδον, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ οὔ, ἤχθω ἀπὸ τοῦ Δ σημείου τῇ ΒΓ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΔΕ, καὶ ἤχθω ἀπὸ τοῦ Α ἐπὶ τὴν ΔΕ κάθετος ἡ ΑΖ, καὶ διὰ τοῦ Ζ σημείου τῇ ΒΓ παράλληλος ἤχθω ἡ ΗΘ. καὶ ἐπεὶ ἡ ΒΓ ἑκατέρᾳ τῶν ΔΑ, ΔΕ πρὸς ὀρθάς ἐστιν, ἡ ΒΓ ἄρα καὶ τῷ διὰ τῶν ΕΔΑ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. καί ἐστιν αὐτῇ παράλληλος ἡ ΗΘ: ἐὰν δὲ ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ μία αὐτῶν ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται: καὶ ἡ ΗΘ ἄρα τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ ὀρθή ἐστιν ἡ ΗΘ. ἅπτεται δὲ αὐτῆς ἡ ΑΖ οὖσα ἐν τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ: ἡ ΗΘ ἄρα ὀρθή ἐστι πρὸς τὴν ΖΑ: ὥστε καὶ ἡ ΖΑ ὀρθή ἐστι πρὸς τὴν ΘΗ. ἔστι δὲ ἡ ΑΖ καὶ πρὸς τὴν ΔΕ ὀρθή: ἡ ΑΖ ἄρα πρὸς ἑκατέραν τῶν ΗΘ, ΔΕ ὀρθή ἐστιν. ἐὰν δὲ εὐθεῖα δυσὶν εὐθείαις τεμνούσαις ἀλλήλας ἐπὶ τῆς τομῆς πρὸς ὀρθὰς ἐπισταθῇ, καὶ τῷ δι᾽ αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται: ἡ ΖΑ ἄρα τῷ διὰ τῶν ΕΔ, ΗΘ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ τῶν ΕΔ, ΗΘ ἐπίπεδόν ἐστι τὸ ὑποκείμενον: ἡ ΑΖ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. ἀπὸ τοῦ ἄρα δοθέντος σημείου μετεώρου τοῦ Α ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετος εὐθεῖα γραμμὴ ἦκται ἡ ΑΖ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",319]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'11.12'"],["Rule","'Text'","'To set up a straight line at right angles to a given plane from a given point in it.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'τῷ δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ δοθέντος σημείου πρὸς ὀρθὰς εὐθεῖαν γραμμὴν ἀναστῆσαι.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",11],["Rule","'Theorem'",8]],["List",["Rule","'Book'",11],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'Let the plane of reference be the given plane, and A the point in it; thus it is required to set up from the point A a straight line at right angles to the plane of reference. Let any elevated point B be conceived, from B let BC be drawn perpendicular to the plane of reference, [XI. 11] and through the point A let AD be drawn parallel to BC. [I. 31] Then, since AD, CB are two parallel straight lines, while one of them, BC, is at right angles to the plane of reference, therefore the remaining one, AD, is also at right angles to the plane of reference. [XI. 8] Therefore AD has been set up at right angles to the given plane from the point A in it.'"],["Rule","'ProofWordCount'",131],["Rule","'GreekProof'","'ἔστω τὸ μὲν δοθὲν ἐπίπεδον τὸ ὑποκείμενον, τὸ δὲ πρὸς αὐτῷ σημεῖον τὸ Α: δεῖ δὴ ἀπὸ τοῦ Α σημείου τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς εὐθεῖαν γραμμὴν ἀναστῆσαι. νενοήσθω τι σημεῖον μετέωρον τὸ Β, καὶ ἀπὸ τοῦ Β ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετος ἤχθω ἡ ΒΓ, καὶ διὰ τοῦ Α σημείου τῇ ΒΓ παράλληλος ἤχθω ἡ ΑΔ. ἐπεὶ οὖν δύο εὐθεῖαι παράλληλοί εἰσιν αἱ ΑΔ, ΓΒ, ἡ δὲ μία αὐτῶν ἡ ΒΓ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, καὶ ἡ λοιπὴ ἄρα ἡ ΑΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τῷ ἄρα δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ σημείου τοῦ Α πρὸς ὀρθὰς ἀνέσταται ἡ ΑΔ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",109]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'11.13'"],["Rule","'Text'","'From the same point two straight lines cannot be set up at right angles to the same plane on the same side.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἀπὸ τοῦ αὐτοῦ σημείου τῷ αὐτῷ ἐπιπέδῳ δύο εὐθεῖαι πρὸς ὀρθὰς οὐκ ἀναστήσονται ἐπὶ τὰ αὐτὰ μέρη.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'For, if possible, from the same point A let the two straight lines AB, AC be set up at right angles to the plane of reference and on the same side, and let a plane be drawn through BA, AC; it will then make, as section through A in the plane of reference, a straight line. [XI. 3] Let it make DAE; therefore the straight lines AB, AC, DAE are in one plane. And, since CA is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3] But DAE meets it and is in the plane of reference; therefore the angle CAE is right. For the same reason the angle BAE is also right; therefore the angle CAE is equal to the angle BAE. And they are in one plane: which is impossible.'"],["Rule","'ProofWordCount'",155],["Rule","'GreekProof'","'εἰ γὰρ δυνατόν, ἀπὸ τοῦ αὐτοῦ σημείου τοῦ Α τῷ ὑποκειμένῳ ἐπιπέδῳ δύο εὐθεῖαι αἱ ΑΒ, ΑΓ πρὸς ὀρθὰς ἀνεστάτωσαν ἐπὶ τὰ αὐτὰ μέρη, καὶ διήχθω τὸ διὰ τῶν ΒΑ, ΑΓ ἐπίπεδον: τομὴν δὴ ποιήσει διὰ τοῦ Α ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν. ποιείτω τὴν ΔΑΕ: αἱ ἄρα ΑΒ, ΑΓ, ΔΑΕ εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ. καὶ ἐπεὶ ἡ ΓΑ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. ἅπτεται δὲ αὐτῆς ἡ ΔΑΕ οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ: ἡ ἄρα ὑπὸ ΓΑΕ γωνία ὀρθή ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΒΑΕ ὀρθή ἐστιν: ἴση ἄρα ἡ ὑπὸ ΓΑΕ τῇ ὑπὸ ΒΑΕ. καί εἰσιν ἐν ἑνὶ ἐπιπέδῳ: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἀπὸ τοῦ αὐτοῦ σημείου τῷ αὐτῷ ἐπιπέδῳ δύο εὐθεῖαι πρὸς ὀρθὰς ἀνασταθήσονται ἐπὶ τὰ αὐτὰ μέρη: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",147]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'11.14'"],["Rule","'Text'","'Planes to which the same straight line is at right angles will be parallel.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'πρὸς ἃ ἐπίπεδα ἡ αὐτὴ εὐθεῖα ὀρθή ἐστιν, παράλληλα ἔσται τὰ ἐπίπεδα.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",17]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Definition'",8]],["List",["Rule","'Book'",11],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'For let any straight line AB be at right angles to each of the planes CD, EF; I say that the planes are parallel. For, if not, they will meet when produced. Let them meet; they will then make, as common section, a straight line. [XI. 3] Let them make GH; let a point K be taken at random on GH, and let AK, BK be joined. Now, since AB is at right angles to the plane EF, therefore AB is also at right angles to BK which is a straight line in the plane EF produced; [XI. Def. 3] therefore the angle ABK is right. For the same reason the angle BAK is also right. Thus, in the triangle ABK, the two angles ABK, BAK are equal to two right angles: which is impossible. [I. 17] Therefore the planes CD, EF will not meet when produced; therefore the planes CD, EF are parallel. [XI. Def. 8] Therefore planes to which the same straight line is at right angles are parallel.'"],["Rule","'ProofWordCount'",171],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ πρὸς ἑκάτερον τῶν ΓΔ, ΕΖ ἐπιπέδων πρὸς ὀρθὰς ἔστω: λέγω, ὅτι παράλληλά ἐστι τὰ ἐπίπεδα. εἰ γὰρ μή, ἐκβαλλόμενα συμπεσοῦνται. συμπιπτέτωσαν: ποιήσουσι δὴ κοινὴν τομὴν εὐθεῖαν. ποιείτωσαν τὴν ΗΘ, καὶ εἰλήφθω ἐπὶ τῆς ΗΘ τυχὸν σημεῖον τὸ Κ, καὶ ἐπεζεύχθωσαν αἱ ΑΚ, ΒΚ. καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς τὸ ΕΖ ἐπίπεδον, καὶ πρὸς τὴν ΒΚ ἄρα εὐθεῖαν οὖσαν ἐν τῷ ΕΖ ἐκβληθέντι ἐπιπέδῳ ὀρθή ἐστιν ἡ ΑΒ: ἡ ἄρα ὑπὸ ΑΒΚ γωνία ὀρθή ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΒΑΚ ὀρθή ἐστιν. τριγώνου δὴ τοῦ ΑΒΚ αἱ δύο γωνίαι αἱ ὑπὸ ΑΒΚ, ΒΑΚ δυσὶν ὀρθαῖς εἰσιν ἴσαι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΓΔ, ΕΖ ἐπίπεδα ἐκβαλλόμενα συμπεσοῦνται: παράλληλα ἄρα ἐστὶ τὰ ΓΔ, ΕΖ ἐπίπεδα. πρὸς ἃ ἐπίπεδα ἄρα ἡ αὐτὴ εὐθεῖα ὀρθή ἐστιν, παράλληλά ἐστι τὰ ἐπίπεδα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",140]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'11.15'"],["Rule","'Text'","'If two straight lines meeting one another be parallel to two straight lines meeting one another, not being in the same plane, the planes through them are parallel.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι, παράλληλά ἐστι τὰ δι᾽ αὐτῶν ἐπίπεδα.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",31]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",4]],["List",["Rule","'Book'",11],["Rule","'Theorem'",9]],["List",["Rule","'Book'",11],["Rule","'Theorem'",11]],["List",["Rule","'Book'",11],["Rule","'Theorem'",14]]]],["Rule","'Proof'","'For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not being in the same plane; I say that the planes produced through AB, BC and DE, EF will not meet one another. For let BG be drawn from the point B perpendicular to the plane through DE, EF [XI. 11], and let it meet the plane at the point G; through G let GH be drawn parallel to ED, and GK parallel to EF. [I. 31] Now, since BG is at right angles to the plane through DE, EF, therefore it will also make right angles with all the straight lines which meet it and are in the plane through DE, EF. [XI. Def. 3] But each of the straight lines GH, GK meets it and is in the plane through DE, EF; therefore each of the angles BGH, BGK is right. And, since BA is parallel to GH, [XI. 9] therefore the angles GBA, BGH are equal to two right angles. [I. 29] But the angle BGH is right; therefore the angle GBA is also right; therefore GB is at right angles to BA. For the same reason GB is also at right angles to BC. Since then the straight line GB is set up at right angles to the two straight lines BA, BC which cut one another, therefore GB is also at right angles to the plane through BA, BC. [XI. 4] But planes to which the same straight line is at right angles are parallel; [XI. 14] therefore the plane through AB, BC is parallel to the plane through DE, EF. Therefore, if two straight lines meeting one another be parallel to two straight lines meeting one another, not in the same plane, the planes through them are parallel.'"],["Rule","'ProofWordCount'",307],["Rule","'GreekProof'","'δύο γὰρ εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΑΒ, ΒΓ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΔΕ, ΕΖ ἔστωσαν μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι: λέγω, ὅτι ἐκβαλλόμενα τὰ διὰ τῶν ΑΒ, ΒΓ, ΔΕ, ΕΖ ἐπίπεδα οὐ συμπεσεῖται ἀλλήλοις. ἤχθω γὰρ ἀπὸ τοῦ Β σημείου ἐπὶ τὸ διὰ τῶν ΔΕ, ΕΖ ἐπίπεδον κάθετος ἡ ΒΗ καὶ συμβαλλέτω τῷ ἐπιπέδῳ κατὰ τὸ Η σημεῖον, καὶ διὰ τοῦ Η τῇ μὲν ΕΔ παράλληλος ἤχθω ἡ ΗΘ, τῇ δὲ ΕΖ ἡ ΗΚ. καὶ ἐπεὶ ἡ ΒΗ ὀρθή ἐστι πρὸς τὸ διὰ τῶν ΔΕ, ΕΖ ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. ἅπτεται δὲ αὐτῆς ἑκατέρα τῶν ΗΘ, ΗΚ οὖσα ἐν τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΒΗΘ, ΒΗΚ γωνιῶν. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΒΑ τῇ ΗΘ, αἱ ἄρα ὑπὸ ΗΒΑ, ΒΗΘ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΒΗΘ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΗΒΑ: ἡ ΗΒ ἄρα τῇ ΒΑ πρὸς ὀρθάς ἐστιν. διὰ τὰ αὐτὰ δὴ ἡ ΗΒ καὶ τῇ ΒΓ ἐστι πρὸς ὀρθάς. ἐπεὶ οὖν εὐθεῖα ἡ ΗΒ δυσὶν εὐθείαις ταῖς ΒΑ, ΒΓ τεμνούσαις ἀλλήλας πρὸς ὀρθὰς ἐφέστηκεν, ἡ ΗΒ ἄρα καὶ τῷ διὰ τῶν ΒΑ, ΒΓ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. διὰ τὰ αὐτὰ δὴ ἡ ΒΗ καὶ τῷ διὰ τῶν ΗΘ, ΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ τῶν ΗΘ, ΗΚ ἐπίπεδόν ἐστι τὸ διὰ τῶν ΔΕ, ΕΖ: ἡ ΒΗ ἄρα τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ ἐστὶ πρὸς ὀρθάς. ἐδείχθη δὲ ἡ ΗΒ καὶ τῷ διὰ τῶν ΑΒ, ΒΓ ἐπιπέδῳ πρὸς ὀρθάς. πρὸς ἃ δὲ ἐπίπεδα ἡ αὐτὴ εὐθεῖα ὀρθή ἐστιν, παράλληλά ἐστι τὰ ἐπίπεδα: παράλληλον ἄρα ἐστὶ τὸ διὰ τῶν ΑΒ, ΒΓ ἐπίπεδον τῷ διὰ τῶν ΔΕ, ΕΖ. ἐὰν ἄρα δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, παράλληλά ἐστι τὰ δι᾽ αὐτῶν ἐπίπεδα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",319]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'11.16'"],["Rule","'Text'","'If two parallel planes be cut by any plane, their common sections are parallel.'"],["Rule","'TextWordCount'",14],["Rule","'GreekText'","'ἐὰν δύο ἐπίπεδα παράλληλα ὑπὸ ἐπιπέδου τινὸς τέμνηται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Definition'",23]],["List",["Rule","'Book'",11],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'For let the two parallel planes AB, CD be cut by the plane EFGH, and let EF, GH be their common sections; I say that EF is parallel to GH. For, if not, EF, GH will, when produced, meet either in the direction of F, H or of E, G. Let them be produced, as in the direction of F, H, and let them, first, meet at K. Now, since EFK is in the plane AB, therefore all the points on EFK are also in the plane AB. [XI. 1] But K is one of the points on the straight line EFK; therefore K is in the plane AB. For the same reason K is also in the plane CD; therefore the planes AB, CD will meet when produced. But they do not meet, because they are, by hypothesis, parallel; therefore the straight lines EF, GH will not meet when produced in the direction of F, H. Similarly we can prove that neither will the straight lines EF, GH meet when produced in the direction of E, G. But straight lines which do not meet in either direction are parallel. [I. Def. 23] Therefore EF is parallel to GH.'"],["Rule","'ProofWordCount'",199],["Rule","'GreekProof'","'δύο γὰρ ἐπίπεδα παράλληλα τὰ ΑΒ, ΓΔ ὑπὸ ἐπιπέδου τοῦ ΕΖΗΘ τεμνέσθω, κοιναὶ δὲ αὐτῶν τομαὶ ἔστωσαν αἱ ΕΖ, ΗΘ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΕΖ τῇ ΗΘ. εἰ γὰρ μή, ἐκβαλλόμεναι αἱ ΕΖ, ΗΘ ἤτοι ἐπὶ τὰ Ζ, Θ μέρη ἢ ἐπὶ τὰ Ε, Η συμπεσοῦνται. ἐκβεβλήσθωσαν ὡς ἐπὶ τὰ Ζ, Θ μέρη καὶ συμπιπτέτωσαν πρότερον κατὰ τὸ Κ. καὶ ἐπεὶ ἡ ΕΖΚ ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ, καὶ πάντα ἄρα τὰ ἐπὶ τῆς ΕΖΚ σημεῖα ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ. ἓν δὲ τῶν ἐπὶ τῆς ΕΖΚ εὐθείας σημείων ἐστὶ τὸ Κ: τὸ Κ ἄρα ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ. διὰ τὰ αὐτὰ δὴ τὸ Κ καὶ ἐν τῷ ΓΔ ἐστιν ἐπιπέδῳ: τὰ ΑΒ, ΓΔ ἄρα ἐπίπεδα ἐκβαλλόμενα συμπεσοῦνται. οὐ συμπίπτουσι δὲ διὰ τὸ παράλληλα ὑποκεῖσθαι: οὐκ ἄρα αἱ ΕΖ, ΗΘ εὐθεῖαι ἐκβαλλόμεναι ἐπὶ τὰ Ζ, Θ μέρη συμπεσοῦνται. ὁμοίως δὴ δείξομεν, ὅτι αἱ ΕΖ, ΗΘ εὐθεῖαι οὐδὲ ἐπὶ τὰ Ε, Η μέρη ἐκβαλλόμεναι συμπεσοῦνται. αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη συμπίπτουσαι παράλληλοί εἰσιν. παράλληλος ἄρα ἐστὶν ἡ ΕΖ τῇ ΗΘ. ἐὰν ἄρα δύο ἐπίπεδα παράλληλα ὑπὸ ἐπιπέδου τινὸς τέμνηται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",190]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'11.17'"],["Rule","'Text'","'If two straight lines be cut by parallel planes, they will be cut in the same ratios.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἐὰν δύο εὐθεῖαι ὑπὸ παραλλήλων ἐπιπέδων τέμνωνται εἰς τοὺς αὐτοὺς λόγους τμηθήσονται.'"],["Rule","'GreekTextWordCount'",12],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]],["List",["Rule","'Book'",11],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'For let the two straight lines AB, CD be cut by the parallel planes GH, KL, MN at the points A, E, B and C, F, D; I say that, as the straight line AE is to EB, so is CF to FD. For let AC, BD, AD be joined, let AD meet the plane KL at the point O, and let EO, OF be joined. Now, since the two parallel planes KL, MN are cut by the plane EBDO, their common sections EO, BD are parallel. [XI. 16] For the same reason, since the two parallel planes GH, KL are cut by the plane AOFC, their common sections AC, OF are parallel. [id.] And, since the straight line EO has been drawn parallel to BD, one of the sides of the triangle ABD, therefore, proportionally, as AE is to EB, so is AO to OD. [VI. 2] Again, since the straight line OF has been drawn parallel to AC, one of the sides of the triangle ADC, proportionally, as AO is to OD, so is CF to FD. [id.] But it was also proved that, as AO is to OD, so is AE to EB; therefore also, as AE is to EB, so is CF to FD. [V. 11]'"],["Rule","'ProofWordCount'",210],["Rule","'GreekProof'","'δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ ὑπὸ παραλλήλων ἐπιπέδων τῶν ΗΘ, ΚΛ, ΜΝ τεμνέσθωσαν κατὰ τὰ Α, Ε, Β, Γ, Ζ, Δ σημεῖα: λέγω, ὅτι ἐστὶν ὡς ἡ ΑΕ εὐθεῖα πρὸς τὴν ΕΒ, οὕτως ἡ ΓΖ πρὸς τὴν ΖΔ. ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΒΔ, ΑΔ, καὶ συμβαλλέτω ἡ ΑΔ τῷ ΚΛ ἐπιπέδῳ κατὰ τὸ Ξ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΕΞ, ΞΖ. καὶ ἐπεὶ δύο ἐπίπεδα παράλληλα τὰ ΚΛ, ΜΝ ὑπὸ ἐπιπέδου τοῦ ΕΒΔΞ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ αἱ ΕΞ, ΒΔ παράλληλοί εἰσιν. διὰ τὰ αὐτὰ δὴ ἐπεὶ δύο ἐπίπεδα παράλληλα τὰ ΗΘ, ΚΛ ὑπὸ ἐπιπέδου τοῦ ΑΞΖΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ αἱ ΑΓ, ΞΖ παράλληλοί εἰσιν. καὶ ἐπεὶ τριγώνου τοῦ ΑΒΔ παρὰ μίαν τῶν πλευρῶν τὴν ΒΔ εὐθεῖα ἦκται ἡ ΕΞ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς ΕΒ, οὕτως ἡ ΑΞ πρὸς ΞΔ. πάλιν ἐπεὶ τριγώνου τοῦ ΑΔΓ παρὰ μίαν τῶν πλευρῶν τὴν ΑΓ εὐθεῖα ἦκται ἡ ΞΖ, ἀνάλογόν ἐστιν ὡς ἡ ΑΞ πρὸς ΞΔ, οὕτως ἡ ΓΖ πρὸς ΖΔ. ἐδείχθη δὲ καὶ ὡς ἡ ΑΞ πρὸς ΞΔ, οὕτως ἡ ΑΕ πρὸς ΕΒ: καὶ ὡς ἄρα ἡ ΑΕ πρὸς ΕΒ, οὕτως ἡ ΓΖ πρὸς ΖΔ. ἐὰν ἄρα δύο εὐθεῖαι ὑπὸ παραλλήλων ἐπιπέδων τέμνωνται, εἰς τοὺς αὐτοὺς λόγους τμηθήσονται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",204]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'11.18'"],["Rule","'Text'","'If a straight line be at right angles to any plane, all the planes through it will also be at right angles to the same plane.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐὰν εὐθεῖα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ πάντα τὰ δι᾽ αὐτῆς ἐπίπεδα τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται.'"],["Rule","'GreekTextWordCount'",20],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",11]],["List",["Rule","'Book'",1],["Rule","'Theorem'",28]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Definition'",4]],["List",["Rule","'Book'",11],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'For let any straight line AB be at right angles to the plane of reference; I say that all the planes through AB are also at right angles to the plane of reference. For let the plane DE be drawn through AB, let CE be the common section of the plane DE and the plane of reference, let a point F be taken at random on CE, and from F let FG be drawn in the plane DE at right angles to CE. [I. 11] Now, since AB is at right angles to the plane of reference, AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] so that it is also at right angles to CE; therefore the angle ABF is right. But the angle GFB is also right; therefore AB is parallel to FG. [I. 28] But AB is at right angles to the plane of reference; therefore FG is also at right angles to the plane of reference. [XI. 8] Now a plane is at right angles to a plane, when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane. [XI. Def. 4] And FG, drawn in one of the planes DE at right angles to CE, the common section of the planes, was proved to be at right angles to the plane of reference; therefore the plane DE is at right angles to the plane of reference. Similarly also it can be proved that all the planes through AB are at right angles to the plane of reference.'"],["Rule","'ProofWordCount'",284],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστω: λέγω, ὅτι καὶ πάντα τὰ διὰ τῆς ΑΒ ἐπίπεδα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. Ἐκβεβλήσθω γὰρ διὰ τῆς ΑΒ ἐπίπεδον τὸ ΔΕ, καὶ ἔστω κοινὴ τομὴ τοῦ ΔΕ ἐπιπέδου καὶ τοῦ ὑποκειμένου ἡ ΓΕ, καὶ εἰλήφθω ἐπὶ τῆς ΓΕ τυχὸν σημεῖον τὸ Ζ, καὶ ἀπὸ τοῦ Ζ τῇ ΓΕ πρὸς ὀρθὰς ἤχθω ἐν τῷ ΔΕ ἐπιπέδῳ ἡ ΖΗ. καὶ ἐπεὶ ἡ ΑΒ πρὸς τὸ ὑποκείμενον ἐπίπεδον ὀρθή ἐστιν, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθή ἐστιν ἡ ΑΒ: ὥστε καὶ πρὸς τὴν ΓΕ ὀρθή ἐστιν: ἡ ἄρα ὑπὸ ΑΒΖ γωνία ὀρθή ἐστιν. ἔστι δὲ καὶ ἡ ὑπὸ ΗΖΒ ὀρθή: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΖΗ. ἡ δὲ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν: καὶ ἡ ΖΗ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. καὶ ἐπίπεδον πρὸς ἐπίπεδον ὀρθόν ἐστιν, ὅταν αἱ τῇ κοινῇ τομῇ τῶν ἐπιπέδων πρὸς ὀρθὰς ἀγόμεναι εὐθεῖαι ἐν ἑνὶ τῶν ἐπιπέδων τῷ λοιπῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν. καὶ τῇ κοινῇ τομῇ τῶν ἐπιπέδων τῇ ΓΕ ἐν ἑνὶ τῶν ἐπιπέδων τῷ ΔΕ πρὸς ὀρθὰς ἀχθεῖσα ἡ ΖΗ ἐδείχθη τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς: τὸ ἄρα ΔΕ ἐπίπεδον ὀρθόν ἐστι πρὸς τὸ ὑποκείμενον. ὁμοίως δὴ δειχθήσεται καὶ πάντα τὰ διὰ τῆς ΑΒ ἐπίπεδα ὀρθὰ τυγχάνοντα πρὸς τὸ ὑποκείμενον ἐπίπεδον. ἐὰν ἄρα εὐθεῖα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ πάντα τὰ δι᾽ αὐτῆς ἐπίπεδα τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",246]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",19]],["Association",["Rule","'VertexLabel'","'11.19'"],["Rule","'Text'","'If two planes which cut one another be at right angles to any plane, their common section will also be at right angles to the same plane.'"],["Rule","'TextWordCount'",27],["Rule","'GreekText'","'ἐὰν δύο ἐπίπεδα τέμνοντα ἄλληλα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ κοινὴ αὐτῶν τομὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Definition'",4]],["List",["Rule","'Book'",11],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'For let the two planes AB, BC be at right angles to the plane of reference, and let BD be their common section; I say that BD is at right angles to the plane of reference. For suppose it is not, and from the point D let DE be drawn in the plane AB at right angles to the straight line AD, and DF in the plane BC at right angles to CD. Now, since the plane AB is at right angles to the plane of reference, and DE has been drawn in the plane AB at right angles to AD, their common section, therefore DE is at right angles to the plane of reference. [XI. Def. 4] Similarly we can prove that DF is also at right angles to the plane of reference. Therefore from the same point D two straight lines have been set up at right angles to the plane of reference on the same side: which is impossible. [XI. 13] Therefore no straight line except the common section DB of the planes AB, BC can be set up from the point D at right angles to the plane of reference.'"],["Rule","'ProofWordCount'",194],["Rule","'GreekProof'","'δύο γὰρ ἐπίπεδα τὰ ΑΒ, ΒΓ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστω, κοινὴ δὲ αὐτῶν τομὴ ἔστω ἡ ΒΔ: λέγω, ὅτι ἡ ΒΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. μὴ γάρ, καὶ ἤχθωσαν ἀπὸ τοῦ Δ σημείου ἐν μὲν τῷ ΑΒ ἐπιπέδῳ τῇ ΑΔ εὐθείᾳ πρὸς ὀρθὰς ἡ ΔΕ, ἐν δὲ τῷ ΒΓ ἐπιπέδῳ τῇ ΓΔ πρὸς ὀρθὰς ἡ ΔΖ. καὶ ἐπεὶ τὸ ΑΒ ἐπίπεδον ὀρθόν ἐστι πρὸς τὸ ὑποκείμενον, καὶ τῇ κοινῇ αὐτῶν τομῇ τῇ ΑΔ πρὸς ὀρθὰς ἐν τῷ ΑΒ ἐπιπέδῳ ἦκται ἡ ΔΕ, ἡ ΔΕ ἄρα ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΔΖ ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον. ἀπὸ τοῦ αὐτοῦ ἄρα σημείου τοῦ Δ τῷ ὑποκειμένῳ ἐπιπέδῳ δύο εὐθεῖαι πρὸς ὀρθὰς ἀνεσταμέναι εἰσὶν ἐπὶ τὰ αὐτὰ μέρη: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ ἀπὸ τοῦ Δ σημείου ἀνασταθήσεται πρὸς ὀρθὰς πλὴν τῆς ΔΒ κοινῆς τομῆς τῶν ΑΒ, ΒΓ ἐπιπέδων. ἐὰν ἄρα δύο ἐπίπεδα τέμνοντα ἄλληλα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ κοινὴ αὐτῶν τομὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",177]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",20]],["Association",["Rule","'VertexLabel'","'11.20'"],["Rule","'Text'","'If a solid angle be contained by three plane angles, any two, taken together in any manner, are greater than the remaining one.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'ἐὰν στερεὰ γωνία ὑπὸ τριῶν γωνιῶν ἐπιπέδων περιέχηται, δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",20]],["List",["Rule","'Book'",1],["Rule","'Theorem'",25]]]],["Rule","'Proof'","'For let the solid angle at A be contained by the three plane angles BAC, CAD, DAB; I say that any two of the angles BAC, CAD, DAB, taken together in any manner, are greater than the remaining one. If now the angles BAC, CAD, DAB are equal to one another, it is manifest that any two are greater than the remaining one. But, if not, let BAC be greater, and on the straight line AB, and at the point A on it, let the angle BAE be constructed, in the plane through BA, AC, equal to the angle DAB; let AE be made equal to AD, and let BEC, drawn across through the point E, cut the straight lines AB, AC at the points B, C; let DB, DC be joined. Now, since DA is equal to AE, and AB is common, two sides are equal to two sides; and the angle DAB is equal to the angle BAE; therefore the base DB is equal to the base BE. [I. 4] And, since the two sides BD, DC are greater than BC, [I. 20] and of these DB was proved equal to BE, therefore the remainder DC is greater than the remainder EC. Now, since DA is equal to AE, and AC is common, and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EAC. [I. 25] But the angle DAB was also proved equal to the angle BAE; therefore the angles DAB, DAC are greater than the angle BAC. Similarly we can prove that the remaining angles also, taken together two and two, are greater than the remaining one.'"],["Rule","'ProofWordCount'",280],["Rule","'GreekProof'","'στερεὰ γὰρ γωνία ἡ πρὸς τῷ Α ὑπὸ τριῶν γωνιῶν ἐπιπέδων τῶν ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ περιεχέσθω: λέγω, ὅτι τῶν ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ γωνιῶν δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι. εἰ μὲν οὖν αἱ ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ γωνίαι ἴσαι ἀλλήλαις εἰσίν, φανερόν, ὅτι δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσιν. εἰ δὲ οὔ, ἔστω μείζων ἡ ὑπὸ ΒΑΓ, καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΔΑΒ γωνίᾳ ἐν τῷ διὰ τῶν ΒΑΓ ἐπιπέδῳ ἴση ἡ ὑπὸ ΒΑΕ, καὶ κείσθω τῇ ΑΔ ἴση ἡ ΑΕ, καὶ διὰ τοῦ Ε σημείου διαχθεῖσα ἡ ΒΕΓ τεμνέτω τὰς ΑΒ, ΑΓ εὐθείας κατὰ τὰ Β, Γ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΔΒ, ΔΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΒ, δύο δυσὶν ἴσαι: καὶ γωνία ἡ ὑπὸ ΔΑΒ γωνίᾳ τῇ ὑπὸ ΒΑΕ ἴση: βάσις ἄρα ἡ ΔΒ βάσει τῇ ΒΕ ἐστιν ἴση. καὶ ἐπεὶ δύο αἱ ΒΔ, ΔΓ τῆς ΒΓ μείζονές εἰσιν, ὧν ἡ ΔΒ τῇ ΒΕ ἐδείχθη ἴση, λοιπὴ ἄρα ἡ ΔΓ λοιπῆς τῆς ΕΓ μείζων ἐστίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΓ, καὶ βάσις ἡ ΔΓ βάσεως τῆς ΕΓ μείζων ἐστίν, γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίας τῆς ὑπὸ ΕΑΓ μείζων ἐστίν. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΔΑΒ τῇ ὑπὸ ΒΑΕ ἴση: αἱ ἄρα ὑπὸ ΔΑΒ, ΔΑΓ τῆς ὑπὸ ΒΑΓ μείζονές εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ λοιπαὶ σύνδυο λαμβανόμεναι τῆς λοιπῆς μείζονές εἰσιν. ἐὰν ἄρα στερεὰ γωνία ὑπὸ τριῶν γωνιῶν ἐπιπέδων περιέχηται, δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",263]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",21]],["Association",["Rule","'VertexLabel'","'11.21'"],["Rule","'Text'","'Any solid angle is contained by plane angles less than four right angles.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'ἅπασα στερεὰ γωνία ὑπὸ ἐλασσόνων ἢ τεσσάρων ὀρθῶν γωνιῶν ἐπιπέδων περιέχεται.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",11],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let the angle at A be a solid angle contained by the plane angles BAC, CAD, DAB; I say that the angles BAC, CAD, DAB are less than four right angles. For let points B, C, D be taken at random on the straight lines AB, AC, AD respectively, and let BC, CD, DB be joined. Now, since the solid angle at B is contained by the three plane angles CBA, ABD, CBD, any two are greater than the remaining one; [XI. 20] therefore the angles CBA, ABD are greater than the angle CBD. For the same reason the angles BCA, ACD are also greater than the angle BCD, and the angles CDA, ADB are greater than the angle CDB; therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles CBD, BCD, CDB. But the three angles CBD, BDC, BCD are equal to two right angles; [I. 32] therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles. And, since the three angles of each of the triangles ABC, ACD, ADB are equal to two right angles, therefore the nine angles of the three triangles, the angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, BAD are equal to six right angles; and of them the six angles ABC, BCA, ACD, CDA, ADB, DBA are greater than two right angles; therefore the remaining three angles BAC, CAD, DAB containing the solid angle are less than four right angles.'"],["Rule","'ProofWordCount'",250],["Rule","'GreekProof'","'ἔστω στερεὰ γωνία ἡ πρὸς τῷ Α περιεχομένη ὑπὸ ἐπιπέδων γωνιῶν τῶν ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ: λέγω, ὅτι αἱ ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ τεσσάρων ὀρθῶν ἐλάσσονές εἰσιν. εἰλήφθω γὰρ ἐφ᾽ ἑκάστης τῶν ΑΒ, ΑΓ, ΑΔ τυχόντα σημεῖα τὰ Β, Γ, Δ, καὶ ἐπεζεύχθωσαν αἱ ΒΓ, ΓΔ, ΔΒ. καὶ ἐπεὶ στερεὰ γωνία ἡ πρὸς τῷ Β ὑπὸ τριῶν γωνιῶν ἐπιπέδων περιέχεται τῶν ὑπὸ ΓΒΑ, ΑΒΔ, ΓΒΔ, δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσιν: αἱ ἄρα ὑπὸ ΓΒΑ, ΑΒΔ τῆς ὑπὸ ΓΒΔ μείζονές εἰσιν. διὰ τὰ αὐτὰ δὴ καὶ αἱ μὲν ὑπὸ ΒΓΑ, ΑΓΔ τῆς ὑπὸ ΒΓΔ μείζονές εἰσιν, αἱ δὲ ὑπὸ ΓΔΑ, ΑΔΒ τῆς ὑπὸ ΓΔΒ μείζονές εἰσιν: αἱ ἓξ ἄρα γωνίαι αἱ ὑπὸ ΓΒΑ, ΑΒΔ, ΒΓΑ, ΑΓΔ, ΓΔΑ, ΑΔΒ τριῶν τῶν ὑπὸ ΓΒΔ, ΒΓΔ, ΓΔΒ μείζονές εἰσιν. ἀλλὰ αἱ τρεῖς αἱ ὑπὸ ΓΒΔ, ΒΔΓ, ΒΓΔ δυσὶν ὀρθαῖς ἴσαι εἰσίν: αἱ ἓξ ἄρα αἱ ὑπὸ ΓΒΑ, ΑΒΔ, ΒΓΑ, ΑΓΔ, ΓΔΑ, ΑΔΒ δύο ὀρθῶν μείζονές εἰσιν. καὶ ἐπεὶ ἑκάστου τῶν ΑΒΓ, ΑΓΔ, ΑΔΒ τριγώνων αἱ τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν, αἱ ἄρα τῶν τριῶν τριγώνων ἐννέα γωνίαι αἱ ὑπὸ ΓΒΑ, ΑΓΒ, ΒΑΓ, ΑΓΔ, ΓΔΑ, ΓΑΔ, ΑΔΒ, ΔΒΑ, ΒΑΔ ἓξ ὀρθαῖς ἴσαι εἰσίν, ὧν αἱ ὑπὸ ΑΒΓ, ΒΓΑ, ΑΓΔ, ΓΔΑ, ΑΔΒ, ΔΒΑ ἓξ γωνίαι δύο ὀρθῶν εἰσι μείζονες: λοιπαὶ ἄρα αἱ ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ τρεῖς γωνίαι περιέχουσαι τὴν στερεὰν γωνίαν τεσσάρων ὀρθῶν ἐλάσσονές εἰσιν. ἅπασα ἄρα στερεὰ γωνία ὑπὸ ἐλασσόνων ἢ τεσσάρων ὀρθῶν γωνιῶν ἐπιπέδων περιέχεται: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",238]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",22]],["Association",["Rule","'VertexLabel'","'11.22'"],["Rule","'Text'","'If there be three plane angles of which two, taken together in any manner, are greater than the remaining one, and they are contained by equal straight lines, it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines.'"],["Rule","'TextWordCount'",48],["Rule","'GreekText'","'ἐὰν ὦσι τρεῖς γωνίαι ἐπίπεδοι, ὧν αἱ δύο τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, περιέχωσι δὲ αὐτὰς ἴσαι εὐθεῖαι, δυνατόν ἐστιν ἐκ τῶν ἐπιζευγνυουσῶν τὰς ἴσας εὐθείας τρίγωνον συστήσασθαι.'"],["Rule","'GreekTextWordCount'",29],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'Let there be three plane angles ABC, DEF, GHK, of which two, taken together in any manner, are greater than the remaining one, namely the angles ABC, DEF greater than the angle GHK, the angles DEF, GHK greater than the angle ABC, and, further, the angles GHK, ABC greater than the angle DEF; let the straight lines AB, BC, DE, EF, GH, HK be equal, and let AC, DF, GK be joined; I say that it is possible to construct a triangle out of straight lines equal to AC, DF, GK, that is, that any two of the straight lines AC, DF, GK are greater than the remaining one. Now, if the angles ABC, DEF, GHK are equal to one another, it is manifest that, AC, DF, GK being equal also, it is possible to construct a triangle out of straight lines equal to AC, DF, GK. But, if not, let them be unequal, and on the straight line HK, and at the point H on it, let the angle KHL be constructed equal to the angle ABC; let HL be made equal to one of the straight lines AB, BC, DE, EF, GH, HK, and let KL, GL be joined. Now, since the two sides AB, BC are equal to the two sides KH, HL, and the angle at B is equal to the angle KHL, therefore the base AC is equal to the base KL. [I. 4] And, since the angles ABC, GHK are greater than the angle DEF, while the angle ABC is equal to the angle KHL, therefore the angle GHL is greater than the angle DEF. And, since the two sides GH, HL are equal to the two sides DE, EF, and the angle GHL is greater than the angle DEF, therefore the base GL is greater than the base DF. [I. 24] But GK, KL are greater than GL. Therefore GK, KL are much greater than DF. But KL is equal to AC; therefore AC, GK are greater than the remaining straight line DF. Similarly we can prove that AC, DF are greater than GK, and further DF, GK are greater than AC. Therefore it is possible to construct a triangle out of straight lines equal to AC, DF, GK.'"],["Rule","'ProofWordCount'",376],["Rule","'GreekProof'","'ἔστωσαν τρεῖς γωνίαι ἐπίπεδοι αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ, ὧν αἱ δύο τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, αἱ μὲν ὑπὸ ΑΒΓ, ΔΕΖ τῆς ὑπὸ ΗΘΚ, αἱ δὲ ὑπὸ ΔΕΖ, ΗΘΚ τῆς ὑπὸ ΑΒΓ, καὶ ἔτι αἱ ὑπὸ ΗΘΚ, ΑΒΓ τῆς ὑπὸ ΔΕΖ, καὶ ἔστωσαν ἴσαι αἱ ΑΒ, ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ εὐθεῖαι, καὶ ἐπεζεύχθωσαν αἱ ΑΓ, ΔΖ, ΗΚ: λέγω, ὅτι δυνατόν ἐστιν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον συστήσασθαι, τουτέστιν ὅτι τῶν ΑΓ, ΔΖ, ΗΚ δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσιν. εἰ μὲν οὖν αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ γωνίαι ἴσαι ἀλλήλαις εἰσίν, φανερόν, ὅτι καὶ τῶν ΑΓ, ΔΖ, ΗΚ ἴσων γινομένων δυνατόν ἐστιν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον συστήσασθαι. εἰ δὲ οὔ, ἔστωσαν ἄνισοι, καὶ συνεστάτω πρὸς τῇ ΘΚ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ τῇ ὑπὸ ΑΒΓ γωνίᾳ ἴση ἡ ὑπὸ ΚΘΛ: καὶ κείσθω μιᾷ τῶν ΑΒ, ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ ἴση ἡ ΘΛ, καὶ ἐπεζεύχθωσαν αἱ ΚΛ, ΗΛ. καὶ ἐπεὶ δύο αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΚΘ, ΘΛ ἴσαι εἰσίν, καὶ γωνία ἡ πρὸς τῷ Β γωνίᾳ τῇ ὑπὸ ΚΘΛ ἴση, βάσις ἄρα ἡ ΑΓ βάσει τῇ ΚΛ ἴση. καὶ ἐπεὶ αἱ ὑπὸ ΑΒΓ, ΗΘΚ τῆς ὑπὸ ΔΕΖ μείζονές εἰσιν, ἴση δὲ ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΚΘΛ, ἡ ἄρα ὑπὸ ΗΘΛ τῆς ὑπὸ ΔΕΖ μείζων ἐστίν. καὶ ἐπεὶ δύο αἱ ΗΘ, ΘΛ δύο ταῖς ΔΕ, ΕΖ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ ΗΘΛ γωνίας τῆς ὑπὸ ΔΕΖ μείζων, βάσις ἄρα ἡ ΗΛ βάσεως τῆς ΔΖ μείζων ἐστίν. ἀλλὰ αἱ ΗΚ, ΚΛ τῆς ΗΛ μείζονές εἰσιν. πολλῷ ἄρα αἱ ΗΚ, ΚΛ τῆς ΔΖ μείζονές εἰσιν. ἴση δὲ ἡ ΚΛ τῇ ΑΓ: αἱ ΑΓ, ΗΚ ἄρα τῆς λοιπῆς τῆς ΔΖ μείζονές εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ μὲν ΑΓ, ΔΖ τῆς ΗΚ μείζονές εἰσιν, καὶ ἔτι αἱ ΔΖ, ΗΚ τῆς ΑΓ μείζονές εἰσιν. δυνατὸν ἄρα ἐστὶν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον συστήσασθαι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",321]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",23]],["Association",["Rule","'VertexLabel'","'11.23'"],["Rule","'Text'","'To construct a solid angle out of three plane angles two of which, taken together in any manner, are greater than the remaining one: thus the three angles must be less than four right angles.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'ἐκ τριῶν γωνιῶν ἐπιπέδων, ὧν αἱ δύο τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, στερεὰν γωνίαν συστήσασθαι: δεῖ δὴ τὰς τρεῖς τεσσάρων ὀρθῶν ἐλάσσονας εἶναι.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",25]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]],["List",["Rule","'Book'",11],["Rule","'Theorem'",12]],["List",["Rule","'Book'",11],["Rule","'Theorem'",22]]]],["Rule","'Proof'","'Let the angles ABC, DEF, GHK be the three given plane angles, and let two of these, taken together in any manner, be greater than the remaining one, while, further, the three are less than four right angles; thus it is required to construct a solid angle out of angles equal to the angles ABC, DEF, GHK. Let AB, BC, DE, EF, GH, HK be cut off equal to one another, and let AC, DF, GK be joined; it is therefore possible to construct a triangle out of straight lines equal to AC, DF, GK. [XI. 22] Let LMN be so constructed that AC is equal to LM, DF to MN, and further GK to NL, let the circle LMN be described about the triangle LMN, let its centre be taken, and let it be O; let LO, MO, NO be joined; I say that AB is greater than LO. For, if not, AB is either equal to LO, or less. First, let it be equal. Then, since AB is equal to LO, while AB is equal to BC, and OL to OM, the two sides AB, BC are equal to the two sides LO, OM respectively; and, by hypothesis, the base AC is equal to the base LM; therefore the angle ABC is equal to the angle LOM. [I. 8] For the same reason the angle DEF is also equal to the angle MON, and further the angle GHK to the angle NOL; therefore the three angles ABC, DEF, GHK are equal to the three angles LOM, MON, NOL. But the three angles LOM, MON, NOL are equal to four right angles; therefore the angles ABC, DEF, GHK are equal to four right angles. But they are also, by hypothesis, less than four right angles: which is absurd. Therefore AB is not equal to LO. I say next that neither is AB less than LO. For, if possible, let it be so, and let OP be made equal to AB, and OQ equal to BC, and let PQ be joined. Then, since AB is equal to BC, OP is also equal to OQ, so that the remainder LP is equal to QM. Therefore LM is parallel to PQ, [VI. 2] and LMO is equiangular with PQO; [I. 29] therefore, as OL is to LM, so is OP to PQ; [VI. 4] and alternately, as LO is to OP, so is LM to PQ. [V. 16] But LO is greater than OP; therefore LM is also greater than PQ. But LM was made equal to AC; therefore AC is also greater than PQ. Since, then, the two sides AB, BC are equal to the two sides PO, OQ, and the base AC is greater than the base PQ, therefore the angle ABC is greater than the angle POQ. [I. 25] Similarly we can prove that the angle DEF is also greater than the angle MON, and the angle GHK greater than the angle NOL. Therefore the three angles ABC, DEF, GHK are greater than the three angles LOM, MON, NOL. But, by hypothesis, the angles ABC, DEF, GHK are less than four right angles; therefore the angles LOM, MON, NOL are much less than four right angles. But they are also equal to four right angles: which is absurd. Therefore AB is not less than LO. And it was proved that neither is it equal; therefore AB is greater than LO. Let then OR be set up from the point O at right angles to the plane of the circle LMN, [XI. 12] and let the square on OR be equal to that area by which the square on AB is greater than the square on LO; [Lemma]let RL, RM, RN be joined. Then, since RO is at right angles to the plane of the circle LMN, therefore RO is also at right angles to each of the straight lines LO, MO, NO. And, since LO is equal to OM, while OR is common and at right angles, therefore the base RL is equal to the base RM. [I. 4] For the same reason RN is also equal to each of the straight lines RL, RM; therefore the three straight lines RL, RM, RN are equal to one another. Next, since by hypothesis the square on OR is equal to that area by which the square on AB is greater than the square on LO, therefore the square on AB is equal to the squares on LO, OR. But the square on LR is equal to the squares on LO, OR, for the angle LOR is right; [I. 47] therefore the square on AB is equal to the square on RL; therefore AB is equal to RL. But each of the straight lines BC, DE, EF, GH, HK is equal to AB, while each of the straight lines RM, RN is equal to RL; therefore each of the straight lines AB, BC, DE, EF, GH, HK is equal to each of the straight lines RL, RM, RN. And, since the two sides LR, RM are equal to the two sides AB, BC, and the base LM is by hypothesis equal to the base AC, therefore the angle LRM is equal to the angle ABC. [I. 8] For the same reason the angle MRN is also equal to the angle DEF, and the angle LRN to the angle GHK. Therefore, out of the three plane angles LRM, MRN, LRN, which are equal to the three given angles ABC, DEF, GHK, the solid angle at R has been constructed, which is contained by the angles LRM, MRN, LRN.'"],["Rule","'ProofWordCount'",937],["Rule","'GreekProof'","'ἔστωσαν αἱ δοθεῖσαι τρεῖς γωνίαι ἐπίπεδοι αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ, ὧν αἱ δύο τῆς λοιπῆς μείζονες ἔστωσαν πάντῃ μεταλαμβανόμεναι, ἔτι δὲ αἱ τρεῖς τεσσάρων ὀρθῶν ἐλάσσονες: δεῖ δὴ ἐκ τῶν ἴσων ταῖς ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ στερεὰν γωνίαν συστήσασθαι. Ἀπειλήφθωσαν ἴσαι αἱ ΑΒ, ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ, καὶ ἐπεζεύχθωσαν αἱ ΑΓ, ΔΖ, ΗΚ: δυνατὸν ἄρα ἐστὶν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον συστήσασθαι. συνεστάτω τὸ ΛΜΝ, ὥστε ἴσην εἶναι τὴν μὲν ΑΓ τῇ ΛΜ, τὴν δὲ ΔΖ τῇ ΜΝ, καὶ ἔτι τὴν ΗΚ τῇ ΝΛ, καὶ περιγεγράφθω περὶ τὸ ΛΜΝ τρίγωνον κύκλος ὁ ΛΜΝ καὶ εἰλήφθω αὐτοῦ τὸ κέντρον καὶ ἔστω τὸ Ξ, καὶ ἐπεζεύχθωσαν αἱ ΛΞ, ΜΞ, ΝΞ: λέγω, ὅτι ἡ ΑΒ μείζων ἐστὶ τῆς ΛΞ. εἰ γὰρ μή, ἤτοι ἴση ἐστὶν ἡ ΑΒ τῇ ΛΞ ἢ ἐλάττων. ἔστω πρότερον ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΛΞ, ἀλλὰ ἡ μὲν ΑΒ τῇ ΒΓ ἐστιν ἴση, ἡ δὲ ΞΛ τῇ ΞΜ, δύο δὴ αἱ ΑΒ, ΒΓ δύο ταῖς ΛΞ, ΞΜ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΑΓ βάσει τῇ ΛΜ ὑπόκειται ἴση: γωνία ἄρα ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΛΞΜ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ μὲν ὑπὸ ΔΕΖ τῇ ὑπὸ ΜΞΝ ἐστιν ἴση, καὶ ἔτι ἡ ὑπὸ ΗΘΚ τῇ ὑπὸ ΝΞΛ: αἱ ἄρα τρεῖς αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ γωνίαι τρισὶ ταῖς ὑπὸ ΛΞΜ, ΜΞΝ, ΝΞΛ εἰσιν ἴσαι. ἀλλὰ αἱ τρεῖς αἱ ὑπὸ ΛΞΜ, ΜΞΝ, ΝΞΛ τέτταρσιν ὀρθαῖς εἰσιν ἴσαι: καὶ αἱ τρεῖς ἄρα αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ τέτταρσιν ὀρθαῖς ἴσαι εἰσίν. ὑπόκεινται δὲ καὶ τεσσάρων ὀρθῶν ἐλάσσονες: ὅπερ ἄτοπον. οὐκ ἄρα ἡ ΑΒ τῇ ΛΞ ἴση ἐστίν. λέγω δή, ὅτι οὐδὲ ἐλάττων ἐστὶν ἡ ΑΒ τῆς ΛΞ. εἰ γὰρ δυνατόν, ἔστω: καὶ κείσθω τῇ μὲν ΑΒ ἴση ἡ ΞΟ, τῇ δὲ ΒΓ ἴση ἡ ΞΠ, καὶ ἐπεζεύχθω ἡ ΟΠ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΒΓ, ἴση ἐστὶ καὶ ἡ ΞΟ τῇ ΞΠ: ὥστε καὶ λοιπὴ ἡ ΛΟ τῇ ΠΜ ἐστιν ἴση. παράλληλος ἄρα ἐστὶν ἡ ΛΜ τῇ ΟΠ, καὶ ἰσογώνιον τὸ ΛΜΞ τῷ ΟΠΞ: ἔστιν ἄρα ὡς ἡ ΞΛ πρὸς ΛΜ, οὕτως ἡ ΞΟ πρὸς ΟΠ: ἐναλλὰξ ὡς ἡ ΛΞ πρὸς ΞΟ, οὕτως ἡ ΛΜ πρὸς ΟΠ. μείζων δὲ ἡ ΛΞ τῆς ΞΟ: μείζων ἄρα καὶ ἡ ΛΜ τῆς ΟΠ. ἀλλὰ ἡ ΛΜ κεῖται τῇ ΑΓ ἴση: καὶ ἡ ΑΓ ἄρα τῆς ΟΠ μείζων ἐστίν. ἐπεὶ οὖν δύο αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΟΞ, ΞΠ ἴσαι εἰσίν, καὶ βάσις ἡ ΑΓ βάσεως τῆς ΟΠ μείζων ἐστίν, γωνία ἄρα ἡ ὑπὸ ΑΒΓ γωνίας τῆς ὑπὸ ΟΞΠ μείζων ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ μὲν ὑπὸ ΔΕΖ τῆς ὑπὸ ΜΞΝ μείζων ἐστίν, ἡ δὲ ὑπὸ ΗΘΚ τῆς ὑπὸ ΝΞΛ. αἱ ἄρα τρεῖς γωνίαι αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ τριῶν τῶν ὑπὸ ΛΞΜ, ΜΞΝ, ΝΞΛ μείζονές εἰσιν. ἀλλὰ αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ τεσσάρων ὀρθῶν ἐλάσσονες ὑπόκεινται: πολλῷ ἄρα αἱ ὑπὸ ΛΞΜ, ΜΞΝ, ΝΞΛ τεσσάρων ὀρθῶν ἐλάσσονές εἰσιν. ἀλλὰ καὶ ἴσαι: ὅπερ ἐστὶν ἄτοπον. οὐκ ἄρα ἡ ΑΒ ἐλάσσων ἐστὶ τῆς ΛΞ. ἐδείχθη δέ, ὅτι οὐδὲ ἴση: μείζων ἄρα ἡ ΑΒ τῆς ΛΞ. ἀνεστάτω δὴ ἀπὸ τοῦ Ξ σημείου τῷ τοῦ ΛΜΝ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΞΡ, καὶ ᾧ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΑΒ τετράγωνον τοῦ ἀπὸ τῆς ΛΞ, ἐκείνῳ ἴσον ἔστω τὸ ἀπὸ τῆς ΞΡ, καὶ ἐπεζεύχθωσαν αἱ ΡΛ, ΡΜ, ΡΝ. καὶ ἐπεὶ ἡ ΡΞ ὀρθή ἐστι πρὸς τὸ τοῦ ΛΜΝ κύκλου ἐπίπεδον, καὶ πρὸς ἑκάστην ἄρα τῶν ΛΞ, ΜΞ, ΝΞ ὀρθή ἐστιν ἡ ΡΞ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΛΞ τῇ ΞΜ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΞΡ, βάσις ἄρα ἡ ΡΛ βάσει τῇ ΡΜ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΡΝ ἑκατέρᾳ τῶν ΡΛ, ΡΜ ἐστιν ἴση: αἱ τρεῖς ἄρα αἱ ΡΛ, ΡΜ, ΡΝ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ᾧ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΛΞ, ἐκείνῳ ἴσον ὑπόκειται τὸ ἀπὸ τῆς ΞΡ, τὸ ἄρα ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΛΞ, ΞΡ. τοῖς δὲ ἀπὸ τῶν ΛΞ, ΞΡ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΛΡ: ὀρθὴ γὰρ ἡ ὑπὸ ΛΞΡ: τὸ ἄρα ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΡΛ: ἴση ἄρα ἡ ΑΒ τῇ ΡΛ. ἀλλὰ τῇ μὲν ΑΒ ἴση ἐστὶν ἑκάστη τῶν ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ, τῇ δὲ ΡΛ ἴση ἑκατέρα τῶν ΡΜ, ΡΝ: ἑκάστη ἄρα τῶν ΑΒ, ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ ἑκάστῃ τῶν ΡΛ, ΡΜ, ΡΝ ἴση ἐστίν. καὶ ἐπεὶ δύο αἱ ΛΡ, ΡΜ δυσὶ ταῖς ΑΒ, ΒΓ ἴσαι εἰσίν, καὶ βάσις ἡ ΛΜ βάσει τῇ ΑΓ ὑπόκειται ἴση, γωνία ἄρα ἡ ὑπὸ ΛΡΜ γωνίᾳ τῇ ὑπὸ ΑΒΓ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ μὲν ὑπὸ ΜΡΝ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση, ἡ δὲ ὑπὸ ΛΡΝ τῇ ὑπὸ ΗΘΚ. ἐκ τριῶν ἄρα γωνιῶν ἐπιπέδων τῶν ὑπὸ ΛΡΜ, ΜΡΝ, ΛΡΝ, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις ταῖς ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ, στερεὰ γωνία συνέσταται ἡ πρὸς τῷ Ρ περιεχομένη ὑπὸ τῶν ΛΡΜ, ΜΡΝ, ΛΡΝ γωνιῶν: ὅπερ ἔδει ποιῆσαι. λῆμμα ὃν δὲ τρόπον, ᾧ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΛΞ, ἐκείνῳ ἴσον λαβεῖν ἔστι τὸ ἀπὸ τῆς ΞΡ, δείξομεν οὕτως. ἐκκείσθωσαν αἱ ΑΒ, ΛΞ εὐθεῖαι, καὶ ἔστω μείζων ἡ ΑΒ, καὶ γεγράφθω ἐπ᾽ αὐτῆς ἡμικύκλιον τὸ ΑΒΓ, καὶ εἰς τὸ ΑΒΓ ἡμικύκλιον ἐνηρμόσθω τῇ ΛΞ εὐθείᾳ μὴ μείζονι οὔσῃ τῆς ΑΒ διαμέτρου ἴση ἡ ΑΓ, καὶ ἐπεζεύχθω ἡ ΓΒ. ἐπεὶ οὖν ἐν ἡμικυκλίῳ τῷ ΑΓΒ γωνία ἐστὶν ἡ ὑπὸ ΑΓΒ, ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΑΓΒ. τὸ ἄρα ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΓ, ΓΒ. ὥστε τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΑΓ μεῖζόν ἐστι τῷ ἀπὸ τῆς ΓΒ. ἴση δὲ ἡ ΑΓ τῇ ΛΞ. τὸ ἄρα ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΛΞ μεῖζόν ἐστι τῷ ἀπὸ τῆς ΓΒ. ἐὰν οὖν τῇ ΒΓ ἴσην τὴν ΞΡ ἀπολάβωμεν, ἔσται τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΛΞ μεῖζον τῷ ἀπὸ τῆς ΞΡ: ὅπερ προέκειτο ποιῆσαι.'"],["Rule","'GreekProofWordCount'",967]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",24]],["Association",["Rule","'VertexLabel'","'11.24'"],["Rule","'Text'","'If a solid be contained by parallel planes, the opposite planes in it are equal and parallelogrammic.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'ἐὰν στερεὸν ὑπὸ παραλλήλων ἐπιπέδων περιέχηται, τὰ ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά ἐστιν.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",11],["Rule","'Theorem'",10]],["List",["Rule","'Book'",11],["Rule","'Theorem'",16]]]],["Rule","'Proof'","'For let the solid CDHG be contained by the parallel planes AC, GF, AH, DF, BF, AE; I say that the opposite planes in it are equal and parallelogrammic. For, since the two parallel planes BG, CE are cut by the plane AC, their common sections are parallel. [XI. 16] Therefore AB is parallel to DC. Again, since the two parallel planes BF, AE are cut by the plane AC, their common sections are parallel. [XI. 16] Therefore BC is parallel to AD. But AB was also proved parallel to DC; therefore AC is a parallelogram. Similarly we can prove that each of the planes DF, FG, GB, BF, AE is a parallelogram. Let AH, DF be joined. Then, since AB is parallel to DC, and BH to CF, the two straight lines AB, BH which meet one another are parallel to the two straight lines DC, CF which meet one another, not in the same plane; therefore they will contain equal angles; [XI. 10] therefore the angle ABH is equal to the angle DCF. And, since the two sides AB, BH are equal to the two sides DC, CF, [I. 34] and the angle ABH is equal to the angle DCF, therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DCF. [I. 4] And the parallelogram BG is double of the triangle ABH, and the parallelogram CE double of the triangle DCF; [I. 34] therefore the parallelogram BG is equal to the parallelogram CE. Similarly we can prove that AC is also equal to GF, and AE to BF.'"],["Rule","'ProofWordCount'",270],["Rule","'GreekProof'","'στερεὸν γὰρ τὸ ΓΔΘΗ ὑπὸ παραλλήλων ἐπιπέδων περιεχέσθω τῶν ΑΓ, ΗΖ, ΑΘ, ΔΖ, ΒΖ, ΑΕ: λέγω, ὅτι τὰ ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά ἐστιν. ἐπεὶ γὰρ δύο ἐπίπεδα παράλληλα τὰ ΒΗ, ΓΕ ὑπὸ ἐπιπέδου τοῦ ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν. παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΔΓ. πάλιν, ἐπεὶ δύο ἐπίπεδα παράλληλα τὰ ΒΖ, ΑΕ ὑπὸ ἐπιπέδου τοῦ ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν. παράλληλος ἄρα ἐστὶν ἡ ΒΓ τῇ ΑΔ. ἐδείχθη δὲ καὶ ἡ ΑΒ τῇ ΔΓ παράλληλος: παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΓ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἕκαστον τῶν ΔΖ, ΖΗ, ΗΒ, ΒΖ, ΑΕ παραλληλόγραμμόν ἐστιν. ἐπεζεύχθωσαν αἱ ΑΘ, ΔΖ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ μὲν ΑΒ τῇ ΔΓ, ἡ δὲ ΒΘ τῇ ΓΖ, δύο δὴ αἱ ΑΒ, ΒΘ ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας τὰς ΔΓ, ΓΖ ἁπτομένας ἀλλήλων εἰσὶν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ: ἴσας ἄρα γωνίας περιέξουσιν: ἴση ἄρα ἡ ὑπὸ ΑΒΘ γωνία τῇ ὑπὸ ΔΓΖ. καὶ ἐπεὶ δύο αἱ ΑΒ, ΒΘ δυσὶ ταῖς ΔΓ, ΓΖ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ ΑΒΘ γωνίᾳ τῇ ὑπὸ ΔΓΖ ἐστιν ἴση, βάσις ἄρα ἡ ΑΘ βάσει τῇ ΔΖ ἐστιν ἴση, καὶ τὸ ΑΒΘ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἴσον ἐστίν. καί ἐστι τοῦ μὲν ΑΒΘ διπλάσιον τὸ ΒΗ παραλληλόγραμμον, τοῦ δὲ ΔΓΖ διπλάσιον τὸ ΓΕ παραλληλόγραμμον: ἴσον ἄρα τὸ ΒΗ παραλληλόγραμμον τῷ ΓΕ παραλληλογράμμῳ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τὸ μὲν ΑΓ τῷ ΗΖ ἐστιν ἴσον, τὸ δὲ ΑΕ τῷ ΒΖ. ἐὰν ἄρα στερεὸν ὑπὸ παραλλήλων ἐπιπέδων περιέχηται, τὰ ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά ἐστιν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",258]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",25]],["Association",["Rule","'VertexLabel'","'11.25'"],["Rule","'Text'","'If a parallelepipedal solid be cut by a plane which is parallel to the opposite planes, then, as the base is to the base, so will the solid be to the solid.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν στερεὸν παραλληλεπίπεδον ἐπιπέδῳ τμηθῇ παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔσται ὡς ἡ βάσις πρὸς τὴν βάσιν, οὕτως τὸ στερεὸν πρὸς τὸ στερεόν.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",5]],["List",["Rule","'Book'",11],["Rule","'Theorem'",24]]]],["Rule","'Proof'","'For let the parallelepipedal solid ABCD be cut by the plane FG which is parallel to the opposite planes RA, DH; I say that, as the base AEFV is to the base EHCF, so is the solid ABFU to the solid EGCD. For let AH be produced in each direction, let any number of straight lines whatever, AK, KL, be made equal to AE, and any number whatever, HM, MN, equal to EH; and let the parallelograms LP, KV, HW, MS and the solids LQ, KR, DM, MT be completed. Then, since the straight lines LK, KA, AE are equal to one another, the parallelograms LP, KV, AF are also equal to one another, KO, KB, AG are equal to one another, and further LX, KQ, AR are equal to one another, for they are opposite. [XI. 24] For the same reason the parallelograms EC, HW, MS are also equal to one another, HG, HI, IN are equal to one another, and further DH, MY, NT are equal to one another. Therefore in the solids LQ, KR, AU three planes are equal to three planes. But the three planes are equal to the three opposite; therefore the three solids LQ, KR, AU are equal to one another. For the same reason the three solids ED, DM, MT are also equal to one another. Therefore, whatever multiple the base LF is of the base AF, the same multiple also is the solid LU of the solid AU. For the same reason, whatever multiple the base NF is of the base FH, the same multiple also is the solid NU of the solid HU. And, if the base LF is equal to the base NF, the solid LU is also equal to the solid NU; if the base LF exceeds the base NF, the solid LU also exceeds the solid NU; and, if one falls short, the other falls short. Therefore, there being four magnitudes, the two bases AF, FH, and the two solids AU, UH, equimultiples have been taken of the base AF and the solid AU, namely the base LF and the solid LU, and equimultiples of the base HF and the solid HU, namely the base NF and the solid NU, and it has been proved that, if the base LF exceeds the base FN, the solid LU also exceeds the solid NU, if the bases are equal, the solids are equal, and if the base falls short, the solid falls short. Therefore, as the base AF is to the base FH, so is the solid AU to the solid UH. [V. Def. 5]'"],["Rule","'ProofWordCount'",435],["Rule","'GreekProof'","'στερεὸν γὰρ παραλληλεπίπεδον τὸ ΑΒΓΔ ἐπιπέδῳ τῷ ΖΗ τετμήσθω παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις τοῖς ΡΑ, ΔΘ: λέγω, ὅτι ἐστὶν ὡς ἡ ΑΕΖΦ βάσις πρὸς τὴν ΕΘΓΖ βάσιν, οὕτως τὸ ΑΒΖΥ στερεὸν πρὸς τὸ ΕΗΓΔ στερεόν. Ἐκβεβλήσθω γὰρ ἡ ΑΘ ἐφ᾽ ἑκάτερα τὰ μέρη, καὶ κείσθωσαν τῇ μὲν ΑΕ ἴσαι ὁσαιδηποτοῦν αἱ ΑΚ, ΚΛ, τῇ δὲ ΕΘ ἴσαι ὁσαιδηποτοῦν αἱ ΘΜ, ΜΝ, καὶ συμπεπληρώσθω τὰ ΛΟ, ΚΦ, ΘΧ, ΜΣ παραλληλόγραμμα καὶ τὰ ΛΠ, ΚΡ, ΔΜ, ΜΤ στερεά. καὶ ἐπεὶ ἴσαι εἰσὶν αἱ ΛΚ, ΚΑ, ΑΕ εὐθεῖαι ἀλλήλαις, ἴσα ἐστὶ καὶ τὰ μὲν ΛΟ, ΚΦ, ΑΖ παραλληλόγραμμα ἀλλήλοις, τὰ δὲ ΚΞ, ΚΒ, ΑΗ ἀλλήλοις καὶ ἔτι τὰ ΛΨ, ΚΠ, ΑΡ ἀλλήλοις: ἀπεναντίον γάρ. διὰ τὰ αὐτὰ δὴ καὶ τὰ μὲν ΕΓ, ΘΧ, ΜΣ παραλληλόγραμμα ἴσα εἰσὶν ἀλλήλοις, τὰ δὲ ΘΗ, ΘΙ, ΙΝ ἴσα εἰσὶν ἀλλήλοις, καὶ ἔτι τὰ ΔΘ, ΜΩ, ΝΤ: τρία ἄρα ἐπίπεδα τῶν ΛΠ, ΚΡ, ΑΥ στερεῶν τρισὶν ἐπιπέδοις ἐστὶν ἴσα. ἀλλὰ τὰ τρία τρισὶ τοῖς ἀπεναντίον ἐστὶν ἴσα: τὰ ἄρα τρία στερεὰ τὰ ΛΠ, ΚΡ, ΑΥ ἴσα ἀλλήλοις ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ τὰ τρία στερεὰ τὰ ΕΔ, ΔΜ, ΜΤ ἴσα ἀλλήλοις ἐστίν: ὁσαπλασίων ἄρα ἐστὶν ἡ ΛΖ βάσις τῆς ΑΖ βάσεως, τοσαυταπλάσιόν ἐστι καὶ τὸ ΛΥ στερεὸν τοῦ ΑΥ στερεοῦ. διὰ τὰ αὐτὰ δὴ ὁσαπλασίων ἐστὶν ἡ ΝΖ βάσις τῆς ΖΘ βάσεως, τοσαυταπλάσιόν ἐστι καὶ τὸ ΝΥ στερεὸν τοῦ ΘΥ στερεοῦ. καὶ εἰ ἴση ἐστὶν ἡ ΛΖ βάσις τῇ ΝΖ βάσει, ἴσον ἐστὶ καὶ τὸ ΛΥ στερεὸν τῷ ΝΥ στερεῷ, καὶ εἰ ὑπερέχει ἡ ΛΖ βάσις τῆς ΝΖ βάσεως, ὑπερέχει καὶ τὸ ΛΥ στερεὸν τοῦ ΝΥ στερεοῦ, καὶ εἰ ἐλλείπει, ἐλλείπει. τεσσάρων δὴ ὄντων μεγεθῶν, δύο μὲν βάσεων τῶν ΑΖ, ΖΘ, δύο δὲ στερεῶν τῶν ΑΥ, ΥΘ, εἴληπται ἰσάκις πολλαπλάσια τῆς μὲν ΑΖ βάσεως καὶ τοῦ ΑΥ στερεοῦ ἥ τε ΛΖ βάσις καὶ τὸ ΛΥ στερεόν, τῆς δὲ ΘΖ βάσεως καὶ τοῦ ΘΥ στερεοῦ ἥ τε ΝΖ βάσις καὶ τὸ ΝΥ στερεόν, καὶ δέδεικται, ὅτι εἰ ὑπερέχει ἡ ΛΖ βάσις τῆς ΖΝ βάσεως, ὑπερέχει καὶ τὸ ΛΥ στερεὸν τοῦ ΝΥ στερεοῦ, καὶ εἰ ἴση, ἴσον, καὶ εἰ ἐλλείπει, ἐλλείπει. ἔστιν ἄρα ὡς ἡ ΑΖ βάσις πρὸς τὴν ΖΘ βάσιν, οὕτως τὸ ΑΥ στερεὸν πρὸς τὸ ΥΘ στερεόν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",365]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",26]],["Association",["Rule","'VertexLabel'","'11.26'"],["Rule","'Text'","'On a given straight line, and at a given point on it, to construct a solid angle equal to a given solid angle.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'πρὸς τῇ δοθείσῃ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῇ δοθείσῃ στερεᾷ γωνίᾳ ἴσην στερεὰν γωνίαν συστήσασθαι.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",11]],["List",["Rule","'Book'",11],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'Let AB be the given straight line, A the given point on it, and the angle at D, contained by the angles EDC, EDF, FDC, the given solid angle; thus it is required to construct on the straight line AB, and at the point A on it, a solid angle equal to the solid angle at D. For let a point F be taken at random on DF, let FG be drawn from F perpendicular to the plane through ED, DC, and let it meet the plane at G, [XI. 11] let DG be joined, let there be constructed on the straight line AB and at the point A on it the angle BAL equal to the angle EDC, and the angle BAK equal to the angle EDG, [I. 23] let AK be made equal to DG, let KH be set up from the point K at right angles to the plane through BA, AL, [XI. 12] let KH be made equal to GF, and let HA be joined; I say that the solid angle at A, contained by the angles BAL, BAH, HAL is equal to the solid angle at D contained by the angles EDC, EDF, FDC. For let AB, DE be cut off equal to one another, and let HB, KB, FE, GE be joined. Then, since FG is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] therefore each of the angles FGD, FGE is right. For the same reason each of the angles HKA, HKB is also right. And, since the two sides KA, AB are equal to the two sides GD, DE respectively, and they contain equal angles, therefore the base KB is equal to the base GE. [I. 4] But KH is also equal to GF, and they contain right angles; therefore HB is also equal to FE. [I. 4] Again, since the two sides AK, KH are equal to the two sides DG, GF, and they contain right angles, therefore the base AH is equal to the base FD. [I. 4] But AB is also equal to DE; therefore the two sides HA, AB are equal to the two sides DF, DE. And the base HB is equal to the base FE; therefore the angle BAH is equal to the angle EDF. [I. 8] For the same reason the angle HAL is also equal to the angle FDC. And the angle BAL is also equal to the angle EDC. Therefore on the straight line AB, and at the point A on it, a solid angle has been constructed equal to the given solid angle at D.'"],["Rule","'ProofWordCount'",457],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ πρὸς αὐτῇ δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα στερεὰ γωνία ἡ πρὸς τῷ δ περιεχομένη ὑπὸ τῶν ὑπὸ ΕΔΓ, ΕΔΖ, ΖΔΓ γωνιῶν ἐπιπέδων: δεῖ δὴ πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ πρὸς τῷ Δ στερεᾷ γωνίᾳ ἴσην στερεὰν γωνίαν συστήσασθαι. εἰλήφθω γὰρ ἐπὶ τῆς ΔΖ τυχὸν σημεῖον τὸ Ζ, καὶ ἤχθω ἀπὸ τοῦ Ζ ἐπὶ τὸ διὰ τῶν ΕΔ, ΔΓ ἐπίπεδον κάθετος ἡ ΖΗ, καὶ συμβαλλέτω τῷ ἐπιπέδῳ κατὰ τὸ Η, καὶ ἐπεζεύχθω ἡ ΔΗ, καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ μὲν ὑπὸ ΕΔΓ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΛ, τῇ δὲ ὑπὸ ΕΔΗ ἴση ἡ ὑπὸ ΒΑΚ, καὶ κείσθω τῇ ΔΗ ἴση ἡ ΑΚ, καὶ ἀνεστάτω ἀπὸ τοῦ Κ σημείου τῷ διὰ τῶν ΒΑΛ ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΚΘ, καὶ κείσθω ἴση τῇ ΗΖ ἡ ΚΘ, καὶ ἐπεζεύχθω ἡ ΘΑ: λέγω, ὅτι ἡ πρὸς τῷ Α στερεὰ γωνία περιεχομένη ὑπὸ τῶν ΒΑΛ, ΒΑΘ, ΘΑΛ γωνιῶν ἴση ἐστὶ τῇ πρὸς τῷ Δ στερεᾷ γωνίᾳ τῇ περιεχομένῃ ὑπὸ τῶν ΕΔΓ, ΕΔΖ, ΖΔΓ γωνιῶν. Ἀπειλήφθωσαν γὰρ ἴσαι αἱ ΑΒ, ΔΕ, καὶ ἐπεζεύχθωσαν αἱ ΘΒ, ΚΒ, ΖΕ, ΗΕ. καὶ ἐπεὶ ἡ ΖΗ ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΖΗΔ, ΖΗΕ γωνιῶν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΘΚΑ, ΘΚΒ γωνιῶν ὀρθή ἐστιν. καὶ ἐπεὶ δύο αἱ ΚΑ, ΑΒ δύο ταῖς ΗΔ, ΔΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΚΒ βάσει τῇ ΗΕ ἴση ἐστίν. ἔστι δὲ καὶ ἡ ΚΘ τῇ ΗΖ ἴση: καὶ γωνίας ὀρθὰς περιέχουσιν: ἴση ἄρα καὶ ἡ ΘΒ τῇ ΖΕ. πάλιν ἐπεὶ δύο αἱ ΑΚ, ΚΘ δυσὶ ταῖς ΔΗ, ΗΖ ἴσαι εἰσίν, καὶ γωνίας ὀρθὰς περιέχουσιν, βάσις ἄρα ἡ ΑΘ βάσει τῇ ΖΔ ἴση ἐστίν. ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΕ ἴση: δύο δὴ αἱ ΘΑ, ΑΒ δύο ταῖς ΔΖ, ΔΕ ἴσαι εἰσίν. καὶ βάσις ἡ ΘΒ βάσει τῇ ΖΕ ἴση: γωνία ἄρα ἡ ὑπὸ ΒΑΘ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΘΑΛ τῇ ὑπὸ ΖΔΓ ἐστιν ἴση ἐπειδήπερ ἐὰν ἀπολάβωμεν ἴσας τὰς ΑΛ, ΔΓ καὶ ἐπιζεύξωμεν τὰς ΚΛ, ΘΛ, ΗΓ, ΖΓ, ἐπεὶ ὅλη ἡ ὑπὸ ΒΑΛ ὅλῃ τῇ ὑπὸ ΕΔΓ ἐστιν ἴση, ὧν ἡ ὑπὸ ΒΑΚ τῇ ὑπὸ ΕΔΗ ὑπόκειται ἴση, λοιπὴ ἄρα ἡ ὑπὸ ΚΑΛ λοιπῇ τῇ ὑπὸ ΗΔΓ ἐστιν ἴση. καὶ ἐπεὶ δύο αἱ ΚΑ, ΑΛ δυσὶ ταῖς ΗΔ, ΔΓ ἴσαι εἰσίν, καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΚΛ βάσει τῇ ΗΓ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΚΘ τῇ ΗΖ ἴση: δύο δὴ αἱ ΛΚ, ΚΘ δυσὶ ταῖς ΓΗ, ΗΖ εἰσιν ἴσαι: καὶ γωνίας ὀρθὰς περιέχουσιν: βάσις ἄρα ἡ ΘΛ βάσει τῇ ΖΓ ἐστιν ἴση. καὶ ἐπεὶ δύο αἱ ΘΑ, ΑΛ δυσὶ ταῖς ΖΔ, ΔΓ εἰσιν ἴσαι, καὶ βάσις ἡ ΘΛ βάσει τῇ ΖΓ ἐστιν ἴση, γωνία ἄρα ἡ ὑπὸ ΘΑΛ γωνίᾳ τῇ ὑπὸ ΖΔΓ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ὑπὸ ΒΑΛ τῇ ὑπὸ ΕΔΓ ἴση. πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ δοθείσῃ στερεᾷ γωνίᾳ τῇ πρὸς τῷ Δ ἴση συνέσταται: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",540]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",27]],["Association",["Rule","'VertexLabel'","'11.27'"],["Rule","'Text'","'On a given straight line to describe a parallelepipedal solid similar and similarly situated to a given parallelepipedal solid.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'ἀπὸ τῆς δοθείσης εὐθείας τῷ δοθέντι στερεῷ παραλληλεπιπέδῳ ὅμοιόν τε καὶ ὁμοίως κείμενον στερεὸν παραλληλεπίπεδον ἀναγράψαι.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",6],["Rule","'Theorem'",12]],["List",["Rule","'Book'",11],["Rule","'Definition'",9]]]],["Rule","'Proof'","'Let AB be the given straight line and CD the given parallelepipedal solid; thus it is required to describe on the given straight line AB a parallelepipedal solid similar and similarly situated to the given parallelepipedal solid CD. For on the straight line AB and at the point A on it let the solid angle, contained by the angles BAH, HAK, KAB, be constructed equal to the solid angle at C, so that the angle BAH is equal to the angle ECF, the angle BAK equal to the angle ECG, and the angle KAH to the angle GCF; and let it be contrived that, as EC is to CG, so is BA to AK, and, as GC is to CF, so is KA to AH. [VI. 12] Therefore also, ex aequali, as EC is to CF, so is BA to AH. [V. 22] Let the parallelogram HB and the solid AL be completed. Now since, as EC is to CG, so is BA to AK, and the sides about the equal angles ECG, BAK are thus proportional, therefore the parallelogram GE is similar to the parallelogram KB. For the same reason the parallelogram KH is also similar to the parallelogram GF, and further FE to HB; therefore three parallelograms of the solid CD are similar to three parallelograms of the solid AL. But the former three are both equal and similar to the three opposite parallelograms, and the latter three are both equal and similar to the three opposite parallelograms; therefore the whole solid CD is similar to the whole solid AL. [XI. Def. 9] Therefore on the given straight line AB there has been described AL similar and similarly situated to the given parallelepipedal solid CD.'"],["Rule","'ProofWordCount'",287],["Rule","'GreekProof'","'ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν στερεὸν παραλληλεπίπεδον τὸ ΓΔ: δεῖ δὴ ἀπὸ τῆς δοθείσης εὐθείας τῆς ΑΒ τῷ δοθέντι στερεῷ παραλληλεπιπέδῳ τῷ ΓΔ ὅμοιόν τε καὶ ὁμοίως κείμενον στερεὸν παραλληλεπίπεδον ἀναγράψαι. συνεστάτω γὰρ πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ πρὸς τῷ Γ στερεᾷ γωνίᾳ ἴση ἡ περιεχομένη ὑπὸ τῶν ΒΑΘ, ΘΑΚ, ΚΑΒ, ὥστε ἴσην εἶναι τὴν μὲν ὑπὸ ΒΑΘ γωνίαν τῇ ὑπὸ ΕΓΖ, τὴν δὲ ὑπὸ ΒΑΚ τῇ ὑπὸ ΕΓΗ, τὴν δὲ ὑπὸ ΚΑΘ τῇ ὑπὸ ΗΓΖ: καὶ γεγονέτω ὡς μὲν ἡ ΕΓ πρὸς τὴν ΓΗ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΚ, ὡς δὲ ἡ ΗΓ πρὸς τὴν ΓΖ, οὕτως ἡ ΚΑ πρὸς τὴν ΑΘ. καὶ δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ ΕΓ πρὸς τὴν ΓΖ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΘ. καὶ συμπεπληρώσθω τὸ ΘΒ παραλληλόγραμμον καὶ τὸ ΑΛ στερεόν. καὶ ἐπεί ἐστιν ὡς ἡ ΕΓ πρὸς τὴν ΓΗ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΚ, καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΕΓΗ, ΒΑΚ αἱ πλευραὶ ἀνάλογόν εἰσιν, ὅμοιον ἄρα ἐστὶ τὸ ΗΕ παραλληλόγραμμον τῷ ΚΒ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν ΚΘ παραλληλόγραμμον τῷ ΗΖ παραλληλογράμμῳ ὅμοιόν ἐστι καὶ ἔτι τὸ ΖΕ τῷ ΘΒ: τρία ἄρα παραλληλόγραμμα τοῦ ΓΔ στερεοῦ τρισὶ παραλληλογράμμοις τοῦ ΑΛ στερεοῦ ὅμοιά ἐστιν. ἀλλὰ τὰ μὲν τρία τρισὶ τοῖς ἀπεναντίον ἴσα τέ ἐστι καὶ ὅμοια, τὰ δὲ τρία τρισὶ τοῖς ἀπεναντίον ἴσα τέ ἐστι καὶ ὅμοια: ὅλον ἄρα τὸ ΓΔ στερεὸν ὅλῳ τῷ ΑΛ στερεῷ ὅμοιόν ἐστιν. ἀπὸ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τῷ δοθέντι στερεῷ παραλληλεπιπέδῳ τῷ ΓΔ ὅμοιόν τε καὶ ὁμοίως κείμενον ἀναγέγραπται τὸ ΑΛ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",270]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",28]],["Association",["Rule","'VertexLabel'","'11.28'"],["Rule","'Text'","'If a parallelepipedal solid be cut by a plane through the diagonals of the opposite planes, the solid will be bisected by the plane.'"],["Rule","'TextWordCount'",24],["Rule","'GreekText'","'ἐὰν στερεὸν παραλληλεπίπεδον ἐπιπέδῳ τμηθῇ κατὰ τὰς διαγωνίους τῶν ἀπεναντίον ἐπιπέδων, δίχα τμηθήσεται τὸ στερεὸν ὑπὸ τοῦ ἐπιπέδου.'"],["Rule","'GreekTextWordCount'",18],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",11],["Rule","'Definition'",10]]]],["Rule","'Proof'","'For let the parallelepipedal solid AB be cut by the plane CDEF through the diagonals CF, DE of opposite planes; I say that the solid AB will be bisected by the plane CDEF. For, since the triangle CGF is equal to the triangle CFB, [I. 34] and ADE to DEH, while the parallelogram CA is also equal to the parallelogram EB, for they are opposite, and GE to CH, therefore the prism contained by the two triangles CGF, ADE and the three parallelograms GE, AC, CE is also equal to the prism contained by the two triangles CFB, DEH and the three parallelograms CH, BE, CE; for they are contained by planes equal both in multitude and in magnitude. [XI. Def. 10] Hence the whole solid AB is bisected by the plane CDEF.'"],["Rule","'ProofWordCount'",133],["Rule","'GreekProof'","'στερεὸν γὰρ παραλληλεπίπεδον τὸ ΑΒ ἐπιπέδῳ τῷ ΓΔΕΖ τετμήσθω κατὰ τὰς διαγωνίους τῶν ἀπεναντίον ἐπιπέδων τὰς ΓΖ, ΔΕ: λέγω, ὅτι δίχα τμηθήσεται τὸ ΑΒ στερεὸν ὑπὸ τοῦ ΓΔΕΖ ἐπιπέδου. ἐπεὶ γὰρ ἴσον ἐστὶ τὸ μὲν ΓΗΖ τρίγωνον τῷ ΓΖΒ τριγώνῳ, τὸ δὲ ΑΔΕ τῷ ΔΕΘ, ἔστι δὲ καὶ τὸ μὲν ΓΑ παραλληλόγραμμον τῷ ΕΒ ἴσον: ἀπεναντίον γάρ: τὸ δὲ ΗΕ τῷ ΓΘ, καὶ τὸ πρίσμα ἄρα τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων τῶν ΓΗΖ, ΑΔΕ, τριῶν δὲ παραλληλογράμμων τῶν ΗΕ, ΑΓ, ΓΕ ἴσον ἐστὶ τῷ πρίσματι τῷ περιεχομένῳ ὑπὸ δύο μὲν τριγώνων τῶν ΓΖΒ, ΔΕΘ, τριῶν δὲ παραλληλογράμμων τῶν ΓΘ, ΒΕ, ΓΕ: ὑπὸ γὰρ ἴσων ἐπιπέδων περιέχονται τῷ τε πλήθει καὶ τῷ μεγέθει. ὥστε ὅλον τὸ ΑΒ στερεὸν δίχα τέτμηται ὑπὸ τοῦ ΓΔΕΖ ἐπιπέδου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",127]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",29]],["Association",["Rule","'VertexLabel'","'11.29'"],["Rule","'Text'","'Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are on the same straight lines, are equal to one another.'"],["Rule","'TextWordCount'",35],["Rule","'GreekText'","'τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",24],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",1],["Rule","'Theorem'",36]]]],["Rule","'Proof'","'Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AG, AF, LM, LN, CD, CE, BH, BK, be on the same straight lines FN, DK; I say that the solid CM is equal to the solid CN. For, since each of the figures CH, CK is a parallelogram, CB is equal to each of the straight lines DH, EK, [I. 34] hence DH is also equal to EK. Let EH be subtracted from each; therefore the remainder DE is equal to the remainder HK. Hence the triangle DCE is also equal to the triangle HBK, [I. 8, 4] and the parallelogram DG to the parallelogram HN. [I. 36] For the same reason the triangle AFG is also equal to the triangle MLN. But the parallelogram CF is equal to the parallelogram BM, and CG to BN, for they are opposite; therefore the prism contained by the two triangles AFG, DCE and the three parallelograms AD, DG, CG is equal to the prism contained by the two triangles MLN, HBK and the three parallelograms BM, HN, BN. Let there be added to each the solid of which the parallelogram AB is the base and GEHM its opposite; therefore the whole parallelepipedal solid CM is equal to the whole parallelepipedal solid CN.'"],["Rule","'ProofWordCount'",228],["Rule","'GreekProof'","'ἔστω ἐπὶ τῆς αὐτῆς βάσεως τῆς ΑΒ στερεὰ παραλληλεπίπεδα τὰ ΓΜ, ΓΝ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΗ, ΑΖ, ΛΜ, ΛΝ, ΓΔ, ΓΕ, ΒΘ, ΒΚ ἐπὶ τῶν αὐτῶν εὐθειῶν ἔστωσαν τῶν ΖΝ, ΔΚ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΓΜ στερεὸν τῷ ΓΝ στερεῷ. ἐπεὶ γὰρ παραλληλόγραμμόν ἐστιν ἑκάτερον τῶν ΓΘ, ΓΚ, ἴση ἐστὶν ἡ ΓΒ ἑκατέρᾳ τῶν ΔΘ, ΕΚ: ὥστε καὶ ἡ ΔΘ τῇ ΕΚ ἐστιν ἴση. κοινὴ ἀφῃρήσθω ἡ ΕΘ: λοιπὴ ἄρα ἡ ΔΕ λοιπῇ τῇ ΘΚ ἐστιν ἴση. ὥστε καὶ τὸ μὲν ΔΓΕ τρίγωνον τῷ ΘΒΚ τριγώνῳ ἴσον ἐστίν, τὸ δὲ ΔΗ παραλληλόγραμμον τῷ ΘΝ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΖΗ τρίγωνον τῷ ΜΛΝ τριγώνῳ ἴσον ἐστίν. ἔστι δὲ καὶ τὸ μὲν ΓΖ παραλληλόγραμμον τῷ ΒΜ παραλληλογράμμῳ ἴσον, τὸ δὲ ΓΗ τῷ ΒΝ: ἀπεναντίον γάρ: καὶ τὸ πρίσμα ἄρα τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων τῶν ΑΖΗ, ΔΓΕ, τριῶν δὲ παραλληλογράμμων τῶν ΑΔ, ΔΗ, ΓΗ ἴσον ἐστὶ τῷ πρίσματι τῷ περιεχομένῳ ὑπὸ δύο μὲν τριγώνων τῶν ΜΛΝ, ΘΒΚ, τριῶν δὲ παραλληλογράμμων τῶν ΒΜ, ΘΝ, ΒΝ. κοινὸν προσκείσθω τὸ στερεόν, οὗ βάσις μὲν τὸ ΑΒ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΗΕΘΜ: ὅλον ἄρα τὸ ΓΜ στερεὸν παραλληλεπίπεδον ὅλῳ τῷ ΓΝ στερεῷ παραλληλεπιπέδῳ ἴσον ἐστίν. τὰ ἄρα ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",227]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",30]],["Association",["Rule","'VertexLabel'","'11.30'"],["Rule","'Text'","'Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are not on the same straight lines, are equal to one another.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Theorem'",29]]]],["Rule","'Proof'","'Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AF, AG, LM, LN, CD, CE, BH, BK, not be on the same straight lines; I say that the solid CM is equal to the solid CN. For let NK, DH be produced and meet one another at R, and further let FM, GE be produced to P, Q; let AO, LP, CQ, BR be joined. Then the solid CM, of which the parallelogram ACBL is the base, and FDHM its opposite, is equal to the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite; for they are on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AF, AO, LM, LP, CD, CQ, BH, BR, are on the same straight lines FP, DR. [XI. 29] But the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite, is equal to the solid CN, of which the parallelogram ACBL is the base and GEKN its opposite; for they are again on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AG, AO, CE, CQ, LN, LP, BK, BR, are on the same straight lines GQ, NR. Hence the solid CM is also equal to the solid CN.'"],["Rule","'ProofWordCount'",243],["Rule","'GreekProof'","'ἔστω ἐπὶ τῆς αὐτῆς βάσεως τῆς ΑΒ στερεὰ παραλληλεπίπεδα τὰ ΓΜ, ΓΝ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΖ, ΑΗ, ΛΜ, ΛΝ, ΓΔ, ΓΕ, ΒΘ, ΒΚ μὴ ἔστωσαν ἐπὶ τῶν αὐτῶν εὐθειῶν: λέγω, ὅτι ἴσον ἐστὶ τὸ ΓΜ στερεὸν τῷ ΓΝ στερεῷ. ἐκβεβλήσθωσαν γὰρ αἱ ΝΚ, ΔΘ καὶ συμπιπτέτωσαν ἀλλήλαις κατὰ τὸ Ρ, καὶ ἔτι ἐκβεβλήσθωσαν αἱ ΖΜ, ΗΕ ἐπὶ τὰ Ο, Π, καὶ ἐπεζεύχθωσαν αἱ ΑΞ, ΛΟ, ΓΠ, ΒΡ. ἴσον δή ἐστι τὸ ΓΜ στερεόν, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΖΔΘΜ, τῷ ΓΟ στερεῷ, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΞΠΡΟ: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΑΓΒΛ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΖ, ΑΞ, ΛΜ, ΛΟ, ΓΔ, ΓΠ, ΒΘ, ΒΡ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΖΟ, ΔΡ. ἀλλὰ τὸ ΓΟ στερεόν, οὗ βάσις μέν ἐστι τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΞΠΡΟ, ἴσον ἐστὶ τῷ ΓΝ στερεῷ, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΗΕΚΝ: ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς ΑΓΒΛ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΗ, ΑΞ, ΓΕ, ΓΠ, ΛΝ, ΛΟ, ΒΚ, ΒΡ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΗΠ, ΝΡ. ὥστε καὶ τὸ ΓΜ στερεὸν ἴσον ἐστὶ τῷ ΓΝ στερεῷ. τὰ ἄρα ἐπὶ τῆς αὐτῆς βάσεως στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",238]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",31]],["Association",["Rule","'VertexLabel'","'11.31'"],["Rule","'Text'","'Parallelepipedal solids which are on equal bases and of the same height are equal to one another.'"],["Rule","'TextWordCount'",17],["Rule","'GreekText'","'τὰ ἐπὶ ἴσων βάσεων ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",23]],["List",["Rule","'Book'",1],["Rule","'Theorem'",35]],["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",11],["Rule","'Definition'",10]],["List",["Rule","'Book'",11],["Rule","'Theorem'",24]],["List",["Rule","'Book'",11],["Rule","'Theorem'",25]],["List",["Rule","'Book'",11],["Rule","'Theorem'",29]],["List",["Rule","'Book'",11],["Rule","'Theorem'",30]]]],["Rule","'Proof'","'Let the parallelepipedal solids AE, CF, of the same height, be on equal bases AB, CD. I say that the solid AE is equal to the solid CF. First, let the sides which stand up, HK, BE, AG, LM, PQ, DF, CO, RS, be at right angles to the bases AB, CD; let the straight line RT be produced in a straight line with CR; on the straight line RT, and at the point R on it, let the angle TRU be constructed equal to the angle ALB, [I. 23] let RT be made equal to AL, and RU equal to LB, and let the base RW and the solid XU be completed. Now, since the two sides TR, RU are equal to the two sides AL, LB, and they contain equal angles, therefore the parallelogram RW is equal and similar to the parallelogram HL. Since again AL is equal to RT, and LM to RS, and they contain right angles, therefore the parallelogram RX is equal and similar to the parallelogram AM. For the same reason LE is also equal and similar to SU; therefore three parallelograms of the solid AE are equal and similar to three parallelograms of the solid XU. But the former three are equal and similar to the three opposite, and the latter three to the three opposite; [XI. 24] therefore the whole parallelepipedal solid AE is equal to the whole parallelepipedal solid XU. [XI. Def. 10] Let DR, WU be drawn through and meet one another at Y, let aTb be drawn through T parallel to DY, let PD be produced to a, and let the solids YX, RI be completed. Then the solid XY, of which the parallelogram RX is the base and Yc its opposite, is equal to the solid XU of which the parallelogram RX is the base and UV its opposite, for they are on the same base RX and of the same height, and the extremities of their sides which stand up, namely RY, RU, Tb, TW, Se, Sd, Xc, XV, are on the same straight lines YW, eV. [XI. 29] But the solid XU is equal to AE: therefore the solid XY is also equal to the solid AE. And, since the parallelogram RUWT is equal to the parallelogram YT for they are on the same base RT and in the same parallels RT, YW, [I. 35] while RUWT is equal to CD, since it is also equal to AB, therefore the parallelogram YT is also equal to CD. But DT is another parallelogram; therefore, as the base CD is to DT, so is YT to DT. [V. 7] And, since the parallelepipedal solid CI has been cut by the plane RF which is parallel to opposite planes, as the base CD is to the base DT, so is the solid CF to the solid RI. [XI. 25] For the same reason, since the parallelepipedal solid YI has been cut by the plane RX which is parallel to opposite planes, as the base YT is to the base TD, so is the solid YX to the solid RI. [XI. 25] But, as the base CD is to DT, so is YT to DT; therefore also, as the solid CF is to the solid RI, so is the solid YX to RI. [V. 11] Therefore each of the solids CF, YX has to RI the same ratio; therefore the solid CF is equal to the solid YX. [V. 9] But YX was proved equal to AE; therefore AE is also equal to CF. Next, let the sides standing up, AG, HK, BE, LM, CN, PQ, DF, RS, not be at right angles to the bases AB, CD; I say again that the solid AE is equal to the solid CF. For from the points K, E, G, M, Q, F, N, S let KO, ET, GU, MV, QW, FX, NY, SI be drawn perpendicular to the plane of reference, and let them meet the plane at the points O, T, U, V, W, X, Y, I, and let OT, OU, UV, TV, WX, WY, YI, IX be joined. Then the solid KV is equal to the solid QI, for they are on the equal bases KM, QS and of the same height, and their sides which stand up are at right angles to their bases. [First part of this Prop.] But the solid KV is equal to the solid AE, and QI to CF; for they are on the same base and of the same height, while the extremities of their sides which stand up are not on the same straight lines. [XI. 30] Therefore the solid AE is also equal to the solid CF.'"],["Rule","'ProofWordCount'",786],["Rule","'GreekProof'","'ἔστω ἐπὶ ἴσων βάσεων τῶν ΑΒ, ΓΔ στερεὰ παραλληλεπίπεδα τὰ ΑΕ, ΓΖ ὑπὸ τὸ αὐτὸ ὕψος: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΕ στερεὸν τῷ ΓΖ στερεῷ. ἔστωσαν δὴ πρότερον αἱ ἐφεστηκυῖαι αἱ ΘΚ, ΒΕ, ΑΗ, ΛΜ, ΟΠ, ΔΖ, ΓΞ, ΡΣ πρὸς ὀρθὰς ταῖς ΑΒ, ΓΔ βάσεσιν, καὶ ἐκβεβλήσθω ἐπ᾽ εὐθείας τῇ ΓΡ εὐθεῖα ἡ ΡΤ, καὶ συνεστάτω πρὸς τῇ ΡΤ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ρ τῇ ὑπὸ ΑΛΒ γωνίᾳ ἴση ἡ ὑπὸ ΤΡΥ, καὶ κείσθω τῇ μὲν ΑΛ ἴση ἡ ΡΤ, τῇ δὲ ΛΒ ἴση ἡ ΡΥ, καὶ συμπεπληρώσθω ἥ τε ΡΧ βάσις καὶ τὸ ΨΥ στερεόν. καὶ ἐπεὶ δύο αἱ ΤΡ, ΡΥ δυσὶ ταῖς ΑΛ, ΛΒ ἴσαι εἰσίν, καὶ γωνίας ἴσας περιέχουσιν, ἴσον ἄρα καὶ ὅμοιον τὸ ΡΧ παραλληλόγραμμον τῷ ΘΛ παραλληλογράμμῳ. καὶ ἐπεὶ πάλιν ἴση μὲν ἡ ΑΛ τῇ ΡΤ, ἡ δὲ ΛΜ τῇ ΡΣ, καὶ γωνίας ὀρθὰς περιέχουσιν, ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΡΨ παραλληλόγραμμον τῷ ΑΜ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΛΕ τῷ ΣΥ ἴσον τέ ἐστι καὶ ὅμοιον: τρία ἄρα παραλληλόγραμμα τοῦ ΑΕ στερεοῦ τρισὶ παραλληλογράμμοις τοῦ ΨΥ στερεοῦ ἴσα τέ ἐστι καὶ ὅμοια. ἀλλὰ τὰ μὲν τρία τρισὶ τοῖς ἀπεναντίον ἴσα τέ ἐστι καὶ ὅμοια, τὰ δὲ τρία τρισὶ τοῖς ἀπεναντίον: ὅλον ἄρα τὸ ΑΕ στερεὸν παραλληλεπίπεδον ὅλῳ τῷ ΨΥ στερεῷ παραλληλεπιπέδῳ ἴσον ἐστίν. διήχθωσαν αἱ ΔΡ, ΧΥ καὶ συμπιπτέτωσαν ἀλλήλαις κατὰ τὸ Ω, καὶ διὰ τοῦ Τ τῇ ΔΩ παράλληλος ἤχθω ἡ #αΤ #5, καὶ ἐκβεβλήσθω ἡ ΟΔ κατὰ τὸ #α, καὶ συμπεπληρώσθω τὰ ΩΨ, ΡΙ στερεά. ἴσον δή ἐστι τὸ ΨΩ στερεόν, οὗ βάσις μέν ἐστι τὸ ΡΨ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ Ω #4, τῷ ΨΥ στερεῷ, οὗ βάσις μὲν τὸ ΡΨ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΥΦ: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΡΨ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΡΩ, ΡΥ, Τ #5, ΤΧ, Σ #2, Σο̂, Ψ #4, ΨΦ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΩΧ, #Φ. ἀλλὰ τὸ ΨΥ στερεὸν τῷ ΑΕ ἐστιν ἴσον: καὶ τὸ ΨΩ ἄρα στερεὸν τῷ ΑΕ στερεῷ ἐστιν ἴσον. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΡΥΧΤ παραλληλόγραμμον τῷ ΩΤ παραλληλογράμμῳ: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΡΤ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΡΤ, ΩΧ: ἀλλὰ τὸ ΡΥΧΤ τῷ ΓΔ ἐστιν ἴσον, ἐπεὶ καὶ τῷ ΑΒ, καὶ τὸ ΩΤ ἄρα παραλληλόγραμμον τῷ ΓΔ ἐστιν ἴσον. ἄλλο δὲ τὸ ΔΤ: ἔστιν ἄρα ὡς ἡ ΓΔ βάσις πρὸς τὴν ΔΤ, οὕτως ἡ ΩΤ πρὸς τὴν ΔΤ. καὶ ἐπεὶ στερεὸν παραλληλεπίπεδον τὸ ΓΙ ἐπιπέδῳ τῷ ΡΖ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ὡς ἡ ΓΔ βάσις πρὸς τὴν ΔΤ βάσιν, οὕτως τὸ ΓΖ στερεὸν πρὸς τὸ ΡΙ στερεόν. διὰ τὰ αὐτὰ δή, ἐπεὶ στερεὸν παραλληλεπίπεδον τὸ ΩΙ ἐπιπέδῳ τῷ ΡΨ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ὡς ἡ ΩΤ βάσις πρὸς τὴν ΤΔ βάσιν, οὕτως τὸ ΩΨ στερεὸν πρὸς τὸ ΡΙ. ἀλλ᾽ ὡς ἡ ΓΔ βάσις πρὸς τὴν ΔΤ, οὕτως ἡ ΩΤ πρὸς τὴν ΔΤ: καὶ ὡς ἄρα τὸ ΓΖ στερεὸν πρὸς τὸ ΡΙ στερεόν, οὕτως τὸ ΩΨ στερεὸν πρὸς τὸ ΡΙ. ἑκάτερον ἄρα τῶν ΓΖ, ΩΨ στερεῶν πρὸς τὸ ΡΙ τὸν αὐτὸν ἔχει λόγον: ἴσον ἄρα ἐστὶ τὸ ΓΖ στερεὸν τῷ ΩΨ στερεῷ. ἀλλὰ τὸ ΩΨ τῷ ΑΕ ἐδείχθη ἴσον: καὶ τὸ ΑΕ ἄρα τῷ ΓΖ ἐστιν ἴσον. μὴ ἔστωσαν δὴ αἱ ἐφεστηκυῖαι αἱ ΑΗ, ΘΚ, ΒΕ, ΛΜ, ΓΝ, ΟΠ, ΔΖ, ΡΣ πρὸς ὀρθὰς ταῖς ΑΒ, ΓΔ βάσεσιν: λέγω πάλιν, ὅτι ἴσον τὸ ΑΕ στερεὸν τῷ ΓΖ στερεῷ. ἤχθωσαν γὰρ ἀπὸ τῶν Κ, Ε, Η, Μ, Π, Ζ, Ν, Σ σημείων ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετοι αἱ ΚΞ, ΕΤ, ΗΥ, ΜΦ, ΠΧ, ΖΨ, ΝΩ, ΣΙ, καὶ συμβαλλέτωσαν τῷ ἐπιπέδῳ κατὰ τὰ Ξ, Τ, Υ, Φ, Χ, Ψ, Ω, Ι σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΞΤ, ΞΥ, ΥΦ, ΤΦ, ΧΨ, ΧΩ, ΩΙ, ΙΨ. ἴσον δή ἐστι τὸ ΚΦ στερεὸν τῷ ΠΙ στερεῷ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΚΜ, ΠΣ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι πρὸς ὀρθάς εἰσι ταῖς βάσεσιν. ἀλλὰ τὸ μὲν ΚΦ στερεὸν τῷ ΑΕ στερεῷ ἐστιν ἴσον, τὸ δὲ ΠΙ τῷ ΓΖ: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὔκ εἰσιν ἐπὶ τῶν αὐτῶν εὐθειῶν. καὶ τὸ ΑΕ ἄρα στερεὸν τῷ ΓΖ στερεῷ ἐστιν ἴσον. τὰ ἄρα ἐπὶ ἴσων βάσεων ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",717]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",32]],["Association",["Rule","'VertexLabel'","'11.32'"],["Rule","'Text'","'Parallelepipedal solids which are of the same height are to one another as their bases.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",45]],["List",["Rule","'Book'",11],["Rule","'Theorem'",25]],["List",["Rule","'Book'",11],["Rule","'Theorem'",31]]]],["Rule","'Proof'","'Let AB, CD be parallelepipedal solids of the same height; I say that the parallelepipedal solids AB, CD are to one another as their bases, that is, that, as the base AE is to the base CF, so is the solid AB to the solid CD. For let FH equal to AE be applied to FG, [I. 45] and on FH as base, and with the same height as that of CD, let the parallelepipedal solid GK be completed. Then the solid AB is equal to the solid GK; for they are on equal bases AE, FH and of the same height. [XI. 31] And, since the parallelepipedal solid CK is cut by the plane DG which is parallel to opposite planes, therefore, as the base CF is to the base FH, so is the solid CD to the solid DH. [XI. 25] But the base FH is equal to the base AE, and the solid GK to the solid AB; therefore also, as the base AE is to the base CF, so is the solid AB to the solid CD.'"],["Rule","'ProofWordCount'",181],["Rule","'GreekProof'","'ἔστω ὑπὸ τὸ αὐτὸ ὕψος στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ: λέγω, ὅτι τὰ ΑΒ, ΓΔ στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις, τουτέστιν ὅτι ἐστὶν ὡς ἡ ΑΕ βάσις πρὸς τὴν ΓΖ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΔ στερεόν. παραβεβλήσθω γὰρ παρὰ τὴν ΖΗ τῷ ΑΕ ἴσον τὸ ΖΘ, καὶ ἀπὸ βάσεως μὲν τῆς ΖΘ, ὕψους δὲ τοῦ αὐτοῦ τῷ ΓΔ στερεὸν παραλληλεπίπεδον συμπεπληρώσθω τὸ ΗΚ. ἴσον δή ἐστι τὸ ΑΒ στερεὸν τῷ ΗΚ στερεῷ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΑΕ, ΖΘ καὶ ὑπὸ τὸ αὐτὸ ὕψος. καὶ ἐπεὶ στερεὸν παραλληλεπίπεδον τὸ ΓΚ ἐπιπέδῳ τῷ ΔΗ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ἄρα ὡς ἡ ΓΖ βάσις πρὸς τὴν ΖΘ βάσιν, οὕτως τὸ ΓΔ στερεὸν πρὸς τὸ ΔΘ στερεόν. ἴση δὲ ἡ μὲν ΖΘ βάσις τῇ ΑΕ βάσει, τὸ δὲ ΗΚ στερεὸν τῷ ΑΒ στερεῷ: ἔστιν ἄρα καὶ ὡς ἡ ΑΕ βάσις πρὸς τὴν ΓΖ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΔ στερεόν. τὰ ἄρα ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",178]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",33]],["Association",["Rule","'VertexLabel'","'11.33'"],["Rule","'Text'","'Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'τὰ ὅμοια στερεὰ παραλληλεπίπεδα πρὸς ἄλληλα ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν.'"],["Rule","'GreekTextWordCount'",13],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",10]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",11],["Rule","'Definition'",10]],["List",["Rule","'Book'",11],["Rule","'Theorem'",24]],["List",["Rule","'Book'",11],["Rule","'Theorem'",32]]]],["Rule","'Proof'","'Let AB, CD be similar parallelepipedal solids, and let AE be the side corresponding to CF; I say that the solid AB has to the solid CD the ratio triplicate of that which AE has to CF. For let EK, EL, EM be produced in a straight line with AE, GE, HE, let EK be made equal to CF, EL equal to FN, and further EM equal to FR, and let the parallelogram KL and the solid KP be completed. Now, since the two sides KE, EL are equal to the two sides CF, FN, while the angle KEL is also equal to the angle CFN, inasmuch as the angle AEG is also equal to the angle CFN because of the similarity of the solids AB, CD, therefore the parallelogram KL is equal <and similar> to the parallelogram CN. For the same reason the parallelogram KM is also equal and similar to CR, and further EP to DF; therefore three parallelograms of the solid KP are equal and similar to three parallelograms of the solid CD. But the former three parallelograms are equal and similar to their opposites, and the latter three to their opposites; [XI. 24] therefore the whole solid KP is equal and similar to the whole solid CD. [XI. Def. 10] Let the parallelogram GK be completed, and on the parallelograms GK, KL as bases, and with the same height as that of AB, let the solids EO, LQ be completed. Then since; owing to the similarity of the solids AB, CD, as AE is to CF, so is EG to FN, and EH to FR, while CF is equal to EK, FN to EL, and FR to EM, therefore, as AE is to EK, so is GE to EL, and HE to EM. But, as AE is to EK, so is AG to the parallelogram GK, as GE is to EL, so is GK to KL, and, as HE is to EM, so is QE to KM; [VI. 1] therefore also, as the parallelogram AG is to GK, so is GK to KL, and QE to KM. But, as AG is to GK, so is the solid AB to the solid EO, as GK is to KL, so is the solid OE to the solid QL, and, as QE is to KM, so is the solid QL to the solid KP; [XI. 32] therefore also, as the solid AB is to EO, so is EO to QL, and QL to KP. But, if four magnitudes be continuously proportional, the first has to the fourth the ratio triplicate of that which it has to the second; [V. Def. 10] therefore the solid AB has to KP the ratio triplicate of that which AB has to EO. But, as AB is to EO, so is the parallelogram AG to GK, and the straight line AE to EK [VI. 1]; hence the solid AB has also to KP the ratio triplicate of that which AE has to EK. But the solid KP is equal to the solid CD, and the straight line EK to CF; therefore the solid AB has also to the solid CD the ratio triplicate of that which the corresponding side of it, AE, has to the corresponding side CF.'"],["Rule","'ProofWordCount'",543],["Rule","'GreekProof'","'ἔστω ὅμοια στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ, ὁμόλογος δὲ ἔστω ἡ ΑΕ τῇ ΓΖ: λέγω, ὅτι τὸ ΑΒ στερεὸν πρὸς τὸ ΓΔ στερεὸν τριπλασίονα λόγον ἔχει, ἤπερ ἡ ΑΕ πρὸς τὴν ΓΖ. ἐκβεβλήσθωσαν γὰρ ἐπ᾽ εὐθείας ταῖς ΑΕ, ΗΕ, ΘΕ αἱ ΕΚ, ΕΛ, ΕΜ, καὶ κείσθω τῇ μὲν ΓΖ ἴση ἡ ΕΚ, τῇ δὲ ΖΝ ἴση ἡ ΕΛ, καὶ ἔτι τῇ ΖΡ ἴση ἡ ΕΜ, καὶ συμπεπληρώσθω τὸ ΚΛ παραλληλόγραμμον καὶ τὸ ΚΟ στερεόν. καὶ ἐπεὶ δύο αἱ ΚΕ, ΕΛ δυσὶ ταῖς ΓΖ, ΖΝ ἴσαι εἰσίν, ἀλλὰ καὶ γωνία ἡ ὑπὸ ΚΕΛ γωνίᾳ τῇ ὑπὸ ΓΖΝ ἐστιν ἴση, ἐπειδήπερ καὶ ἡ ὑπὸ ΑΕΗ τῇ ὑπὸ ΓΖΝ ἐστιν ἴση διὰ τὴν ὁμοιότητα τῶν ΑΒ, ΓΔ στερεῶν, ἴσον ἄρα ἐστὶ καὶ ὅμοιον τὸ ΚΛ παραλληλόγραμμον τῷ ΓΝ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν ΚΜ παραλληλόγραμμον ἴσον ἐστὶ καὶ ὅμοιον τῷ ΓΡ παραλληλογράμμῳ καὶ ἔτι τὸ ΕΟ τῷ ΔΖ: τρία ἄρα παραλληλόγραμμα τοῦ ΚΟ στερεοῦ τρισὶ παραλληλογράμμοις τοῦ ΓΔ στερεοῦ ἴσα ἐστὶ καὶ ὅμοια. ἀλλὰ τὰ μὲν τρία τρισὶ τοῖς ἀπεναντίον ἴσα ἐστὶ καὶ ὅμοια, τὰ δὲ τρία τρισὶ τοῖς ἀπεναντίον ἴσα ἐστὶ καὶ ὅμοια: ὅλον ἄρα τὸ ΚΟ στερεὸν ὅλῳ τῷ ΓΔ στερεῷ ἴσον ἐστὶ καὶ ὅμοιον. συμπεπληρώσθω τὸ ΗΚ παραλληλόγραμμον, καὶ ἀπὸ βάσεων μὲν τῶν ΗΚ, ΚΛ παραλληλογράμμων, ὕψους δὲ τοῦ αὐτοῦ τῷ ΑΒ στερεὰ συμπεπληρώσθω τὰ ΕΞ, ΛΠ. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν ΑΒ, ΓΔ στερεῶν ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΝ, καὶ ἡ ΕΘ πρὸς τὴν ΖΡ, ἴση δὲ ἡ μέν ΓΖ τῇ ΕΚ, ἡ δὲ ΖΝ τῇ ΕΛ, ἡ δὲ ΖΡ τῇ ΕΜ, ἔστιν ἄρα ὡς ἡ ΑΕ πρὸς τὴν ΕΚ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΛ καὶ ἡ ΘΕ πρὸς τὴν ΕΜ. ἀλλ᾽ ὡς μὲν ἡ ΑΕ πρὸς τὴν ΕΚ, οὕτως τὸ ΑΗ παραλληλόγραμμον πρὸς τὸ ΗΚ παραλληλόγραμμον, ὡς δὲ ἡ ΗΕ πρὸς τὴν ΕΛ, οὕτως τὸ ΗΚ πρὸς τὸ ΚΛ, ὡς δὲ ἡ ΘΕ πρὸς ΕΜ, οὕτως τὸ ΠΕ πρὸς τὸ ΚΜ: καὶ ὡς ἄρα τὸ ΑΗ παραλληλόγραμμον πρὸς τὸ ΗΚ, οὕτως τὸ ΗΚ πρὸς τὸ ΚΛ καὶ τὸ ΠΕ πρὸς τὸ ΚΜ. ἀλλ᾽ ὡς μὲν τὸ ΑΗ πρὸς τὸ ΗΚ, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΕΞ στερεόν, ὡς δὲ τὸ ΗΚ πρὸς τὸ ΚΛ, οὕτως τὸ ΞΕ στερεὸν πρὸς τὸ ΠΛ στερεόν, ὡς δὲ τὸ ΠΕ πρὸς τὸ ΚΜ, οὕτως τὸ ΠΛ στερεὸν πρὸς τὸ ΚΟ στερεόν: καὶ ὡς ἄρα τὸ ΑΒ στερεὸν πρὸς τὸ ΕΞ, οὕτως τὸ ΕΞ πρὸς τὸ ΠΛ καὶ τὸ ΠΛ πρὸς τὸ ΚΟ. ἐὰν δὲ τέσσαρα μεγέθη κατὰ τὸ συνεχὲς ἀνάλογον ᾖ, τὸ πρῶτον πρὸς τὸ τέταρτον τριπλασίονα λόγον ἔχει ἤπερ πρὸς τὸ δεύτερον: τὸ ΑΒ ἄρα στερεὸν πρὸς τὸ ΚΟ τριπλασίονα λόγον ἔχει ἤπερ τὸ ΑΒ πρὸς τὸ ΕΞ. ἀλλ᾽ ὡς τὸ ΑΒ πρὸς τὸ ΕΞ, οὕτως τὸ ΑΗ παραλληλόγραμμον πρὸς τὸ ΗΚ καὶ ἡ ΑΕ εὐθεῖα πρὸς τὴν ΕΚ: ὥστε καὶ τὸ ΑΒ στερεὸν πρὸς τὸ ΚΟ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΑΕ πρὸς τὴν ΕΚ. ἴσον δὲ τὸ μὲν ΚΟ στερεὸν τῷ ΓΔ στερεῷ, ἡ δὲ ΕΚ εὐθεῖα τῇ ΓΖ: καὶ τὸ ΑΒ ἄρα στερεὸν πρὸς τὸ ΓΔ στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος αὐτοῦ πλευρὰ ἡ ΑΕ πρὸς τὴν ὁμόλογον πλευρὰν τὴν ΓΖ. τὰ ἄρα ὅμοια στερεὰ παραλληλεπίπεδα ἐν τριπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, ἔσται ὡς ἡ πρώτη πρὸς τὴν τετάρτην, οὕτω τὸ ἀπὸ τῆς πρώτης στερεὸν παραλληλεπίπεδον πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον, ἐπείπερ καὶ ἡ πρώτη πρὸς τὴν τετάρτην τριπλασίονα λόγον ἔχει ἤπερ πρὸς τὴν δευτέραν.'"],["Rule","'GreekProofWordCount'",597]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",34]],["Association",["Rule","'VertexLabel'","'11.34'"],["Rule","'Text'","'In equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'τῶν ἴσων στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν: καὶ ὧν στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἴσα ἐστὶν ἐκεῖνα.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",11],["Rule","'Theorem'",25]],["List",["Rule","'Book'",11],["Rule","'Theorem'",29]],["List",["Rule","'Book'",11],["Rule","'Theorem'",30]],["List",["Rule","'Book'",11],["Rule","'Theorem'",31]],["List",["Rule","'Book'",11],["Rule","'Theorem'",32]]]],["Rule","'Proof'","'Let AB, CD be equal parallelepipedal solids; I say that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB. First, let the sides which stand up, namely AG, EF, LB, HK, CM, NO, PD, QR, be at right angles to their bases; I say that, as the base EH is to the base NQ, so is CM to AG. If now the base EH is equal to the base NQ, while the solid AB is also equal to the solid CD, CM will also be equal to AG. For parallelepipedal solids of the same height are to one another as the bases; [XI. 32] and, as the base EH is to NQ, so will CM be to AG, and it is manifest that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights. Next, let the base EH not be equal to the base NQ, but let EH be greater. Now the solid AB is equal to the solid CD; therefore CM is also greater than AG. Let then CT be made equal to AG, and let the parallelepipedal solid VC be completed on NQ as base and with CT as height. Now, since the solid AB is equal to the solid CD, and CV is outside them, while equals have to the same the same ratio, [V. 7] therefore, as the solid AB is to the solid CV, so is the solid CD to the solid CV. But, as the solid AB is to the solid CV, so is the base EH to the base NQ, for the solids AB, CV are of equal height; [XI. 32] and, as the solid CD is to the solid CV, so is the base MQ to the base TQ [XI. 25] and CM to CT [VI. 1]; therefore also, as the base EH is to the base NQ, so is MC to CT. But CT is equal to AG; therefore also, as the base EH is to the base NQ, so is MC to AG. Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights. Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD. Let the sides which stand up be again at right angles to the bases. Now, if the base EH is equal to the base NQ, and, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB, therefore the height of the solid CD is also equal to the height of the solid AB. But parallelepipedal solids on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid AB is equal to the solid CD. Next, let the base EH not be equal to the base NQ, but let EH be greater; therefore the height of the solid CD is also greater than the height of the solid AB, that is, CM is greater than AG. Let CT be again made equal to AG, and let the solid CV be similarly completed. Since, as the base EH is to the base NQ, so is MC to AG, while AG is equal to CT, therefore, as the base EH is to the base NQ, so is CM to CT. But, as the base EH is to the base NQ, so is the solid AB to the solid CV, for the solids AB, CV are of equal height; [XI. 32] and, as CM is to CT, so is the base MQ to the base QT [VI. 1] and the solid CD to the solid CV. [XI. 25] Therefore also, as the solid AB is to the solid CV, so is the solid CD to the solid CV; therefore each of the solids AB, CD has to CV the same ratio. Therefore the solid AB is equal to the solid CD. [V. 9] Now let the sides which stand up, FE, BL, GA, HK, ON, DP, MC, RQ, not be at right angles to their bases; let perpendiculars be drawn from the points F, G, B, K, O, M, D, R to the planes through EH, NQ, and let them meet the planes at S, T, U, V, W, X, Y, a, and let the solids FV, Oa be completed; I say that, in this case too, if the solids AB, CD are equal, the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB. Since the solid AB is equal to the solid CD, while AB is equal to BT, for they are on the same base FK and of the same height; [XI. 29, 30] and the solid CD is equal to DX, for they are again on the same base RO and of the same height; [id.] therefore the solid BT is also equal to the solid DX. Therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT. [Part 1.] But the base FK is equal to the base EH, and the base OR to the base NQ; therefore, as the base EH is to the base NQ, so is the height of the solid DX to the height of the solid BT. But the solids DX, BT and the solids DC, BA have the same heights respectively; therefore, as the base EH is to the base NQ, so is the height of the solid DC to the height of the solid AB. Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights. Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD. For, with the same construction, since, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB, while the base EH is equal to the base FK, and NQ to OR, therefore, as the base FK is to the base OR, so is the height of the solid CD to the height of the solid AB. But the solids AB, CD and BT, DX have the same heights respectively; therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT. Therefore in the parallelepipedal solids BT, DX the bases are reciprocally proportional to the heights; therefore the solid BT is equal to the solid DX. [Part 1.] But BT is equal to BA, for they are on the same base FK and of the same height; [XI. 29, 30] and the solid DX is equal to the solid DC. [id.] Therefore the solid AB is also equal to the solid CD.'"],["Rule","'ProofWordCount'",1267],["Rule","'GreekProof'","'ἔστω ἴσα στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ: λέγω, ὅτι τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, καί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος. ἔστωσαν γὰρ πρότερον αἱ ἐφεστηκυῖαι αἱ ΑΗ, ΕΖ, ΛΒ, ΘΚ, ΓΜ, ΝΞ, ΟΔ, ΠΡ πρὸς ὀρθὰς ταῖς βάσεσιν αὐτῶν: λέγω, ὅτι ἐστὶν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΓΜ πρὸς τὴν ΑΗ. εἰ μὲν οὖν ἴση ἐστιν ἡ ΕΘ βάσις τῇ ΝΠ βάσει, ἔστι δὲ καὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ ἴσον, ἔσται καὶ ἡ ΓΜ τῇ ΑΗ ἴση. τὰ γὰρ ὑπὸ τὸ αὐτὸ ὕψος στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις εἰ γὰρ τῶν ΕΘ, ΝΠ βάσεων ἴσων οὐσῶν μὴ εἴη τὰ ΑΗ, ΓΜ ὕψη ἴσα, οὐδ᾽ ἄρα τὸ ΑΒ στερεὸν ἴσον ἔσται τῷ ΓΔ. ὑπόκειται δὲ ἴσον: οὐκ ἄρα ἄνισόν ἐστι τὸ ΓΜ ὕψος τῷ ΑΗ ὕψει: ἴσον ἄρα. καὶ ἔσται ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ, οὕτως ἡ ΓΜ πρὸς τὴν ΑΗ, καὶ φανερόν, ὅτι τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. μὴ ἔστω δὴ ἴση ἡ ΕΘ βάσις τῇ ΝΠ βάσει, ἀλλ᾽ ἔστω μείζων ἡ ΕΘ. ἔστι δὲ καὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ ἴσον: μείζων ἄρα ἐστὶ καὶ ἡ ΓΜ τῆς ΑΗ εἰ γὰρ μή, οὐδ᾽ ἄρα πάλιν τὰ ΑΒ, ΓΔ στερεὰ ἴσα ἔσται: ὑπόκειται δὲ ἴσα. κείσθω οὖν τῇ ΑΗ ἴση ἡ ΓΤ, καὶ συμπεπληρώσθω ἀπὸ βάσεως μὲν τῆς ΝΠ, ὕψους δὲ τοῦ ΓΤ, στερεὸν παραλληλεπίπεδον τὸ ΦΓ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ, ἔξωθεν δὲ τὸ ΓΦ, τὰ δὲ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν. ἀλλ᾽ ὡς μὲν τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν: ἰσοϋψῆ γὰρ τὰ ΑΒ, ΓΦ στερεά: ὡς δὲ τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως ἡ ΜΠ βάσις πρὸς τὴν ΤΠ βάσιν καὶ ἡ ΓΜ πρὸς τὴν ΓΤ: καὶ ὡς ἄρα ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΜΓ πρὸς τὴν ΓΤ. ἴση δὲ ἡ ΓΤ τῇ ΑΗ: καὶ ὡς ἄρα ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΜΓ πρὸς τὴν ΑΗ. τῶν ΑΒ, ΓΔ ἄρα στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. πάλιν δὴ τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντιπεπονθέτωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ. ἔστωσαν γὰρ πάλιν αἱ ἐφεστηκυῖαι πρὸς ὀρθὰς ταῖς βάσεσιν, καὶ εἰ μὲν ἴση ἐστὶν ἡ ΕΘ βάσις τῇ ΝΠ βάσει, καί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος, ἴσον ἄρα ἐστὶ καὶ τὸ τοῦ ΓΔ στερεοῦ ὕψος τῷ τοῦ ΑΒ στερεοῦ ὕψει. τὰ δὲ ἐπὶ ἴσων βάσεων στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν: ἴσον ἄρα ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ. μὴ ἔστω δὴ ἡ ΕΘ βάσις τῇ ΝΠ βάσει ἴση, ἀλλ᾽ ἔστω μείζων ἡ ΕΘ: μεῖζον ἄρα ἐστὶ καὶ τὸ τοῦ ΓΔ στερεοῦ ὕψος τοῦ τοῦ ΑΒ στερεοῦ ὕψους, τουτέστιν ἡ ΓΜ τῆς ΑΗ. κείσθω τῇ ΑΗ ἴση πάλιν ἡ ΓΤ, καὶ συμπεπληρώσθω ὁμοίως τὸ ΓΦ στερεόν. ἐπεί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΜΓ πρὸς τὴν ΑΗ, ἴση δὲ ἡ ΑΗ τῇ ΓΤ, ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΓΜ πρὸς τὴν ΓΤ. ἀλλ᾽ ὡς μὲν ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν: ἰσοϋψῆ γάρ ἐστι τὰ ΑΒ, ΓΦ στερεά: ὡς δὲ ἡ ΓΜ πρὸς τὴν ΓΤ, οὕτως ἥ τε ΜΠ βάσις πρὸς τὴν ΠΤ βάσιν καὶ τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν. καὶ ὡς ἄρα τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν: ἑκάτερον ἄρα τῶν ΑΒ, ΓΔ πρὸς τὸ ΓΦ τὸν αὐτὸν ἔχει λόγον. ἴσον ἄρα ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ ὅπερ ἔδει δεῖξαι. μὴ ἔστωσαν δὴ αἱ ἐφεστηκυῖαι αἱ ΖΕ, ΒΛ, ΗΑ, ΘΚ, ΞΝ, ΔΟ, ΜΓ, ΡΠ πρὸς ὀρθὰς ταῖς βάσεσιν αὐτῶν, καὶ ἤχθωσαν ἀπὸ τῶν Ζ, Η, Β, Κ, Ξ, Μ, Δ, Ρ σημείων ἐπὶ τὰ διὰ τῶν ΕΘ, ΝΠ ἐπίπεδα κάθετοι καὶ συμβαλλέτωσαν τοῖς ἐπιπέδοις κατὰ τὰ Σ, Τ, Υ, Φ, Χ, Ψ, Ω, #2, καὶ συμπεπληρώσθω τὰ ΖΦ, ΞΩ στερεά: λέγω, ὅτι καὶ οὕτως ἴσων ὄντων τῶν ΑΒ, ΓΔ στερεῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, καί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος. ἐπεὶ ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ, ἀλλὰ τὸ μὲν ΑΒ τῷ ΒΤ ἐστιν ἴσον: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΖΚ καὶ ὑπὸ τὸ αὐτὸ ὕψος ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν: τὸ δὲ ΓΔ στερεὸν τῷ ΔΨ ἐστιν ἴσον: ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς ΡΞ καὶ ὑπὸ τὸ αὐτὸ ὕψος ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν: καὶ τὸ ΒΤ ἄρα στερεὸν τῷ ΔΨ στερεῷ ἴσον ἐστίν τῶν δὲ ἴσων στερεῶν παραλληλεπιπέδων, ὧν τὰ ὕψη πρὸς ὀρθάς ἐστι ταῖς βάσεσιν αὐτῶν, ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ στερεοῦ ὕψος. ἴση δὲ ἡ μὲν ΖΚ βάσις τῇ ΕΘ βάσει, ἡ δὲ ΞΡ βάσις τῇ ΝΠ βάσει: ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ στερεοῦ ὕψος. τὰ δ᾽ αὐτὰ ὕψη ἐστὶ τῶν ΔΨ, ΒΤ στερεῶν καὶ τῶν ΔΓ, ΒΑ: ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΔΓ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος. τῶν ΑΒ, ΓΔ ἄρα στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. πάλιν δὴ τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντιπεπονθέτωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος, ἴση δὲ ἡ μὲν ΕΘ βάσις τῇ ΖΚ βάσει, ἡ δὲ ΝΠ τῇ ΞΡ, ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος. τὰ δ᾽ αὐτὰ ὕψη ἐστὶ τῶν ΑΒ, ΓΔ στερεῶν καὶ τῶν ΒΤ, ΔΨ: ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ στερεοῦ ὕψος. τῶν ΒΤ, ΔΨ ἄρα στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν ὧν δὲ στερεῶν παραλληλεπιπέδων τὰ ὕψη πρὸς ὀρθάς ἐστι ταῖς βάσεσιν αὐτῶν, ἀντιπεπόνθασι δὲ αἱ βάσεις τοῖς ὕψεσιν, ἴσα ἐστὶν ἐκεῖνα: ἴσον ἄρα ἐστὶ τὸ ΒΤ στερεὸν τῷ ΔΨ στερεῷ. ἀλλὰ τὸ μὲν ΒΤ τῷ ΒΑ ἴσον ἐστίν: ἐπί τε γὰρ τῆς αὐτῆς βάσεως εἰσι τῆς ΖΚ καὶ ὑπὸ τὸ αὐτὸ ὕψος ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν. τὸ δὲ ΔΨ στερεὸν τῷ ΔΓ στερεῷ ἴσον ἐστίν ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς ΞΡ καὶ ὑπὸ τὸ αὐτὸ ὕψος καὶ οὐκ ἐν ταῖς αὐταῖς εὐθείαις. καὶ τὸ ΑΒ ἄρα στερεὸν τῷ ΓΔ στερεῷ ἐστιν ἴσον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",1262]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",35]],["Association",["Rule","'VertexLabel'","'11.35'"],["Rule","'Text'","'If there be two equal plane angles, and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points be taken at random and perpendiculars be drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines be joined to the vertices of the original angles, they will contain, with the elevated straight lines, equal angles.'"],["Rule","'TextWordCount'",84],["Rule","'GreekText'","'ἐὰν ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι, ἐπὶ δὲ τῶν κορυφῶν αὐτῶν μετέωροι εὐθεῖαι ἐπισταθῶσιν ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, ἐπὶ δὲ τῶν μετεώρων ληφθῇ τυχόντα σημεῖα, καὶ ἀπ᾽ αὐτῶν ἐπὶ τὰ ἐπίπεδα, ἐν οἷς εἰσιν αἱ ἐξ ἀρχῆς γωνίαι, κάθετοι ἀχθῶσιν, ἀπὸ δὲ τῶν γενομένων σημείων ἐν τοῖς ἐπιπέδοις ἐπὶ τὰς ἐξ ἀρχῆς γωνίας ἐπιζευχθῶσιν εὐθεῖαι, ἴσας γωνίας περιέξουσι μετὰ τῶν μετεώρων.'"],["Rule","'GreekTextWordCount'",68],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",1],["Rule","'Theorem'",48]],["List",["Rule","'Book'",11],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'Let the angles BAC, EDF be two equal rectilineal angles, and from the points A, D let the elevated straight lines AG, DM be set up containing, with the original straight lines, equal angles respectively, namely, the angle MDE to the angle GAB and the angle MDF to the angle GAC, let points G, M be taken at random on AG, DM, let GL, MN be drawn from the points G, M perpendicular to the planes through BA, AC and ED, DF, and let them meet the planes at L, N, and let LA, ND be joined; I say that the angle GAL is equal to the angle MDN. Let AH be made equal to DM, and let HK be drawn through the point H parallel to GL. But GL is perpendicular to the plane through BA, AC; therefore HK is also perpendicular to the plane through. BA, AC. [XI. 8] From the points K, N let KC, NF, KB, NE be drawn perpendicular to the straight lines AC, DF, AB, DE, and let HC, CB, MF, FE be joined. Since the square on HA is equal to the squares on HK, KA, and the squares on KC, CA are equal to the square on KA, [I. 47] therefore the square on HA is also equal to the squares on HK, KC, CA. But the square on HC is equal to the squares on HK, KC; [I. 47] therefore the square on HA is equal to the squares on HC, CA. Therefore the angle HCA is right. [I. 48] For the same reason the angle DFM is also right. Therefore the angle ACH is equal to the angle DFM. But the angle HAC is also equal to the angle MDF. Therefore MDF, HAC are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that subtending one of the equal angles, that is, HA equal to MD; therefore they will also have the remaining sides equal to the remaining sides respectively. [I. 26] Therefore AC is equal to DF. Similarly we can prove that AB is also equal to DE. Since then AC is equal to DF, and AB to DE, the two sides CA, AB are equal to the two sides FD, DE. But the angle CAB is also equal to the angle FDE; therefore the base BC is equal to the base EF, the triangle to the triangle, and the remaining angles to the remaining angles; [I. 4] therefore the angle ACB is equal to the angle DFE. But the right angle ACK is also equal to the right angle DFN; therefore the remaining angle BCK is also equal to the remaining angle EFN. For the same reason the angle CBK is also equal to the angle FEN. Therefore BCK, EFN are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is, BC equal to EF; therefore they will also have the remaining sides equal to the remaining sides. [I. 26] Therefore CK is equal to FN. But AC is also equal to DF; therefore the two sides AC, CK are equal to the two sides DF, FN; and they contain right angles. Therefore the base AK is equal to the base DN. [I. 4] And, since AH is equal to DM, the square on AH is also equal to the square on DM. But the squares on AK, KH are equal to the square on AH, for the angle AKH is right; [I. 47] and the squares on DN, NM are equal to the square on DM, for the angle DNM is right; [I. 47] therefore the squares on AK, KH are equal to the squares on DN, NM; and of these the square on AK is equal to the square on DN; therefore the remaining square on KH is equal to the square on NM; therefore HK is equal to MN. And, since the two sides HA, AK are equal to the two sides MD, DN respectively, and the base HK was proved equal to the base MN, therefore the angle HAK is equal to the angle MDN. [I. 8]'"],["Rule","'ProofWordCount'",710],["Rule","'GreekProof'","'ἔστωσαν δύο γωνίαι εὐθύγραμμοι ἴσαι αἱ ὑπὸ ΒΑΓ, ΕΔΖ, ἀπὸ δὲ τῶν Α, Δ σημείων μετέωροι εὐθεῖαι ἐφεστάτωσαν αἱ ΑΗ, ΔΜ ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, τὴν μὲν ὑπὸ ΜΔΕ τῇ ὑπὸ ΗΑΒ, τὴν δὲ ὑπὸ ΜΔΖ τῇ ὑπὸ ΗΑΓ, καὶ εἰλήφθω ἐπὶ τῶν ΑΗ, ΔΜ τυχόντα σημεῖα τὰ Η, Μ, καὶ ἤχθωσαν ἀπὸ τῶν Η, Μ σημείων ἐπὶ τὰ διὰ τῶν ΒΑΓ, ΕΔΖ ἐπίπεδα κάθετοι αἱ ΗΛ, ΜΝ, καὶ συμβαλλέτωσαν τοῖς ἐπιπέδοις κατὰ τὰ Ν, Λ, καὶ ἐπεζεύχθωσαν αἱ ΛΑ, ΝΔ: λέγω, ὅτι ἴση ἐστὶν ἡ ὑπὸ ΗΑΛ γωνία τῇ ὑπὸ ΜΔΝ γωνίᾳ. κείσθω τῇ ΔΜ ἴση ἡ ΑΘ, καὶ ἤχθω διὰ τοῦ Θ σημείου τῇ ΗΛ παράλληλος ἡ ΘΚ. ἡ δὲ ΗΛ κάθετός ἐστιν ἐπὶ τὸ διὰ τῶν ΒΑΓ ἐπίπεδον: καὶ ἡ ΘΚ ἄρα κάθετός ἐστιν ἐπὶ τὸ διὰ τῶν ΒΑΓ ἐπίπεδον. ἤχθωσαν ἀπὸ τῶν Κ, Ν σημείων ἐπὶ τὰς ΑΒ, ΑΓ, ΔΖ, ΔΕ εὐθείας κάθετοι αἱ ΚΓ, ΝΖ, ΚΒ, ΝΕ, καὶ ἐπεζεύχθωσαν αἱ ΘΓ, ΓΒ, ΜΖ, ΖΕ. ἐπεὶ τὸ ἀπὸ τῆς ΘΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΑ, τῷ δὲ ἀπὸ τῆς ΚΑ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΚΓ, ΓΑ, καὶ τὸ ἀπὸ τῆς ΘΑ ἄρα ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΓ, ΓΑ. τοῖς δὲ ἀπὸ τῶν ΘΚ, ΚΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΘΓ: τὸ ἄρα ἀπὸ τῆς ΘΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΓ, ΓΑ. ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΘΓΑ γωνία. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΔΖΜ γωνία ὀρθή ἐστιν. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΓΘ γωνία τῇ ὑπὸ ΔΖΜ. ἔστι δὲ καὶ ἡ ὑπὸ ΘΑΓ τῇ ὑπὸ ΜΔΖ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΜΔΖ, ΘΑΓ δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν τὴν ΘΑ τῇ ΜΔ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ. ἴση ἄρα ἐστὶν ἡ ΑΓ τῇ ΔΖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΑΒ τῇ ΔΕ ἐστιν ἴση οὕτως: ἐπεζεύχθωσαν αἱ ΘΒ, ΜΕ. καὶ ἐπεὶ τὸ ἀπὸ τῆς ΑΘ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΚ, ΚΘ, τῷ δὲ ἀπὸ τῆς ΑΚ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΚ, τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΚ, ΚΘ ἴσα ἐστὶ τῷ ἀπὸ ΑΘ. ἀλλὰ τοῖς ἀπὸ τῶν ΒΚ, ΚΘ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΘ: ὀρθὴ γὰρ ἡ ὑπὸ ΘΚΒ γωνία διὰ τὸ καὶ τὴν ΘΚ κάθετον εἶναι ἐπὶ τὸ ὑποκείμενον ἐπίπεδον: τὸ ἄρα ἀπὸ τῆς ΑΘ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΒ, ΒΘ: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΑΒΘ γωνία. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΔΕΜ γωνία ὀρθή ἐστιν. ἔστι δὲ καὶ ἡ ὑπὸ ΒΑΘ γωνία τῇ ὑπὸ ΕΔΜ ἴση: ὑπόκεινται γάρ: καὶ ἔστιν ἡ ΑΘ τῇ ΔΜ ἴση: ἴση ἄρα ἐστὶ καὶ ἡ ΑΒ τῇ ΔΕ. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΓ τῇ ΔΖ, ἡ δὲ ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΓΑ, ΑΒ δυσὶ ταῖς ΖΔ, ΔΕ ἴσαι εἰσίν. ἀλλὰ καὶ γωνία ἡ ὑπὸ ΓΑΒ γωνίᾳ τῇ ὑπὸ ΖΔΕ ἐστιν ἴση: βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστὶ καὶ τὸ τρίγωνον τῷ τριγώνῳ καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις: ἴση ἄρα ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἔστι δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΓΚ ὀρθῇ τῇ ὑπὸ ΔΖΝ ἴση: καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΓΚ λοιπῇ τῇ ὑπὸ ΕΖΝ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΒΚ τῇ ὑπὸ ΖΕΝ ἐστιν ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΒΓΚ, ΕΖΝ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΒΓ τῇ ΕΖ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν. ἴση ἄρα ἐστὶν ἡ ΓΚ τῇ ΖΝ. ἔστι δὲ καὶ ἡ ΑΓ τῇ ΔΖ ἴση: δύο δὴ αἱ ΑΓ, ΓΚ δυσὶ ταῖς ΔΖ, ΖΝ ἴσαι εἰσίν: καὶ ὀρθὰς γωνίας περιέχουσιν. βάσις ἄρα ἡ ΑΚ βάσει τῇ ΔΝ ἴση ἐστίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΘ τῇ ΔΜ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΘ τῷ ἀπὸ τῆς ΔΜ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΘ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΚ, ΚΘ: ὀρθὴ γὰρ ἡ ὑπὸ ΑΚΘ: τῷ δὲ ἀπὸ τῆς ΔΜ ἴσα τὰ ἀπὸ τῶν ΔΝ, ΝΜ: ὀρθὴ γὰρ ἡ ὑπὸ ΔΝΜ: τὰ ἄρα ἀπὸ τῶν ΑΚ, ΚΘ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΔΝ, ΝΜ, ὧν τὸ ἀπὸ τῆς ΑΚ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΝ: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΚΘ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΝΜ: ἴση ἄρα ἡ ΘΚ τῇ ΜΝ. καὶ ἐπεὶ δύο αἱ ΘΑ, ΑΚ δυσὶ ταῖς ΜΔ, ΔΝ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ βάσις ἡ ΘΚ βάσει τῇ ΜΝ ἐδείχθη ἴση, γωνία ἄρα ἡ ὑπὸ ΘΑΚ γωνίᾳ τῇ ὑπὸ ΜΔΝ ἐστιν ἴση. ἐὰν ἄρα ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι καὶ τὰ ἑξῆς τῆς προτάσεως ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι, ἐπισταθῶσι δὲ ἐπ᾽ αὐτῶν μετέωροι εὐθεῖαι ἴσαι ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, αἱ ἀπ᾽ αὐτῶν κάθετοι ἀγόμεναι ἐπὶ τὰ ἐπίπεδα, ἐν οἷς εἰσιν αἱ ἐξ ἀρχῆς γωνίαι, ἴσαι ἀλλήλαις εἰσίν. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",836]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",36]],["Association",["Rule","'VertexLabel'","'11.36'"],["Rule","'Text'","'If three straight lines be proportional, the parallelepipedal solid formed out of the three is equal to the parallelepipedal solid on the mean which is equilateral, but equiangular with the aforesaid solid.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ἐκ τῶν τριῶν στερεὸν παραλληλεπίπεδον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης στερεῷ παραλληλεπιπέδῳ ἰσοπλεύρῳ μέν, ἰσογωνίῳ δὲ τῷ προειρημένῳ.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",14]],["List",["Rule","'Book'",11],["Rule","'Theorem'",31]],["List",["Rule","'Book'",11],["Rule","'Theorem'",35]]]],["Rule","'Proof'","'Let A, B, C be three straight lines in proportion, so that, as A is to B, so is B to C; I say that the solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid. Let there be set out the solid angle at E contained by the angles DEG, GEF, FED, let each of the straight lines DE, GE, EF be made equal to B, and let the parallelepipedal solid EK be completed, let LM be made equal to A, and on the straight line LM, and at the point L on it, let there be constructed a solid angle equal to the solid angle at E, namely that contained by NLO, OLM, MLN; let LO be made equal to B, and LN equal to C. Now, since, as A is to B, so is B to C, while A is equal to LM, B to each of the straight lines LO, ED, and C to LN, therefore, as LM is to EF, so is DE to LN. Thus the sides about the equal angles NLM, DEF are reciprocally proportional; therefore the parallelogram MN is equal to the parallelogram DF. [VI. 14] And, since the angles DEF, NLM are two plane rectilineal angles, and on them the elevated straight lines LO, EG are set up which are equal to one another and contain equal angles with the original straight lines respectively, therefore the perpendiculars drawn from the points G, O to the planes through NL, LM and DE, EF are equal to one another; [XI. 35, Por.]hence the solids LH, EK are of the same height. But parallelepipedal solids on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid HL is equal to the solid EK. And LH is the solid formed out of A, B, C, and EK the solid on B; therefore the parallelepipedal solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid.'"],["Rule","'ProofWordCount'",353],["Rule","'GreekProof'","'ἔστωσαν τρεῖς εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ: λέγω, ὅτι τὸ ἐκ τῶν Α, Β, Γ στερεὸν ἴσον ἐστὶ τῷ ἀπὸ τῆς Β στερεῷ ἰσοπλεύρῳ μέν, ἰσογωνίῳ δὲ τῷ προειρημένῳ. Ἐκκείσθω στερεὰ γωνία ἡ πρὸς τῷ Ε περιεχομένη ὑπὸ τῶν ὑπὸ ΔΕΗ, ΗΕΖ, ΖΕΔ, καὶ κείσθω τῇ μὲν Β ἴση ἑκάστη τῶν ΔΕ, ΗΕ, ΕΖ, καὶ συμπεπληρώσθω τὸ ΕΚ στερεὸν παραλληλεπίπεδον, τῇ δὲ Α ἴση ἡ ΛΜ, καὶ συνεστάτω πρὸς τῇ ΛΜ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Λ τῇ πρὸς τῷ Ε στερεᾷ γωνίᾳ ἴση στερεὰ γωνία ἡ περιεχομένη ὑπὸ τῶν ΝΛΞ, ΞΛΜ, ΜΛΝ, καὶ κείσθω τῇ μὲν Β ἴση ἡ ΛΞ, τῇ δὲ Γ ἴση ἡ ΛΝ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ, ἴση δὲ ἡ μὲν Α τῇ ΛΜ, ἡ δὲ Β ἑκατέρᾳ τῶν ΛΞ, ΕΔ, ἡ δὲ Γ τῇ ΛΝ, ἔστιν ἄρα ὡς ἡ ΛΜ πρὸς τὴν ΕΖ, οὕτως ἡ ΔΕ πρὸς τὴν ΛΝ. καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΝΛΜ, ΔΕΖ αἱ πλευραὶ ἀντιπεπόνθασιν: ἴσον ἄρα ἐστὶ τὸ ΜΝ παραλληλόγραμμον τῷ ΔΖ παραλληλογράμμῳ. καὶ ἐπεὶ δύο γωνίαι ἐπίπεδοι εὐθύγραμμοι ἴσαι εἰσὶν αἱ ὑπὸ ΔΕΖ, ΝΛΜ, καὶ ἐπ᾽ αὐτῶν μετέωροι εὐθεῖαι ἐφεστᾶσιν αἱ ΛΞ, ΕΗ ἴσαι τε ἀλλήλαις καὶ ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, αἱ ἄρα ἀπὸ τῶν Η, Ξ σημείων κάθετοι ἀγόμεναι ἐπὶ τὰ διὰ τῶν ΝΛΜ, ΔΕΖ ἐπίπεδα ἴσαι ἀλλήλαις εἰσίν: ὥστε τὰ ΛΘ, ΕΚ στερεὰ ὑπὸ τὸ αὐτὸ ὕψος ἐστίν. τὰ δὲ ἐπὶ ἴσων βάσεων στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν: ἴσον ἄρα ἐστὶ τὸ ΘΛ στερεὸν τῷ ΕΚ στερεῷ. καί ἐστι τὸ μὲν ΛΘ τὸ ἐκ τῶν Α, Β, Γ στερεόν, τὸ δὲ ΕΚ τὸ ἀπὸ τῆς Β στερεόν: τὸ ἄρα ἐκ τῶν Α, Β, Γ στερεὸν παραλληλεπίπεδον ἴσον ἐστὶ τῷ ἀπὸ τῆς Β στερεῷ ἰσοπλεύρῳ μέν, ἰσογωνίῳ δὲ τῷ προειρημένῳ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",324]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",37]],["Association",["Rule","'VertexLabel'","'11.37'"],["Rule","'Text'","'If four straight lines be proportional, the parallelepipedal solids on them which are similar and similarly described will also be proportional; and, if the parallelepipedal solids on them which are similar and similarly described be proportional, the straight lines will themselves also be proportional.'"],["Rule","'TextWordCount'",44],["Rule","'GreekText'","'ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπ᾽ αὐτῶν στερεὰ παραλληλεπίπεδα ὅμοιά τε καὶ ὁμοίως ἀναγραφόμενα ἀνάλογον ἔσται: καὶ ἐὰν τὰ ἀπ᾽ αὐτῶν στερεὰ παραλληλεπίπεδα ὅμοιά τε καὶ ὁμοίως ἀναγραφόμενα ἀνάλογον ᾖ, καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται.'"],["Rule","'GreekTextWordCount'",40],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Theorem'",33]]]],["Rule","'Proof'","'Let AB, CD, EF, GH be four straight lines in proportion, so that, as AB is to CD, so is EF to GH; and let there be described on AB, CD, EF, GH the similar and similarly situated parallelepipedal solids KA, LC, ME, NG; I say that, as KA is to LC, so is ME to NG. For, since the parallelepipedal solid KA is similar to LC, therefore KA has to LC the ratio triplicate of that which AB has to CD. [XI. 33] For the same reason ME also has to NG the ratio triplicate of that which EF has to GH. And, as AB is to CD, so is EF to GH. Therefore also, as AK is to LC, so is ME to NG. Next, as the solid AK is to the solid LC, so let the solid ME be to the solid NG; I say that, as the straight line AB is to CD, so is EF to GH. For since, again, KA has to LC the ratio triplicate of that which AB has to CD, [XI. 33] and ME also has to NG the ratio triplicate of that which EF has to GH, and, as KA is to LC, so is ME to NG, therefore also, as AB is to CD, so is EF to GH.'"],["Rule","'ProofWordCount'",220],["Rule","'GreekProof'","'ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, ΕΖ, ΗΘ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, καὶ ἀναγεγράφθωσαν ἀπὸ τῶν ΑΒ, ΓΔ, ΕΖ, ΗΘ ὅμοιά τε καὶ ὁμοίως κείμενα στερεὰ παραλληλεπίπεδα τὰ ΚΑ, ΛΓ, ΜΕ, ΝΗ: λέγω, ὅτι ἐστὶν ὡς τὸ ΚΑ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ πρὸς τὸ ΝΗ. ἐπεὶ γὰρ ὅμοιόν ἐστι τὸ ΚΑ στερεὸν παραλληλεπίπεδον τῷ ΛΓ, τὸ ΚΑ ἄρα πρὸς τὸ ΛΓ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΓΔ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΜΕ πρὸς τὸ ΝΗ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΗΘ. καί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. καὶ ὡς ἄρα τὸ ΑΚ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ πρὸς τὸ ΝΗ. ἀλλὰ δὴ ἔστω ὡς τὸ ΑΚ στερεὸν πρὸς τὸ ΛΓ στερεόν, οὕτως τὸ ΜΕ στερεὸν πρὸς τὸ ΝΗ: λέγω, ὅτι ἐστὶν ὡς ἡ ΑΒ εὐθεῖα πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. ἐπεὶ γὰρ πάλιν τὸ ΚΑ πρὸς τὸ ΛΓ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΓΔ, ἔχει δὲ καὶ τὸ ΜΕ πρὸς τὸ ΝΗ τριπλασίονα λόγον ἤπερ ἡ ΕΖ πρὸς τὴν ΗΘ, καί ἐστιν ὡς τὸ ΚΑ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ πρὸς τὸ ΝΗ, καὶ ὡς ἄρα ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. ἐὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσι καὶ τὰ ἑξῆς τῆς προτάσεως: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",237]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",38]],["Association",["Rule","'VertexLabel'","'11.38'"],["Rule","'Text'","'If the sides of the opposite planes of a cube be bisected, and planes be carried through the points of section, the common section of the planes and the diameter of the cube bisect one another.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν κύβου τῶν ἀπεναντίον ἐπιπέδων αἱ πλευραὶ δίχα τμηθῶσιν, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβληθῇ, ἡ κοινὴ τομὴ τῶν ἐπιπέδων καὶ ἡ τοῦ κύβου διάμετρος δίχα τέμνουσιν ἀλλήλας.'"],["Rule","'GreekTextWordCount'",28],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",14]],["List",["Rule","'Book'",1],["Rule","'Theorem'",15]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",33]],["List",["Rule","'Book'",11],["Rule","'Theorem'",9]]]],["Rule","'Proof'","'For let the sides of the opposite planes CF, AH of the cube AF be bisected at the points K, L, M, N, O, Q, P, R, and through the points of section let the planes KN, OR be carried; let US be the common section of the planes, and DG the diameter of the cube AF. I say that UT is equal to TS, and DT to TG. For let DU, UE, BS, SG be joined. Then, since DO is parallel to PE, the alternate angles DOU, UPE are equal to one another. [I. 29] And, since DO is equal to PE, and OU to UP, and they contain equal angles, therefore the base DU is equal to the base UE, the triangle DOU is equal to the triangle PUE, and the remaining angles are equal to the remaining angles; [I. 4] therefore the angle OUD is equal to the angle PUE. For this reason DUE is a straight line. [I. 14] For the same reason, BSG is also a straight line, and BS is equal to SG. Now, since CA is equal and parallel to DB, while CA is also equal and parallel to EG, therefore DB is also equal and parallel to EG. [XI. 9] And the straight lines DE, BG join their extremities; therefore DE is parallel to BG. [I. 33] Therefore the angle EDT is equal to the angle BGT, for they are alternate; [I. 29] and the angle DTU is equal to the angle GTS. [I. 15] Therefore DTU, GTS are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is, DU equal to GS, for they are the halves of DE, BG; therefore they will also have the remaining sides equal to the remaining sides. [I. 26] Therefore DT is equal to TG, and UT to TS.'"],["Rule","'ProofWordCount'",321],["Rule","'GreekProof'","'κύβου γὰρ τοῦ ΑΖ τῶν ἀπεναντίον ἐπιπέδων τῶν ΓΖ, ΑΘ αἱ πλευραὶ δίχα τετμήσθωσαν κατὰ τὰ Κ, Λ, Μ, Ν, Ξ, Π, Ο, Ρ σημεῖα, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβεβλήσθω τὰ ΚΝ, ΞΡ, κοινὴ δὲ τομὴ τῶν ἐπιπέδων ἔστω ἡ ΥΣ, τοῦ δὲ ΑΖ κύβου διαγώνιος ἡ ΔΗ. λέγω, ὅτι ἴση ἐστὶν ἡ μὲν ΥΤ τῇ ΤΣ, ἡ δὲ ΔΤ τῇ ΤΗ. ἐπεζεύχθωσαν γὰρ αἱ ΔΥ, ΥΕ, ΒΣ, ΣΗ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΔΞ τῇ ΟΕ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΔΞΥ, ΥΟΕ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΔΞ τῇ ΟΕ, ἡ δὲ ΞΥ τῇ ΥΟ, καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΔΥ τῇ ΥΕ ἐστιν ἴση, καὶ τὸ ΔΞΥ τρίγωνον τῷ ΟΥΕ τριγώνῳ ἐστὶν ἴσον καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι: ἴση ἄρα ἡ ὑπὸ ΞΥΔ γωνία τῇ ὑπὸ ΟΥΕ γωνίᾳ. διὰ δὴ τοῦτο εὐθεῖά ἐστιν ἡ ΔΥΕ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΒΣΗ εὐθεῖά ἐστιν, καὶ ἴση ἡ ΒΣ τῇ ΣΗ. καὶ ἐπεὶ ἡ ΓΑ τῇ ΔΒ ἴση ἐστὶ καὶ παράλληλος, ἀλλὰ ἡ ΓΑ καὶ τῇ ΕΗ ἴση τέ ἐστι καὶ παράλληλος, καὶ ἡ ΔΒ ἄρα τῇ ΕΗ ἴση τέ ἐστι καὶ παράλληλος. καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ΔΕ, ΒΗ: παράλληλος ἄρα ἐστὶν ἡ ΔΕ τῇ ΒΗ. ἴση ἄρα ἡ μὲν ὑπὸ ΕΔΤ γωνία τῇ ὑπὸ ΒΗΤ: ἐναλλὰξ γάρ: ἡ δὲ ὑπὸ ΔΤΥ τῇ ὑπὸ ΗΤΣ. δύο δὴ τρίγωνά ἐστι τὰ ΔΤΥ, ΗΤΣ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν τὴν ΔΥ τῇ ΗΣ: ἡμίσειαι γάρ εἰσι τῶν ΔΕ, ΒΗ: καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει. ἴση ἄρα ἡ μὲν ΔΤ τῇ ΤΗ, ἡ δὲ ΥΤ τῇ ΤΣ. ἐὰν ἄρα κύβου τῶν ἀπεναντίον ἐπιπέδων αἱ πλευραὶ δίχα τμηθῶσιν, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβληθῇ, ἡ κοινὴ τομὴ τῶν ἐπιπέδων καὶ ἡ τοῦ κύβου διάμετρος δίχα τέμνουσιν ἀλλήλας: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",319]]],["Rule",["Association",["Rule","'Book'",11],["Rule","'Theorem'",39]],["Association",["Rule","'VertexLabel'","'11.39'"],["Rule","'Text'","'If there be two prisms of equal height, and one have a parallelogram as base and the other a triangle, and if the parallelogram be double of the triangle, the prisms will be equal.'"],["Rule","'TextWordCount'",34],["Rule","'GreekText'","'ἐὰν ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῃ βάσιν παραλληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ παραλληλόγραμμον τοῦ τριγώνου, ἴσα ἔσται τὰ πρίσματα.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",11],["Rule","'Theorem'",28]],["List",["Rule","'Book'",11],["Rule","'Theorem'",31]]]],["Rule","'Proof'","'Let ABCDEF, GHKLMN be two prisms of equal height, let one have the parallelogram AF as base, and the other the triangle GHK, and let the parallelogram AF be double of the triangle GHK; I say that the prism ABCDEF is equal to the prism GHKLMN. For let the solids AO, GP be completed. Since the parallelogram AF is double of the triangle GHK, while the parallelogram HK is also double of the triangle GHK, [I. 34] therefore the parallelogram AF is equal to the parallelogram HK. But parallelepipedal solids which are on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid AO is equal to the solid GP. And the prism ABCDEF is half of the solid AO, and the prism GHKLMN is half of the solid GP; [XI. 28] therefore the prism ABCDEF is equal to the prism GHKLMN.'"],["Rule","'ProofWordCount'",149],["Rule","'GreekProof'","'ἔστω δύο πρίσματα ἰσοϋψῆ τὰ ΑΒΓΔΕΖ, ΗΘΚΛ ΜΝ, καὶ τὸ μὲν ἐχέτω βάσιν τὸ ΑΖ παραλληλόγραμμον, τὸ δὲ τὸ ΗΘΚ τρίγωνον, διπλάσιον δὲ ἔστω τὸ ΑΖ παραλληλόγραμμον τοῦ ΗΘΚ τριγώνου: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔΕΖ πρίσμα τῷ ΗΘΚΛΜΝ πρίσματι. συμπεπληρώσθω γὰρ τὰ ΑΞ, ΗΟ στερεά. ἐπεὶ διπλάσιόν ἐστι τὸ ΑΖ παραλληλόγραμμον τοῦ ΗΘΚ τριγώνου, ἔστι δὲ καὶ τὸ ΘΚ παραλληλόγραμμον διπλάσιον τοῦ ΗΘΚ τριγώνου, ἴσον ἄρα ἐστὶ τὸ ΑΖ παραλληλόγραμμον τῷ ΘΚ παραλληλογράμμῳ. τὰ δὲ ἐπὶ ἴσων βάσεων ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν: ἴσον ἄρα ἐστὶ τὸ ΑΞ στερεὸν τῷ ΗΟ στερεῷ. καί ἐστι τοῦ μὲν ΑΞ στερεοῦ ἥμισυ τὸ ΑΒΓΔΕΖ πρίσμα, τοῦ δὲ ΗΟ στερεοῦ ἥμισυ τὸ ΗΘΚΛΜΝ πρίσμα: ἴσον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ πρίσμα τῷ ΗΘΚΛΜΝ πρίσματι. ἐὰν ἄρα ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῃ βάσιν παραλληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ παραλληλόγραμμον τοῦ τριγώνου, ἴσα ἐστὶ τὰ πρίσματα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",155]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'12.1'"],["Rule","'Text'","'Similar polygons inscribed in circles are to one another as the squares on the diameters.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'τὰ ἐν τοῖς κύκλοις ὅμοια πολύγωνα πρὸς ἄλληλά ἐστιν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",3],["Rule","'Theorem'",27]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",6]],["List",["Rule","'Book'",6],["Rule","'Theorem'",20]]]],["Rule","'Proof'","'Let ABC, FGH be circles, let ABCDE, FGHKL be similar polygons inscribed in them, and let BM, GN be diameters of the circles; I say that, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL. For let BE, AM, GL, FN be joined. Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and, as BA is to AE, so is GF to FL. [VI. Def. I] Thus BAE, GFL are two triangles which have one angle equal to one angle, namely the angle BAE to the angle GFL, and the sides about the equal angles proportional; therefore the triangle ABE is equiangular with the triangle FGL. [VI. 6] Therefore the angle AEB is equal to the angle FLG. But the angle AEB is equal to the angle AMB, for they stand on the same circumference; [III. 27] and the angle FLG to the angle FNG; therefore the angle AMB is also equal to the angle FNG. But the right angle BAM is also equal to the right angle GFN; [III. 31] therefore the remaining angle is equal to the remaining angle. [I. 32] Therefore the triangle ABM is equiangular with the triangle FGN. Therefore, proportionally, as BM is to GN, so is BA to GF. [VI. 4] But the ratio of the square on BM to the square on GN is duplicate of the ratio of BM to GN, and the ratio of the polygon ABCDE to the polygon FGHKL is duplicate of the ratio of BA to GF; [VI. 20] therefore also, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL.'"],["Rule","'ProofWordCount'",295],["Rule","'GreekProof'","'ἔστωσαν κύκλοι οἱ ΑΒΓ, ΖΗΘ, καὶ ἐν αὐτοῖς ὅμοια πολύγωνα ἔστω τὰ ΑΒΓΔΕ, ΖΗΘΚΛ, διάμετροι δὲ τῶν κύκλων ἔστωσαν αἱ ΒΜ, ΗΝ: λέγω, ὅτι ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΜ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, οὕτως τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΑΜ, ΗΛ, ΖΝ. καὶ ἐπεὶ ὅμοιον τὸ ΑΒΓΔΕ πολύγωνον τῷ ΖΗΘΚΛ πολυγώνῳ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΒΑΕ γωνία τῇ ὑπὸ ΗΖΛ, καί ἐστιν ὡς ἡ ΒΑ πρὸς τὴν ΑΕ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΛ. δύο δὴ τρίγωνά ἐστι τὰ ΒΑΕ, ΗΖΛ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα τὴν ὑπὸ ΒΑΕ τῇ ὑπὸ ΗΖΛ, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΗΛ τριγώνῳ. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ γωνία τῇ ὑπὸ ΖΛΗ. ἀλλ᾽ ἡ μὲν ὑπὸ ΑΕΒ τῇ ὑπὸ ΑΜΒ ἐστιν ἴση: ἐπὶ γὰρ τῆς αὐτῆς περιφερείας βεβήκασιν: ἡ δὲ ὑπὸ ΖΛΗ τῇ ὑπὸ ΖΝΗ: καὶ ἡ ὑπὸ ΑΜΒ ἄρα τῇ ὑπὸ ΖΝΗ ἐστιν ἴση. ἔστι δὲ καὶ ὀρθὴ ἡ ὑπὸ ΒΑΜ ὀρθῇ τῇ ὑπὸ ΗΖΝ ἴση: καὶ ἡ λοιπὴ ἄρα τῇ λοιπῇ ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΜ τρίγωνον τῷ ΖΗΝ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΜ πρὸς τὴν ΗΝ, οὕτως ἡ ΒΑ πρὸς τὴν ΗΖ. ἀλλὰ τοῦ μὲν τῆς ΒΜ πρὸς τὴν ΗΝ λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τῆς ΒΜ τετραγώνου πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, τοῦ δὲ τῆς ΒΑ πρὸς τὴν ΗΖ διπλασίων ἐστὶν ὁ τοῦ ΑΒΓΔΕ πολυγώνου πρὸς τὸ ΖΗ ΘΚΛ πολύγωνον: καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΒΜ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, οὕτως τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. τὰ ἄρα ἐν τοῖς κύκλοις ὅμοια πολύγωνα πρὸς ἄλληλά ἐστιν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",291]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'12.2'"],["Rule","'Text'","'Circles are to one another as the squares on the diameters.'"],["Rule","'TextWordCount'",11],["Rule","'GreekText'","'οἱ κύκλοι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",12],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let ABCD, EFGH be circles, and BD, FH their diameters; I say that, as the circle ABCD is to the circle EFGH, so is the square on BD to the square on FH. For, if the square on BD is not to the square on FH as the circle ABCD is to the circle EFGH, then, as the square on BD is to the square on FH, so will the circle ABCD be either to some less area than the circle EFGH, or to a greater. First, let it be in that ratio to a less area S. Let the square EFGH be inscribed in the circle EFGH; then the inscribed square is greater than the half of the circle EFGH, inasmuch as, if through the points E, F, G, H we draw tangents to the circle, the square EFGH is half the square circumscribed about the circle, and the circle is less than the circumscribed square; hence the inscribed square EFGH is greater than the half of the circle EFGH. Let the circumferences EF, FG, GH, HE be bisected at the points K, L, M, N, and let EK, KF, FL, LG, GM, MH, HN, NE be joined; therefore each of the triangles EKF, FLG, GMH, HNE is also greater than the half of the segment of the circle about it, inasmuch as, if through the points K, L, M, N we draw tangents to the circle and complete the parallelograms on the straight lines EF, FG, GH, HE, each of the triangles EKF, FLG, GMH, HNE will be half of the parallelogram about it, while the segment about it is less than the parallelogram; hence each of the triangles EKF, FLG, GMH, HNE is greater than the half of the segment of the circle about it. Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, we shall leave some segments of the circle which will be less than the excess by which the circle EFGH exceeds the area S. For it was proved in the first theorem of the tenth book that, if two unequal magnitudes be set out, and if from the greater there be subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out. Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, NE be less than the excess by which the circle EFGH exceeds the area S. Therefore the remainder, the polygon EKFLGMHN, is greater than the area S. Let there be inscribed, also, in the circle ABCD the polygon AOBPCQDR similar to the polygon EKFLGMHN; therefore, as the square on BD is to the square on FH, so is the polygon AOBPCQDR to the polygon EKFLGMHN. [XII. 1] But, as the square on BD is to the square on FH, so also is the circle ABCD to the area S; therefore also, as the circle ABCD is to the area S, so is the polygon AOBPCQDR to the polygon EKFLGMHN; [V. 11] therefore, alternately, as the circle ABCD is to the polygon inscribed in it, so is the area S to the polygon EKFLGMHN. [V. 16] But the circle ABCD is greater than the polygon inscribed in it; therefore the area S is also greater than the polygon EKFLGMHN. But it is also less: which is impossible. Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area less than the circle EFGH. Similarly we can prove that neither is the circle EFGH to any area less than the circle ABCD as the square on FH is to the square on BD. I say next that neither is the circle ABCD to any area greater than the circle EFGH as the square on BD is to the square on FH. For, if possible, let it be in that ratio to a greater area S. Therefore, inversely, as the square on FH is to the square on DB, so is the area S to the circle ABCD. But, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD; therefore also, as the square on FH is to the square on BD, so is the circle EFGH to some area less than the circle ABCD: [V. 11] which was proved impossible. Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area greater than the circle EFGH. And it was proved that neither is it in that ratio to any area less than the circle EFGH; therefore, as the square on BD is to the square on FH, so is the circle ABCD to the circle EFGH.'"],["Rule","'ProofWordCount'",833],["Rule","'GreekProof'","'ἔστωσαν κύκλοι οἱ ΑΒΓΔ, ΕΖΗΘ, διάμετροι δὲ αὐτῶν ἔστωσαν αἱ ΒΔ, ΖΘ: λέγω, ὅτι ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ τετράγωνον. εἰ γὰρ μή ἐστιν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ, οὕτως τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, ἔσται ὡς τὸ ἀπὸ τῆς ΒΔ πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος ἤτοι πρὸς ἔλασσόν τι τοῦ ΕΖΗΘ κύκλου χωρίον ἢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ Σ. καὶ ἐγγεγράφθω εἰς τὸν ΕΖΗΘ κύκλον τετράγωνον τὸ ΕΖΗΘ: τὸ δὴ ἐγγεγραμμένον τετράγωνον μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ ΕΖΗΘ κύκλου, ἐπειδήπερ ἐὰν διὰ τῶν ε, Ζ, Η, Θ σημείων ἐφαπτομένας εὐθείας τοῦ κύκλου ἀγάγωμεν, τοῦ περιγραφομένου περὶ τὸν κύκλον τετραγώνου ἥμισύ ἐστι τὸ ΕΖΗΘ τετράγωνον, τοῦ δὲ περιγραφέντος τετραγώνου ἐλάττων ἐστὶν ὁ κύκλος: ὥστε τὸ ΕΖΗΘ ἐγγεγραμμένον τετράγωνον μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ ΕΖΗΘ κύκλου. τετμήσθωσαν δίχα αἱ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι κατὰ τὰ Κ, Λ, Μ, Ν σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΕΚ, ΚΖ, ΖΛ, ΛΗ, ΗΜ, ΜΘ, ΘΝ, ΝΕ: καὶ ἕκαστον ἄρα τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ καθ᾽ ἑαυτὸ τμήματος τοῦ κύκλου, ἐπειδήπερ ἐὰν διὰ τῶν Κ, Λ, Μ, Ν σημείων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν καὶ ἀναπληρώσωμεν τὰ ἐπὶ τῶν ΕΖ, ΖΗ, ΗΘ, ΘΕ εὐθειῶν παραλληλόγραμμα, ἕκαστον τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων ἥμισυ ἔσται τοῦ καθ᾽ ἑαυτὸ παραλληλογράμμου, ἀλλὰ τὸ καθ᾽ ἑαυτὸ τμῆμα ἔλαττόν ἐστι τοῦ παραλληλογράμμου: ὥστε ἕκαστον τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ καθ᾽ ἑαυτὸ τμήματος τοῦ κύκλου. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κύκλου, ἃ ἔσται ἐλάσσονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ ΕΖΗΘ κύκλος τοῦ Σ χωρίου. ἐδείχθη γὰρ ἐν τῷ πρώτῳ θεωρήματι τοῦ δεκάτου βιβλίου, ὅτι δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους. λελείφθω οὖν, καὶ ἔστω τὰ ἐπὶ τῶν ΕΚ, ΚΖ, ΖΛ, ΛΗ, ΗΜ, ΜΘ, ΘΝ, ΝΕ τμήματα τοῦ ΕΖΗΘ κύκλου ἐλάττονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ ΕΖΗΘ κύκλος τοῦ Σ χωρίου. λοιπὸν ἄρα τὸ ΕΚΖΛΗ ΜΘΝ πολύγωνον μεῖζόν ἐστι τοῦ Σ χωρίου. ἐγγεγράφθω καὶ εἰς τὸν ΑΒΓΔ κύκλον τῷ ΕΚΖΛΗΜΘΝ πολυγώνῳ ὅμοιον πολύγωνον τὸ ΑΞΒΟΓΠΔΡ: ἔστιν ἄρα ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ τετράγωνον, οὕτως τὸ ΑΞΒΟΓΠΔΡ πολύγωνον πρὸς τὸ ΕΚΖΛ ΗΜΘΝ πολύγωνον. ἀλλὰ καὶ ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸ Σ χωρίον: καὶ ὡς ἄρα ὁ ΑΒΓΔ κύκλος πρὸς τὸ Σ χωρίον, οὕτως τὸ ΑΞΒΟΓΠΔΡ πολύγωνον πρὸς τὸ ΕΚΖΛΗΜΘΝ πολύγωνον: ἐναλλὰξ ἄρα ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸ ἐν αὐτῷ πολύγωνον, οὕτως τὸ Σ χωρίον πρὸς τὸ ΕΚΖΛΗΜΘΝ πολύγωνον. μείζων δὲ ὁ ΑΒΓΔ κύκλος τοῦ ἐν αὐτῷ πολυγώνου: μεῖζον ἄρα καὶ τὸ Σ χωρίον τοῦ ΕΚΖΛΗΜΘΝ πολυγώνου. ἀλλὰ καὶ ἔλαττον: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς ἔλασσόν τι τοῦ ΕΖΗΘ κύκλου χωρίον. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ὡς τὸ ἀπὸ ΖΘ πρὸς τὸ ἀπὸ ΒΔ, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλασσόν τι τοῦ ΑΒΓΔ κύκλου χωρίον. λέγω δή, ὅτι οὐδὲ ὡς τὸ ἀπὸ τῆς ΒΔ πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς μεῖζόν τι τοῦ ΕΖΗΘ κύκλου χωρίον. εἰ γὰρ δυνατόν, ἔστω πρὸς μεῖζον τὸ Σ. ἀνάπαλιν ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΖΘ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΔΒ, οὕτως τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον. ἀλλ᾽ ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον: καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΖΘ πρὸς τὸ ἀπὸ τῆς ΒΔ, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλασσόν τι τοῦ ΑΒΓΔ κύκλου χωρίον: ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς μεῖζόν τι τοῦ ΕΖΗΘ κύκλου χωρίον. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἔλασσον: ἔστιν ἄρα ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον. οἱ ἄρα κύκλοι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα: ὅπερ ἔδει δεῖξαι. λῆμμα λέγω δή, ὅτι τοῦ Σ χωρίου μείζονος ὄντος τοῦ ΕΖΗΘ κύκλου ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον. γεγονέτω γὰρ ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον. λέγω, ὅτι ἔλαττόν ἐστι τὸ Τ χωρίον τοῦ ΑΒΓΔ κύκλου. ἐπεὶ γάρ ἐστιν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον, ἐναλλάξ ἐστιν ὡς τὸ Σ χωρίον πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸ Τ χωρίον. μεῖζον δὲ τὸ Σ χωρίον τοῦ ΕΖΗΘ κύκλου: μείζων ἄρα καὶ ὁ ΑΒΓΔ κύκλος τοῦ Τ χωρίου. ὥστε ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",859]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'12.3'"],["Rule","'Text'","'Any pyramid which has a triangular base is divided into two pyramids equal and similar to one another, similar to the whole and having triangular bases, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.'"],["Rule","'TextWordCount'",44],["Rule","'GreekText'","'πᾶσα πυραμὶς τρίγωνον ἔχουσα βάσιν διαιρεῖται εἰς δύο πυραμίδας ἴσας τε καὶ ὁμοίας ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ τριγώνους ἐχούσας βάσεις καὶ εἰς δύο πρίσματα ἴσα: καὶ τὰ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος.'"],["Rule","'GreekTextWordCount'",38],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",29]],["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]],["List",["Rule","'Book'",11],["Rule","'Definition'",10]],["List",["Rule","'Book'",11],["Rule","'Theorem'",10]],["List",["Rule","'Book'",11],["Rule","'Theorem'",39]]]],["Rule","'Proof'","'Let there be a pyramid of which the triangle ABC is the base and the point D the vertex; I say that the pyramid ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid. For let AB, BC, CA, AD, DB, DC be bisected at the points E, F, G, H, K, L, and let HE, EG, GH, HK, KL, LH, KF, FG be joined. Since AE is equal to EB, and AH to DH, therefore EH is parallel to DB. [VI. 2] For the same reason HK is also parallel to AB. Therefore HEBK is a parallelogram; therefore HK is equal to EB. [I. 34] But EB is equal to EA; therefore AE is also equal to HK. But AH is also equal to HD; therefore the two sides EA, AH are equal to the two sides KH, HD respectively, and the angle EAH is equal to the angle KHD; therefore the base EH is equal to the base KD. [I. 4] Therefore the triangle AEH is equal and similar to the triangle HKD. For the same reason the triangle AHG is also equal and similar to the triangle HLD. Now, since two straight lines EH, HG meeting one another are parallel to two straight lines KD, DL meeting one another, and are not in the same plane, they will contain equal angles. [XI. 10] Therefore the angle EHG is equal to the angle KDL. And, since the two straight lines EH, HG are equal to the two KD, DL respectively, and the angle EHG is equal to the angle KDL, therefore the base EG is equal to the base KL; [I. 4] therefore the triangle EHG is equal and similar to the triangle KDL. For the same reason the triangle AEG is also equal and similar to the triangle HKL. Therefore the pyramid of which the triangle AEG is the base and the point H the vertex is equal and similar to the pyramid of which the triangle HKL is the base and the point D the vertex. [XI. Def. 10] And, since HK has been drawn parallel to AB, one of the sides of the triangle ADB, the triangle ADB is equiangular to the triangle DHK, [I. 29] and they have their sides proportional; therefore the triangle ADB is similar to the triangle DHK. [VI. Def. 1] For the same reason the triangle DBC is also similar to the triangle DKL, and the triangle ADC to the triangle DLH. Now, since the two straight lines BA, AC meeting one another are parallel to the two straight lines KH, HL meeting one another, not in the same plane, they will contain equal angles. [XI. 10] Therefore the angle BAC is equal to the angle KHL. And, as BA is to AC, so is KH to HL; therefore the triangle ABC is similar to the triangle HKL. Therefore also the pyramid of which the triangle ABC is the base and the point D the vertex is similar to the pyramid of which the triangle HKL is the base and the point D the vertex. But the pyramid of which the triangle HKL is the base and the point D the vertex was proved similar to the pyramid of which the triangle AEG is the base and the point H the vertex. Therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD. Next, since BF is equal to FC, the parallelogram EBFG is double of the triangle GFC. And since, if there be two prisms of equal height, and one have a parallelogram as base, and the other a triangle, and if the parallelogram be double of the triangle, the prisms are equal, [XI. 39] therefore the prism contained by the two triangles BKF, EHG, and the three parallelograms EBFG, EBKH, HKFG is equal to the prism contained by the two triangles GFC, HKL and the three parallelograms KFCL, LCGH, HKFG. And it is manifest that each of the prisms, namely that in which the parallelogram EBFG is the base and the straight line HK is its opposite, and that in which the triangle GFC is the base and the triangle HKL its opposite, is greater than each of the pyramids of which the triangles AEG, HKL are the bases and the points H, D the vertices, inasmuch as, if we join the straight lines EF, EK, the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle EBF is the base and the point K the vertex. But the pyramid of which the triangle EBF is the base and the point K the vertex is equal to the pyramid of which the triangle AEG is the base and the point H the vertex; for they are contained by equal and similar planes. Hence also the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle AEG is the base and the point H the vertex. But the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is equal to the prism in which the triangle GFC is the base and the triangle HKL its opposite, and the pyramid of which the triangle AEG is the base and the point H the vertex is equal to the pyramid of which the triangle HKL is the base and the point D the vertex. Therefore the said two prisms are greater than the said two pyramids of which the triangles AEG, HKL are the bases and the points H, D the vertices. Therefore the whole pyramid, of which the triangle ABC is the base and the point D the vertex, has been divided into two pyramids equal to one another and into two equal prisms, and the two prisms are greater than the half of the whole pyramid.'"],["Rule","'ProofWordCount'",1024],["Rule","'GreekProof'","'ἔστω πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον: λέγω, ὅτι ἡ ΑΒΓΔ πυραμὶς διαιρεῖται εἰς δύο πυραμίδας ἴσας ἀλλήλαις τριγώνους βάσεις ἐχούσας καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα: καὶ τὰ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος. τετμήσθωσαν γὰρ αἱ ΑΒ, ΒΓ, ΓΑ, ΑΔ, ΔΒ, ΔΓ δίχα κατὰ τὰ Ε, Ζ, Η, Θ, Κ, Λ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΘΕ, ΕΗ, ΗΘ, ΘΚ, ΚΛ, ΛΘ, ΚΖ, ΖΗ. ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΒ, ἡ δὲ ΑΘ τῇ ΔΘ, παράλληλος ἄρα ἐστὶν ἡ ΕΘ τῇ ΔΒ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΚ τῇ ΑΒ παράλληλός ἐστιν. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΕ ΒΚ: ἴση ἄρα ἐστὶν ἡ ΘΚ τῇ ΕΒ. ἀλλὰ ἡ ΕΒ τῇ ΕΑ ἐστιν ἴση: καὶ ἡ ΑΕ ἄρα τῇ ΘΚ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΑΘ τῇ ΘΔ ἴση: δύο δὴ αἱ ΕΑ, ΑΘ δυσὶ ταῖς ΚΘ, ΘΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΕΑΘ γωνίᾳ τῇ ὑπὸ ΚΘΔ ἴση: βάσις ἄρα ἡ ΕΘ βάσει τῇ ΚΔ ἐστιν ἴση. ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΑΕΘ τρίγωνον τῷ ΘΚΔ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΘΗ τρίγωνον τῷ ΘΛΔ τριγώνῳ ἴσον τέ ἐστι καὶ ὅμοιον. καὶ ἐπεὶ δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΕΘ, ΘΗ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΚΔ, ΔΛ εἰσιν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι, ἴσας γωνίας περιέξουσιν. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΕΘΗ γωνία τῇ ὑπὸ ΚΔΛ γωνίᾳ. καὶ ἐπεὶ δύο εὐθεῖαι αἱ ΕΘ, ΘΗ δυσὶ ταῖς ΚΔ, ΔΛ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΕΘΗ γωνίᾳ τῇ ὑπὸ ΚΔΛ ἐστιν ἴση, βάσις ἄρα ἡ ΕΗ βάσει τῇ ΚΛ ἐστιν ἴση: ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΕΘΗ τρίγωνον τῷ ΚΔΛ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΕΗ τρίγωνον τῷ ΘΚΛ τριγώνῳ ἴσον τε καὶ ὅμοιόν ἐστιν. ἡ ἄρα πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον, ἴση καὶ ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. καὶ ἐπεὶ τριγώνου τοῦ ΑΔΒ παρὰ μίαν τῶν πλευρῶν τὴν ΑΒ ἦκται ἡ ΘΚ, ἰσογώνιόν ἐστι τὸ ΑΔΒ τρίγωνον τῷ ΔΘΚ τριγώνῳ, καὶ τὰς πλευρὰς ἀνάλογον ἔχουσιν: ὅμοιον ἄρα ἐστὶ τὸ ΑΔΒ τρίγωνον τῷ ΔΘΚ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν ΔΒΓ τρίγωνον τῷ ΔΚΛ τριγώνῳ ὅμοιόν ἐστιν, τὸ δὲ ΑΔΓ τῷ ΔΛΘ. καὶ ἐπεὶ δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΒΑ, ΑΓ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΚΘ, ΘΛ εἰσιν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἴσας γωνίας περιέξουσιν. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΚΘΛ. καί ἐστιν ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΚΘ πρὸς τὴν ΘΛ: ὅμοιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΘΚΛ τριγώνῳ. καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. ἀλλὰ πυραμίς, ἧς βάσις μὲν ἐστι τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ὁμοία ἐδείχθη πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον ὥστε καὶ πυραμίς, ἧς βάσις μὲν τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μὲν τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον. ἑκατέρα ἄρα τῶν ΑΕΗΘ, ΘΚΛΔ πυραμίδων ὁμοία ἐστὶ τῇ ὅλῃ τῇ ΑΒΓΔ πυραμίδι. _ Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΖ τῇ ΖΓ, διπλάσιόν ἐστι τὸ ΕΒΖΗ παραλληλόγραμμον τοῦ ΗΖΓ τριγώνου. καὶ ἐπεί, ἐὰν ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῃ βάσιν παραλληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ παραλληλόγραμμον τοῦ τριγώνου, ἴσα ἐστὶ τὰ πρίσματα, ἴσον ἄρα ἐστὶ τὸ πρίσμα τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων τῶν ΒΚΖ, ΕΘΗ, τριῶν δὲ παραλληλογράμμων τῶν ΕΒΖΗ, ΕΒΚΘ, ΘΚΖΗ τῷ πρίσματι τῷ περιεχομένῳ ὑπὸ δύο μὲν τριγώνων τῶν ΗΖΓ, ΘΚΛ, τριῶν δὲ παραλληλογράμμων τῶν ΚΖΓΛ, ΛΓΗΘ, ΘΚΖΗ. καὶ φανερόν, ὅτι ἑκάτερον τῶν πρισμάτων, οὗ τε βάσις τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, καὶ οὗ βάσις τὸ ΗΖΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΘΚΛ τρίγωνον, μεῖζόν ἐστιν ἑκατέρας τῶν πυραμίδων, ὧν βάσεις μὲν τὰ ΑΕΗ, ΘΚΛ τρίγωνα, κορυφαὶ δὲ τὰ Θ, Δ σημεῖα, ἐπειδήπερ καὶ ἐὰν ἐπιζεύξωμεν τὰς ΕΖ, ΕΚ εὐθείας, τὸ μὲν πρίσμα, οὗ βάσις τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, μεῖζόν ἐστι τῆς πυραμίδος, ἧς βάσις τὸ ΕΒΖ τρίγωνον, κορυφὴ δὲ τὸ Κ σημεῖον. ἀλλ᾽ ἡ πυραμίς, ἧς βάσις τὸ ΕΒΖ τρίγωνον, κορυφὴ δὲ τὸ Κ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον: ὑπὸ γὰρ ἴσων καὶ ὁμοίων ἐπιπέδων περιέχονται. ὥστε καὶ τὸ πρίσμα, οὗ βάσις μὲν τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, μεῖζόν ἐστι πυραμίδος, ἧς βάσις μὲν τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον. ἴσον δὲ τὸ μὲν πρίσμα, οὗ βάσις τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, τῷ πρίσματι, οὗ βάσις μὲν τὸ ΗΖΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΘΚΛ τρίγωνον: ἡ δὲ πυραμίς, ἧς βάσις τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. τὰ ἄρα εἰρημένα δύο πρίσματα μείζονά ἐστι τῶν εἰρημένων δύο πυραμίδων, ὧν βάσεις μὲν τὰ ΑΕΗ, ΘΚΛ τρίγωνα, κορυφαὶ δὲ τὰ Θ, Δ σημεῖα. ἡ ἄρα ὅλη πυραμίς, ἧς βάσις τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, διῄρηται εἴς τε δύο πυραμίδας ἴσας ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα, καὶ τὰ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",900]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'12.4'"],["Rule","'Text'","'If there be two pyramids of the same height which have triangular bases, and cach of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms, then, as the base of the one pyramid is to the base of the other pyramid, so will all the prisms in the one pyramid be to all the prisms, being equal in multitude, in the other pyramid.'"],["Rule","'TextWordCount'",74],["Rule","'GreekText'","'ἐὰν ὦσι δύο πυραμίδες ὑπὸ τὸ αὐτὸ ὕψος τριγώνους ἔχουσαι βάσεις, διαιρεθῇ δὲ ἑκατέρα αὐτῶν εἴς τε δύο πυραμίδας ἴσας ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα, ἔσται ὡς ἡ τῆς μιᾶς πυραμίδος βάσις πρὸς τὴν τῆς ἑτέρας πυραμίδος βάσιν, οὕτως τὰ ἐν τῇ μιᾷ πυραμίδι πρίσματα πάντα πρὸς τὰ ἐν τῇ ἑτέρᾳ πυραμίδι πρίσματα πάντα ἰσοπληθῆ.'"],["Rule","'GreekTextWordCount'",60],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",12]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",6],["Rule","'Theorem'",22]],["List",["Rule","'Book'",11],["Rule","'Theorem'",39]],["List",["Rule","'Book'",12],["Rule","'Theorem'",3]]]],["Rule","'Proof'","'Let there be two pyramids of the same height which have the triangular bases ABC, DEF, and vertices the points G, H, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; [XII. 3] I say that, as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH, For, since BO is equal to OC, and AL to LC, therefore LO is parallel to AB, and the triangle ABC is similar to the triangle LOC. For the same reason the triangle DEF is also similar to the triangle RVF. And, since BC is double of CO, and EF of FV, therefore, as BC is to CO, so is EF to FV. And on BC, CO are described the similar and similarly situated rectilineal figures ABC, LOC, and on EF, FV the similar and similarly situated figures DEF, RVF; therefore, as the triangle ABC is to the triangle LOC, so is the triangle DEF to the triangle RVF; [VI. 22] therefore, alternately, as the triangle ABC is to the triangle DEF, so is the triangle LOC to the triangle RVF. [V. 16] But, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite; [Lemma following]therefore also, as the triangle ABC is to the triangle DEF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite. But, as the said prisms are to one another, so is the prism in which the parallelogram KBOL is the base and the straight line PM its opposite to the prism in which the parallelogram QEVR is the base and the straight line ST its opposite. [XI. 39; cf. XII. 3] Therefore also the two prisms, that in which the parallelogram KBOL is the base and PM its opposite, and that in which the triangle LOC is the base and PMN its opposite, are to the prisms in which QEVR is the base and the straight line ST its opposite and in which the triangle RVF is the base and STU its opposite in the same ratio [V. 12] Therefore also, as the base ABC is to the base DEF, so are the said two prisms to the said two prisms. And similarly, if the pyramids PMNG, STUH be divided into two prisms and two pyramids, as the base PMN is to the base STU, so will the two prisms in the pyramid PMNG be to the two prisms in the pyramid STUH. But, as the base PMN is to the base STU, so is the base ABC to the base DEF; for the triangles PMN, STU are equal to the triangles LOC, RVF respectively. Therefore also, as the base ABC is to the base DEF, so are the four prisms to the four prisms. And similarly also, if we divide the remaining pyramids into two pyramids and into two prisms, then, as the base ABC is to the base DEF, so will all the prisms in the pyramid ABCG be to all the prisms, being equal in multitude, in the pyramid DEFH.'"],["Rule","'ProofWordCount'",582],["Rule","'GreekProof'","'ἔστωσαν δύο πυραμίδες ὑπὸ τὸ αὐτὸ ὕψος τριγώνους ἔχουσαι βάσεις τὰς ΑΒΓ, ΔΕΖ, κορυφὰς δὲ τὰ Η, Θ σημεῖα, καὶ διῃρήσθω ἑκατέρα αὐτῶν εἴς τε δύο πυραμίδας ἴσας ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα: λέγω, ὅτι ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ ἐν τῇ ΑΒΓΗ πυραμίδι πρίσματα πάντα πρὸς τὰ ἐν τῇ ΔΕΖΘ πυραμίδι πρίσματα ἰσοπληθῆ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ μὲν ΒΞ τῇ ΞΓ, ἡ δὲ ΑΛ τῇ ΛΓ, παράλληλος ἄρα ἐστὶν ἡ ΛΞ τῇ ΑΒ καὶ ὅμοιον τὸ ΑΒΓ τρίγωνον τῷ ΛΞΓ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΔΕΖ τρίγωνον τῷ ΡΦΖ τριγώνῳ ὅμοιόν ἐστιν. καὶ ἐπεὶ διπλασίων ἐστὶν ἡ μὲν ΒΓ τῆς ΓΞ, ἡ δὲ ΕΖ τῆς ΖΦ, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΞ, οὕτως ἡ ΕΖ πρὸς τὴν ΖΦ. καὶ ἀναγέγραπται ἀπὸ μὲν τῶν ΒΓ, ΓΞ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΑΒΓ, ΛΞΓ, ἀπὸ δὲ τῶν ΕΖ, ΖΦ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΔΕΖ, ΡΦΖ. ἔστιν ἄρα ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΛΞΓ τρίγωνον, οὕτως τὸ ΔΕΖ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον: ἐναλλὰξ ἄρα ἐστὶν ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον, οὕτως τὸ ΛΞΓ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον. ἀλλ᾽ ὡς τὸ ΛΞΓ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον, οὕτως τὸ πρίσμα, οὗ βάσις μὲν ἐστι τὸ ΛΞΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΡΦΖ τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ: καὶ ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον, οὕτως τὸ πρίσμα, οὗ βάσις μὲν τὸ ΛΞΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΡΦΖ τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ. ὡς δὲ τὰ εἰρημένα πρίσματα πρὸς ἄλληλα, οὕτως τὸ πρίσμα, οὗ βάσις μὲν τὸ ΚΒΞΛ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΟΜ εὐθεῖα, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΠΕΦΡ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΣΤ εὐθεῖα. καὶ τὰ δύο ἄρα πρίσματα, οὗ τε βάσις μὲν τὸ ΚΒΞΛ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΟΜ, καὶ οὗ βάσις μὲν τὸ ΛΞΓ, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὰ πρίσματα, οὗ τε βάσις μὲν τὸ ΠΕΦΡ, ἀπεναντίον δὲ ἡ ΣΤ εὐθεῖα, καὶ οὗ βάσις μὲν τὸ ΡΦΖ τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ. καὶ ὡς ἄρα ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ εἰρημένα δύο πρίσματα πρὸς τὰ εἰρημένα δύο πρίσματα. καὶ ὁμοίως, ἐὰν διαιρεθῶσιν αἱ ΟΜΝΗ, ΣΤΥΘ πυραμίδες εἴς τε δύο πρίσματα καὶ δύο πυραμίδας, ἔσται ὡς ἡ ΟΜΝ βάσις πρὸς τὴν ΣΤΥ βάσιν, οὕτως τὰ ἐν τῇ ΟΜ ΝΗ πυραμίδι δύο πρίσματα πρὸς τὰ ἐν τῇ ΣΤΥΘ πυραμίδι δύο πρίσματα. ἀλλ᾽ ὡς ἡ ΟΜΝ βάσις πρὸς τὴν ΣΤΥ βάσιν, οὕτως ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν: ἴσον γὰρ ἑκάτερον τῶν ΟΜΝ, ΣΤΥ τριγώνων ἑκατέρῳ τῶν ΛΞΓ, ΡΦΖ. καὶ ὡς ἄρα ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ τέσσαρα πρίσματα πρὸς τὰ τέσσαρα πρίσματα. ὁμοίως δὲ κἂν τὰς ὑπολειπομένας πυραμίδας διέλωμεν εἴς τε δύο πυραμίδας καὶ εἰς δύο πρίσματα, ἔσται ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ ἐν τῇ ΑΒ ΓΗ πυραμίδι πρίσματα πάντα πρὸς τὰ ἐν τῇ ΔΕΖΘ πυραμίδι πρίσματα πάντα ἰσοπληθῆ: ὅπερ ἔδει δεῖξαι. λῆμμα ὅτι δέ ἐστιν ὡς τὸ ΛΞΓ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον, οὕτως τὸ πρίσμα, οὗ βάσις τὸ ΛΞΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΡΦΖ τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ, οὕτω δεικτέον. ἐπὶ γὰρ τῆς αὐτῆς καταγραφῆς νενοήσθωσαν ἀπὸ τῶν Η, Θ κάθετοι ἐπὶ τὰ ΑΒΓ, ΔΕΖ ἐπίπεδα, ἴσαι δηλαδὴ τυγχάνουσαι διὰ τὸ ἰσοϋψεῖς ὑποκεῖσθαι τὰς πυραμίδας. καὶ ἐπεὶ δύο εὐθεῖαι ἥ τε ΗΓ καὶ ἡ ἀπὸ τοῦ Η κάθετος ὑπὸ παραλλήλων ἐπιπέδων τῶν ΑΒΓ, ΟΜΝ τέμνονται, εἰς τοὺς αὐτοὺς λόγους τμηθήσονται. καὶ τέτμηται ἡ ΗΓ δίχα ὑπὸ τοῦ ΟΜΝ ἐπιπέδου κατὰ τὸ Ν: καὶ ἡ ἀπὸ τοῦ Η ἄρα κάθετος ἐπὶ τὸ ΑΒΓ ἐπίπεδον δίχα τμηθήσεται ὑπὸ τοῦ ΟΜΝ ἐπιπέδου. διὰ τὰ αὐτὰ δὴ καὶ ἡ ἀπὸ τοῦ Θ κάθετος ἐπὶ τὸ ΔΕΖ ἐπίπεδον δίχα τμηθήσεται ὑπὸ τοῦ ΣΤΥ ἐπιπέδου. καί εἰσιν ἴσαι αἱ ἀπὸ τῶν Η, Θ κάθετοι ἐπὶ τὰ ΑΒΓ, ΔΕΖ ἐπίπεδα: ἴσαι ἄρα καὶ αἱ ἀπὸ τῶν ΟΜΝ, ΣΤΥ τριγώνων ἐπὶ τὰ ΑΒΓ, ΔΕΖ κάθετοι. ἰσοϋψῆ ἄρα ἐστὶ τὰ πρίσματα, ὧν βάσεις μέν εἰσι τὰ ΛΞΓ, ΡΦΖ τρίγωνα, ἀπεναντίον δὲ τὰ ΟΜΝ, ΣΤΥ. ὥστε καὶ τὰ στερεὰ παραλληλεπίπεδα τὰ ἀπὸ τῶν εἰρημένων πρισμάτων ἀναγραφόμενα ἰσοϋψῆ καὶ πρὸς ἄλληλα εἰσὶν ὡς αἱ βάσεις: καὶ τὰ ἡμίση ἄρα ἐστὶν ὡς ἡ ΛΞΓ βάσις πρὸς τὴν ΡΦΖ βάσιν, οὕτως τὰ εἰρημένα πρίσματα πρὸς ἄλληλα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",739]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'12.5'"],["Rule","'Text'","'Pyramids which are of the same height and have triangular bases are to one another as the bases.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'αἱ ὑπὸ τὸ αὐτὸ ὕψος οὖσαι πυραμίδες καὶ τριγώνους ἔχουσαι βάσεις πρὸς ἀλλήλας εἰσὶν ὡς αἱ βάσεις.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",2]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",1]],["List",["Rule","'Book'",12],["Rule","'Theorem'",2]],["List",["Rule","'Book'",12],["Rule","'Theorem'",3]],["List",["Rule","'Book'",12],["Rule","'Theorem'",4]]]],["Rule","'Proof'","'Let there be pyramids of the same height, of which the triangles ABC, DEF are the bases and the points G, H the vertices; I say that, as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH. For, if the pyramid ABCG is not to the pyramid DEFH as the base ABC is to the base DEF, then, as the base ABC is to the base DEF, so will the pyramid ABCG be either to some solid less than the pyramid DEFH or to a greater. Let it, first, be in that ratio to a less solid W, and let the pyramid DEFH be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; then the two prisms are greater than the half of the whole pyramid. [XII. 3] Again, let the pyramids arising from the division be similarly divided, and let this be done continually until there are left over from the pyramid DEFH some pyramids which are less than the excess by which the pyramid DEFH exceeds the solid W. [X. I] Let such be left, and let them be, for the sake of argument, DQRS, STUH; therefore the remainders, the prisms in the pyramid DEFH, are greater than the solid W. Let the pyramid ABCG also be divided similarly, and a similar number of times, with the pyramid DEFH; therefore, as the base ABC is to the base DEF, so are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH. [XII. 4] But, as the base ABC is to the base DEF, so also is the pyramid ABCG to the solid W; therefore also, as the pyramid ABCG is to the solid W, so are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH; [V. II]therefore, alternately, as the pyramid ABCG is to the prisms in it, so is the solid W to the prisms in the pyramid DEFH. [V. 16] But the pyramid ABCG is greater than the prisms in it; therefore the solid W is also greater than the prisms in the pyramid DEFH. But it is also less: which is impossible. Therefore the prism ABCG is not to any solid less than the pyramid DEFH as the base ABC is to the base DEF. Similarly it can be proved that neither is the pyramid DEFH to any solid less than the pyramid ABCG as the base DEF is to the base ABC. I say next that neither is the pyramid ABCG to any solid greater than the pyramid DEFH as the base ABC is to the base DEF. For, if possible, let it be in that ratio to a greater solid W; therefore, inversely, as the base DEF is to the base ABC, so is the solid W to the pyramid ABCG. But, as the solid W is to the solid ABCG, so is the pyramid DEFH to some solid less than the pyramid ABCG, as was before proved; [XII. 2, Lemma] therefore also, as the base DEF is to the base ABC, so is the pyramid DEFH to some solid less than the pyramid ABCG: [V. II]which was proved absurd. Therefore the pyramid ABCG is not to any solid greater than the pyramid DEFH as the base ABC is to the base DEF. But it was proved that neither is it in that ratio to a less solid. Therefore, as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH.'"],["Rule","'ProofWordCount'",600],["Rule","'GreekProof'","'ἔστωσαν ὑπὸ τὸ αὐτὸ ὕψος πυραμίδες, ὧν βάσεις μὲν τὰ ΑΒΓ, ΔΕΖ τρίγωνα, κορυφαὶ δὲ τὰ Η, Θ σημεῖα: λέγω, ὅτι ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα. εἰ γὰρ μή ἐστιν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα, ἔσται ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς ἤτοι πρὸς ἔλασσόν τι τῆς ΔΕΖΘ πυραμίδος στερεὸν ἢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ Χ, καὶ διῃρήσθω ἡ ΔΕΖΘ πυραμὶς εἴς τε δύο πυραμίδας ἴσας ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα: τὰ δὴ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος. καὶ πάλιν αἱ ἐκ τῆς διαιρέσεως γινόμεναι πυραμίδες ὁμοίως διῃρήσθωσαν, καὶ τοῦτο ἀεὶ γινέσθω, ἕως οὗ λειφθῶσί τινες πυραμίδες ἀπὸ τῆς ΔΕΖΘ πυραμίδος, αἵ εἰσιν ἐλάττονες τῆς ὑπεροχῆς, ᾗ ὑπερέχει ἡ ΔΕ ΖΘ πυραμὶς τοῦ Χ στερεοῦ. λελείφθωσαν καὶ ἔστωσαν λόγου ἕνεκεν αἱ ΔΠΡΣ, ΣΤΥΘ: λοιπὰ ἄρα τὰ ἐν τῇ ΔΕ ΖΘ πυραμίδι πρίσματα μείζονά ἐστι τοῦ Χ στερεοῦ. διῃρήσθω καὶ ἡ ΑΒΓΗ πυραμὶς ὁμοίως καὶ ἰσοπληθῶς τῇ ΔΕΖΘ πυραμίδι: ἔστιν ἄρα ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ ἐν τῇ ΑΒΓΗ πυραμίδι πρίσματα πρὸς τὰ ἐν τῇ ΔΕΖΘ πυραμίδι πρίσματα. ἀλλὰ καὶ ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς τὸ Χ στερεόν: καὶ ὡς ἄρα ἡ ΑΒΓΗ πυραμὶς πρὸς τὸ Χ στερεόν, οὕτως τὰ ἐν τῇ ΑΒΓΗ πυραμίδι πρίσματα πρὸς τὰ ἐν τῇ ΔΕΖΘ πυραμίδι πρίσματα: ἐναλλὰξ ἄρα ὡς ἡ ΑΒΓΗ πυραμὶς πρὸς τὰ ἐν αὐτῇ πρίσματα, οὕτως τὸ Χ στερεὸν πρὸς τὰ ἐν τῇ ΔΕΖΘ πυραμίδι πρίσματα. μείζων δὲ ἡ ΑΒΓΗ πυραμὶς τῶν ἐν αὐτῇ πρισμάτων: μεῖζον ἄρα καὶ τὸ Χ στερεὸν τῶν ἐν τῇ ΔΕΖΘ πυραμίδι πρισμάτων. ἀλλὰ καὶ ἔλαττον: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς ἔλασσόν τι τῆς ΔΕΖΘ πυραμίδος στερεόν. ὁμοίως δὴ δειχθήσεται, ὅτι οὐδὲ ὡς ἡ ΔΕΖ βάσις πρὸς τὴν ΑΒΓ βάσιν, οὕτως ἡ ΔΕΖΘ πυραμὶς πρὸς ἔλαττόν τι τῆς ΑΒΓΗ πυραμίδος στερεόν. λέγω δή, ὅτι οὐκ ἔστιν οὐδὲ ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς μεῖζόν τι τῆς ΔΕΖΘ πυραμίδος στερεόν. εἰ γὰρ δυνατόν, ἔστω πρὸς μεῖζον τὸ Χ: ἀνάπαλιν ἄρα ἐστὶν ὡς ἡ ΔΕΖ βάσις πρὸς τὴν ΑΒΓ βάσιν, οὕτως τὸ Χ στερεὸν πρὸς τὴν ΑΒΓΗ πυραμίδα. ὡς δὲ τὸ Χ στερεὸν πρὸς τὴν ΑΒΓΗ πυραμίδα, οὕτως ἡ ΔΕΖΘ πυραμὶς πρὸς ἔλασσόν τι τῆς ΑΒΓΗ πυραμίδος, ὡς ἔμπροσθεν ἐδείχθη: καὶ ὡς ἄρα ἡ ΔΕΖ βάσις πρὸς τὴν ΑΒΓ βάσιν, οὕτως ἡ ΔΕΖΘ πυραμὶς πρὸς ἔλασσόν τι τῆς ΑΒΓΗ πυραμίδος: ὅπερ ἄτοπον ἐδείχθη. οὐκ ἄρα ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς μεῖζόν τι τῆς ΔΕΖΘ πυραμίδος στερεόν. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἔλασσον. ἔστιν ἄρα ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",497]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'12.6'"],["Rule","'Text'","'Pyramids which are of the same height and have polygonal bases are to one another as the bases.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'αἱ ὑπὸ τὸ αὐτὸ ὕψος οὖσαι πυραμίδες καὶ πολυγώνους ἔχουσαι βάσεις πρὸς ἀλλήλας εἰσὶν ὡς αἱ βάσεις.'"],["Rule","'GreekTextWordCount'",17],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",18]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",12],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'Let there be pyramids of the same height of which the polygons ABCDE, FGHKL are the bases and the points M, N the vertices; I say that, as the base ABCDE is to the base FGHKL, so is the pyramid ABCDEM to the pyramid FGHKLN. For let AC, AD, FH, FK be joined. Since then ABCM, ACDM are two pyramids which have triangular bases and equal height, they are to one another as the bases; [XII. 5] therefore, as the base ABC is to the base ACD, so is the pyramid ABCM to the pyramid ACDM. And, componendo, as the base ABCD is to the base ACD, so is the pyramid ABCDM to the pyramid ACDM. [V. 18] But also, as the base ACD is to the base ADE, so is the pyramid ACDM to the pyramid ADEM. [XII. 5] Therefore, ex aequali, as the base ABCD is to the base ADE, so is the pyramid ABCDM to the pyramid ADEM. [V. 22] And again componendo, as the base ABCDE is to the base ADE, so is the pyramid ABCDEM to the pyramid ADEM. [V. 18] Similarly also it can be proved that, as the base FGHKL is to the base FGH, so is the pyramid FGHKLN to the pyramid FGHN. And, since ADEM, FGHN are two pyramids which have triangular bases and equal height, therefore, as the base ADE is to the base FGH, so is the pyramid ADEM to the pyramid FGHN. [XII. 5] But, as the base ADE is to the base ABCDE, so was the pyramid ADEM to the pyramid ABCDEM. Therefore also, ex aequali, as the base ABCDE is to the base FGH, so is the pyramid ABCDEM to the pyramid FGHN. [V. 22] But further, as the base FGH is to the base FGHKL, so also was the pyramid FGHN to the pyramid FGHKLN. Therefore also, ex aequali, as the base ABCDE is to the base FGHKL, so is the pyramid ABCDEM to the pyramid FGHKLN. [V. 22]'"],["Rule","'ProofWordCount'",334],["Rule","'GreekProof'","'ἔστωσαν ὑπὸ τὸ αὐτὸ ὕψος πυραμίδες, ὧν αἱ βάσεις μὲν τὰ ΑΒΓΔΕ, ΖΗΘΚΛ πολύγωνα, κορυφαὶ δὲ τὰ Μ, ν σημεῖα: λέγω, ὅτι ἐστὶν ὡς ἡ ΑΒΓΔΕ βάσις πρὸς τὴν ΖΗΘΚΛ βάσιν, οὕτως ἡ ΑΒΓΔΕΜ πυραμὶς πρὸς τὴν ΖΗΘΚΛΝ πυραμίδα. ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΑΔ, ΖΘ, ΖΚ. ἐπεὶ οὖν δύο πυραμίδες εἰσὶν αἱ ΑΒΓΜ, ΑΓΔΜ τριγώνους ἔχουσαι βάσεις καὶ ὕψος ἴσον, πρὸς ἀλλήλας εἰσὶν ὡς αἱ βάσεις: ἔστιν ἄρα ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΑΓΔ βάσιν, οὕτως ἡ ΑΒΓΜ πυραμὶς πρὸς τὴν ΑΓΔΜ πυραμίδα. καὶ συνθέντι ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΑΓΔ βάσιν, οὕτως ἡ ΑΒΓΔΜ πυραμὶς πρὸς τὴν ΑΓΔΜ πυραμίδα. ἀλλὰ καὶ ὡς ἡ ΑΓΔ βάσις πρὸς τὴν ΑΔΕ βάσιν, οὕτως ἡ ΑΓΔΜ πυραμὶς πρὸς τὴν ΑΔΕΜ πυραμίδα. δι᾽ ἴσου ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΑΔΕ βάσιν, οὕτως ἡ ΑΒΓΔΜ πυραμὶς πρὸς τὴν ΑΔΕΜ πυραμίδα. καὶ συνθέντι πάλιν, ὡς ἡ ΑΒΓΔΕ βάσις πρὸς τὴν ΑΔΕ βάσιν, οὕτως ἡ ΑΒΓΔΕΜ πυραμὶς πρὸς τὴν ΑΔΕΜ πυραμίδα. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ὡς ἡ ΖΗΘΚΛ βάσις πρὸς τὴν ΖΗΘ βάσιν, οὕτως καὶ ἡ ΖΗΘΚΛΝ πυραμὶς πρὸς τὴν ΖΗΘΝ πυραμίδα. καὶ ἐπεὶ δύο πυραμίδες εἰσὶν αἱ ΑΔ ΕΜ, ΖΗΘΝ τριγώνους ἔχουσαι βάσεις καὶ ὕψος ἴσον, ἔστιν ἄρα ὡς ἡ ΑΔΕ βάσις πρὸς τὴν ΖΗΘ βάσιν, οὕτως ἡ ΑΔΕΜ πυραμὶς πρὸς τὴν ΖΗΘΝ πυραμίδα. ἀλλ᾽ ὡς ἡ ΑΔΕ βάσις πρὸς τὴν ΑΒΓΔΕ βάσιν, οὕτως ἦν ἡ ΑΔ ΕΜ πυραμὶς πρὸς τὴν ΑΒΓΔΕΜ πυραμίδα. καὶ δι᾽ ἴσου ἄρα ὡς ἡ ΑΒΓΔΕ βάσις πρὸς τὴν ΖΗΘ βάσιν, οὕτως ἡ ΑΒΓΔΕΜ πυραμὶς πρὸς τὴν ΖΗΘΝ πυραμίδα. ἀλλὰ μὴν καὶ ὡς ἡ ΖΗΘ βάσις πρὸς τὴν ΖΗΘΚΛ βάσιν, οὕτως ἦν καὶ ἡ ΖΗΘΝ πυραμὶς πρὸς τὴν ΖΗΘΚΛΝ πυραμίδα. καὶ δι᾽ ἴσου ἄρα ὡς ἡ ΑΒΓΔΕ βάσις πρὸς τὴν ΖΗΘΚΛ βάσιν, οὕτως ἡ ΑΒΓΔΕΜ πυραμὶς πρὸς τὴν ΖΗΘΚΛΝ πυραμίδα: ὅπερ ἔδει δεῖξαι. πᾶν πρίσμα τρίγωνον ἔχον βάσιν διαιρεῖται εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις τριγώνους βάσεις ἐχούσας. ἔστω πρίσμα, οὗ βάσις μὲν τὸ ΑΒΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΔΕΖ: λέγω, ὅτι τὸ ΑΒΓΔΕΖ πρίσμα διαιρεῖται εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις τριγώνους ἐχούσας βάσεις. ἐπεζεύχθωσαν γὰρ αἱ ΒΔ, ΕΓ, ΓΔ. ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΑΒΕΔ, διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΒΔ, ἴσον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΕΒΔ τριγώνῳ: καὶ ἡ πυραμὶς ἄρα, ἧς βάσις μὲν τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΔΕΒ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον. ἀλλὰ ἡ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΔΕΒ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἡ αὐτή ἐστι πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον: ὑπὸ γὰρ τῶν αὐτῶν ἐπιπέδων περιέχεται. καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. πάλιν, ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΖΓΒΕ, διάμετρος δέ ἐστιν αὐτοῦ ἡ ΓΕ, ἴσον ἐστὶ τὸ ΓΕΖ τρίγωνον τῷ ΓΒΕ τριγώνῳ. καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΒΓΕ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΓΖ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. ἡ δὲ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΒΓΕ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ἴση ἐδείχθη πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον: καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΓΕΖ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μὲν ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον: διῄρηται ἄρα τὸ ΑΒΓΔΕΖ πρίσμα εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις τριγώνους ἐχούσας βάσεις. καὶ ἐπεὶ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἡ αὐτή ἐστι πυραμίδι, ἧς βάσις τὸ ΓΑΒ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον: ὑπὸ γὰρ τῶν αὐτῶν ἐπιπέδων περιέχονται: ἡ δὲ πυραμίς, ἧς βάσις τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, τρίτον ἐδείχθη τοῦ πρίσματος, οὗ βάσις τὸ ΑΒΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΔΕΖ, καὶ ἡ πυραμὶς ἄρα, ἧς βάσις τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, τρίτον ἐστὶ τοῦ πρίσματος τοῦ ἔχοντος βάσιν τὴν αὐτὴν τὸ ΑΒΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΔΕΖ. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι πᾶσα πυραμὶς τρίτον μέρος ἐστὶ τοῦ πρίσματος τοῦ τὴν αὐτὴν βάσιν ἔχοντος αὐτῇ καὶ ὕψος ἴσον ἐπειδήπερ κἂν ἕτερόν τι σχῆμα εὐθύγραμμον ἔχῃ ἡ βάσις τοῦ πρίσματος, τοιοῦτο καὶ τὸ ἀπεναντίον, καὶ διαιρεῖται εἰς πρίσματα τρίγωνα ἔχοντα τὰς βάσεις καὶ τὰ ἀπεναντίον, καὶ ὡς ἡ ὅλη βάσις πρὸς ἕκαστον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",735]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'12.7'"],["Rule","'Text'","'Any prism which has a triangular base is divided into three pyramids equal to one another which have triangular bases.'"],["Rule","'TextWordCount'",20],["Rule","'GreekText'","'πᾶν πρίσμα τρίγωνον ἔχον βάσιν διαιρεῖται εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις τριγώνους βάσεις ἐχούσας.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",12],["Rule","'Theorem'",5]]]],["Rule","'Proof'","'Let there be a prism in which the triangle ABC is the base and DEF its opposite; I say that the prism ABCDEF is divided into three pyramids equal to one another, which have triangular bases. For let BD, EC, CD be joined. Since ABED is a parallelogram, and BD is its diameter, therefore the triangle ABD is equal to the triangle EBD; [I. 34] therefore also the pyramid of which the triangle ABD is the base and the point C the vertex is equal to the pyramid of which the triangle DEB is the base and the point C the vertex. [XII. 5] But the pyramid of which the triangle DEB is the base and the point C the vertex is the same with the pyramid of which the triangle EBC is the base and the point D the vertex; for they are contained by the same planes. Therefore the pyramid of which the triangle ABD is the base and the point C the vertex is also equal to the pyramid of which the triangle EBC is the base and the point D the vertex. Again, since FCBE is a parallelogram, and CE is its diameter, the triangle CEF is equal to the triangle CBE. [I. 34] Therefore also the pyramid of which the triangle BCE is the base and the point D the vertex is equal to the pyramid of which the triangle ECF is the base and the point D the vertex. [XII. 5] But the pyramid of which the triangle BCE is the base and the point D the vertex was proved equal to the pyramid of which the triangle ABD is the base and the point C the vertex; therefore also the pyramid of which the triangle CEF is the base and the point D the vertex is equal to the pyramid of which the triangle ABD is the base and the point C the vertex; therefore the prism ABCDEF has been divided into three pyramids equal to one another which have triangular bases. And, since the pyramid of which the triangle ABD is the base and the point C the vertex is the same with the pyramid of which the triangle CAB is the base and the point D the vertex, for they are contained by the same planes, while the pyramid of which the triangle ABD is the base and the point C the vertex was proved to be a third of the prism in which the triangle ABC is the base and DEF its opposite, therefore also the pyramid of which the triangle ABC is the base and the point D the vertex is a third of the prism which has the same base, the triangle ABC, and DEF as its opposite.'"],["Rule","'ProofWordCount'",460],["Rule","'GreekProof'","'Ἔστω πρίσμα, οὗ βάσις μὲν τὸ ΑΒΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΔΕΖ:  λέγω, ὅτι τὸ ΑΒΓΔΕΖ πρίσμα διαιρεῖται εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις τριγώνους ἐχούσας βάσεις. Ἐπεζεύχθωσαν γὰρ αἱ ΒΔ, ΕΓ, ΓΔ. ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΑΒΕΔ, διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΒΔ, ἴσον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΕΒΔ τριγώνῳ: καὶ ἡ πυραμὶς ἄρα, ἧς βάσις μὲν τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΔΕΒ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον.  ἀλλὰ ἡ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΔΕΒ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἡ αὐτή ἐστι πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον: ὑπὸ γὰρ τῶν αὐτῶν ἐπιπέδων περιέχεται.  καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. πάλιν, ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΖΓΒΕ, διάμετρος δέ ἐστιν αὐτοῦ ἡ ΓΕ, ἴσον ἐστὶ τὸ ΓΕΖ τρίγωνον τῷ ΓΒΕ τριγώνῳ.  καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΒΓΕ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΓΖ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. ἡ δὲ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΒΓΕ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ἴση ἐδείχθη πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον: καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΓΕΖ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μὲν [ἐστι] τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον: διῄρηται ἄρα τὸ ΑΒΓΔΕΖ πρίσμα εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις τριγώνους ἐχούσας βάσεις. Καὶ ἐπεὶ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἡ αὐτή ἐστι πυραμίδι, ἧς βάσις τὸ ΓΑΒ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον: ὑπὸ γὰρ τῶν αὐτῶν ἐπιπέδων περιέχονται: ἡ δὲ πυραμίς, ἧς βάσις τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, τρίτον ἐδείχθη τοῦ πρίσματος, οὗ βάσις τὸ ΑΒΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΔΕΖ,  καὶ ἡ πυραμὶς ἄρα, ἧς βάσις τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, τρίτον ἐστὶ τοῦ πρίσματος τοῦ ἔχοντος βάσιν τὴν αὐτὴν τὸ ΑΒΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΔΕΖ.'"],["Rule","'GreekProofWordCount'",363]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'12.8'"],["Rule","'Text'","'Similar pyramids which have triangular bases are in the triplicate ratio of their corresponding sides.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'αἱ ὅμοιαι πυραμίδες καὶ τριγώνους ἔχουσαι βάσεις ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",11],["Rule","'Theorem'",24]],["List",["Rule","'Book'",11],["Rule","'Theorem'",28]],["List",["Rule","'Book'",11],["Rule","'Theorem'",33]],["List",["Rule","'Book'",12],["Rule","'Theorem'",7]]]],["Rule","'Proof'","'Let there be similar and similarly situated pyramids of which the triangles ABC, DEF, are the bases and the points G, H the vertices; I say that the pyramid ABCG has to the pyramid DEFH the ratio triplicate of that which BC has to EF. For let the parallelepipedal solids BGML, EHQP be completed. Now, since the pyramid ABCG is similar to the pyramid DEFH, therefore the angle ABC is equal to the angle DEF, the angle GBC to the angle HEF, and the angle ABG to the angle DEH; and, as AB is to DE, so is BC to EF, and BG to EH. And since, as AB is to DE, so is BC to EF, and the sides are proportional about equal angles, therefore the parallelogram BM is similar to the parallelogram EQ. For the same reason BN is also similar to ER, and BK to EO; therefore the three parallelograms MB, BK, BN are similar to the three EQ, EO, ER. But the three parallelograms MB, BK, BN are equal and similar to their three opposites, and the three EQ, EO, ER are equal and similar to their three opposites. [XI. 24] Therefore the solids BGML, EHQP are contained by similar planes equal in multitude. Therefore the solid BGML is similar to the solid EHQP. But similar parallelepipedal solids are in the triplicate ratio of their corresponding sides. [XI. 33] Therefore the solid BGML has to the solid EHQP the ratio triplicate of that which the corresponding side BC has to the corresponding side EF. But, as the solid BGML is to the solid EHQP, so is the pyramid ABCG to the pyramid DEFH, inasmuch as the pyramid is a sixth part of the solid, because the prism which is half of the parallelepipedal solid [XI. 28] is also triple of the pyramid. [XII. 7] Therefore the pyramid ABCG also has to the pyramid DEFH the ratio triplicate of that which BC has to EF.'"],["Rule","'ProofWordCount'",328],["Rule","'GreekProof'","'ἔστωσαν ὅμοιαι καὶ ὁμοίως κείμεναι πυραμίδες, ὧν βάσεις μέν εἰσι τὰ ΑΒΓ, ΔΕΖ τρίγωνα, κορυφαὶ δὲ τὰ Η, θ σημεῖα: λέγω, ὅτι ἡ ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. συμπεπληρώσθω γὰρ τὰ ΒΗΜΛ, ΕΘΠΟ στερεὰ παραλληλεπίπεδα. καὶ ἐπεὶ ὁμοία ἐστὶν ἡ ΑΒΓΗ πυραμὶς τῇ ΔΕΖΘ πυραμίδι, ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ γωνίᾳ, ἡ δὲ ὑπὸ ΗΒΓ τῇ ὑπὸ ΘΕΖ, ἡ δὲ ὑπὸ ΑΒΗ τῇ ὑπὸ ΔΕΘ, καί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΔΕ, οὕτως ἡ ΒΓ πρὸς τὴν ΕΖ, καὶ ἡ ΒΗ πρὸς τὴν ΕΘ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΔΕ, οὕτως ἡ ΒΓ πρὸς τὴν ΕΖ, καὶ περὶ ἴσας γωνίας αἱ πλευραὶ ἀνάλογόν εἰσιν, ὅμοιον ἄρα ἐστὶ τὸ ΒΜ παραλληλόγραμμον τῷ ΕΠ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν ΒΝ τῷ ΕΡ ὅμοιόν ἐστι, τὸ δὲ ΒΚ τῷ ΕΞ: τὰ τρία ἄρα τὰ ΜΒ, ΒΚ, ΒΝ τρισὶ τοῖς ΕΠ, ΕΞ, ΕΡ ὅμοιά ἐστιν. ἀλλὰ τὰ μὲν τρία τὰ ΜΒ, ΒΚ, ΒΝ τρισὶ τοῖς ἀπεναντίον ἴσα τε καὶ ὅμοιά ἐστιν, τὰ δὲ τρία τὰ ΕΠ, ΕΞ, ΕΡ τρισὶ τοῖς ἀπεναντίον ἴσα τε καὶ ὅμοιά ἐστιν. τὰ ΒΗΜΛ, ΕΘΠΟ ἄρα στερεὰ ὑπὸ ὁμοίων ἐπιπέδων ἴσων τὸ πλῆθος περιέχεται. ὅμοιον ἄρα ἐστὶ τὸ ΒΗΜΛ στερεὸν τῷ ΕΘΠΟ στερεῷ. τὰ δὲ ὅμοια στερεὰ παραλληλεπίπεδα ἐν τριπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν. τὸ ΒΗΜΛ ἄρα στερεὸν πρὸς τὸ ΕΘΠΟ στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ ἡ ΒΓ πρὸς τὴν ὁμόλογον πλευρὰν τὴν ΕΖ. ὡς δὲ τὸ ΒΗΜΛ στερεὸν πρὸς τὸ ΕΘΠΟ στερεόν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα, ἐπειδήπερ ἡ πυραμὶς ἕκτον μέρος ἐστὶ τοῦ στερεοῦ διὰ τὸ καὶ τὸ πρίσμα ἥμισυ ὂν τοῦ στερεοῦ παραλληλεπιπέδου τριπλάσιον εἶναι τῆς πυραμίδος. καὶ ἡ ΑΒΓΗ ἄρα πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι καὶ αἱ πολυγώνους ἔχουσαι βάσεις ὅμοιαι πυραμίδες πρὸς ἀλλήλας ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. διαιρεθεισῶν γὰρ αὐτῶν εἰς τὰς ἐν αὐταῖς πυραμίδας τριγώνους βάσεις ἐχούσας τῷ καὶ τὰ ὅμοια πολύγωνα τῶν βάσεων εἰς ὅμοια τρίγωνα διαιρεῖσθαι καὶ ἴσα τῷ πλήθει καὶ ὁμόλογα τοῖς ὅλοις ἔσται ὡς ἡ ἐν τῇ ἑτέρᾳ μία πυραμὶς τρίγωνον ἔχουσα βάσιν πρὸς τὴν ἐν τῇ ἑτέρᾳ μίαν πυραμίδα τρίγωνον ἔχουσαν βάσιν, οὕτως καὶ ἅπασαι αἱ ἐν τῇ ἑτέρᾳ πυραμίδι πυραμίδες τριγώνους ἔχουσαι βάσεις πρὸς τὰς ἐν τῇ ἑτέρᾳ πυραμίδι πυραμίδας τριγώνους βάσεις ἐχούσας, τουτέστιν αὐτὴ ἡ πολύγωνον βάσιν ἔχουσα πυραμὶς πρὸς τὴν πολύγωνον βάσιν ἔχουσαν πυραμίδα. ἡ δὲ τρίγωνον βάσιν ἔχουσα πυραμὶς πρὸς τὴν τρίγωνον βάσιν ἔχουσαν ἐν τριπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν: καὶ ἡ πολύγωνον ἄρα βάσιν ἔχουσα πρὸς τὴν ὁμοίαν βάσιν ἔχουσαν τριπλασίονα λόγον ἔχει ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν.'"],["Rule","'GreekProofWordCount'",457]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'12.9'"],["Rule","'Text'","'In equal pyramids which have triangular bases the bases are reciprocally proportional to the heights; and those pyramids in which the bases are reciprocally proportional to the heights are equal.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'τῶν ἴσων πυραμίδων καὶ τριγώνους βάσεις ἐχουσῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν: καὶ ὧν πυραμίδων τριγώνους βάσεις ἐχουσῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἴσαι εἰσὶν ἐκεῖναι.'"],["Rule","'GreekTextWordCount'",26],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",34]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",11],["Rule","'Theorem'",34]]]],["Rule","'Proof'","'For let there be equal pyramids which have the triangular bases ABC, DEF and vertices the points G, H; I say that in the pyramids ABCG, DEFH the bases are reciprocally proportional to the heights, that is, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG. For let the parallelepipedal solids BGML, EHQP be completed. Now, since the pyramid ABCG is equal to the pyramid DEFH, and the solid BGML is six times the pyramid ABCG, and the solid EHQP six times the pyramid DEFH, therefore the solid BGML is equal to the solid EHQP. But in equal parallelepipedal solids the bases are reciprocally proportional to the heights; [XI. 34] therefore, as the base BM is to the base EQ, so is the height of the solid EHQP to the height of the solid BGML. But, as the base BM is to EQ, so is the triangle ABC to the triangle DEF. [I. 34] Therefore also, as the triangle ABC is to the triangle DEF, so is the height of the solid EHQP to the height of the solid BGML. [V. 11] But the height of the solid EHQP is the same with the height of the pyramid DEFH, and the height of the solid BGML is the same with the height of the pyramid ABCG, therefore, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG. Therefore in the pyramids ABCG, DEFH the bases are reciprocally proportional to the heights. Next, in the pyramids ABCG, DEFH let the bases be reciprocally proportional to the heights; that is, as the base ABC is to the base DEF, so let the height of the pyramid DEFH be to the height of the pyramid ABCG; I say that the pyramid ABCG is equal to the pyramid DEFH. For, with the same construction, since, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG, while, as the base ABC is to the base DEF, so is the parallelogram BM to the parallelogram EQ, therefore also, as the parallelogram BM is to the parallelogram EQ, so is the height of the pyramid DEFH to the height of the pyramid ABCG. [V. 11] But the height of the pyramid DEFH is the same with the height of the parallelepiped EHQP, and the height of the pyramid ABCG is the same with the height of the parallelepiped BGML; therefore, as the base BM is to the base EQ, so is the height of the parallelepiped EHQP to the height of the parallelepiped BGML. But those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal; [XI. 34] therefore the parallelepipedal solid BGML is equal to the parallelepipedal solid EHQP. And the pyramid ABCG is a sixth part of BGML, and the pyramid DEFH a sixth part of the parallelepiped EHQP; therefore the pyramid ABCG is equal to the pyramid DEFH. Therefore etc'"],["Rule","'ProofWordCount'",523],["Rule","'GreekProof'","'ἔστωσαν γὰρ ἴσαι πυραμίδες τριγώνους βάσεις ἔχουσαι τὰς ΑΒΓ, ΔΕΖ, κορυφὰς δὲ τὰ Η, Θ σημεῖα: λέγω, ὅτι τῶν ΑΒΓΗ, ΔΕΖΘ πυραμίδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, καί ἐστιν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς τὸ τῆς ΑΒΓΗ πυραμίδος ὕψος. συμπεπληρώσθω γὰρ τὰ ΒΗΜΛ, ΕΘΠΟ στερεὰ παραλληλεπίπεδα. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒΓΗ πυραμὶς τῇ ΔΕΖΘ πυραμίδι, καί ἐστι τῆς μὲν ΑΒΓΗ πυραμίδος ἑξαπλάσιον τὸ ΒΗ ΜΛ στερεόν, τῆς δὲ ΔΕΖΘ πυραμίδος ἑξαπλάσιον τὸ ΕΘΠΟ στερεόν, ἴσον ἄρα ἐστὶ τὸ ΒΗΜΛ στερεὸν τῷ ΕΘΠΟ στερεῷ. τῶν δὲ ἴσων στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν: ἔστιν ἄρα ὡς ἡ ΒΜ βάσις πρὸς τὴν ΕΠ βάσιν, οὕτως τὸ τοῦ ΕΘΠΟ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΗΜΛ στερεοῦ ὕψος. ἀλλ᾽ ὡς ἡ ΒΜ βάσις πρὸς τὴν ΕΠ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον. καὶ ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον, οὕτως τὸ τοῦ ΕΘΠΟ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΗΜΛ στερεοῦ ὕψος. ἀλλὰ τὸ μὲν τοῦ ΕΘΠΟ στερεοῦ ὕψος τὸ αὐτό ἐστι τῷ τῆς ΔΕΖΘ πυραμίδος ὕψει, τὸ δὲ τοῦ ΒΗΜΛ στερεοῦ ὕψος τὸ αὐτό ἐστι τῷ τῆς ΑΒΓΗ πυραμίδος ὕψει: ἔστιν ἄρα ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς τὸ τῆς ΑΒΓΗ πυραμίδος ὕψος. τῶν ΑΒΓΗ, ΔΕΖΘ ἄρα πυραμίδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. ἀλλὰ δὴ τῶν ΑΒΓΗ, ΔΕΖΘ πυραμίδων ἀντιπεπονθέτωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς τὸ τῆς ΑΒΓΗ πυραμίδος ὕψος: λέγω, ὅτι ἴση ἐστὶν ἡ ΑΒΓΗ πυραμὶς τῇ ΔΕΖΘ πυραμίδι. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς τὸ τῆς ΑΒΓΗ πυραμίδος ὕψος, ἀλλ᾽ ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὸ ΒΜ παραλληλόγραμμον πρὸς τὸ ΕΠ παραλληλόγραμμον, καὶ ὡς ἄρα τὸ ΒΜ παραλληλόγραμμον πρὸς τὸ ΕΠ παραλληλόγραμμον, οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς τὸ τῆς ΑΒΓΗ πυραμίδος ὕψος. ἀλλὰ τὸ μὲν τῆς ΔΕΖΘ πυραμίδος ὕψος τὸ αὐτό ἐστι τῷ τοῦ ΕΘΠΟ παραλληλεπιπέδου ὕψει, τὸ δὲ τῆς ΑΒΓΗ πυραμίδος ὕψος τὸ αὐτό ἐστι τῷ τοῦ ΒΗΜΛ παραλληλεπιπέδου ὕψει: ἔστιν ἄρα ὡς ἡ ΒΜ βάσις πρὸς τὴν ΕΠ βάσιν, οὕτως τὸ τοῦ ΕΘΠΟ παραλληλεπιπέδου ὕψος πρὸς τὸ τοῦ ΒΗΜΛ παραλληλεπιπέδου ὕψος. ὧν δὲ στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἴσα ἐστὶν ἐκεῖνα: ἴσον ἄρα ἐστὶ τὸ ΒΗΜΛ στερεὸν παραλληλεπίπεδον τῷ ΕΘ ΠΟ στερεῷ παραλληλεπιπέδῳ. καί ἐστι τοῦ μὲν ΒΗΜΛ ἕκτον μέρος ἡ ΑΒΓΗ πυραμίς, τοῦ δὲ ΕΘΠΟ παραλληλεπιπέδου ἕκτον μέρος ἡ ΔΕΖΘ πυραμίς: ἴση ἄρα ἡ ΑΒΓΗ πυραμὶς τῇ ΔΕΖΘ πυραμίδι. τῶν ἄρα ἴσων πυραμίδων καὶ τριγώνους βάσεις ἐχουσῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν: καὶ ὧν πυραμίδων τριγώνους βάσεις ἐχουσῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἴσαι εἰσὶν ἐκεῖναι: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",467]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'12.10'"],["Rule","'Text'","'Any cone is a third part of the cylinder which has the same base with it and equal height.'"],["Rule","'TextWordCount'",19],["Rule","'GreekText'","'πᾶς κῶνος κυλίνδρου τρίτον μέρος ἐστὶ τοῦ τὴν αὐτὴν βάσιν ἔχοντος αὐτῷ καὶ ὕψος ἴσον.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",4],["Rule","'Theorem'",6]],["List",["Rule","'Book'",4],["Rule","'Theorem'",7]],["List",["Rule","'Book'",10],["Rule","'Theorem'",1]],["List",["Rule","'Book'",11],["Rule","'Theorem'",32]],["List",["Rule","'Book'",12],["Rule","'Theorem'",2]],["List",["Rule","'Book'",12],["Rule","'Theorem'",7]]]],["Rule","'Proof'","'For let a cone have the same base, namely the circle ABCD, with a cylinder and equal height; I say that the cone is a third part of the cylinder, that is, that the cylinder is triple of the cone. For if the cylinder is not triple of the cone, the cylinder will be either greater than triple or less than triple of the cone. First let it be greater than triple, and let the square ABCD be inscribed in the circle ABCD; [IV. 6] then the square ABCD is greater than the half of the circle ABCD. From the square ABCD let there be set up a prism of equal height with the cylinder. Then the prism so set up is greater than the half of the cylinder, inasmuch as, if we also circumscribe a square about the circle ABCD [IV. 7], the square inscribed in the circle ABCD is half of that circumscribed about it, and the solids set up from them are parallelepipedal prisms of equal height, while parallelepipedal solids which are of the same height are to one another as their bases; [XI. 32] therefore also the prism set up on the square ABCD is half of the prism set up from the square circumscribed about the circle ABCD; [cf. XI. 28, or XII. 6 and 7] and the cylinder is less than the prism set up from the square circumscribed about the circle ABCD; therefore the prism set up from the square ABCD and of equal height with the cylinder is greater than the half of the cylinder. Let the circumferences AB, BC, CD, DA be bisected at the points E, F, G, H, and let AE, EB, BF, FC, CG, GD, DH, HA be joined; then each of the triangles AEB, BFC, CGD, DHA is greater than the half of that segment of the circle ABCD which is about it, as we proved before. [XII. 2] On each of the triangles AEB, BFC, CGD, DHA let prisms be set up of equal height with the cylinder; then each of the prisms so set up is greater than the half part of that segment of the cylinder which is about it, inasmuch as, if we draw through the points E, F, G, H parallels to AB, BC, CD, DA, complete the parallelograms on AB, BC, CD, DA, and set up from them parallelepipedal solids of equal height with the cylinder, the prisms on the triangles AEB, BFC, CGD, DHA are halves of the several solids set up; and the segments of the cylinder are less than the parallelepipedal solids set up; hence also the prisms on the triangles AEB, BFC, CGD, DHA are greater than the half of the segments of the cylinder about them. Thus, bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles prisms of equal height with the cylinder, and doing this continually, we shall leave some segments of the cylinder which will be less than the excess by which the cylinder exceeds the triple of the cone. [X. 1] Let such segments be left, and let them be AE, EB, BF, FC, CG, GD, DH, HA; therefore the remainder, the prism of which the polygon AEBFCGDH is the base and the height is the same as that of the cylinder, is greater than triple of the cone. But the prism of which the polygon AEBFCGDH is the base and the height the same as that of the cylinder is triple of the pyramid of which the polygon AEBFCGDH is the base and the vertex is the same as that of the cone; [XII. 7] therefore also the pyramid of which the polygon AEBFCGDH is the base and the vertex is the same as that of the cone is greater than the cone which has the circle ABCD as base. But it is also less, for it is enclosed by it: which is impossible. Therefore the cylinder is not greater than triple of the cone. I say next that neither is the cylinder less than triple of the cone, For, if possible, let the cylinder be less than triple of the cone, therefore, inversely, the cone is greater than a third part of the cylinder. Let the square ABCD be inscribed in the circle ABCD; therefore the square ABCD is greater than the half of the circle ABCD. Now let there be set up from the square ABCD a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone, seeing that, as we proved before, if we circumscribe a square about the circle, the square ABCD will be half of the square circumscribed about the circle, and if we set up from the squares parallelepipedal solids of equal height with the cone, which are also called prisms, the solid set up from the square ABCD will be half of that set up from the square circumscribed about the circle; for they are to one another as their bases. [XI. 32] Hence also the thirds of them are in that ratio; therefore also the pyramid of which the square ABCD is the base is half of the pyramid set up from the square circumscribed about the circle. And the pyramid set up from the square about the circle is greater than the cone, for it encloses it. Therefore the pyramid of which the square ABCD is the base and the vertex is the same with that of the cone is greater than the half of the cone. Let the circumferences AB, BC, CD, DA be bisected at the points E, F, G, H, and let AE, EB, BF, FC, CG, GD, DH, HA be joined; therefore also each of the triangles AEB, BFC, CGD, DHA is greater than the half part of that segment of the circle ABCD which is about it. Now, on each of the triangles AEB, BFC, CGD, DHA let pyramids be set up which have the same vertex as the cone; therefore also each of the pyramids so set up is, in the same manner, greater than the half part of that segment of the cone which is about it. Thus, by bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles a pyramid which has the same vertex as the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone exceeds the third part of the cylinder. [X. 1] Let such be left, and let them be the segments on AE, EB, BF, FC, CG, GD, DH, HA; therefore the remainder, the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone, is greater than a third part of the cylinder. But the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone is a third part of the prism of which the polygon AEBFCGDH is the base and the height is the same with that of the cylinder; therefore the prism of which the polygon AEBFCGDH is the base and the height is the same with that of the cylinder is greater than the cylinder of which the circle ABCD is the base. But it is also less, for it is enclosed by it: which is impossible. Therefore the cylinder is not less than triple of the cone. But it was proved that neither is it greater than triple; therefore the cylinder is triple of the cone; hence the cone is a third part of the cylinder.'"],["Rule","'ProofWordCount'",1288],["Rule","'GreekProof'","'ἐχέτω γὰρ κῶνος κυλίνδρῳ βάσιν τε τὴν αὐτὴν τὸν ΑΒΓΔ κύκλον καὶ ὕψος ἴσον: λέγω, ὅτι ὁ κῶνος τοῦ κυλίνδρου τρίτον ἐστὶ μέρος, τουτέστιν ὅτι ὁ κύλινδρος τοῦ κώνου τριπλασίων ἐστίν. εἰ γὰρ μή ἐστιν ὁ κύλινδρος τοῦ κώνου τριπλασίων, ἔσται ὁ κύλινδρος τοῦ κώνου ἤτοι μείζων ἢ τριπλασίων ἢ ἐλάσσων ἢ τριπλασίων. ἔστω πρότερον μείζων ἢ τριπλασίων, καὶ ἐγγεγράφθω εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον τὸ ΑΒΓΔ: τὸ δὴ ΑΒΓΔ τετράγωνον μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ ΑΒΓΔ κύκλου. καὶ ἀνεστάτω ἀπὸ τοῦ ΑΒΓΔ τετραγώνου πρίσμα ἰσουψὲς τῷ κυλίνδρῳ. τὸ δὴ ἀνιστάμενον πρίσμα μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ κυλίνδρου, ἐπειδήπερ κἂν περὶ τὸν ΑΒΓΔ κύκλον τετράγωνον περιγράψωμεν, τὸ ἐγγεγραμμένον εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον ἥμισύ ἐστι τοῦ περιγεγραμμένου: καί ἐστι τὰ ἀπ᾽ αὐτῶν ἀνιστάμενα στερεὰ παραλληλεπίπεδα πρίσματα ἰσοϋψῆ: τὰ δὲ ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις: καὶ τὸ ἐπὶ τοῦ ΑΒΓΔ ἄρα τετραγώνου ἀνασταθὲν πρίσμα ἥμισύ ἐστι τοῦ ἀνασταθέντος πρίσματος ἀπὸ τοῦ περὶ τὸν ΑΒΓΔ κύκλον περιγραφέντος τετραγώνου: καί ἐστιν ὁ κύλινδρος ἐλάττων τοῦ πρίσματος τοῦ ἀνασταθέντος ἀπὸ τοῦ περὶ τὸν ΑΒΓΔ κύκλον περιγραφέντος τετραγώνου: τὸ ἄρα πρίσμα τὸ ἀνασταθὲν ἀπὸ τοῦ ΑΒΓΔ τετραγώνου ἰσοϋψὲς τῷ κυλίνδρῳ μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ κυλίνδρου. τετμήσθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΑ περιφέρειαι δίχα κατὰ τὰ Ε, Ζ, Η, Θ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ: καὶ ἕκαστον ἄρα τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ καθ᾽ ἑαυτὸ τμήματος τοῦ ΑΒΓΔ κύκλου, ὡς ἔμπροσθεν ἐδείκνυμεν. ἀνεστάτω ἐφ᾽ ἑκάστου τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων πρίσματα ἰσουψῆ τῷ κυλίνδρῳ: καὶ ἕκαστον ἄρα τῶν ἀνασταθέντων πρισμάτων μεῖζόν ἐστιν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾽ ἑαυτὸ τμήματος τοῦ κυλίνδρου, ἐπειδήπερ ἐὰν διὰ τῶν Ε, Ζ, Η, θ σημείων παραλλήλους ταῖς ΑΒ, ΒΓ, ΓΔ, ΔΑ ἀγάγωμεν, καὶ συμπληρώσωμεν τὰ ἐπὶ τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ παραλληλόγραμμα, καὶ ἀπ᾽ αὐτῶν ἀναστήσωμεν στερεὰ παραλληλεπίπεδα ἰσοϋψῆ τῷ κυλίνδρῳ, ἑκάστου τῶν ἀνασταθέντων ἡμίση ἐστὶ τὰ πρίσματα τὰ ἐπὶ τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων: καί ἐστι τὰ τοῦ κυλίνδρου τμήματα ἐλάττονα τῶν ἀνασταθέντων στερεῶν παραλληλεπιπέδων: ὥστε καὶ τὰ ἐπὶ τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῶν καθ᾽ ἑαυτὰ τοῦ κυλίνδρου τμημάτων. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐφ᾽ ἑκάστου τῶν τριγώνων πρίσματα ἰσοϋψῆ τῷ κυλίνδρῳ καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κυλίνδρου, ἃ ἔσται ἐλάττονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ κύλινδρος τοῦ τριπλασίου τοῦ κώνου. λελείφθω, καὶ ἔστω τὰ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ: λοιπὸν ἄρα τὸ πρίσμα, οὗ βάσις μὲν τὸ ΑΕΒΖ ΓΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, μεῖζόν ἐστιν ἢ τριπλάσιον τοῦ κώνου. ἀλλὰ τὸ πρίσμα, οὗ βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, τριπλάσιόν ἐστι τῆς πυραμίδος, ἧς βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ: καὶ ἡ πυραμὶς ἄρα, ἧς βάσις μὲν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, μείζων ἐστὶ τοῦ κώνου τοῦ βάσιν ἔχοντος τὸν ΑΒ ΓΔ κύκλον. ἀλλὰ καὶ ἐλάττων: ἐμπεριέχεται γὰρ ὑπ᾽ αὐτοῦ: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐστὶν ὁ κύλινδρος τοῦ κώνου μείζων ἢ τριπλάσιος. λέγω δή, ὅτι οὐδὲ ἐλάττων ἐστὶν ἢ τριπλάσιος ὁ κύλινδρος τοῦ κώνου. εἰ γὰρ δυνατόν, ἔστω ἐλάττων ἢ τριπλάσιος ὁ κύλινδρος τοῦ κώνου: ἀνάπαλιν ἄρα ὁ κῶνος τοῦ κυλίνδρου μείζων ἐστὶν ἢ τρίτον μέρος. ἐγγεγράφθω δὴ εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον τὸ ΑΒΓΔ: τὸ ΑΒΓΔ ἄρα τετράγωνον μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ ΑΒΓΔ κύκλου. καὶ ἀνεστάτω ἀπὸ τοῦ ΑΒΓΔ τετραγώνου πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ κώνῳ: ἡ ἄρα ἀνασταθεῖσα πυραμὶς μείζων ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ κώνου, ἐπειδήπερ, ὡς ἔμπροσθεν ἐδείκνυμεν, ὅτι ἐὰν περὶ τὸν κύκλον τετράγωνον περιγράψωμεν, ἔσται τὸ ΑΒΓΔ τετράγωνον ἥμισυ τοῦ περὶ τὸν κύκλον περιγεγραμμένου τετραγώνου: καὶ ἐὰν ἀπὸ τῶν τετραγώνων στερεὰ παραλληλεπίπεδα ἀναστήσωμεν ἰσοϋψῆ τῷ κώνῳ, ἃ καὶ καλεῖται πρίσματα, ἔσται τὸ ἀνασταθὲν ἀπὸ τοῦ ΑΒΓΔ τετραγώνου ἥμισυ τοῦ ἀνασταθέντος ἀπὸ τοῦ περὶ τὸν κύκλον περιγραφέντος τετραγώνου: πρὸς ἄλληλα γάρ εἰσιν ὡς αἱ βάσεις. ὥστε καὶ τὰ τρίτα: καὶ πυραμὶς ἄρα, ἧς βάσις τὸ ΑΒΓΔ τετράγωνον, ἥμισύ ἐστι τῆς πυραμίδος τῆς ἀνασταθείσης ἀπὸ τοῦ περὶ τὸν κύκλον περιγραφέντος τετραγώνου. καί ἐστι μείζων ἡ πυραμὶς ἡ ἀνασταθεῖσα ἀπὸ τοῦ περὶ τὸν κύκλον τετραγώνου τοῦ κώνου: ἐμπεριέχει γὰρ αὐτόν. ἡ ἄρα πυραμὶς, ἧς βάσις τὸ ΑΒΓΔ τετράγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, μείζων ἐστὶν ἢ τὸ ἥμισυ τοῦ κώνου. τετμήσθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΑ περιφέρειαι δίχα κατὰ τὰ Ε, Ζ, Η, Θ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ: καὶ ἕκαστον ἄρα τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾽ ἑαυτὸ τμήματος τοῦ ΑΒΓΔ κύκλου. καὶ ἀνεστάτωσαν ἐφ᾽ ἑκάστου τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων πυραμίδες τὴν αὐτὴν κορυφὴν ἔχουσαι τῷ κώνῳ: καὶ ἑκάστη ἄρα τῶν ἀνασταθεισῶν πυραμίδων κατὰ τὸν αὐτὸν τρόπον μείζων ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾽ ἑαυτὴν τμήματος τοῦ κώνου. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐφ᾽ ἑκάστου τῶν τριγώνων πυραμίδα τὴν αὐτὴν κορυφὴν ἔχουσαν τῷ κώνῳ καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κώνου, ἃ ἔσται ἐλάττονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ κῶνος τοῦ τρίτου μέρους τοῦ κυλίνδρου. λελείφθω, καὶ ἔστω τὰ ἐπὶ τῶν ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ: λοιπὴ ἄρα ἡ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, μείζων ἐστὶν ἢ τρίτον μέρος τοῦ κυλίνδρου. ἀλλ᾽ ἡ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΕΒΖΓ ΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, τρίτον ἐστὶ μέρος τοῦ πρίσματος, οὗ βάσις μέν ἐστι τὸ ΑΕΒΖΓ ΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ: τὸ ἄρα πρίσμα, οὗ βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, μεῖζόν ἐστι τοῦ κυλίνδρου, οὗ βάσις ἐστὶν ὁ ΑΒΓΔ κύκλος. ἀλλὰ καὶ ἔλαττον: ἐμπεριέχεται γὰρ ὑπ᾽ αὐτοῦ: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ὁ κύλινδρος τοῦ κώνου ἐλάττων ἐστὶν ἢ τριπλάσιος. ἐδείχθη δέ, ὅτι οὐδὲ μείζων ἢ τριπλάσιος: τριπλάσιος ἄρα ὁ κύλινδρος τοῦ κώνου: ὥστε ὁ κῶνος τρίτον ἐστὶ μέρος τοῦ κυλίνδρου. πᾶς ἄρα κῶνος κυλίνδρου τρίτον μέρος ἐστὶ τοῦ τὴν αὐτὴν βάσιν ἔχοντος αὐτῷ καὶ ὕψος ἴσον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",1033]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'12.11'"],["Rule","'Text'","'Cones and cylinders which are of the same height are to one another as their bases.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'οἱ ὑπὸ τὸ αὐτὸ ὕψος ὄντες κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις.'"],["Rule","'GreekTextWordCount'",15],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",1]],["List",["Rule","'Book'",12],["Rule","'Theorem'",1]],["List",["Rule","'Book'",12],["Rule","'Theorem'",2]],["List",["Rule","'Book'",12],["Rule","'Theorem'",6]],["List",["Rule","'Book'",12],["Rule","'Theorem'",10]]]],["Rule","'Proof'","'Let there be cones and cylinders of the same height, let the circles ABCD, EFGH be their bases, KL, MN their axes and AC, EG the diameters of their bases; I say that, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN. For, if not, then, as the circle ABCD is to the circle EFGH, so will the cone AL be either to some solid less than the cone EN or to a greater. First, let it be in that ratio to a less solid O, and let the solid X be equal to that by which the solid O is less than the cone EN; therefore the cone EN is equal to the solids O, X. Let the square EFGH be inscribed in the circle EFGH; therefore the square is greater than the half of the circle. Let there be set up from the square EFGH a pyramid of equal height with the cone; therefore the pyramid so set up is greater than the half of the cone, inasmuch as, if we circumscribe a square about the circle, and set up from it a pyramid of equal height with the cone, the inscribed pyramid is half of the circumscribed pyramid, for they are to one another as their bases, [XII. 6] while the cone is less than the circumscribed pyramid. Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let HP, PE, EQ, QF, FR, RG, GS, SH be joined. Therefore each of the triangles HPE, EQF, FRG, GSH is greater than the half of that segment of the circle which is about it. On each of the triangles HPE, EQF, FRG, GSH let there be set up a pyramid of equal height with the cone; therefore, also, each of the pyramids so set up is greater than the half of that segment of the cone which is about it. Thus, bisecting the circumferences which are left, joining straight lines, setting up on each of the triangles pyramids of equal height with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the solid X. [X. 1] Let such be left, and let them be the segments on HP, PE, EQ, QF, FR, RG, GS, SH; therefore the remainder, the pyramid of which the polygon HPEQFRGS is the base and the height the same with that of the cone, is greater than the solid O. Let there also be inscribed in the circle ABCD the polygon DTAUBVCW similar and similarly situated to the polygon HPEQFRGS, and on it let a pyramid be set up of equal height with the cone AL. Since then, as the square on AC is to the square on EG, so is the polygon DTAUBVCW to the polygon HPEQFRGS, [XII. 1] while, as the square on AC is to the square on EG, so is the circle ABCD to the circle EFGH, [XII. 2] therefore also, as the circle ABCD is to the circle EFGH, so is the polygon DTAUBVCW to the polygon HPEQFRGS. But, as the circle ABCD is to the circle EFGH, so is the cone AL to the solid O, and, as the polygon DTAUBVCW is to the polygon HPEQFRGS, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex. [XII. 6] Therefore also, as the cone AL is to the solid O, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex; [V. 11] therefore, alternately, as the cone AL is to the pyramid in it, so is the solid O to the pyramid in the cone EN. [V. 16] But the cone AL is greater than the pyramid in it; therefore the solid O is also greater than the pyramid in the cone EN. But it is also less: which is absurd. Therefore the cone AL is not to any solid less than the cone EN as the circle ABCD is to the circle EFGH. Similarly we can prove that neither is the cone EN to any solid less than the cone AL as the circle EFGH is to the circle ABCD. I say next that neither is the cone AL to any solid greater than the cone EN as the circle ABCD is to the circle EFGH. For, if possible, let it be in that ratio to a greater solid O; therefore, inversely, as the circle EFGH is to the circle ABCD, so is the solid O to the cone AL. But, as the solid O is to the cone AL, so is the cone EN to some solid less than the cone AL; therefore also, as the circle EFGH is to the circle ABCD, so is the cone EN to some solid less than the cone AL: which was proved impossible. Therefore the cone AL is not to any solid greater than the cone EN as the circle ABCD is to the circle EFGH. But it was proved that neither is it in this ratio to a less solid; therefore, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN. But, as the cone is to the cone, so is the cylinder to the cylinder, for each is triple of each; [XII. 10] Therefore also, as the circle ABCD is to the circle EFGH, so are the cylinders on them which are of equal height.'"],["Rule","'ProofWordCount'",961],["Rule","'GreekProof'","'ἔστωσαν ὑπὸ τὸ αὐτὸ ὕψος κῶνοι καὶ κύλινδροι, ὧν βάσεις μὲν εἰσιν οἱ ΑΒΓΔ, ΕΖΗΘ κύκλοι, ἄξονες δὲ οἱ ΚΛ, ΜΝ, διάμετροι δὲ τῶν βάσεων αἱ ΑΓ, ΕΗ: λέγω, ὅτι ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς τὸν ΕΝ κῶνον. εἰ γὰρ μή, ἔσται ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος ἤτοι πρὸς ἔλασσόν τι τοῦ ΕΝ κώνου στερεὸν ἢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ Ξ, καὶ ᾧ ἔλασσόν ἐστι τὸ Ξ στερεὸν τοῦ ΕΝ κώνου, ἐκείνῳ ἴσον ἔστω τὸ Ψ στερεόν: ὁ ΕΝ κῶνος ἄρα ἴσος ἐστὶ τοῖς Ξ, Ψ στερεοῖς. ἐγγεγράφθω εἰς τὸν ΕΖΗΘ κύκλον τετράγωνον τὸ ΕΖΗΘ: τὸ ἄρα τετράγωνον μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ κύκλου. ἀνεστάτω ἀπὸ τοῦ ΕΖ ΗΘ τετραγώνου πυραμὶς ἰσοϋψὴς τῷ κώνῳ: ἡ ἄρα ἀνασταθεῖσα πυραμὶς μείζων ἐστὶν ἢ τὸ ἥμισυ τοῦ κώνου, ἐπειδήπερ ἐὰν περιγράψωμεν περὶ τὸν κύκλον τετράγωνον, καὶ ἀπ᾽ αὐτοῦ ἀναστήσωμεν πυραμίδα ἰσοϋψῆ τῷ κώνῳ, ἡ ἐγγραφεῖσα πυραμὶς ἥμισύ ἐστι τῆς περιγραφείσης: πρὸς ἀλλήλας γάρ εἰσιν ὡς αἱ βάσεις: ἐλάττων δὲ ὁ κῶνος τῆς περιγραφείσης πυραμίδος. τετμήσθωσαν αἱ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι δίχα κατὰ τὰ Ο, Π, Ρ, Σ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΘΟ, ΟΕ, ΕΠ, ΠΖ, ΖΡ, ΡΗ, ΗΣ, ΣΘ. ἕκαστον ἄρα τῶν ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΣΘ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ καθ᾽ ἑαυτὸ τμήματος τοῦ κύκλου. ἀνεστάτω ἐφ᾽ ἑκάστου τῶν ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΣΘ τριγώνων πυραμὶς ἰσοϋψὴς τῷ κώνῳ: καὶ ἑκάστη ἄρα τῶν ἀνασταθεισῶν πυραμίδων μείζων ἐστὶν ἢ τὸ ἥμισυ τοῦ καθ᾽ ἑαυτὴν τμήματος τοῦ κώνου. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐπὶ ἑκάστου τῶν τριγώνων πυραμίδας ἰσοϋψεῖς τῷ κώνῳ καὶ ἀεὶ τοῦτο ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κώνου, ἃ ἔσται ἐλάσσονα τοῦ Ψ στερεοῦ. λελείφθω, καὶ ἔστω τὰ ἐπὶ τῶν ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΣΘ: λοιπὴ ἄρα ἡ πυραμίς, ἧς βάσις τὸ ΘΟΕΠΖΡΗΣ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κώνῳ, μείζων ἐστὶ τοῦ Ξ στερεοῦ. ἐγγεγράφθω καὶ εἰς τὸν ΑΒΓΔ κύκλον τῷ ΘΟΕΠΖΡΗΣ πολυγώνῳ ὅμοιόν τε καὶ ὁμοίως κείμενον πολύγωνον τὸ ΔΤΑΥΒ ΦΓΧ, καὶ ἀνεστάτω ἐπ᾽ αὐτοῦ πυραμὶς ἰσοϋψὴς τῷ ΑΛ κώνῳ. ἐπεὶ οὖν ἐστιν ὡς τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ἀπὸ τῆς ΕΗ, οὕτως τὸ ΔΤΑΥΒΦΓΧ πολύγωνον πρὸς τὸ ΘΟΕ ΠΖΡΗΣ πολύγωνον, ὡς δὲ τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ἀπὸ τῆς ΕΗ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, καὶ ὡς ἄρα ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως τὸ ΔΤΑΥΒΦΓΧ πολύγωνον πρὸς τὸ ΘΟΕΠΖΡΗΣ πολύγωνον. ὡς δὲ ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς τὸ Ξ στερεόν, ὡς δὲ τὸ ΔΤΑΥΒΦΓΧ πολύγωνον πρὸς τὸ ΘΟΕΠΖ ΡΗΣ πολύγωνον, οὕτως ἡ πυραμίς, ἧς βάσις μὲν τὸ ΔΤΑΥΒΦΓΧ πολύγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς τὴν πυραμίδα, ἧς βάσις μὲν τὸ ΘΟΕΠΖΡΗΣ πολύγωνον, κορυφὴ δὲ τὸ Ν σημεῖον. καὶ ὡς ἄρα ὁ ΑΛ κῶνος πρὸς τὸ Ξ στερεόν, οὕτως ἡ πυραμίς, ἧς βάσις μὲν τὸ ΔΤΑΥΒΦΓΧ πολύγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς τὴν πυραμίδα, ἧς βάσις μὲν τὸ ΘΟΕΠΖΡΗΣ πολύγωνον, κορυφὴ δὲ τὸ Ν σημεῖον: ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ ΑΛ κῶνος πρὸς τὴν ἐν αὐτῷ πυραμίδα, οὕτως τὸ Ξ στερεὸν πρὸς τὴν ἐν τῷ ΕΝ κώνῳ πυραμίδα. μείζων δὲ ὁ ΑΛ κῶνος τῆς ἐν αὐτῷ πυραμίδος: μεῖζον ἄρα καὶ τὸ Ξ στερεὸν τῆς ἐν τῷ ΕΝ κώνῳ πυραμίδος. ἀλλὰ καὶ ἔλασσον: ὅπερ ἄτοπον. οὐκ ἄρα ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς ἔλασσόν τι τοῦ ΕΝ κώνου στερεόν. ὁμοίως δὴ δείξομεν, ὅτι οὐδέ ἐστιν ὡς ὁ ΕΖΗΘ κύκλος πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΝ κῶνος πρὸς ἔλασσόν τι τοῦ ΑΛ κώνου στερεόν. λέγω δή, ὅτι οὐδέ ἐστιν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς μεῖζόν τι τοῦ ΕΝ κώνου στερεόν. εἰ γὰρ δυνατόν, ἔστω πρὸς μεῖζον τὸ Ξ: ἀνάπαλιν ἄρα ἐστὶν ὡς ὁ ΕΖΗΘ κύκλος πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως τὸ Ξ στερεὸν πρὸς τὸν ΑΛ κῶνον. ἀλλ᾽ ὡς τὸ Ξ στερεὸν πρὸς τὸν ΑΛ κῶνον, οὕτως ὁ ΕΝ κῶνος πρὸς ἔλασσόν τι τοῦ ΑΛ κώνου στερεόν: καὶ ὡς ἄρα ὁ ΕΖΗΘ κύκλος πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΝ κῶνος πρὸς ἔλασσόν τι τοῦ ΑΛ κώνου στερεόν: ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς μεῖζόν τι τοῦ ΕΝ κώνου στερεόν. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἔλασσον: ἔστιν ἄρα ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς τὸν ΕΝ κῶνον. ἀλλ᾽ ὡς ὁ κῶνος πρὸς τὸν κῶνον, ὁ κύλινδρος πρὸς τὸν κύλινδρον: τριπλασίων γὰρ ἑκάτερος ἑκατέρου. καὶ ὡς ἄρα ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως οἱ ἐπ᾽ αὐτῶν ἰσοϋψεῖς τοῖς κώνοις κύλινδροι. οἱ ἄρα ὑπὸ τὸ αὐτὸ ὕψος ὄντες κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",797]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'12.12'"],["Rule","'Text'","'Similar cones and cylinders are to one another in the triplicate ratio of the diameters in their bases.'"],["Rule","'TextWordCount'",18],["Rule","'GreekText'","'οἱ ὅμοιοι κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ἐν ταῖς βάσεσι διαμέτρων.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",4],["Rule","'Theorem'",6]],["List",["Rule","'Book'",5],["Rule","'Theorem'",12]],["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",5],["Rule","'Theorem'",22]],["List",["Rule","'Book'",6],["Rule","'Definition'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",5]],["List",["Rule","'Book'",6],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",1]],["List",["Rule","'Book'",11],["Rule","'Definition'",9]],["List",["Rule","'Book'",11],["Rule","'Definition'",24]],["List",["Rule","'Book'",12],["Rule","'Theorem'",8]],["List",["Rule","'Book'",12],["Rule","'Theorem'",10]]]],["Rule","'Proof'","'Let there be similar cones and cylinders, let the circles ABCD, EFGH be their bases, BD, FH the diameters of the bases, and KL, MN the axes of the cones and cylinders; I say that the cone of which the circle ABCD is the base and the point L the vertex has to the cone of which the circle EFGH is the base and the point N the vertex the ratio triplicate of that which BD has to FH. For, if the cone ABCDL has not to the cone EFGHN the ratio triplicate of that which BD has to FH, the cone ABCDL will have that triplicate ratio either to some solid less than the cone EFGHN or to a greater. First, let it have that triplicate ratio to a less solid O. Let the square EFGH be inscribed in the circle EFGH; [IV. 6] therefore the square EFGH is greater than the half of the circle EFGH. Now let there be set up on the square EFGH a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone. Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let EP, PF, FQ, QG, GR, RH, HS, SE be joined. Therefore each of the triangles EPF, FQG, GRH, HSE is also greater than the half part of that segment of the circle EFGH which is about it. Now on each of the triangles EPF, FQG, GRH, HSE let a pyramid be set up having the same vertex with the cone; therefore each of the pyramids so set up is also greater than the half part of that segment of the cone which is about it. Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids having the same vertex with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone EFGHN exceeds the solid O. [X. 1] Let such be left, and let them be the segments on EP, PF, FQ, QG, GR, RH, HS, SE; therefore the remainder, the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex, is greater than the solid O. Let there be also inscribed in the circle ABCD the polygon ATBUCVDW similar and similarly situated to the polygon EPFQGRHS, and let there be set up on the polygon ATBUCVDW a pyramid having the same vertex with the cone; of the triangles containing the pyramid of which the polygon ATBUCVDW is the base and the point L the vertex let LBT be one, and of the triangles containing the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex let NFP be one; and let KT, MP be joined. Now, since the cone ABCDL is similar to the cone EFGHN, therefore, as BD is to FH, so is the axis KL to the axis MN. [XI. Def. 24] But, as BD is to FH, so is BK to FM; therefore also, as BK is to FM, so is KL to MN. And, alternately, as BK is to KL, so is FM to MN. [V. 16] And the sides are proportional about equal angles, namely the angles BKL, FMN; therefore the triangle BKL is similar to the triangle FMN. [VI. 6] Again, since, as BK is to KT, so is FM to MP, and they are about equal angles, namely the angles BKT, FMP, inasmuch as, whatever part the angle BKT is of the four right angles at the centre K, the same part also is the angle FMP of the four right angles at the centre M; since then the sides are proportional about equal angles, therefore the triangle BKT is similar to the triangle FMP. [VI. 6] Again, since it was proved that, as BK is to KL, so is FM to MN, while BK is equal to KT, and FM to PM, therefore, as TK is to KL, so is PM to MN; and the sides are proportional about equal angles, namely the angles TKL, PMN, for they are right; therefore the triangle LKT is similar to the triangle NMP. [VI. 6] And since, owing to the similarity of the triangles LKB, NMF, as LB is to BK, so is NF to FM, and, owing to the similarity of the triangles BKT, FMP, as KB is to BT, so is MF to FP, therefore, ex aequali, as LB is to BT, so is NF to FP. [V. 22] Again since, owing to the similarity of the triangles LTK, NPM, as LT is to TK, so is NP to PM, and, owing to the similarity of the triangles TKB, PMF, as KT is to TB, so is MP to PF; therefore, ex aequali, as LT is to TB, so is NP to PF. [V. 22] But it was also proved that, as TB is to BL, so is PF to FN. Therefore, ex aequali, as TL is to LB, so is PN to NF. [V. 22] Therefore in the triangles LTB, NPF the sides are proportional; therefore the triangles LTB, NPF are equiangular; [VI. 5] hence they are also similar. [VI. Def. I] Therefore the pyramid of which the triangle BKT is the base and the point L the vertex is also similar to the pyramid of which the triangle FMP is the base and the point N the vertex, for they are contained by similar planes equal in multitude. [XI. Def. 9] But similar pyramids which have triangular bases are to one another in the triplicate ratio of their corresponding sides. [XII. 8] Therefore the pyramid BKTL has to the pyramid FMPN the ratio triplicate of that which BK has to FM. Similarly, by joining straight lines from A, W, D, V, C, U to K, and from E, S, H, R, G, Q to M, and setting up on each of the triangles pyramids which have the same vertex with the cones, we can prove that each of the similarly arranged pyramids will also have to each similarly arranged pyramid the ratio triplicate of that which the corresponding side BK has to the corresponding side FM, that is, which BD has to FH. And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore also, as the pyramid BKTL is to the pyramid FMPN, so is the whole pyramid of which the polygon ATBUCVDW is the base and the point L the vertex to the whole pyramid of which the polygon EPFQGRHS is the base and the point N the vertex; hence also the pyramid of which ATBUCVDW is the base and the point L the vertex has to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex the ratio triplicate of that which BD has to FH. But, by hypothesis, the cone of which the circle ABCD  is the base and the point L the vertex has also to the solid O the ratio triplicate of that which BD has to FH; therefore, as the cone of which the circle ABCD is the base and the point L the vertex is to the solid O, so is the pyramid of which the polygon ATBUCVDW is the base and L the vertex to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex; therefore, alternately, as the cone of which the circle ABCD is the base and L the vertex is to the pyramid contained in it of which the polygon ATBUCVDW is the base and L the vertex, so is the solid O to the pyramid of which the polygon EPFQGRHS is the base and N the vertex. [V. 16] But the said cone is greater than the pyramid in it; for it encloses it. Therefore the solid O is also greater than the pyramid of which the polygon EPFQGRHS is the base and N the vertex. But it is also less: which is impossible. Therefore the cone of which the circle ABCD is the base and L the vertex has not to any solid less than the cone of which the circle EFGH is the base and the point N the vertex the ratio triplicate of that which BD has to FH: Similarly we can prove that neither has the cone EFGHN to any solid less than the cone ABCDL the ratio triplicate of that which FH has to BD. I say next that neither has the cone ABCDL to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH. For, if possible, let it have that ratio to a greater solid O. Therefore, inversely, the solid O has to the cone ABCDL the ratio triplicate of that which FH has to BD. But, as the solid O is to the cone ABCDL, so is the cone EFGHN to some solid less than the cone ABCDL. Therefore the cone EFGHN also has to some solid less than the cone ABCDL the ratio triplicate of that which FH has to BD: which was proved impossible. Therefore the cone ABCDL has not to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH. But it was proved that neither has it this ratio to a less solid than the cone EFGHN. Therefore the cone ABCDL has to the cone EFGHN the ratio triplicate of that which BD has to FH. But, as the cone is to the cone, so is the cylinder to the cylinder, for the cylinder which is on the same base as the cone and of equal height with it is triple of the cone; [XII. 10] therefore the cylinder also has to the cylinder the ratio triplicate of that which BD has to FH.'"],["Rule","'ProofWordCount'",1679],["Rule","'GreekProof'","'ἔστωσαν ὅμοιοι κῶνοι καὶ κύλινδροι, ὧν βάσεις μὲν οἱ ΑΒΓΔ, ΕΖΗΘ κύκλοι, διάμετροι δὲ τῶν βάσεων αἱ ΒΔ, ΖΘ, ἄξονες δὲ τῶν κώνων καὶ κυλίνδρων οἱ ΚΛ, ΜΝ: λέγω, ὅτι ὁ κῶνος, οὗ βάσις μὲν ἐστιν ὁ ΑΒΓΔ κύκλος, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς τὸν κῶνον, οὗ βάσις μὲν ἐστιν ὁ ΕΖΗΘ κύκλος, κορυφὴ δὲ τὸ Ν σημεῖον, τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. εἰ γὰρ μὴ ἔχει ὁ ΑΒΓΔΛ κῶνος πρὸς τὸν ΕΖΗΘΝ κῶνον τριπλασίονα λόγον ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ, ἕξει ὁ ΑΒΓΔΛ κῶνος ἢ πρὸς ἔλασσόν τι τοῦ ΕΖΗΘΝ κώνου στερεὸν τριπλασίονα λόγον ἢ πρὸς μεῖζον. ἐχέτω πρότερον πρὸς ἔλασσον τὸ Ξ, καὶ ἐγγεγράφθω εἰς τὸν ΕΖΗΘ κύκλον τετράγωνον τὸ ΕΖΗΘ: τὸ ἄρα ΕΖΗΘ τετράγωνον μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ ΕΖΗΘ κύκλου. καὶ ἀνεστάτω ἐπὶ τοῦ ΕΖΗΘ τετραγώνου πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ κώνῳ: ἡ ἄρα ἀνασταθεῖσα πυραμὶς μείζων ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ κώνου. τετμήσθωσαν δὴ αἱ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι δίχα κατὰ τὰ Ο, Π, Ρ, Σ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΕΟ, ΟΖ, ΖΠ, ΠΗ, ΗΡ, ΡΘ, ΘΣ, ΣΕ. καὶ ἕκαστον ἄρα τῶν ΕΟΖ, ΖΠΗ, ΗΡΘ, ΘΣΕ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾽ ἑαυτὸ τμήματος τοῦ ΕΖΗΘ κύκλου. καὶ ἀνεστάτω ἐφ᾽ ἑκάστου τῶν ΕΟΖ, ΖΠΗ, ΗΡΘ, ΘΣΕ τριγώνων πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ κώνῳ: καὶ ἑκάστη ἄρα τῶν ἀνασταθεισῶν πυραμίδων μείζων ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾽ ἑαυτὴν τμήματος τοῦ κώνου. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐφ᾽ ἑκάστου τῶν τριγώνων πυραμίδας τὴν αὐτὴν κορυφὴν ἐχούσας τῷ κώνῳ καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κώνου, ἃ ἔσται ἐλάσσονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ ΕΖΗΘΝ κῶνος τοῦ Ξ στερεοῦ. λελείφθω, καὶ ἔστω τὰ ἐπὶ τῶν ΕΟ, ΟΖ, ΖΠ, ΠΗ, ΗΡ, ΡΘ, ΘΣ, ΣΕ: λοιπὴ ἄρα ἡ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν σημεῖον, μείζων ἐστὶ τοῦ Ξ στερεοῦ. ἐγγεγράφθω καὶ εἰς τὸν ΑΒΓΔ κύκλον τῷ ΕΟΖΠΗΡΘΣ πολυγώνῳ ὅμοιόν τε καὶ ὁμοίως κείμενον πολύγωνον τὸ ΑΤΒΥΓΦΔΧ, καὶ ἀνεστάτω ἐπὶ τοῦ ΑΤΒΥΓΦΔΧ πολυγώνου πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ κώνῳ, καὶ τῶν μὲν περιεχόντων τὴν πυραμίδα, ἧς βάσις μέν ἐστι τὸ ΑΤΒΥ ΓΦΔΧ πολύγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, ἓν τρίγωνον ἔστω τὸ ΛΒΤ, τῶν δὲ περιεχόντων τὴν πυραμίδα, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν σημεῖον, ἓν τρίγωνον ἔστω τὸ ΝΖΟ, καὶ ἐπεζεύχθωσαν αἱ ΚΤ, ΜΟ. καὶ ἐπεὶ ὅμοιός ἐστιν ὁ ΑΒΓΔΛ κῶνος τῷ ΕΖΗΘΝ κώνῳ, ἔστιν ἄρα ὡς ἡ ΒΔ πρὸς τὴν ΖΘ, οὕτως ὁ ΚΛ ἄξων πρὸς τὸν ΜΝ ἄξονα. ὡς δὲ ἡ ΒΔ πρὸς τὴν ΖΘ, οὕτως ἡ ΒΚ πρὸς τὴν ΖΜ: καὶ ὡς ἄρα ἡ ΒΚ πρὸς τὴν ΖΜ, οὕτως ἡ ΚΛ πρὸς τὴν ΜΝ. καὶ ἐναλλὰξ ὡς ἡ ΒΚ πρὸς τὴν ΚΛ, οὕτως ἡ ΖΜ πρὸς τὴν ΜΝ. καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΒΚΛ, ΖΜΝ αἱ πλευραὶ ἀνάλογόν εἰσιν: ὅμοιον ἄρα ἐστὶ τὸ ΒΚΛ τρίγωνον τῷ ΖΜΝ τριγώνῳ. πάλιν, ἐπεί ἐστιν ὡς ἡ ΒΚ πρὸς τὴν ΚΤ, οὕτως ἡ ΖΜ πρὸς τὴν ΜΟ, καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΒΚΤ, ΖΜΟ, ἐπειδήπερ, ὃ μέρος ἐστὶν ἡ ὑπὸ ΒΚΤ γωνία τῶν πρὸς τῷ Κ κέντρῳ τεσσάρων ὀρθῶν, τὸ αὐτὸ μέρος ἐστὶ καὶ ἡ ὑπὸ ΖΜΟ γωνία τῶν πρὸς τῷ Μ κέντρῳ τεσσάρων ὀρθῶν: ἐπεὶ οὖν περὶ ἴσας γωνίας αἱ πλευραὶ ἀνάλογόν εἰσιν, ὅμοιον ἄρα ἐστὶ τὸ ΒΚΤ τρίγωνον τῷ ΖΜΟ τριγώνῳ. πάλιν, ἐπεὶ ἐδείχθη ὡς ἡ ΒΚ πρὸς τὴν ΚΛ, οὕτως ἡ ΖΜ πρὸς τὴν ΜΝ, ἴση δὲ ἡ μὲν ΒΚ τῇ ΚΤ, ἡ δὲ ΖΜ τῇ ΟΜ, ἔστιν ἄρα ὡς ἡ ΤΚ πρὸς τὴν ΚΛ, οὕτως ἡ ΟΜ πρὸς τὴν ΜΝ. καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΤΚΛ, ΟΜΝ: ὀρθαὶ γάρ: αἱ πλευραὶ ἀνάλογόν εἰσιν: ὅμοιον ἄρα ἐστὶ τὸ ΛΚΤ τρίγωνον τῷ ΝΜΟ τριγώνῳ. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν ΛΚΒ, ΝΜΖ τριγώνων ἐστὶν ὡς ἡ ΛΒ πρὸς τὴν ΒΚ, οὕτως ἡ ΝΖ πρὸς τὴν ΖΜ, διὰ δὲ τὴν ὁμοιότητα τῶν ΒΚΤ, ΖΜΟ τριγώνων ἐστὶν ὡς ἡ ΚΒ πρὸς τὴν ΒΤ, οὕτως ἡ ΜΖ πρὸς τὴν ΖΟ, δι᾽ ἴσου ἄρα ὡς ἡ ΛΒ πρὸς τὴν ΒΤ, οὕτως ἡ ΝΖ πρὸς τὴν ΖΟ. πάλιν, ἐπεὶ διὰ τὴν ὁμοιότητα τῶν ΛΤΚ, ΝΟΜ τριγώνων ἐστὶν ὡς ἡ ΛΤ πρὸς τὴν ΤΚ, οὕτως ἡ ΝΟ πρὸς τὴν ΟΜ, διὰ δὲ τὴν ὁμοιότητα τῶν ΤΚΒ, ΟΜΖ τριγώνων ἐστὶν ὡς ἡ ΚΤ πρὸς τὴν ΤΒ, οὕτως ἡ ΜΟ πρὸς τὴν ΟΖ, δι᾽ ἴσου ἄρα ὡς ἡ ΛΤ πρὸς τὴν ΤΒ, οὕτως ἡ ΝΟ πρὸς τὴν ΟΖ. ἐδείχθη δὲ καὶ ὡς ἡ ΤΒ πρὸς τὴν ΒΛ, οὕτως ἡ ΟΖ πρὸς τὴν ΖΝ. δι᾽ ἴσου ἄρα ὡς ἡ ΤΛ πρὸς τὴν ΛΒ, οὕτως ἡ ΟΝ πρὸς τὴν ΝΖ. τῶν ΛΤΒ, ΝΟΖ ἄρα τριγώνων ἀνάλογόν εἰσιν αἱ πλευραί: ἰσογώνια ἄρα ἐστὶ τὰ ΛΤΒ, ΝΟΖ τρίγωνα: ὥστε καὶ ὅμοια. καὶ πυραμὶς ἄρα, ἧς βάσις μὲν τὸ ΒΚΤ τρίγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μὲν τὸ ΖΜΟ τρίγωνον, κορυφὴ δὲ τὸ Ν σημεῖον: ὑπὸ γὰρ ὁμοίων ἐπιπέδων περιέχονται ἴσων τὸ πλῆθος. αἱ δὲ ὅμοιαι πυραμίδες καὶ τριγώνους ἔχουσαι βάσεις ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ἡ ἄρα ΒΚΤΛ πυραμὶς πρὸς τὴν ΖΜΟΝ πυραμίδα τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΚ πρὸς τὴν ΖΜ. ὁμοίως δὴ ἐπιζευγνύντες ἀπὸ τῶν Α, Χ, Δ, Φ, Γ, Υ ἐπὶ τὸ Κ εὐθείας καὶ ἀπὸ τῶν Ε, Σ, Θ, Ρ, Η, Π ἐπὶ τὸ Μ καὶ ἀνιστάντες ἐφ᾽ ἑκάστου τῶν τριγώνων πυραμίδας τὴν αὐτὴν κορυφὴν ἐχούσας τοῖς κώνοις δείξομεν, ὅτι καὶ ἑκάστη τῶν ὁμοταγῶν πυραμίδων πρὸς ἑκάστην ὁμοταγῆ πυραμίδα τριπλασίονα λόγον ἕξει ἤπερ ἡ ΒΚ ὁμόλογος πλευρὰ πρὸς τὴν ΖΜ ὁμόλογον πλευράν, τουτέστιν ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. καὶ ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ἔστιν ἄρα καὶ ὡς ἡ ΒΚΤΛ πυραμὶς πρὸς τὴν ΖΜΟΝ πυραμίδα, οὕτως ἡ ὅλη πυραμίς, ἧς βάσις τὸ ΑΤΒΥΓΦΔΧ πολύγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς τὴν ὅλην πυραμίδα, ἧς βάσις μὲν τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν σημεῖον: ὥστε καὶ πυραμίς, ἧς βάσις μὲν τὸ ΑΤΒΥΓΦΔΧ, κορυφὴ δὲ τὸ Λ, πρὸς τὴν πυραμίδα, ἧς βάσις μὲν τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν σημεῖον, τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. ὑπόκειται δὲ καὶ ὁ κῶνος, οὗ βάσις μὲν ὁ ΑΒΓΔ κύκλος, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς τὸ Ξ στερεὸν τριπλασίονα λόγον ἔχων ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ: ἔστιν ἄρα ὡς ὁ κῶνος, οὗ βάσις μέν ἐστιν ὁ ΑΒΓΔ κύκλος, κορυφὴ δὲ τὸ Λ, πρὸς τὸ Ξ στερεόν, οὕτως ἡ πυραμίς, ἧς βάσις μὲν τὸ ΑΤΒΥΓΦΔΧ πολύγωνον, κορυφὴ δὲ τὸ Λ, πρὸς τὴν πυραμίδα, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν: ἐναλλὰξ ἄρα, ὡς ὁ κῶνος, οὗ βάσις μὲν ὁ ΑΒΓΔ κύκλος, κορυφὴ δὲ τὸ Λ, πρὸς τὴν ἐν αὐτῷ πυραμίδα, ἧς βάσις μὲν τὸ ΑΤΒΥΓΦΔΧ πολύγωνον, κορυφὴ δὲ τὸ Λ, οὕτως τὸ Ξ στερεὸν πρὸς τὴν πυραμίδα, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν. μείζων δὲ ὁ εἰρημένος κῶνος τῆς ἐν αὐτῷ πυραμίδος: ἐμπεριέχει γὰρ αὐτήν. μεῖζον ἄρα καὶ τὸ Ξ στερεὸν τῆς πυραμίδος, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν. ἀλλὰ καὶ ἔλαττον: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ὁ κῶνος, οὗ βάσις ὁ ΑΒΓΔ κύκλος, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς ἔλαττόν τι τοῦ κώνου στερεόν, οὗ βάσις μὲν ὁ ΕΖΗΘ κύκλος, κορυφὴ δὲ τὸ Ν σημεῖον, τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ὁ ΕΖΗΘΝ κῶνος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔΛ κώνου στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΖΘ πρὸς τὴν ΒΔ. λέγω δή, ὅτι οὐδὲ ὁ ΑΒΓΔΛ κῶνος πρὸς μεῖζόν τι τοῦ ΕΖΗΘΝ κώνου στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. εἰ γὰρ δυνατόν, ἐχέτω πρὸς μεῖζον τὸ Ξ. ἀνάπαλιν ἄρα τὸ Ξ στερεὸν πρὸς τὸν ΑΒΓΔΛ κῶνον τριπλασίονα λόγον ἔχει ἤπερ ἡ ΖΘ πρὸς τὴν ΒΔ. ὡς δὲ τὸ Ξ στερεὸν πρὸς τὸν ΑΒΓΔΛ κῶνον, οὕτως ὁ ΕΖΗΘΝ κῶνος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔΛ κώνου στερεόν. καὶ ὁ ΕΖΗΘΝ ἄρα κῶνος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔΛ κώνου στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΖΘ πρὸς τὴν ΒΔ: ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα ὁ ΑΒΓΔΛ κῶνος πρὸς μεῖζόν τι τοῦ ΕΖΗΘΝ κώνου στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἔλαττον. ὁ ΑΒΓΔΛ ἄρα κῶνος πρὸς τὸν ΕΖΗΘΝ κῶνον τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. ὡς δὲ ὁ κῶνος πρὸς τὸν κῶνον, ὁ κύλινδρος πρὸς τὸν κύλινδρον: τριπλάσιος γὰρ ὁ κύλινδρος τοῦ κώνου ὁ ἐπὶ τῆς αὐτῆς βάσεως τῷ κώνῳ καὶ ἰσοϋψὴς αὐτῷ. καὶ ὁ κύλινδρος ἄρα πρὸς τὸν κύλινδρον τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. οἱ ἄρα ὅμοιοι κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ἐν ταῖς βάσεσι διαμέτρων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",1454]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'12.13'"],["Rule","'Text'","'If a cylinder be cut by a plane which is parallel to its opposite planes, then, as the cylinder is to the cylinder, so will the axis be to the axis.'"],["Rule","'TextWordCount'",31],["Rule","'GreekText'","'ἐὰν κύλινδρος ἐπιπέδῳ τμηθῇ παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔσται ὡς ὁ κύλινδρος πρὸς τὸν κύλινδρον, οὕτως ὁ ἄξων πρὸς τὸν ἄξονα.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Definition'",5]],["List",["Rule","'Book'",12],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'For let the cylinder AD be cut by the plane GH which is parallel to the opposite planes AB, CD, and let the plane GH meet the axis at the point K; I say that, as the cylinder BG is to the cylinder GD, so is the axis EK to the axis KF. For let the axis EF be produced in both directions to the points L, M, and let there be set out any number whatever of axes EN, NL equal to the axis EK, and any number whatever FO, OM equal to FK; and let the cylinder PW on the axis LM be conceived of which the circles PQ, VW are the bases. Let planes be carried through the points N, O parallel to AB, CD and to the bases of the cylinder PW, and let them produce the circles RS, TU about the centres N, O. Then, since the axes LN, NE, EK are equal to one another, therefore the cylinders QR, RB, BG are to one another as their bases. [XII. 11] But the bases are equal; therefore the cylinders QR, RB, BG are also equal to one another. Since then the axes LN, NE, EK are equal to one another, and the cylinders QR, RB, BG are also equal to one another, and the multitude of the former is equal to the multitude of the latter, therefore, whatever multiple the axis KL is of the axis EK, the same multiple also will the cylinder QG be of the cylinder GB. For the same reason, whatever multiple the axis MK is of the axis KF, the same multiple also is the cylinder WG of the cylinder GD. And, if the axis KL is equal to the axis KM, the cylinder QG will also be equal to the cylinder GW, if the axis is greater than the axis, the cylinder will also be greater than the cylinder, and if less, less. Thus, there being four magnitudes, the axes EK, KF and the cylinders BG, GD, there have been taken equimultiples of the axis EK and of the cylinder BG, namely the axis LK and the cylinder QG, and equimultiples of the axis KF and of the cylinder GD, namely the axis KM and the cylinder GW; and it has been proved that, if the axis KL is in excess of the axis KM, the cylinder QG is also in excess of the cylinder GW, if equal, equal, and if less, less. Therefore, as the axis EK is to the axis KF, so is the cylinder BG to the cylinder GD. [V. Def. 5]'"],["Rule","'ProofWordCount'",436],["Rule","'GreekProof'","'κύλινδρος γὰρ ὁ ΑΔ ἐπιπέδῳ τῷ ΗΘ τετμήσθω παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις τοῖς ΑΒ, ΓΔ, καὶ συμβαλλέτω τῷ ἄξονι τὸ ΗΘ ἐπίπεδον κατὰ τὸ Κ σημεῖον: λέγω, ὅτι ἐστὶν ὡς ὁ ΒΗ κύλινδρος πρὸς τὸν ΗΔ κύλινδρον, οὕτως ὁ ΕΚ ἄξων πρὸς τὸν ΚΖ ἄξονα. Ἐκβεβλήσθω γὰρ ὁ ΕΖ ἄξων ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ τὰ Λ, Μ σημεῖα, καὶ ἐκκείσθωσαν τῷ ΕΚ ἄξονι ἴσοι ὁσοιδηποτοῦν οἱ ΕΝ, ΝΛ, τῷ δὲ ΖΚ ἴσοι ὁσοιδηποτοῦν οἱ ΖΞ, ΞΜ, καὶ νοείσθω ὁ ἐπὶ τοῦ ΛΜ ἄξονος κύλινδρος ὁ ΟΧ, οὗ βάσεις οἱ ΟΠ, ΦΧ κύκλοι. καὶ ἐκβεβλήσθω διὰ τῶν Ν, Ξ σημείων ἐπίπεδα παράλληλα τοῖς ΑΒ, ΓΔ καὶ ταῖς βάσεσι τοῦ ΟΧ κυλίνδρου καὶ ποιείτωσαν τοὺς ΡΣ, ΤΥ κύκλους περὶ τὰ Ν, Ξ κέντρα. καὶ ἐπεὶ οἱ ΛΝ, ΝΕ, ΕΚ ἄξονες ἴσοι εἰσὶν ἀλλήλοις, οἱ ἄρα ΠΡ, ΡΒ, ΒΗ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις. ἴσαι δέ εἰσιν αἱ βάσεις: ἴσοι ἄρα καὶ οἱ ΠΡ, ΡΒ, ΒΗ κύλινδροι ἀλλήλοις. ἐπεὶ οὖν οἱ ΛΝ, ΝΕ, ΕΚ ἄξονες ἴσοι εἰσὶν ἀλλήλοις, εἰσὶ δὲ καὶ οἱ ΠΡ, ΡΒ, ΒΗ κύλινδροι ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῷ πλήθει, ὁσαπλασίων ἄρα ὁ ΚΛ ἄξων τοῦ ΕΚ ἄξονος, τοσαυταπλασίων ἔσται καὶ ὁ ΠΗ κύλινδρος τοῦ ΗΒ κυλίνδρου. διὰ τὰ αὐτὰ δὴ καὶ ὁσαπλασίων ἐστὶν ὁ ΜΚ ἄξων τοῦ ΚΖ ἄξονος, τοσαυταπλασίων ἐστὶ καὶ ὁ ΧΗ κύλινδρος τοῦ ΗΔ κυλίνδρου. καὶ εἰ μὲν ἴσος ἐστὶν ὁ ΚΛ ἄξων τῷ ΚΜ ἄξονι, ἴσος ἔσται καὶ ὁ ΠΗ κύλινδρος τῷ ΗΧ κυλίνδρῳ, εἰ δὲ μείζων ὁ ἄξων τοῦ ἄξονος, μείζων καὶ ὁ κύλινδρος τοῦ κυλίνδρου, καὶ εἰ ἐλάσσων, ἐλάσσων. τεσσάρων δὴ μεγεθῶν ὄντων, ἀξόνων μὲν τῶν ΕΚ, ΚΖ, κυλίνδρων δὲ τῶν ΒΗ, ΗΔ, εἴληπται ἰσάκις πολλαπλάσια, τοῦ μὲν ΕΚ ἄξονος καὶ τοῦ ΒΗ κυλίνδρου ὅ τε ΛΚ ἄξων καὶ ὁ ΠΗ κύλινδρος, τοῦ δὲ ΚΖ ἄξονος καὶ τοῦ ΗΔ κυλίνδρου ὅ τε ΚΜ ἄξων καὶ ὁ ΗΧ κύλινδρος, καὶ δέδεικται, ὅτι εἰ ὑπερέχει ὁ ΚΛ ἄξων τοῦ ΚΜ ἄξονος, ὑπερέχει καὶ ὁ ΠΗ κύλινδρος τοῦ ΗΧ κυλίνδρου, καὶ εἰ ἴσος, ἴσος, καὶ εἰ ἐλάσσων, ἐλάσσων. ἔστιν ἄρα ὡς ὁ ΕΚ ἄξων πρὸς τὸν ΚΖ ἄξονα, οὕτως ὁ ΒΗ κύλινδρος πρὸς τὸν ΗΔ κύλινδρον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",360]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'12.14'"],["Rule","'Text'","'Cones and cylinders which are on equal bases are to one another as their heights.'"],["Rule","'TextWordCount'",15],["Rule","'GreekText'","'οἱ ἐπὶ ἴσων βάσεων ὄντες κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ὕψη.'"],["Rule","'GreekTextWordCount'",14],["Rule","'References'",["List",["List",["Rule","'Book'",12],["Rule","'Theorem'",10]],["List",["Rule","'Book'",12],["Rule","'Theorem'",11]],["List",["Rule","'Book'",12],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'For let EB, FD be cylinders on equal bases, the circles AB, CD; I say that, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL. For let the axis KL be produced to the point N, let LN be made equal to the axis GH, and let the cylinder CM be conceived about LN as axis. Since then the cylinders EB, CM are of the same height, they are to one another as their bases. [XII. 11] But the bases are equal to one another; therefore the cylinders EB, CM are also equal. And, since the cylinder FM has been cut by the plane CD which is parallel to its opposite planes, therefore, as the cylinder CM is to the cylinder FD, so is the axis LN to the axis KL. [XII. 13] But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL. But, as the cylinder EB is to the cylinder FD, so is the cone ABG to the cone CDK. [XII. 10] Therefore also, as the axis GH is to the axis KL, so is the cone ABG to the cone CDK and the cylinder EB to the cylinder FD.'"],["Rule","'ProofWordCount'",228],["Rule","'GreekProof'","'ἔστωσαν γὰρ ἐπὶ ἴσων βάσεων τῶν ΑΒ, ΓΔ κύκλων κύλινδροι οἱ ΕΒ, ΖΔ: λέγω, ὅτι ἐστὶν ὡς ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΗΘ ἄξων πρὸς τὸν ΚΛ ἄξονα. Ἐκβεβλήσθω γὰρ ὁ ΚΛ ἄξων ἐπὶ τὸ Ν σημεῖον, καὶ κείσθω τῷ ΗΘ ἄξονι ἴσος ὁ ΛΝ, καὶ περὶ ἄξονα τὸν ΛΝ κύλινδρος νενοήσθω ὁ ΓΜ. ἐπεὶ οὖν οἱ ΕΒ, ΓΜ κύλινδροι ὑπὸ τὸ αὐτὸ ὕψος εἰσίν, πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις. ἴσαι δέ εἰσιν αἱ βάσεις ἀλλήλαις: ἴσοι ἄρα εἰσὶ καὶ οἱ ΕΒ, ΓΜ κύλινδροι. καὶ ἐπεὶ κύλινδρος ὁ ΖΜ ἐπιπέδῳ τέτμηται τῷ ΓΔ παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ἄρα ὡς ὁ ΓΜ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΛΝ ἄξων πρὸς τὸν ΚΛ ἄξονα. ἴσος δέ ἐστιν ὁ μὲν ΓΜ κύλινδρος τῷ ΕΒ κυλίνδρῳ, ὁ δὲ ΛΝ ἄξων τῷ ΗΘ ἄξονι: ἔστιν ἄρα ὡς ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΗΘ ἄξων πρὸς τὸν ΚΛ ἄξονα. ὡς δὲ ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΑΒΗ κῶνος πρὸς τὸν ΓΔΚ κῶνον. καὶ ὡς ἄρα ὁ ΗΘ ἄξων πρὸς τὸν ΚΛ ἄξονα, οὕτως ὁ ΑΒΗ κῶνος πρὸς τὸν ΓΔΚ κῶνον καὶ ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",202]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'12.15'"],["Rule","'Text'","'In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'τῶν ἴσων κώνων καὶ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν: καὶ ὧν κώνων καὶ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἴσοι εἰσὶν ἐκεῖνοι.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",7]],["List",["Rule","'Book'",5],["Rule","'Theorem'",9]],["List",["Rule","'Book'",5],["Rule","'Theorem'",11]],["List",["Rule","'Book'",12],["Rule","'Theorem'",11]],["List",["Rule","'Book'",12],["Rule","'Theorem'",13]]]],["Rule","'Proof'","'Let there be equal cones and cylinders of which the circles ABCD, EFGH are the bases; let AC, EG be the diameters of the bases, and KL, MN the axes, which are also the heights of the cones or cylinders; let the cylinders AO, EP be completed. I say that in the cylinders AO, EP the bases are reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so is the height MN to the height KL. For the height LK is either equal to the height MN or not equal. First, let it be equal. Now the cylinder AO is also equal to the cylinder EP. But cones and cylinders which are of the same height are to one another as their bases; [XII. 11] therefore the base ABCD is also equal to the base EFGH. Hence also, reciprocally, as the base ABCD is to the base EFGH, so is the height MN to the height KL. Next, let the height LK not be equal to MN, but let MN be greater; from the height MN let QN be cut off equal to KL, through the point Q let the cylinder EP be cut by the plane TUS parallel to the planes of the circles EFGH, RP, and let the cylinder ES be conceived erected from the circle EFGH as base and with height NQ. Now, since the cylinder AO is equal to the cylinder EP, therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 7] But, as the cylinder AO is to the cylinder ES, so is the base ABCD to the base EFGH, for the cylinders AO, ES are of the same height; [XII. 11] and, as the cylinder EP is to the cylinder ES, so is the height MN to the height QN, for the cylinder EP has been cut by a plane which is parallel to its opposite planes. [XII. 13] Therefore also, as the base ABCD is to the base EFGH, so is the height MN to the height QN. [V. 11] But the height QN is equal to the height KL; therefore, as the base ABCD is to the base EFGH, so is the height MN to the height KL. Therefore in the cylinders AO, EP the bases are reciprocally proportional to the heights. Next, in the cylinders AO, EP let the bases be reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so let the height MN be to the height KL; I say that the cylinder AO is equal to the cylinder EP. For, with the same construction, since, as the base ABCD is to the base EFGH, so is the height MN to the height KL, while the height KL is equal to the height QN, therefore, as the base ABCD is to the base EFGH, so is the height MN to the height QN   But, as the base ABCD is to the base EFGH, so is the cylinder AO to the cylinder ES, for they are of the same height; [XII. 11] and, as the height MN is to QN, so is the cylinder EP to the cylinder ES; [XII. 13] therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 11] Therefore the cylinder AO is equal to the cylinder EP. [V. 9] And the same is true for the cones also.'"],["Rule","'ProofWordCount'",588],["Rule","'GreekProof'","'ἔστωσαν ἴσοι κῶνοι καὶ κύλινδροι, ὧν βάσεις μὲν οἱ ΑΒΓΔ, ΕΖΗΘ κύκλοι, διάμετροι δὲ αὐτῶν αἱ ΑΓ, ΕΗ, ἄξονες δὲ οἱ ΚΛ, ΜΝ, οἵτινες καὶ ὕψη εἰσὶ τῶν κώνων ἢ κυλίνδρων, καὶ συμπεπληρώσθωσαν οἱ ΑΞ, ΕΟ κύλινδροι. λέγω, ὅτι τῶν ΑΞ, ΕΟ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, καί ἐστιν ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος. τὸ γὰρ ΛΚ ὕψος τῷ ΜΝ ὕψει ἤτοι ἴσον ἐστὶν ἢ οὔ. ἔστω πρότερον ἴσον. ἔστι δὲ καὶ ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ ἴσος. οἱ δὲ ὑπὸ τὸ αὐτὸ ὕψος ὄντες κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις: ἴση ἄρα καὶ ἡ ΑΒΓΔ βάσις τῇ ΕΖΗΘ βάσει. ὥστε καὶ ἀντιπέπονθεν, ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος. ἀλλὰ δὴ μὴ ἔστω τὸ ΛΚ ὕψος τῷ ΜΝ ἴσον, ἀλλ᾽ ἔστω μεῖζον τὸ ΜΝ, καὶ ἀφῃρήσθω ἀπὸ τοῦ ΜΝ ὕψους τῷ ΚΛ ἴσον τὸ ΠΝ, καὶ διὰ τοῦ Π σημείου τετμήσθω ὁ ΕΟ κύλινδρος ἐπιπέδῳ τῷ ΤΥΣ παραλλήλῳ τοῖς τῶν ΕΖΗΘ, ΡΟ κύκλων ἐπιπέδοις, καὶ ἀπὸ βάσεως μὲν τοῦ ΕΖΗΘ κύκλου, ὕψους δὲ τοῦ ΝΠ κύλινδρος νενοήσθω ὁ ΕΣ. καὶ ἐπεὶ ἴσος ἐστὶν ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ, ἔστιν ἄρα ὡς ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον, οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον. ἀλλ᾽ ὡς μὲν ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον, οὕτως ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ: ὑπὸ γὰρ τὸ αὐτὸ ὕψος εἰσὶν οἱ ΑΞ, ΕΣ κύλινδροι: ὡς δὲ ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος: ὁ γὰρ ΕΟ κύλινδρος ἐπιπέδῳ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις. ἔστιν ἄρα καὶ ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος. ἴσον δὲ τὸ ΠΝ ὕψος τῷ ΚΛ ὕψει: ἔστιν ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος. τῶν ἄρα ΑΞ, ΕΟ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. ἀλλὰ δὴ τῶν ΑΞ, ΕΟ κυλίνδρων ἀντιπεπονθέτωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος: λέγω, ὅτι ἴσος ἐστὶν ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ. τῶν γὰρ αὐτῶν κατασκευασθέντων ἐπεί ἐστιν ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος, ἴσον δὲ τὸ ΚΛ ὕψος τῷ ΠΝ ὕψει, ἔστιν ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος. ἀλλ᾽ ὡς μὲν ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον: ὑπὸ γὰρ τὸ αὐτὸ ὕψος εἰσίν: ὡς δὲ τὸ ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος, οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον: ἔστιν ἄρα ὡς ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον, οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ. ἴσος ἄρα ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ. ὡσαύτως δὲ καὶ ἐπὶ τῶν κώνων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",497]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'12.16'"],["Rule","'Text'","'Given two circles about the same centre, to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle.'"],["Rule","'TextWordCount'",29],["Rule","'GreekText'","'δύο κύκλων περὶ τὸ αὐτὸ κέντρον ὄντων εἰς τὸν μείζονα κύκλον πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον ἐγγράψαι μὴ ψαῦον τοῦ ἐλάσσονος κύκλου.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",3],["Rule","'Theorem'",3]],["List",["Rule","'Book'",3],["Rule","'Theorem'",16]],["List",["Rule","'Book'",10],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let ABCD, EFGH be the two given circles about the same centre K; thus it is required to inscribe in the greater circle ABCD an equilateral polygon with an even number of sides which does not touch the circle EFGH. For let the straight line BKD be drawn through the centre K, and from the point G let GA be drawn at right angles to the straight line BD and carried through to C; therefore AC touches the circle EFGH. [III. 16] Then, bisecting the circumference BAD, bisecting the half of it, and doing this continually, we shall leave a circumference less than AD. [X. 1] Let such be left, and let it be LD; from L let LM be drawn perpendicular to BD and carried through to N, and let LD, DN be joined; therefore LD is equal to DN. [III. 3, I. 4] Now, since LN is parallel to AC, and AC touches the circle EFGH, therefore LN does not touch the circle EFGH; therefore LD, DN are far from touching the circle EFGH. If then we fit into the circle ABCD straight lines equal to the straight line LD and placed continuously, there will be inscribed in the circle ABCD an equilateral polygon with an even number of sides which does not touch the lesser circle EFGH.'"],["Rule","'ProofWordCount'",220],["Rule","'GreekProof'","'ἔστωσαν οἱ δοθέντες δύο κύκλοι οἱ ΑΒΓΔ, ΕΖΗΘ περὶ τὸ αὐτὸ κέντρον τὸ Κ: δεῖ δὴ εἰς τὸν μείζονα κύκλον τὸν ΑΒΓΔ πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον ἐγγράψαι μὴ ψαῦον τοῦ ΕΖΗΘ κύκλου. ἤχθω γὰρ διὰ τοῦ Κ κέντρου εὐθεῖα ἡ ΒΚΔ, καὶ ἀπὸ τοῦ Η σημείου τῇ ΒΔ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ ΗΑ καὶ διήχθω ἐπὶ τὸ Γ: ἡ ΑΓ ἄρα ἐφάπτεται τοῦ ΕΖΗΘ κύκλου. τέμνοντες δὴ τὴν ΒΑΔ περιφέρειαν δίχα καὶ τὴν ἡμίσειαν αὐτῆς δίχα καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομεν περιφέρειαν ἐλάσσονα τῆς ΑΔ. λελείφθω, καὶ ἔστω ἡ ΛΔ, καὶ ἀπὸ τοῦ Λ ἐπὶ τὴν ΒΔ κάθετος ἤχθω ἡ ΛΜ καὶ διήχθω ἐπὶ τὸ Ν, καὶ ἐπεζεύχθωσαν αἱ ΛΔ, ΔΝ: ἴση ἄρα ἐστὶν ἡ ΛΔ τῇ ΔΝ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΛΝ τῇ ΑΓ, ἡ δὲ ΑΓ ἐφάπτεται τοῦ ΕΖΗΘ κύκλου, ἡ ΛΝ ἄρα οὐκ ἐφάπτεται τοῦ ΕΖΗΘ κύκλου: πολλῷ ἄρα αἱ ΛΔ, ΔΝ οὐκ ἐφάπτονται τοῦ ΕΖΗΘ κύκλου. ἐὰν δὴ τῇ ΛΔ εὐθείᾳ ἴσας κατὰ τὸ συνεχὲς ἐναρμόσωμεν εἰς τὸν ΑΒΓΔ κύκλον, ἐγγραφήσεται εἰς τὸν ΑΒΓΔ κύκλον πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον μὴ ψαῦον τοῦ ἐλάσσονος κύκλου τοῦ ΕΖΗΘ: ὅπερ ἔδει ποιῆσαι.'"],["Rule","'GreekProofWordCount'",187]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'12.17'"],["Rule","'Text'","'Given two spheres about the same centre, to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'δύο σφαιρῶν περὶ τὸ αὐτὸ κέντρον οὐσῶν εἰς τὴν μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγράψαι μὴ ψαῦον τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",1],["Rule","'Theorem'",33]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Theorem'",27]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",6],["Rule","'Theorem'",2]],["List",["Rule","'Book'",6],["Rule","'Theorem'",8]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Definition'",4]],["List",["Rule","'Book'",11],["Rule","'Definition'",14]],["List",["Rule","'Book'",11],["Rule","'Theorem'",2]],["List",["Rule","'Book'",11],["Rule","'Theorem'",6]],["List",["Rule","'Book'",11],["Rule","'Theorem'",7]],["List",["Rule","'Book'",11],["Rule","'Theorem'",9]],["List",["Rule","'Book'",11],["Rule","'Theorem'",11]],["List",["Rule","'Book'",11],["Rule","'Theorem'",18]]]],["Rule","'Proof'","'Let two spheres be conceived about the same centre A; thus it is required to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface. Let the spheres be cut by any plane through the centre; then the sections will be circles, inasmuch as the sphere was produced by the diameter remaining fixed and the semicircle being carried round it; [XI. Def. 14] hence, in whatever position we conceive the semicircle to be, the plane carried through it will produce a circle on the circumference of the sphere. And it is manifest that this circle is the greatest possible, inasmuch as the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere. Let then BCDE be the circle in the greater sphere, and FGH the circle in the lesser sphere; let two diameters in them, BD, CE, be drawn at right angles to one another; then, given the two circles BCDE, FGH about the same centre, let there be inscribed in the greater circle BCDE an equilateral polygon with an even number of sides which does not touch the lesser circle FGH, let BK, KL, LM   ME be its sides in the quadrant BE. let KA be joined and carried through to N, let AO be set up from the point A at right angles to the plane of the circle BCDE, and let it meet the surface of the sphere at O, and through AO and each of the straight lines BD, KN let planes be carried; they will then make greatest circles on the surface of the sphere, for the reason stated. Let them make such, and in them let BOD, KON be the semicircles on BD, KN. Now, since OA is at right angles to the plane of the circle BCDE, therefore all the planes through OA are also at right angles to the plane of the circle BCDE; [XI. 18] hence the semicircles BOD, KON are also at right angles to the plane of the circle BCDE. And, since the semicircles BED, BOD, KON are equal, for they are on the equal diameters BD, KN, therefore the quadrants BE, BO, KO are also equal to one another. Therefore there are as many straight lines in the quadrants BO, KO equal to the straight lines BK, KL, LM, ME as there are sides of the polygon in the quadrant BE. Let them be inscribed, and let them be BP, PQ, QR, RO and KS, ST, TU, UO, let SP, TQ, UR be joined, and from P, S let perpendiculars be drawn to the plane of the circle BCDE; [XI. 11] these will fall on BD, KN, the common sections of the planes, inasmuch as the planes of BOD, KON are also at right angles to the plane of the circle BCDE. [cf. XI. Def. 4] Let them so fall, and let them be PV, SW, and let WV be joined. Now since, in the equal semicircles BOD, KON, equal straight lines BP, KS have been cut off, and the perpendiculars PV, SW have been drawn, therefore PV is equal to SW, and BV to KW. [III. 27, I. 26] But the whole BA is also equal to the whole KA; therefore the remainder VA is also equal to the remainder WA; therefore, as BV is to VA, so is KW to WA; therefore WV is parallel to KB. [VI. 2] And, since each of the straight lines PV, SW is at right angles to the plane of the circle BCDE, therefore PV is parallel to SW. [XI. 6] But it was also proved equal to it; therefore WV, SP are also equal and parallel. [I. 33] And, since WV is parallel to SP, while WV is parallel to KB, therefore SP is also parallel to KB. [XI. 9] And BP, KS join their extremities; therefore the quadrilateral KBPS is in one plane, inasmuch as, if two straight lines be parallel, and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallels. [XI. 7] For the same reason each of the quadrilaterals SPQT, TQRU is also in one plane. But the triangle URO is also in one plane. [XI. 2] If then we conceive straight lines joined from the points P, S, Q, T, R, U to A, there will be constructed a certain polyhedral solid figure between the circumferences BO, KO, consisting of pyramids of which the quadrilaterals KBPS, SPQT, TQRU and the triangle URO are the bases and the point A the vertex. And, if we make the same construction in the case of each of the sides KL, LM, ME as in the case of BK, and further in the case of the remaining three quadrants, there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids, of which the said quadrilaterals and the triangle URO, and the others corresponding to them, are the bases and the point A the vertex. I say that the said polyhedron will not touch the lesser sphere at the surface on which the circle FGH is. Let AX be drawn from the point A perpendicular to the plane of the quadrilateral KBPS, and let it meet the plane at the point X; [XI. 11] let XB, XK be joined. Then, since AX is at right angles to the plane of the quadrilateral KBPS, therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. [XI. Def. 3] Therefore AX is at right angles to each of the straight lines BX, XK. And, since AB is equal to AK, the square on AB is also equal to the square on AK. And the squares on AX, XB are equal to the square on AB, for the angle at X is right; [I. 47] and the squares on AX, XK are equal to the square on AK. [id.] Therefore the squares on AX, XB are equal to the squares on AX, XK. Let the square on AX be subtracted from each; therefore the remainder, the square on BX, is equal to the remainder, the square on XK; therefore BX is equal to XK. Similarly we can prove that the straight lines joined from X to P, S are equal to each of the straight lines BX, XK. Therefore the circle described with centre X and distance one of the straight lines XB, XK will pass through P, S also, and KBPS will be a quadrilateral in a circle. Now, since KB is greater than WV, while WV is equal to SP, therefore KB is greater than SP. But KB is equal to each of the straight lines KS, BP; therefore each of the straight lines KS, BP is greater than SP. And, since KBPS is a quadrilateral in a circle, and KB, BP, KS are equal, and PS less, and BX is the radius of the circle, therefore the square on KB is greater than double of the square on BX. Let KZ be drawn from K perpendicular to BV. Then, since BD is less than double of DZ, and, as BD is to DZ, so is the rectangle DB, BZ to the rectangle DZ, ZB, if a square be described upon BZ and the parallelogram on ZD be completed, then the rectangle DB, BZ is also less than double of the rectangle DZ, ZB. And, if KD be joined, the rectangle DB, BZ is equal to the square on BK, and the rectangle DZ, ZB equal to the square on KZ; [III. 31, VI. 8 and Por.] therefore the square on KB is less than double of the square on KZ. But the square on KB is greater than double of the square on BX; therefore the square on KZ is greater than the square on BX. And, since BA is equal to KA, the square on BA is equal to the square on AK. And the squares on BX, XA are equal to the square on BA, and the squares on KZ, ZA equal to the square on KA; [I. 47] therefore the squares on BX, XA are equal to the squares on KZ, ZA, and of these the square on KZ is greater than the square on BX; therefore the remainder, the square on ZA, is less than the square on XA. Therefore AX is greater than AZ; therefore AX is much greater than AG. And AX is the perpendicular on one base of the polyhedron, and AG on the surface of the lesser sphere; hence the polyhedron will not touch the lesser sphere on its surface. Therefore, given two spheres about the same centre, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface.'"],["Rule","'ProofWordCount'",1508],["Rule","'GreekProof'","'Νενοήσθωσαν δύο σφαῖραι περὶ τὸ αὐτὸ κέντρον τὸ Α: δεῖ δὴ εἰς τὴν μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγράψαι μὴ ψαῦον τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν. τετμήσθωσαν αἱ σφαῖραι ἐπιπέδῳ τινὶ διὰ τοῦ κέντρου: ἔσονται δὴ αἱ τομαὶ κύκλοι, ἐπειδήπερ μενούσης τῆς διαμέτρου καὶ περιφερομένου τοῦ ἡμικυκλίου ἐγίγνετο ἡ σφαῖρα: ὥστε καὶ καθ᾽ οἵας ἂν θέσεως ἐπινοήσωμεν τὸ ἡμικύκλιον, τὸ δι᾽ αὐτοῦ ἐκβαλλόμενον ἐπίπεδον ποιήσει ἐπὶ τῆς ἐπιφανείας τῆς σφαίρας κύκλον. καὶ φανερόν, ὅτι καὶ μέγιστον, ἐπειδήπερ ἡ διάμετρος τῆς σφαίρας, ἥτις ἐστὶ καὶ τοῦ ἡμικυκλίου διάμετρος δηλαδὴ καὶ τοῦ κύκλου, μείζων ἐστὶ πασῶν τῶν εἰς τὸν κύκλον ἢ τὴν σφαῖραν διαγομένων εὐθειῶν. ἔστω οὖν ἐν μὲν τῇ μείζονι σφαίρᾳ κύκλος ὁ ΒΓΔΕ, ἐν δὲ τῇ ἐλάσσονι σφαίρᾳ κύκλος ὁ ΖΗΘ, καὶ ἤχθωσαν αὐτῶν δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ ΒΔ, ΓΕ, καὶ δύο κύκλων περὶ τὸ αὐτὸ κέντρον ὄντων τῶν ΒΓΔΕ, ΖΗΘ εἰς τὸν μείζονα κύκλον τὸν ΒΓΔΕ πολύγωνον ἰσόπλευρον καὶ ἀρτιόπλευρον ἐγγεγράφθω μὴ ψαῦον τοῦ ἐλάσσονος κύκλου τοῦ ΖΗΘ, οὗ πλευραὶ ἔστωσαν ἐν τῷ ΒΕ τεταρτημορίῳ αἱ ΒΚ, ΚΛ, ΛΜ, ΜΕ, καὶ ἐπιζευχθεῖσα ἡ ΚΑ διήχθω ἐπὶ τὸ Ν, καὶ ἀνεστάτω ἀπὸ τοῦ Α σημείου τῷ τοῦ ΒΓΔΕ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΑΞ καὶ συμβαλλέτω τῇ ἐπιφανείᾳ τῆς σφαίρας κατὰ τὸ Ξ, καὶ διὰ τῆς ΑΞ καὶ ἑκατέρας τῶν ΒΔ, ΚΝ ἐπίπεδα ἐκβεβλήσθω: ποιήσουσι δὴ διὰ τὰ εἰρημένα ἐπὶ τῆς ἐπιφανείας τῆς σφαίρας μεγίστους κύκλους. ποιείτωσαν, ὧν ἡμικύκλια ἔστω ἐπὶ τῶν ΒΔ, ΚΝ διαμέτρων τὰ ΒΞΔ, ΚΞΝ. καὶ ἐπεὶ ἡ ΞΑ ὀρθή ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον, καὶ πάντα ἄρα τὰ διὰ τῆς ΞΑ ἐπίπεδά ἐστιν ὀρθὰ πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον: ὥστε καὶ τὰ ΒΞΔ, ΚΞΝ ἡμικύκλια ὀρθά ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον. καὶ ἐπεὶ ἴσα ἐστὶ τὰ ΒΕΔ, ΒΞΔ, ΚΞΝ ἡμικύκλια: ἐπὶ γὰρ ἴσων εἰσὶ διαμέτρων τῶν ΒΔ, ΚΝ: ἴσα ἐστὶ καὶ τὰ ΒΕ, ΒΞ, ΚΞ τεταρτημόρια ἀλλήλοις. ὅσαι ἄρα εἰσὶν ἐν τῷ ΒΕ τεταρτημορίῳ πλευραὶ τοῦ πολυγώνου, τοσαῦταί εἰσι καὶ ἐν τοῖς ΒΞ, ΚΞ τεταρτημορίοις ἴσαι ταῖς ΒΚ, ΚΛ, ΛΜ, ΜΕ εὐθείαις. ἐγγεγράφθωσαν καὶ ἔστωσαν αἱ ΒΟ, ΟΠ, ΠΡ, ΡΞ, ΚΣ, ΣΤ, ΤΥ, ΥΞ, καὶ ἐπεζεύχθωσαν αἱ ΣΟ, ΤΠ, ΥΡ, καὶ ἀπὸ τῶν Ο, Σ ἐπὶ τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον κάθετοι ἤχθωσαν: πεσοῦνται δὴ ἐπὶ τὰς κοινὰς τομὰς τῶν ἐπιπέδων τὰς ΒΔ, ΚΝ, ἐπειδήπερ καὶ τὰ τῶν ΒΞΔ, ΚΞΝ ἐπίπεδα ὀρθά ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον. πιπτέτωσαν, καὶ ἔστωσαν αἱ ΟΦ, ΣΧ, καὶ ἐπεζεύχθω ἡ ΧΦ. καὶ ἐπεὶ ἐν ἴσοις ἡμικυκλίοις τοῖς ΒΞΔ, ΚΞΝ ἴσαι ἀπειλημμέναι εἰσὶν αἱ ΒΟ, ΚΣ, καὶ κάθετοι ἠγμέναι εἰσὶν αἱ ΟΦ, ΣΧ, ἴση ἄρα ἐστὶν ἡ μὲν ΟΦ τῇ ΣΧ, ἡ δὲ ΒΦ τῇ ΚΧ. ἔστι δὲ καὶ ὅλη ἡ ΒΑ ὅλῃ τῇ ΚΑ ἴση: καὶ λοιπὴ ἄρα ἡ ΦΑ λοιπῇ τῇ ΧΑ ἐστιν ἴση: ἔστιν ἄρα ὡς ἡ ΒΦ πρὸς τὴν ΦΑ, οὕτως ἡ ΚΧ πρὸς τὴν ΧΑ: παράλληλος ἄρα ἐστὶν ἡ ΧΦ τῇ ΚΒ. καὶ ἐπεὶ ἑκατέρα τῶν ΟΦ, ΣΧ ὀρθή ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον, παράλληλος ἄρα ἐστὶν ἡ ΟΦ τῇ ΣΧ. ἐδείχθη δὲ αὐτῇ καὶ ἴση: καὶ αἱ ΧΦ, ΣΟ ἄρα ἴσαι εἰσὶ καὶ παράλληλοι. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΧΦ τῇ ΣΟ, ἀλλὰ ἡ ΧΦ τῇ ΚΒ ἐστι παράλληλος, καὶ ἡ ΣΟ ἄρα τῇ ΚΒ ἐστι παράλληλος. καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΒΟ, ΚΣ: τὸ ΚΒΟΣ ἄρα τετράπλευρον ἐν ἑνί ἐστιν ἐπιπέδῳ, ἐπειδήπερ, ἐὰν ὦσι δύο εὐθεῖαι παράλληλοι, καὶ ἐφ᾽ ἑκατέρας αὐτῶν ληφθῇ τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις. διὰ τὰ αὐτὰ δὴ καὶ ἑκάτερον τῶν ΣΟΠΤ, ΤΠΡΥ τετραπλεύρων ἐν ἑνί ἐστιν ἐπιπέδῳ. ἔστι δὲ καὶ τὸ ΥΡΞ τρίγωνον ἐν ἑνὶ ἐπιπέδῳ. ἐὰν δὴ νοήσωμεν ἀπὸ τῶν Ο, Σ, Π, Τ, Ρ, Υ σημείων ἐπὶ τὸ Α ἐπιζευγνυμένας εὐθείας, συσταθήσεταί τι σχῆμα στερεὸν πολύεδρον μεταξὺ τῶν ΒΞ, ΚΞ περιφερειῶν ἐκ πυραμίδων συγκείμενον, ὧν βάσεις μὲν τὰ ΚΒΟΣ, ΣΟΠΤ, ΤΠΡΥ τετράπλευρα καὶ τὸ ΥΡΞ τρίγωνον, κορυφὴ δὲ τὸ Α σημεῖον. ἐὰν δὲ καὶ ἐπὶ ἑκάστης τῶν ΚΛ, ΛΜ, ΜΕ πλευρῶν καθάπερ ἐπὶ τῆς ΒΚ τὰ αὐτὰ κατασκευάσωμεν καὶ ἔτι ἐπὶ τῶν λοιπῶν τριῶν τεταρτημορίων, συσταθήσεταί τι σχῆμα πολύεδρον ἐγγεγραμμένον εἰς τὴν σφαῖραν πυραμίσι περιεχόμενον, ὧν βάσεις μὲν τὰ εἰρημένα τετράπλευρα καὶ τὸ ΥΡΞ τρίγωνον καὶ τὰ ὁμοταγῆ αὐτοῖς, κορυφὴ δὲ τὸ Α σημεῖον. λέγω, ὅτι τὸ εἰρημένον πολύεδρον οὐκ ἐφάψεται τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν, ἐφ᾽ ἧς ἐστιν ὁ ΖΗΘ κύκλος. ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὸ τοῦ ΚΒΟΣ τετραπλεύρου ἐπίπεδον κάθετος ἡ ΑΨ καὶ συμβαλλέτω τῷ ἐπιπέδῳ κατὰ τὸ Ψ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΨΒ, ΨΚ. καὶ ἐπεὶ ἡ ΑΨ ὀρθή ἐστι πρὸς τὸ τοῦ ΚΒΟΣ τετραπλεύρου ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ τοῦ τετραπλεύρου ἐπιπέδῳ ὀρθή ἐστιν. ἡ ΑΨ ἄρα ὀρθή ἐστι πρὸς ἑκατέραν τῶν ΒΨ, ΨΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΑΚ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΒ τῷ ἀπὸ τῆς ΑΚ. καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΒ ἴσα τὰ ἀπὸ τῶν ΑΨ, ΨΒ: ὀρθὴ γὰρ ἡ πρὸς τῷ Ψ: τῷ δὲ ἀπὸ τῆς ΑΚ ἴσα τὰ ἀπὸ τῶν ΑΨ, ΨΚ. τὰ ἄρα ἀπὸ τῶν ΑΨ, ΨΒ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΑΨ, ΨΚ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΑΨ: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΒΨ λοιπῷ τῷ ἀπὸ τῆς ΨΚ ἴσον ἐστίν: ἴση ἄρα ἡ ΒΨ τῇ ΨΚ. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ἀπὸ τοῦ Ψ ἐπὶ τὰ Ο, Σ ἐπιζευγνύμεναι εὐθεῖαι ἴσαι εἰσὶν ἑκατέρᾳ τῶν ΒΨ, ΨΚ. ὁ ἄρα κέντρῳ τῷ Ψ καὶ διαστήματι ἑνὶ τῶν ΨΒ, ΨΚ γραφόμενος κύκλος ἥξει καὶ διὰ τῶν Ο, Σ, καὶ ἔσται ἐν κύκλῳ τὸ ΚΒΟΣ τετράπλευρον. καὶ ἐπεὶ μείζων ἐστὶν ἡ ΚΒ τῆς ΧΦ, ἴση δὲ ἡ ΧΦ τῇ ΣΟ, μείζων ἄρα ἡ ΚΒ τῆς ΣΟ. ἴση δὲ ἡ ΚΒ ἑκατέρᾳ τῶν ΚΣ, ΒΟ: καὶ ἑκατέρα ἄρα τῶν ΚΣ, ΒΟ τῆς ΣΟ μείζων ἐστίν. καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΚΒΟΣ, καὶ ἴσαι αἱ ΚΒ, ΒΟ, ΚΣ, καὶ ἐλάττων ἡ ΟΣ, καὶ ἐκ τοῦ κέντρου τοῦ κύκλου ἐστὶν ἡ ΒΨ, τὸ ἄρα ἀπὸ τῆς ΚΒ τοῦ ἀπὸ τῆς ΒΨ μεῖζόν ἐστιν ἢ διπλάσιον. ἤχθω ἀπὸ τοῦ Κ ἐπὶ τὴν ΒΦ κάθετος ἡ ΚΩ. καὶ ἐπεὶ ἡ ΒΔ τῆς ΔΩ ἐλάττων ἐστὶν ἢ διπλῆ, καί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΩ, οὕτως τὸ ὑπὸ τῶν ΔΒ, ΒΩ πρὸς τὸ ὑπὸ τῶν ΔΩ, ΩΒ, ἀναγραφομένου ἀπὸ τῆς ΒΩ τετραγώνου καὶ συμπληρουμένου τοῦ ἐπὶ τῆς ΩΔ παραλληλογράμμου καὶ τὸ ὑπὸ ΔΒ, ΒΩ ἄρα τοῦ ὑπὸ ΔΩ, ΩΒ ἔλαττόν ἐστιν ἢ διπλάσιον. καί ἐστι τῆς ΚΔ ἐπιζευγνυμένης τὸ μὲν ὑπὸ ΔΒ, ΒΩ ἴσον τῷ ἀπὸ τῆς ΒΚ, τὸ δὲ ὑπὸ τῶν ΔΩ, ΩΒ ἴσον τῷ ἀπὸ τῆς ΚΩ: τὸ ἄρα ἀπὸ τῆς ΚΒ τοῦ ἀπὸ τῆς ΚΩ ἔλασσόν ἐστιν ἢ διπλάσιον. ἀλλὰ τὸ ἀπὸ τῆς ΚΒ τοῦ ἀπὸ τῆς ΒΨ μεῖζόν ἐστιν ἢ διπλάσιον: μεῖζον ἄρα τὸ ἀπὸ τῆς ΚΩ τοῦ ἀπὸ τῆς ΒΨ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΑ τῇ ΚΑ, ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΑ τῷ ἀπὸ τῆς ΑΚ. καί ἐστι τῷ μὲν ἀπὸ τῆς ΒΑ ἴσα τὰ ἀπὸ τῶν ΒΨ, ΨΑ, τῷ δὲ ἀπὸ τῆς ΚΑ ἴσα τὰ ἀπὸ τῶν ΚΩ, ΩΑ: τὰ ἄρα ἀπὸ τῶν ΒΨ, ΨΑ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΚΩ, ΩΑ, ὧν τὸ ἀπὸ τῆς ΚΩ μεῖζον τοῦ ἀπὸ τῆς ΒΨ: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΩΑ ἔλασσόν ἐστι τοῦ ἀπὸ τῆς ΨΑ. μείζων ἄρα ἡ ΑΨ τῆς ΑΩ: πολλῷ ἄρα ἡ ΑΨ μείζων ἐστὶ τῆς ΑΗ. καί ἐστιν ἡ μὲν ΑΨ ἐπὶ μίαν τοῦ πολυέδρου βάσιν, ἡ δὲ ΑΗ ἐπὶ τὴν τῆς ἐλάσσονος σφαίρας ἐπιφάνειαν: ὥστε τὸ πολύεδρον οὐ ψαύσει τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν. δύο ἄρα σφαιρῶν περὶ τὸ αὐτὸ κέντρον οὐσῶν εἰς τὴν μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγέγραπται μὴ ψαῦον τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν: ὅπερ ἔδει ποιῆσαι. Πόρισμα ἐὰν δὲ καὶ εἰς ἑτέραν σφαῖραν τῷ ἐν τῇ ΒΓΔΕ σφαίρᾳ στερεῷ πολυέδρῳ ὅμοιον στερεὸν πολύεδρον ἐγγραφῇ, τὸ ἐν τῇ ΒΓΔΕ σφαίρᾳ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ ἑτέρᾳ σφαίρᾳ στερεὸν πολύεδρον τριπλασίονα λόγον ἔχει, ἤπερ ἡ τῆς ΒΓΔΕ σφαίρας διάμετρος πρὸς τὴν τῆς ἑτέρας σφαίρας διάμετρον. διαιρεθέντων γὰρ τῶν στερεῶν εἰς τὰς ὁμοιοπληθεῖς καὶ ὁμοιοταγεῖς πυραμίδας ἔσονται αἱ πυραμίδες ὅμοιαι. αἱ δὲ ὅμοιαι πυραμίδες πρὸς ἀλλήλας ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν: ἡ ἄρα πυραμίς, ἧς βάσις μέν ἐστι τὸ ΚΒΟΣ τετράπλευρον, κορυφὴ δὲ τὸ Α σημεῖον, πρὸς τὴν ἐν τῇ ἑτέρᾳ σφαίρᾳ ὁμοιοταγῆ πυραμίδα τριπλασίονα λόγον ἔχει, ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἤπερ ἡ ΑΒ ἐκ τοῦ κέντρου τῆς σφαίρας τῆς περὶ κέντρον τὸ Α πρὸς τὴν ἐκ τοῦ κέντρου τῆς ἑτέρας σφαίρας. ὁμοίως καὶ ἑκάστη πυραμὶς τῶν ἐν τῇ περὶ κέντρον τὸ Α σφαίρᾳ πρὸς ἑκάστην ὁμοταγῆ πυραμίδα τῶν ἐν τῇ ἑτέρᾳ σφαίρᾳ τριπλασίονα λόγον ἕξει, ἤπερ ἡ ΑΒ πρὸς τὴν ἐκ τοῦ κέντρου τῆς ἑτέρας σφαίρας. καὶ ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ὥστε ὅλον τὸ ἐν τῇ περὶ κέντρον τὸ Α σφαίρᾳ στερεὸν πολύεδρον πρὸς ὅλον τὸ ἐν τῇ ἑτέρᾳ σφαίρᾳ στερεὸν πολύεδρον τριπλασίονα λόγον ἕξει, ἤπερ ἡ ΑΒ πρὸς τὴν ἐκ τοῦ κέντρου τῆς ἑτέρας σφαίρας, τουτέστιν ἤπερ ἡ ΒΔ διάμετρος πρὸς τὴν τῆς ἑτέρας σφαίρας διάμετρον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",1504]]],["Rule",["Association",["Rule","'Book'",12],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'12.18'"],["Rule","'Text'","'Spheres are to one another in the triplicate ratio of their respective diameters.'"],["Rule","'TextWordCount'",13],["Rule","'GreekText'","'αἱ σφαῖραι πρὸς ἀλλήλας ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ἰδίων διαμέτρων.'"],["Rule","'GreekTextWordCount'",11],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",16]],["List",["Rule","'Book'",12],["Rule","'Theorem'",2]],["List",["Rule","'Book'",12],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'Let the spheres ABC, DEF be conceived, and let BC, EF be their diameters; I say that the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF. For, if the sphere ABC has not to the sphere DEF the ratio triplicate of that which BC has to EF, then the sphere ABC will have either to some less sphere than the sphere DEF, or to a greater, the ratio triplicate of that which BC has to EF. First, let it have that ratio to a less sphere GHK, let DEF be conceived about the same centre with GHK, let there be inscribed in the greater sphere DEF a polyhedral solid which does not touch the lesser sphere GHK at its surface, [XII. 17] and let there also be inscribed in the sphere ABC a polyhedral solid similar to the polyhedral solid in the sphere DEF; therefore the polyhedral solid in ABC has to the polyhedral solid in DEF the ratio triplicate of that which BC has to EF. [XII. 17] But the sphere ABC also has to the sphere GHK the ratio triplicate of that which BC has to EF; therefore, as the sphere ABC is to the sphere GHK, so is the polyhedral solid in the sphere ABC to the polyhedral solid in the sphere DEF; and, alternately, as the sphere ABC is to the polyhedron in it, so is the sphere GHK to the polyhedral solid in the sphere DEF. [V. 16] But the sphere ABC is greater than the polyhedron in it; therefore the sphere GHK is also greater than the polyhedron in the sphere DEF. But it is also less, for it is enclosed by it. Therefore the sphere ABC has not to a less sphere than the sphere DEF the ratio triplicate of that which the diameter BC has to EF. Similarly we can prove that neither has the sphere DEF to a less sphere than the sphere ABC the ratio triplicate of that which EF has to BC. I say next that neither has the sphere ABC to any greater sphere than the sphere DEF the ratio triplicate of that which BC has to EF. For, if possible, let it have that ratio to a greater, LMN; therefore, inversely, the sphere LMN has to the sphere ABC the ratio triplicate of that which the diameter EF has to the diameter BC. But, inasmuch as LMN is greater than DEF, therefore, as the sphere LMN is to the sphere ABC, so is the sphere DEF to some less sphere than the sphere ABC, as was before proved. [XII. 2, Lemma] Therefore the sphere DEF also has to some less sphere than the sphere ABC the ratio triplicate of that which EF has to BC: which was proved impossible. Therefore the sphere ABC has not to any sphere greater than the sphere DEF the ratio triplicate of that which BC has to EF. But it was proved that neither has it that ratio to a less sphere. Therefore the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF.'"],["Rule","'ProofWordCount'",528],["Rule","'GreekProof'","'Νενοήσθωσαν σφαῖραι αἱ ΑΒΓ, ΔΕΖ, διάμετροι δὲ αὐτῶν αἱ ΒΓ, ΕΖ: λέγω, ὅτι ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. εἰ γὰρ μὴ ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ, ἕξει ἄρα ἡ ΑΒΓ σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἢ πρὸς μείζονα ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἐχέτω πρότερον πρὸς ἐλάσσονα τὴν ΗΘΚ, καὶ νενοήσθω ἡ ΔΕΖ τῇ ΗΘΚ περὶ τὸ αὐτὸ κέντρον, καὶ ἐγγεγράφθω εἰς τὴν μείζονα σφαῖραν τὴν ΔΕΖ στερεὸν πολύεδρον μὴ ψαῦον τῆς ἐλάσσονος σφαίρας τῆς ΗΘΚ κατὰ τὴν ἐπιφάνειαν, ἐγγεγράφθω δὲ καὶ εἰς τὴν ΑΒΓ σφαῖραν τῷ ἐν τῇ ΔΕΖ σφαίρᾳ στερεῷ πολυέδρῳ ὅμοιον στερεὸν πολύεδρον: τὸ ἄρα ἐν τῇ ΑΒΓ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ ΔΕΖ στερεὸν πολύεδρον τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἔχει δὲ καὶ ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΗΘΚ σφαῖραν τριπλασίονα λόγον ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ: ἔστιν ἄρα ὡς ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΗΘΚ σφαῖραν, οὕτως τὸ ἐν τῇ ΑΒΓ σφαίρᾳ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ ΔΕΖ σφαίρᾳ στερεὸν πολύεδρον: ἐναλλὰξ ἄρα ὡς ἡ ΑΒΓ σφαῖρα πρὸς τὸ ἐν αὐτῇ πολύεδρον, οὕτως ἡ ΗΘΚ σφαῖρα πρὸς τὸ ἐν τῇ ΔΕΖ σφαίρᾳ στερεὸν πολύεδρον. μείζων δὲ ἡ ΑΒΓ σφαῖρα τοῦ ἐν αὐτῇ πολυέδρου: μείζων ἄρα καὶ ἡ ΗΘΚ σφαῖρα τοῦ ἐν τῇ ΔΕΖ σφαίρᾳ πολυέδρου. ἀλλὰ καὶ ἐλάττων: ἐμπεριέχεται γὰρ ὑπ᾽ αὐτοῦ. οὐκ ἄρα ἡ ΑΒΓ σφαῖρα πρὸς ἐλάσσονα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ διάμετρος πρὸς τὴν ΕΖ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἡ ΔΕΖ σφαῖρα πρὸς ἐλάσσονα τῆς ΑΒΓ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΒΓ. λέγω δή, ὅτι οὐδὲ ἡ ΑΒΓ σφαῖρα πρὸς μείζονά τινα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. εἰ γὰρ δυνατόν, ἐχέτω πρὸς μείζονα τὴν ΛΜΝ: ἀνάπαλιν ἄρα ἡ ΛΜΝ σφαῖρα πρὸς τὴν ΑΒΓ σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ διάμετρος πρὸς τὴν ΒΓ διάμετρον. ὡς δὲ ἡ ΛΜΝ σφαῖρα πρὸς τὴν ΑΒΓ σφαῖραν, οὕτως ἡ ΔΕΖ σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΑΒΓ σφαίρας, ἐπειδήπερ μείζων ἐστὶν ἡ ΛΜΝ τῆς ΔΕΖ, ὡς ἔμπροσθεν ἐδείχθη. καὶ ἡ ΔΕΖ ἄρα σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΑΒΓ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΒΓ: ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα ἡ ΑΒΓ σφαῖρα πρὸς μείζονά τινα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἐλάσσονα. ἡ ἄρα ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",432]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",1]],["Association",["Rule","'VertexLabel'","'13.1'"],["Rule","'Text'","'If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square on the half.'"],["Rule","'TextWordCount'",32],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ μεῖζον τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Definition'",3]],["List",["Rule","'Book'",6],["Rule","'Theorem'",1]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'For let the straight line AB be cut in extreme and mean ratio at the point C, and let AC be the greater segment; let the straight line AD be produced in a straight line with CA, and let AD be made half of AB; I say that the square on CD is five times the square on AD. For let the squares AE, DF be described on AB, DC, and let the figure in DF be drawn; let FC be carried through to G. Now, since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17] And CE is the rectangle AB, BC, and FH the square on AC; therefore CE is equal to FH. And, since BA is double of AD, while BA is equal to KA, and AD to AH, therefore KA is also double of AH. But, as KA is to AH, so is CK to CH; [VI. 1] therefore CK is double of CH. But LH, HC are also double of CH. Therefore KC is equal to LH, HC. But CE was also proved equal to HF; therefore the whole square AE is equal to the gnomon MNO. And, since BA is double of AD, the square on BA is quadruple of the square on AD, that is, AE is quadruple of DH. But AE is equal to the gnomon MNO; therefore the gnomon MNO is also quadruple of AP; therefore the whole DF is five times AP. And DF is the square on DC, and AP the square on DA; therefore the square on CD is five times the square on DA.'"],["Rule","'ProofWordCount'",286],["Rule","'GreekProof'","'εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τετμήσθω κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα τὸ ΑΓ, καὶ ἐκβεβλήσθω ἐπ᾽ εὐθείας τῇ ΓΑ εὐθεῖα ἡ ΑΔ, καὶ κείσθω τῆς ΑΒ ἡμίσεια ἡ ΑΔ: λέγω, ὅτι πενταπλάσιόν ἐστι τὸ ἀπὸ τῆς ΓΔ τοῦ ἀπὸ τῆς ΔΑ. ἀναγεγράφθωσαν γὰρ ἀπὸ τῶν ΑΒ, ΔΓ τετράγωνα τὰ ΑΕ, ΔΖ, καὶ καταγεγράφθω ἐν τῷ ΔΖ τὸ σχῆμα, καὶ διήχθω ἡ ΖΓ ἐπὶ τὸ Η. καὶ ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Γ, τὸ ἄρα ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΑΒΓ τὸ ΓΕ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΖΘ: ἴσον ἄρα τὸ ΓΕ τῷ ΖΘ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΒΑ τῆς ΑΔ, ἴση δὲ ἡ μὲν ΒΑ τῇ ΚΑ, ἡ δὲ ΑΔ τῇ ΑΘ, διπλῆ ἄρα καὶ ἡ ΚΑ τῆς ΑΘ. ὡς δὲ ἡ ΚΑ πρὸς τὴν ΑΘ, οὕτως τὸ ΓΚ πρὸς τὸ ΓΘ: διπλάσιον ἄρα τὸ ΓΚ τοῦ ΓΘ. εἰσὶ δὲ καὶ τὰ ΛΘ, ΘΓ διπλάσια τοῦ ΓΘ. ἴσον ἄρα τὸ ΚΓ τοῖς ΛΘ, ΘΓ. ἐδείχθη δὲ καὶ τὸ ΓΕ τῷ ΘΖ ἴσον: ὅλον ἄρα τὸ ΑΕ τετράγωνον ἴσον ἐστὶ τῷ ΜΝΞ γνώμονι. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΒΑ τῆς ΑΔ, τετραπλάσιόν ἐστι τὸ ἀπὸ τῆς ΒΑ τοῦ ἀπὸ τῆς ΑΔ, τουτέστι τὸ ΑΕ τοῦ ΔΘ. ἴσον δὲ τὸ ΑΕ τῷ ΜΝΞ γνώμονι: καὶ ὁ ΜΝΞ ἄρα γνώμων τετραπλάσιός ἐστι τοῦ ΑΟ: ὅλον ἄρα τὸ ΔΖ πενταπλάσιόν ἐστι τοῦ ΑΟ. καί ἐστι τὸ μὲν ΔΖ τὸ ἀπὸ τῆς ΔΓ, τὸ δὲ ΑΟ τὸ ἀπὸ τῆς ΔΑ: τὸ ἄρα ἀπὸ τῆς ΓΔ πενταπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΔΑ. ἐὰν ἄρα εὐθεῖα ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ μεῖζον τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",298]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",2]],["Association",["Rule","'VertexLabel'","'13.2'"],["Rule","'Text'","'If the square on a straight line be five times the square on a segment of it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line.'"],["Rule","'TextWordCount'",44],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ τμήματος ἑαυτῆς πενταπλάσιον δύνηται, τῆς διπλασίας τοῦ εἰρημένου τμήματος ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμα τὸ λοιπὸν μέρος ἐστὶ τῆς ἐξ ἀρχῆς εὐθείας.'"],["Rule","'GreekTextWordCount'",28],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'For let the square on the straight line AB be five times the square on the segment AC of it, and let CD be double of AC; I say that, when CD is cut in extreme and mean ratio, the greater segment is CB. Let the squares AF, CG be described on AB, CD respectively, let the figure in AF be drawn, and let BE be drawn through. Now, since the square on BA is five times the square on AC, AF is five times AH. Therefore the gnomon MNO is quadruple of AH. And, since DC is double of CA, therefore the square on DC is quadruple of the square on CA, that is, CG is quadruple of AH. But the gnomon MNO was also proved quadruple of AH; therefore the gnomon MNO is equal to CG. And, since DC is double of CA, while DC is equal to CK, and AC to CH, therefore KB is also double of BH. [VI. 1] But LH, HB are also double of HB; therefore KB is equal to LH, HB. But the whole gnomon MNO was also proved equal to the whole CG; therefore the remainder HF is equal to BG. And BG is the rectangle CD, DB, for CD is equal to DG; and HF is the square on CB; therefore the rectangle CD, DB is equal to the square on CB. Therefore, as DC is to CB, so is CB to BD. But DC is greater than CB; therefore CB is also greater than BD. Therefore, when the straight line CD is cut in extreme and mean ratio, CB is the greater segment.'"],["Rule","'ProofWordCount'",274],["Rule","'GreekProof'","'εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ τμήματος ἑαυτῆς τοῦ ΑΓ πενταπλάσιον δυνάσθω, τῆς δὲ ΑΓ διπλῆ ἔστω ἡ ΓΔ: λέγω, ὅτι τῆς ΓΔ ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΓΒ. Ἀναγεγράφθω γὰρ ἀφ᾽ ἑκατέρας τῶν ΑΒ, ΓΔ τετράγωνα τὰ ΑΖ, ΓΗ, καὶ καταγεγράφθω ἐν τῷ ΑΖ τὸ σχῆμα, καὶ διήχθω ἡ ΒΕ. καὶ ἐπεὶ πενταπλάσιόν ἐστι τὸ ἀπὸ τῆς ΒΑ τοῦ ἀπὸ τῆς ΑΓ, πενταπλάσιόν ἐστι τὸ ΑΖ τοῦ ΑΘ. τετραπλάσιος ἄρα ὁ ΜΝΞ γνώμων τοῦ ΑΘ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΔΓ τῆς ΓΑ, τετραπλάσιον ἄρα ἐστὶ τὸ ἀπὸ ΔΓ τοῦ ἀπὸ ΓΑ, τουτέστι τὸ ΓΗ τοῦ ΑΘ. ἐδείχθη δὲ καὶ ὁ ΜΝΞ γνώμων τετραπλάσιος τοῦ ΑΘ: ἴσος ἄρα ὁ ΜΝΞ γνώμων τῷ ΓΗ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΔΓ τῆς ΓΑ, ἴση δὲ ἡ μὲν ΔΓ τῇ ΓΚ, ἡ δὲ ΑΓ τῇ ΓΘ διπλῆ ἄρα καὶ ἡ ΚΓ τῆς ΓΘ, διπλάσιον ἄρα καὶ τὸ ΚΒ τοῦ ΒΘ. εἰσὶ δὲ καὶ τὰ ΛΘ, ΘΒ τοῦ ΘΒ διπλάσια: ἴσον ἄρα τὸ ΚΒ τοῖς ΛΘ, ΘΒ. ἐδείχθη δὲ καὶ ὅλος ὁ ΜΝΞ γνώμων ὅλῳ τῷ ΓΗ ἴσος: καὶ λοιπὸν ἄρα τὸ ΘΖ τῷ ΒΗ ἐστιν ἴσον. καί ἐστι τὸ μὲν ΒΗ τὸ ὑπὸ τῶν ΓΔΒ: ἴση γὰρ ἡ ΓΔ τῇ ΔΗ: τὸ δὲ ΘΖ τὸ ἀπὸ τῆς ΓΒ: τὸ ἄρα ὑπὸ τῶν ΓΔΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΒ. ἔστιν ἄρα ὡς ἡ ΔΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΓΒ πρὸς τὴν ΒΔ. μείζων δὲ ἡ ΔΓ τῆς ΓΒ: μείζων ἄρα καὶ ἡ ΓΒ τῆς ΒΔ. τῆς ΓΔ ἄρα εὐθείας ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΓΒ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμήματος ἑαυτῆς πενταπλάσιον δύνηται, τῆς διπλασίας τοῦ εἰρημένου τμήματος ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμα τὸ λοιπὸν μέρος ἐστὶ τῆς ἐξ ἀρχῆς εὐθείας: ὅπερ ἔδει δεῖξαι. λῆμμα ὅτι δὲ ἡ διπλῆ τῆς ΑΓ μείζων ἐστὶ τῆς ΒΓ, οὕτως δεικτέον. εἰ γὰρ μή, ἔστω, εἰ δυνατόν, ἡ ΒΓ διπλῆ τῆς ΓΑ. τετραπλάσιον ἄρα τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΑ: πενταπλάσια ἄρα τὰ ἀπὸ τῶν ΒΓ, ΓΑ τοῦ ἀπὸ τῆς ΓΑ. ὑπόκειται δὲ καὶ τὸ ἀπὸ τῆς ΒΑ πενταπλάσιον τοῦ ἀπὸ τῆς ΓΑ: τὸ ἄρα ἀπὸ τῆς ΒΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΓ, ΓΑ: ὅπερ ἀδύνατον. οὐκ ἄρα ἡ ΓΒ διπλασία ἐστὶ τῆς ΑΓ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἡ ἐλάττων τῆς ΓΒ διπλασίων ἐστὶ τῆς ΓΑ: πολλῷ γὰρ μεῖζον τὸ ἄτοπον. ἡ ἄρα τῆς ΑΓ διπλῆ μείζων ἐστὶ τῆς ΓΒ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",405]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",3]],["Association",["Rule","'VertexLabel'","'13.3'"],["Rule","'Text'","'If a straight line be cut in extreme and mean ratio, the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ ἔλασσον τμῆμα προσλαβὸν τὴν ἡμίσειαν τοῦ μείζονος τμήματος πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τοῦ μείζονος τμήματος τετραγώνου.'"],["Rule","'GreekTextWordCount'",27],["Rule","'References'",["List"]],["Rule","'Proof'","'For let any straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AC be bisected at D; I say that the square on BD is five times the square on DC. For let the square AE be described on AB, and let the figure be drawn double. Since AC is double of DC, therefore the square on AC is quadruple of the square on DC, that is, RS is quadruple of FG. And, since the rectangle AB, BC is equal to the square on AC, and CE is the rectangle AB, BC, therefore CE is equal to RS. But RS is quadruple of FG; therefore CE is also quadruple of FG. Again, since AD is equal to DC, HK is also equal to KF. Hence the square GF is also equal to the square HL. Therefore GK is equal to KL, that is, MN to NE; hence MF is also equal to FE. But MF is equal to CG; therefore CG is also equal to FE. Let CN be added to each; therefore the gnomon OPQ is equal to CE. But CE was proved quadruple of GF; therefore the gnomon OPQ is also quadruple of the square FG. Therefore the gnomon OPQ and the square FG are five times FG. But the gnomon OPQ and the square FG are the square DN. And DN is the square on DB, and GF the square on DC. Therefore the square on DB is five times the square on DC.'"],["Rule","'ProofWordCount'",260],["Rule","'GreekProof'","'εὐθεῖα γάρ τις ἡ ΑΒ ἄκρον καὶ μέσον λόγον τετμήσθω κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα τὸ ΑΓ, καὶ τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Δ: λέγω, ὅτι πενταπλάσιόν ἐστι τὸ ἀπὸ τῆς ΒΔ τοῦ ἀπὸ τῆς ΔΓ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΕ, καὶ καταγεγράφθω διπλοῦν τὸ σχῆμα. ἐπεὶ διπλῆ ἐστιν ἡ ΑΓ τῆς ΔΓ, τετραπλάσιον ἄρα τὸ ἀπὸ τῆς ΑΓ τοῦ ἀπὸ τῆς ΔΓ, τουτέστι τὸ ΡΣ τοῦ ΖΗ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ, καί ἐστι τὸ ὑπὸ τῶν ΑΒΓ τὸ ΓΕ, τὸ ἄρα ΓΕ ἴσον ἐστὶ τῷ ΡΣ. τετραπλάσιον δὲ τὸ ΡΣ τοῦ ΖΗ: τετραπλάσιον ἄρα καὶ τὸ ΓΕ τοῦ ΖΗ. πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΓ, ἴση ἐστὶ καὶ ἡ ΘΚ τῇ ΚΖ. ὥστε καὶ τὸ ΗΖ τετράγωνον ἴσον ἐστὶ τῷ ΘΛ τετραγώνῳ. ἴση ἄρα ἡ ΗΚ τῇ ΚΛ, τουτέστιν ἡ ΜΝ τῇ ΝΕ: ὥστε καὶ τὸ ΜΖ τῷ ΖΕ ἐστιν ἴσον. ἀλλὰ τὸ ΜΖ τῷ ΓΗ ἐστιν ἴσον: καὶ τὸ ΓΗ ἄρα τῷ ΖΕ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΓΝ: ὁ ἄρα ΞΟΠ γνώμων ἴσος ἐστὶ τῷ ΓΕ. ἀλλὰ τὸ ΓΕ τετραπλάσιον ἐδείχθη τοῦ ΗΖ: καὶ ὁ ΞΟΠ ἄρα γνώμων τετραπλάσιός ἐστι τοῦ ΖΗ τετραγώνου. ὁ ΞΟΠ ἄρα γνώμων καὶ τὸ ΖΗ τετράγωνον πενταπλάσιός ἐστι τοῦ ΖΗ. ἀλλὰ ὁ ΞΟΠ γνώμων καὶ τὸ ΖΗ τετράγωνόν ἐστι τὸ ΔΝ. καί ἐστι τὸ μὲν ΔΝ τὸ ἀπὸ τῆς ΔΒ, τὸ δὲ ΗΖ τὸ ἀπὸ τῆς ΔΓ. τὸ ἄρα ἀπὸ τῆς ΔΒ πενταπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΔΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",256]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",4]],["Association",["Rule","'VertexLabel'","'13.4'"],["Rule","'Text'","'If a straight line be cut in extreme and mean ratio, the square on the whole and the square on the lesser segment together are triple of the square on the greater segment.'"],["Rule","'TextWordCount'",33],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ ἀπὸ τῆς ὅλης καὶ τοῦ ἐλάσσονος τμήματος, τὰ συναμφότερα τετράγωνα, τριπλάσιά ἐστι τοῦ ἀπὸ τοῦ μείζονος τμήματος τετραγώνου.'"],["Rule","'GreekTextWordCount'",27],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Definition'",3]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'Let AB be a straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that the squares on AB, BC are triple of the square on CA. For let the square ADEB be described on AB, and let the figure be drawn. Since then AB has been cut in extreme and mean ratio at C, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17] And AK is the rectangle AB, BC, and HG the square on AC; therefore AK is equal to HG. And, since AF is equal to FE, let CK be added to each; therefore the whole AK is equal to the whole CE; therefore AK, CE are double of AK. But AK, CE are the gnomon LMN and the square CK; therefore the gnomon LMN and the square CK are double of AK. But, further, AK was also proved equal to HG; therefore the gnomon LMN and the squares CK, HG are triple of the square HG. And the gnomon LMN and the squares CK, HG are the whole square AE and CK, which are the squares on AB, BC, while HG is the square on AC. Therefore the squares on AB, BC are triple of the square on AC.'"],["Rule","'ProofWordCount'",229],["Rule","'GreekProof'","'ἔστω εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω ἄκρον καὶ μέσον λόγον κατὰ τὸ Γ, καὶ ἔστω μεῖζον τμῆμα τὸ ΑΓ: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΓΑ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἡ ΑΒ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Γ, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΑΓ, τὸ ἄρα ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΑΒΓ τὸ ΑΚ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΘΗ: ἴσον ἄρα ἐστὶ τὸ ΑΚ τῷ ΘΗ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΖ τῷ ΖΕ, κοινὸν προσκείσθω τὸ ΓΚ: ὅλον ἄρα τὸ ΑΚ ὅλῳ τῷ ΓΕ ἐστιν ἴσον: τὰ ἄρα ΑΚ, ΓΕ τοῦ ΑΚ ἐστι διπλάσια. ἀλλὰ τὰ ΑΚ, ΓΕ ὁ ΛΜΝ γνώμων ἐστὶ καὶ τὸ ΓΚ τετράγωνον: ὁ ἄρα ΛΜΝ γνώμων καὶ τὸ ΓΚ τετράγωνον διπλάσιά ἐστι τοῦ ΑΚ. ἀλλὰ μὴν καὶ τὸ ΑΚ τῷ ΘΗ ἐδείχθη ἴσον: ὁ ἄρα ΛΜΝ γνώμων καὶ τὸ ΓΚ τετράγωνον διπλάσιά ἐστι τοῦ ΘΗ: ὥστε ὁ ΛΜΝ γνώμων καὶ τὰ ΓΚ, ΘΗ τετράγωνα τριπλάσιά ἐστι τοῦ ΘΗ τετραγώνου. καί ἐστιν ὁ μὲν ΛΜΝ γνώμων καὶ τὰ ΓΚ, ΘΗ τετράγωνα ὅλον τὸ ΑΕ καὶ τὸ ΓΚ, ἅπερ ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα, τὸ δὲ ΗΘ τὸ ἀπὸ τῆς ΑΓ τετράγωνον. τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΑΓ τετραγώνου: ὅπερ ἔδει δεῖξαι. ἐὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, καὶ προστεθῇ αὐτῇ ἴση τῷ μείζονι τμήματι, ἡ ὅλη εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ἐξ ἀρχῆς εὐθεῖα. εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τετμήσθω κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα ἡ ΑΓ, καὶ τῇ ΑΓ ἴση κείσθω ἡ ΑΔ. λέγω, ὅτι ἡ ΔΒ εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Α, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ἐξ ἀρχῆς εὐθεῖα ἡ ΑΒ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΕ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Γ, τὸ ἄρα ὑπὸ ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ ΑΓ. καί ἐστι τὸ μὲν ὑπὸ ΑΒΓ τὸ ΓΕ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΓΘ: ἴσον ἄρα τὸ ΓΕ τῷ ΘΓ. ἀλλὰ τῷ μὲν ΓΕ ἴσον ἐστὶ τὸ ΘΕ, τῷ δὲ ΘΓ ἴσον τὸ ΔΘ: καὶ τὸ ΔΘ ἄρα ἴσον ἐστὶ τῷ ΘΕ κοινὸν προσκείσθω τὸ ΘΒ. ὅλον ἄρα τὸ ΔΚ ὅλῳ τῷ ΑΕ ἐστιν ἴσον. καί ἐστι τὸ μὲν ΔΚ τὸ ὑπὸ τῶν ΒΔ, ΔΑ: ἴση γὰρ ἡ ΑΔ τῇ ΔΛ: τὸ δὲ ΑΕ τὸ ἀπὸ τῆς ΑΒ: τὸ ἄρα ὑπὸ τῶν ΒΔΑ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ. ἔστιν ἄρα ὡς ἡ ΔΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΔ. μείζων δὲ ἡ ΔΒ τῆς ΒΑ: μείζων ἄρα καὶ ἡ ΒΑ τῆς ΑΔ. ἡ ἄρα ΔΒ ἄκρον καὶ μέσον λόγον τέμηται κατὰ τὸ Α, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΑΒ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",488]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",5]],["Association",["Rule","'VertexLabel'","'13.5'"],["Rule","'Text'","'If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment.'"],["Rule","'TextWordCount'",46],["Rule","'GreekText'","'ἐὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, καὶ προστεθῇ αὐτῇ ἴση τῷ μείζονι τμήματι, ἡ ὅλη εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ἐξ ἀρχῆς εὐθεῖα.'"],["Rule","'GreekTextWordCount'",32],["Rule","'References'",["List",["List",["Rule","'Book'",5],["Rule","'Theorem'",14]],["List",["Rule","'Book'",6],["Rule","'Definition'",3]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'For let the straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AD be equal to AC. I say that the straight line DB has been cut in extreme and mean ratio at A, and the original straight line AB is the greater segment. For let the square AE be described on AB, and let the figure be drawn. Since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17] And CE is the rectangle AB, BC, and CH the square on AC; therefore CE is equal to HC. But HE is equal to CE, and DH is equal to HC; therefore DH is also equal to HE. Therefore the whole DK is equal to the whole AE. And DK is the rectangle BD, DA, for AD is equal to DL; and AE is the square on AB; therefore the rectangle BD, DA is equal to the square on AB. Therefore, as DB is to BA, so is BA to AD. [VI. 17] And DB is greater than BA; therefore BA is also greater than AD. [V. 14] Therefore DB has been cut in extreme and mean ratio at A, and AB is the greater segment.'"],["Rule","'ProofWordCount'",227],["Rule","'GreekProof'","'Εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τετμήσθω κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα ἡ ΑΓ, καὶ τῇ ΑΓ ἴση [κείσθω] ἡ ΑΔ. λέγω, ὅτι ἡ ΔΒ εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Α, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ἐξ ἀρχῆς εὐθεῖα ἡ ΑΒ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΕ, καὶ καταγεγράφθω τὸ σχῆμα.  ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Γ, τὸ ἄρα ὑπὸ ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ ΑΓ.  καί ἐστι τὸ μὲν ὑπὸ ΑΒΓ τὸ ΓΕ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΓΘ: ἴσον ἄρα τὸ ΓΕ τῷ ΘΓ. ἀλλὰ τῷ μὲν ΓΕ ἴσον ἐστὶ τὸ ΘΕ, τῷ δὲ ΘΓ ἴσον τὸ ΔΘ:  καὶ τὸ ΔΘ ἄρα ἴσον ἐστὶ τῷ ΘΕ [κοινὸν προσκείσθω τὸ ΘΒ]. ὅλον ἄρα τὸ ΔΚ ὅλῳ τῷ ΑΕ ἐστιν ἴσον.  καί ἐστι τὸ μὲν ΔΚ τὸ ὑπὸ τῶν ΒΔ, ΔΑ: ἴση γὰρ ἡ ΑΔ τῇ ΔΛ: τὸ δὲ ΑΕ τὸ ἀπὸ τῆς ΑΒ: τὸ ἄρα ὑπὸ τῶν ΒΔΑ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ. ἔστιν ἄρα ὡς ἡ ΔΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΔ.  μείζων δὲ ἡ ΔΒ τῆς ΒΑ: μείζων ἄρα καὶ ἡ ΒΑ τῆς ΑΔ. Ἡ ἄρα ΔΒ ἄκρον καὶ μέσον λόγον τέμηται κατὰ τὸ Α, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΑΒ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",221]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",6]],["Association",["Rule","'VertexLabel'","'13.6'"],["Rule","'Text'","'If a rational straight line be cut in extreme and mean ratio, each of the segments is the irrational straight line called apotome.'"],["Rule","'TextWordCount'",23],["Rule","'GreekText'","'ἐὰν εὐθεῖα ῥητὴ ἄκρον καὶ μέσον λόγον τμηθῇ, ἑκάτερον τῶν τμημάτων ἄλογός ἐστιν ἡ καλουμένη ἀποτομή.'"],["Rule","'GreekTextWordCount'",16],["Rule","'References'",["List",["List",["Rule","'Book'",6],["Rule","'Definition'",3]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",10],["Rule","'Definition'",1.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",6]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",97]],["List",["Rule","'Book'",13],["Rule","'Theorem'",1]]]],["Rule","'Proof'","'Let AB be a rational straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that each of the straight lines AC, CB is the irrational straight line called apotome. For let BA be produced, and let AD be made half of BA. Since then the straight line AB has been cut in extreme and mean ratio, and to the greater segment AC is added AD which is half of AB, therefore the square on CD is five times the square on DA. [XIII. 1] Therefore the square on CD has to the square on DA the ratio which a number has to a number; therefore the square on CD is commensurable with the square on DA. [X. 6] But the square on DA is rational, for DA is rational, being half of AB which is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is also rational. And, since the square on CD has not to the square on DA the ratio which a square number has to a square number, therefore CD is incommensurable in length with DA; [X. 9] therefore CD, DA are rational straight lines commensurable in square only; therefore AC is an apotome. [X. 73] Again, since AB has been cut in extreme and mean ratio, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17] Therefore the square on the apotome AC, if applied to the rational straight line AB, produces BC as breadth. But the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome; [X. 97] therefore CB is a first apotome. And CA was also proved to be an apotome.'"],["Rule","'ProofWordCount'",306],["Rule","'GreekProof'","'ἔστω εὐθεῖα ῥητὴ ἡ ΑΒ καὶ τετμήσθω ἄκρον καὶ μέσον λόγον κατὰ τὸ Γ, καὶ ἔστω μεῖζον τμῆμα ἡ ΑΓ: λέγω, ὅτι ἑκατέρα τῶν ΑΓ, ΓΒ ἄλογός ἐστιν ἡ καλουμένη ἀποτομή. Ἐκβεβλήσθω γὰρ ἡ ΒΑ, καὶ κείσθω τῆς ΒΑ ἡμίσεια ἡ ΑΔ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΒ τέτμηται ἄκρον καὶ μέσον λόγον κατὰ τὸ Γ, καὶ τῷ μείζονι τμήματι τῷ ΑΓ πρόσκειται ἡ ΑΔ ἡμίσεια οὖσα τῆς ΑΒ, τὸ ἄρα ἀπὸ ΓΔ τοῦ ἀπὸ ΔΑ πενταπλάσιόν ἐστιν. τὸ ἄρα ἀπὸ ΓΔ πρὸς τὸ ἀπὸ ΔΑ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν: σύμμετρον ἄρα τὸ ἀπὸ ΓΔ τῷ ἀπὸ ΔΑ. ῥητὸν δὲ τὸ ἀπὸ ΔΑ: ῥητὴ γὰρ ἐστιν ἡ ΔΑ ἡμίσεια οὖσα τῆς ΑΒ ῥητῆς οὔσης: ῥητὸν ἄρα καὶ τὸ ἀπὸ ΓΔ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΓΔ. καὶ ἐπεὶ τὸ ἀπὸ ΓΔ πρὸς τὸ ἀπὸ ΔΑ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ἀσύμμετρος ἄρα μήκει ἡ ΓΔ τῇ ΔΑ: αἱ ΓΔ, ΔΑ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀποτομὴ ἄρα ἐστὶν ἡ ΑΓ. πάλιν, ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΑΓ, τὸ ἄρα ὑπὸ ΑΒ, ΒΓ τῷ ἀπὸ ΑΓ ἴσον ἐστίν. τὸ ἄρα ἀπὸ τῆς ΑΓ ἀποτομῆς παρὰ τὴν ΑΒ ῥητὴν παραβληθὲν πλάτος ποιεῖ τὴν ΒΓ. τὸ δὲ ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν πρώτην: ἀποτομὴ ἄρα πρώτη ἐστὶν ἡ ΓΒ. ἐδείχθη δὲ καὶ ἡ ΓΑ ἀποτομή. ἐὰν ἄρα εὐθεῖα ῥητὴ ἄκρον καὶ μέσον λόγον τμηθῇ, ἑκάτερον τῶν τμημάτων ἄλογός ἐστιν ἡ καλουμένη ἀποτομή: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",250]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",7]],["Association",["Rule","'VertexLabel'","'13.7'"],["Rule","'Text'","'If three angles of an equilateral pentagon, taken either in order or not in order, be equal, the pentagon will be equiangular.'"],["Rule","'TextWordCount'",22],["Rule","'GreekText'","'ἐὰν πενταγώνου ἰσοπλεύρου αἱ τρεῖς γωνίαι ἤτοι αἱ κατὰ τὸ ἑξῆς ἢ αἱ μὴ κατὰ τὸ ἑξῆς ἴσαι ὦσιν, ἰσογώνιον ἔσται τὸ πεντάγωνον.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",1],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'For in the equilateral pentagon ABCDE let, first, three angles taken in order, those at A, B, C, be equal to one another; I say that the pentagon ABCDE is equiangular. For let AC, BE, FD be joined. Now, since the two sides CB, BA are equal to the two sides BA, AE respectively, and the angle CBA is equal to the angle BAE, therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [I. 4] that is, the angle BCA to the angle BEA, and the angle ABE to the angle CAB; hence the side AF is also equal to the side BF. [I. 6] But the whole AC was also proved equal to the whole BE; therefore the remainder FC is also equal to the remainder FE. But CD is also equal to DE. Therefore the two sides FC, CD are equal to the two sides FE, ED; and the base FD is common to them; therefore the angle FCD is equal to the angle FED. [I. 8] But the angle BCA was also proved equal to the angle AEB; therefore the whole angle BCD is also equal to the whole angle AED. But, by hypothesis, the angle BCD is equal to the angles at A, B; therefore the angle AED is also equal to the angles at A, B. Similarly we can prove that the angle CDE is also equal to the angles at A, B, C; therefore the pentagon ABCDE is equiangular. Next, let the given equal angles not be angles taken in order, but let the angles at the points A, C, D be equal; I say that in this case too the pentagon ABCDE is equiangular. For let BD be joined. Then, since the two sides BA, AE are equal to the two sides BC, CD, and they contain equal angles, therefore the base BE is equal to the base BD, the triangle ABE is equal to the triangle BCD, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle AEB is equal to the angle CDB. But the angle BED is also equal to the angle BDE, since the side BE is also equal to the side BD. [I. 5] Therefore the whole angle AED is equal to the whole angle CDE. But the angle CDE is, by hypothesis, equal to the angles at A, C; therefore the angle AED is also equal to the angles at A, C. For the same reason the angle ABC is also equal to the angles at A, C, D. Therefore the pentagon ABCDE is equiangular.'"],["Rule","'ProofWordCount'",469],["Rule","'GreekProof'","'πενταγώνου γὰρ ἰσοπλεύρου τοῦ ΑΒΓΔΕ αἱ τρεῖς γωνίαι πρότερον αἱ κατὰ τὸ ἑξῆς αἱ πρὸς τοῖς Α, Β, Γ ἴσαι ἀλλήλαις ἔστωσαν: λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓΔΕ πεντάγωνον. ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΒΕ, ΖΔ. καὶ ἐπεὶ δύο αἱ ΓΒ, ΒΑ δυσὶ ταῖς ΒΑ, ΑΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΓΒΑ γωνίᾳ τῇ ὑπὸ ΒΑΕ ἐστιν ἴση, βάσις ἄρα ἡ ΑΓ βάσει τῇ ΒΕ ἐστιν ἴση, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΕ τριγώνῳ ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΒΓΑ τῇ ὑπὸ ΒΕΑ, ἡ δὲ ὑπὸ ΑΒΕ τῇ ὑπὸ ΓΑΒ: ὥστε καὶ πλευρὰ ἡ ΑΖ πλευρᾷ τῇ ΒΖ ἐστιν ἴση. ἐδείχθη δὲ καὶ ὅλη ἡ ΑΓ ὅλῃ τῇ ΒΕ ἴση: καὶ λοιπὴ ἄρα ἡ ΖΓ λοιπῇ τῇ ΖΕ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΓΔ τῇ ΔΕ ἴση. δύο δὴ αἱ ΖΓ, ΓΔ δυσὶ ταῖς ΖΕ, ΕΔ ἴσαι εἰσίν: καὶ βάσις αὐτῶν κοινὴ ἡ ΖΔ: γωνία ἄρα ἡ ὑπὸ ΖΓΔ γωνίᾳ τῇ ὑπὸ ΖΕΔ ἐστιν ἴση. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΒΓΑ τῇ ὑπὸ ΑΕΒ ἴση: καὶ ὅλη ἄρα ἡ ὑπὸ ΒΓΔ ὅλῃ τῇ ὑπὸ ΑΕΔ ἴση. ἀλλ᾽ ἡ ὑπὸ ΒΓΔ ἴση ὑπόκειται ταῖς πρὸς τοῖς Α, Β γωνίαις: καὶ ἡ ὑπὸ ΑΕΔ ἄρα ταῖς πρὸς τοῖς Α, Β γωνίαις ἴση ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ὑπὸ ΓΔΕ γωνία ἴση ἐστὶ ταῖς πρὸς τοῖς Α, Β, Γ γωνίαις: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. ἀλλὰ δὴ μὴ ἔστωσαν ἴσαι αἱ κατὰ τὸ ἑξῆς γωνίαι, ἀλλ᾽ ἔστωσαν ἴσαι αἱ πρὸς τοῖς Α, Γ, Δ σημείοις: λέγω, ὅτι καὶ οὕτως ἰσογώνιόν ἐστι τὸ ΑΒΓΔΕ πεντάγωνον. ἐπεζεύχθω γὰρ ἡ ΒΔ. καὶ ἐπεὶ δύο αἱ ΒΑ, ΑΕ δυσὶ ταῖς ΒΓ, ΓΔ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΒΕ βάσει τῇ ΒΔ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ ΒΓΔ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ γωνία τῇ ὑπὸ ΓΔΒ. ἔστι δὲ καὶ ἡ ὑπὸ ΒΕΔ γωνία τῇ ὑπὸ ΒΔΕ ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΒΕ πλευρᾷ τῇ ΒΔ ἐστιν ἴση. καὶ ὅλη ἄρα ἡ ὑπὸ ΑΕΔ γωνία ὅλῃ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση. ἀλλὰ ἡ ὑπὸ ΓΔΕ ταῖς πρὸς τοῖς Α, Γ γωνίαις ὑπόκειται ἴση: καὶ ἡ ὑπὸ ΑΕΔ ἄρα γωνία ταῖς πρὸς τοῖς Α, Γ ἴση ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΒΓ ἴση ἐστὶ ταῖς πρὸς τοῖς Α, Γ, Δ γωνίαις. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",423]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",8]],["Association",["Rule","'VertexLabel'","'13.8'"],["Rule","'Text'","'If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'ἐὰν πενταγώνου ἰσοπλεύρου καὶ ἰσογωνίου τὰς κατὰ τὸ ἑξῆς δύο γωνίας ὑποτείνωσιν εὐθεῖαι, ἄκρον καὶ μέσον λόγον τέμνουσιν ἀλλήλας, καὶ τὰ μείζονα αὐτῶν τμήματα ἴσα ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ.'"],["Rule","'GreekTextWordCount'",30],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",6]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",3],["Rule","'Theorem'",28]],["List",["Rule","'Book'",4],["Rule","'Theorem'",14]],["List",["Rule","'Book'",5],["Rule","'Theorem'",14]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",33]]]],["Rule","'Proof'","'For in the equilateral and equiangular pentagon ABCDE let the straight lines AC, BE, cutting one another at the point H, subtend two angles taken in order, the angles at A, B; I say that each of them has been cut in extreme and mean ratio at the point H, and their greater segments are equal to the side of the pentagon. For let the circle ABCDE be circumscribed about the pentagon ABCDE. [IV. 14] Then, since the two straight lines EA, AB are equal to the two AB, BC, and they contain equal angles, therefore the base BE is equal to the base AC, the triangle ABE is equal to the triangle ABC, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. [I. 4] Therefore the angle BAC is equal to the angle ABE; therefore the angle AHE is double of the angle BAH. [I. 32] But the angle EAC is also double of the angle BAC, inasmuch as the circumference EDC is also double of the circumference CB; [III. 28, VI. 33] therefore the angle HAE is equal to the angle AHE; hence the straight line HE is also equal to EA, that is, to AB. [I. 6] And, since the straight line BA is equal to AE, the angle ABE is also equal to the angle AEB. [I. 5] But the angle ABE was proved equal to the angle BAH; therefore the angle BEA is also equal to the angle BAH. And the angle ABE is common to the two triangles ABE and ABH; therefore the remaining angle BAE is equal to the remaining angle AHB; [I. 32] therefore the triangle ABE is equiangular with the triangle ABH; therefore, proportionally, as EB is to BA, so is AB to BH. [VI. 4] But BA is equal to EH; therefore, as BE is to EH, so is EH to HB. And BE is greater than EH; therefore EH is also greater than HB. [V. 14] Therefore BE has been cut in extreme and mean ratio at H, and the greater segment HE is equal to the side of the pentagon. Similarly we can prove that AC has also been cut in extreme and mean ratio at H, and its greater segment CH is equal to the side of the pentagon.'"],["Rule","'ProofWordCount'",392],["Rule","'GreekProof'","'πενταγώνου γὰρ ἰσοπλεύρου καὶ ἰσογωνίου τοῦ ΑΒΓ ΔΕ δύο γωνίας τὰς κατὰ τὸ ἑξῆς τὰς πρὸς τοῖς Α, Β ὑποτεινέτωσαν εὐθεῖαι αἱ ΑΓ, ΒΕ τέμνουσαι ἀλλήλας κατὰ τὸ Θ σημεῖον: λέγω, ὅτι ἑκατέρα αὐτῶν ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Θ σημεῖον, καὶ τὰ μείζονα αὐτῶν τμήματα ἴσα ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ. περιγεγράφθω γὰρ περὶ τὸ ΑΒΓΔΕ πεντάγωνον κύκλος ὁ ΑΒΓΔΕ. καὶ ἐπεὶ δύο εὐθεῖαι αἱ ΕΑ, ΑΒ δυσὶ ταῖς ΑΒ, ΒΓ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΒΕ βάσει τῇ ΑΓ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ ΑΒΓ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΑΒΕ: διπλῆ ἄρα ἡ ὑπὸ ΑΘΕ τῆς ὑπὸ ΒΑΘ. ἔστι δὲ καὶ ἡ ὑπὸ ΕΑΓ τῆς ὑπὸ ΒΑΓ διπλῆ, ἐπειδήπερ καὶ περιφέρεια ἡ ΕΔΓ περιφερείας τῆς ΓΒ ἐστι διπλῆ: ἴση ἄρα ἡ ὑπὸ ΘΑΕ γωνία τῇ ὑπὸ ΑΘΕ: ὥστε καὶ ἡ ΘΕ εὐθεῖα τῇ ΕΑ, τουτέστι τῇ ΑΒ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΑ εὐθεῖα τῇ ΑΕ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΒΕ τῇ ὑπὸ ΑΕΒ. ἀλλὰ ἡ ὑπὸ ΑΒΕ τῇ ὑπὸ ΒΑΘ ἐδείχθη ἴση: καὶ ἡ ὑπὸ ΒΕΑ ἄρα τῇ ὑπὸ ΒΑΘ ἐστιν ἴση. καὶ κοινὴ τῶν δύο τριγώνων τοῦ τε ΑΒΕ καὶ τοῦ ΑΒΘ ἐστιν ἡ ὑπὸ ΑΒΕ: λοιπὴ ἄρα ἡ ὑπὸ ΒΑΕ γωνία λοιπῇ τῇ ὑπὸ ΑΘΒ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΑΒΘ τριγώνῳ: ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΕΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΘ. ἴση δὲ ἡ ΒΑ τῇ ΕΘ: ὡς ἄρα ἡ ΒΕ πρὸς τὴν ΕΘ, οὕτως ἡ ΕΘ πρὸς τὴν ΘΒ. μείζων δὲ ἡ ΒΕ τῆς ΕΘ: μείζων ἄρα καὶ ἡ ΕΘ τῆς ΘΒ. ἡ ΒΕ ἄρα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Θ, καὶ τὸ μεῖζον τμῆμα τὸ ΘΕ ἴσον ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΑΓ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Θ, καὶ τὸ μεῖζον αὐτῆς τμῆμα ἡ ΓΘ ἴσον ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",350]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",9]],["Association",["Rule","'VertexLabel'","'13.9'"],["Rule","'Text'","'If the side of the hexagon and that of the decagon inscribed in the same circle be added together, the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon.'"],["Rule","'TextWordCount'",41],["Rule","'GreekText'","'ἐὰν ἡ τοῦ ἑξαγώνου πλευρὰ καὶ ἡ τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων συντεθῶσιν, ἡ ὅλη εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ τοῦ ἑξαγώνου πλευρά.'"],["Rule","'GreekTextWordCount'",34],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",4],["Rule","'Theorem'",15]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",33]]]],["Rule","'Proof'","'Let ABC be a circle; of the figures inscribed in the circle ABC let BC be the side of a decagon, CD that of a hexagon, and let them be in a straight line; I say that the whole straight line BD has been cut in extreme and mean ratio, and CD is its greater segment. For let the centre of the circle, the point E, be taken, let EB, EC, ED be joined, and let BE be carried through to A. Since BC is the side of an equilateral decagon, therefore the circumference ACB is five times the circumference BC; therefore the circumference AC is quadruple of CB. But, as the circumference AC is to CB, so is the angle AEC to the angle CEB; [VI. 33] therefore the angle AEC is quadruple of the angle CEB. And, since the angle EBC is equal to the angle ECB, [I. 5] therefore the angle AEC is double of the angle ECB. [I. 32] And, since the straight line EC is equal to CD, for each of them is equal to the side of the hexagon inscribed in the circle ABC, [IV. 15] the angle CED is also equal to the angle CDE; [I. 5] therefore the angle ECB is double of the angle EDC. [I. 32] But the angle AEC was proved double of the angle ECB; therefore the angle AEC is quadruple of the angle EDC. But the angle AEC was also proved quadruple of the angle BEC; therefore the angle EDC is equal to the angle BEC. But the angle EBD is common to the two triangles BEC and BED; therefore the remaining angle BED is also equal to the remaining angle ECB; [I. 32] therefore the triangle EBD is equiangular with the triangle EBC. Therefore, proportionally, as DB is to BE, so is EB to BC. [VI. 4] But EB is equal to CD. Therefore, as BD is to DC, so is DC to CB. And BD is greater than DC; therefore DC is also greater than CB. Therefore the straight line BD has been cut in extreme and mean ratio, and DC is its greater segment.'"],["Rule","'ProofWordCount'",360],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓ, καὶ τῶν εἰς τὸν ΑΒΓ κύκλον ἐγγραφομένων σχημάτων, δεκαγώνου μὲν ἔστω πλευρὰ ἡ ΒΓ, ἑξαγώνου δὲ ἡ ΓΔ, καὶ ἔστωσαν ἐπ᾽ εὐθείας: λέγω, ὅτι ἡ ὅλη εὐθεῖα ἡ ΒΔ ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΓΔ. εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ε σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΕΒ, ΕΓ, ΕΔ, καὶ διήχθω ἡ ΒΕ ἐπὶ τὸ Α. ἐπεὶ δεκαγώνου ἰσοπλεύρου πλευρά ἐστιν ἡ ΒΓ, πενταπλασίων ἄρα ἡ ΑΓΒ περιφέρεια τῆς ΒΓ περιφερείας: τετραπλασίων ἄρα ἡ ΑΓ περιφέρεια τῆς ΓΒ. ὡς δὲ ἡ ΑΓ περιφέρεια πρὸς τὴν ΓΒ, οὕτως ἡ ὑπὸ ΑΕΓ γωνία πρὸς τὴν ὑπὸ ΓΕΒ: τετραπλασίων ἄρα ἡ ὑπὸ ΑΕΓ τῆς ὑπὸ ΓΕΒ. καὶ ἐπεὶ ἴση ἡ ὑπὸ ΕΒΓ γωνία τῇ ὑπὸ ΕΓΒ, ἡ ἄρα ὑπὸ ΑΕΓ γωνία διπλασία ἐστὶ τῆς ὑπὸ ΕΓΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΕΓ εὐθεῖα τῇ ΓΔ: ἑκατέρα γὰρ αὐτῶν ἴση ἐστὶ τῇ τοῦ ἑξαγώνου πλευρᾷ τοῦ εἰς τὸν ΑΒΓ κύκλον ἐγγραφομένου: ἴση ἐστὶ καὶ ἡ ὑπὸ ΓΕΔ γωνία τῇ ὑπὸ ΓΔΕ γωνίᾳ: διπλασία ἄρα ἡ ὑπὸ ΕΓΒ γωνία τῆς ὑπὸ ΕΔΓ. ἀλλὰ τῆς ὑπὸ ΕΓΒ διπλασία ἐδείχθη ἡ ὑπὸ ΑΕΓ: τετραπλασία ἄρα ἡ ὑπὸ ΑΕΓ τῆς ὑπὸ ΕΔΓ. ἐδείχθη δὲ καὶ τῆς ὑπὸ ΒΕΓ τετραπλασία ἡ ὑπὸ ΑΕΓ: ἴση ἄρα ἡ ὑπὸ ΕΔΓ τῇ ὑπὸ ΒΕΓ. κοινὴ δὲ τῶν δύο τριγώνων, τοῦ τε ΒΕΓ καὶ τοῦ ΒΕΔ, ἡ ὑπὸ ΕΒΔ γωνία: καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΕΔ τῇ ὑπὸ ΕΓΒ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΕΒΔ τρίγωνον τῷ ΕΒΓ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΕΒ πρὸς τὴν ΒΓ. ἴση δὲ ἡ ΕΒ τῇ ΓΔ. ἔστιν ἄρα ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΔΓ πρὸς τὴν ΓΒ. μείζων δὲ ἡ ΒΔ τῆς ΔΓ: μείζων ἄρα καὶ ἡ ΔΓ τῆς ΓΒ. ἡ ΒΔ ἄρα εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Γ, καὶ τὸ μεῖζον τμῆμα αὐτῆς ἐστιν ἡ ΔΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",321]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",10]],["Association",["Rule","'VertexLabel'","'13.10'"],["Rule","'Text'","'If an equilateral pentagon be inscribed in a circle, the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle.'"],["Rule","'TextWordCount'",39],["Rule","'GreekText'","'ἐὰν εἰς κύκλον πεντάγωνον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ πενταγώνου πλευρὰ δύναται τήν τε τοῦ ἑξαγώνου καὶ τὴν τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων.'"],["Rule","'GreekTextWordCount'",25],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",5]],["List",["Rule","'Book'",1],["Rule","'Theorem'",26]],["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",2],["Rule","'Theorem'",2]],["List",["Rule","'Book'",3],["Rule","'Theorem'",26]],["List",["Rule","'Book'",4],["Rule","'Theorem'",15]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",6],["Rule","'Theorem'",33]]]],["Rule","'Proof'","'Let ABCDE be a circle, and let the equilateral pentagon ABCDE be inscribed in the circle ABCDE. I say that the square on the side of the pentagon ABCDE is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle ABCDE. For let the centre of the circle, the point F, be taken, let AF be joined and carried through to the point G, let FB be joined, let FH be drawn from F perpendicular to AB and be carried through to K, let AK, KB be joined, let FL be again drawn from F perpendicular to AK, and be carried through to M, and let KN be joined. Since the circumference ABCG is equal to the circumference AEDG, and in them ABC is equal to AED, therefore the remainder, the circumference CG, is equal to the remainder GD. But CD belongs to a pentagon; therefore CG belongs to a decagon. And, since FA is equal to FB, and FH is perpendicular, therefore the angle AFK is also equal to the angle KFB. [I. 5, I. 26] Hence the circumference AK is also equal to KB; [III. 26] therefore the circumference AB is double of the circumference BK; therefore the straight line AK is a side of a decagon. For the same reason AK is also double of KM. Now, since the circumference AB is double of the circumference BK, while the circumference CD is equal to the circumference AB, therefore the circumference CD is also double of the circumference BK. But the circumference CD is also double of CG; therefore the circumference CG is equal to the circumference BK. But BK is double of KM, since KA is so also; therefore CG is also double of KM. But, further, the circumference CB is also double of the circumference BK, for the circumference CB is equal to BA. Therefore the whole circumference GB is also double of BM; hence the angle GFB is also double of the angle BFM. [VI. 33] But the angle GFB is also double of the angle FAB, for the angle FAB is equal to the angle ABF. Therefore the angle BFN is also equal to the angle FAB. But the angle ABF is common to the two triangles ABF and BFN; therefore the remaining angle AFB is equal to the remaining angle BNF; [I. 32] therefore the triangle ABF is equiangular with the triangle BFN. Therefore, proportionally, as the straight line AB is to BF, so is FB to BN; [VI. 4] therefore the rectangle AB, BN is equal to the square on BF. [VI. 17] Again, since AL is equal to LK, while LN is common and at right angles, therefore the base KN is equal to the base AN; [I. 4] therefore the angle LKN is also equal to the angle LAN. But the angle LAN is equal to the angle KBN; therefore the angle LKN is also equal to the angle KBN. And the angle at A is common to the two triangles AKB and AKN. Therefore the remaining angle AKB is equal to the remaining angle KNA; [I. 32] therefore the triangle KBA is equiangular with the triangle KNA. Therefore, proportionally, as the straight line BA is to AK, so is KA to AN; [VI. 4] therefore the rectangle BA, AN is equal to the square on AK. [VI. 17] But the rectangle AB, BN was also proved equal to the square on BF; therefore the rectangle AB, BN together with the rectangle BA, AN, that is, the square on BA [II. 2], is equal to the square on BF together with the square on AK. And BA is a side of the pentagon, BF of the hexagon [IV. 15], and AK of the decagon.'"],["Rule","'ProofWordCount'",634],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓΔΕ, καὶ εἰς τὸν ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρον ἐγγεγράφθω τὸ ΑΒΓΔΕ. λέγω, ὅτι ἡ τοῦ ΑΒΓΔΕ πενταγώνου πλευρὰ δύναται τήν τε τοῦ ἑξαγώνου καὶ τὴν τοῦ δεκαγώνου πλευρὰν τῶν εἰς τὸν ΑΒΓΔΕ κύκλον ἐγγραφομένων. εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ζ σημεῖον, καὶ ἐπιζευχθεῖσα ἡ ΑΖ διήχθω ἐπὶ τὸ Η σημεῖον, καὶ ἐπεζεύχθω ἡ ΖΒ, καὶ ἀπὸ τοῦ Ζ ἐπὶ τὴν ΑΒ κάθετος ἤχθω ἡ ΖΘ, καὶ διήχθω ἐπὶ τὸ Κ, καὶ ἐπεζεύχθωσαν αἱ ΑΚ, ΚΒ, καὶ πάλιν ἀπὸ τοῦ Ζ ἐπὶ τὴν ΑΚ κάθετος ἤχθω ἡ ΖΛ, καὶ διήχθω ἐπὶ τὸ Μ, καὶ ἐπεζεύχθω ἡ ΚΝ. ἐπεὶ ἴση ἐστὶν ἡ ΑΒΓΗ περιφέρεια τῇ ΑΕΔΗ περιφερείᾳ, ὧν ἡ ΑΒΓ τῇ ΑΕΔ ἐστιν ἴση, λοιπὴ ἄρα ἡ ΓΗ περιφέρεια λοιπῇ τῇ ΗΔ ἐστιν ἴση. πενταγώνου δὲ ἡ ΓΔ: δεκαγώνου ἄρα ἡ ΓΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΖΑ τῇ ΖΒ, καὶ κάθετος ἡ ΖΘ, ἴση ἄρα καὶ ἡ ὑπὸ ΑΖΚ γωνία τῇ ὑπὸ ΚΖΒ. ὥστε καὶ περιφέρεια ἡ ΑΚ τῇ ΚΒ ἐστιν ἴση: διπλῆ ἄρα ἡ ΑΒ περιφέρεια τῆς ΒΚ περιφερείας: δεκαγώνου ἄρα πλευρά ἐστιν ἡ ΑΚ εὐθεῖα. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΑΚ τῆς ΚΜ ἐστι διπλῆ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΒ περιφέρεια τῆς ΒΚ περιφερείας, ἴση δὲ ἡ ΓΔ περιφέρεια τῇ ΑΒ περιφερείᾳ, διπλῆ ἄρα καὶ ἡ ΓΔ περιφέρεια τῆς ΒΚ περιφερείας. ἔστι δὲ ἡ ΓΔ περιφέρεια καὶ τῆς ΓΗ διπλῆ: ἴση ἄρα ἡ ΓΗ περιφέρεια τῇ ΒΚ περιφερείᾳ. ἀλλὰ ἡ ΒΚ τῆς ΚΜ ἐστι διπλῆ, ἐπεὶ καὶ ἡ ΚΑ: καὶ ἡ ΓΗ ἄρα τῆς ΚΜ ἐστι διπλῆ. ἀλλὰ μὴν καὶ ἡ ΓΒ περιφέρεια τῆς ΒΚ περιφερείας ἐστὶ διπλῆ: ἴση γὰρ ἡ ΓΒ περιφέρεια τῇ ΒΑ. καὶ ὅλη ἄρα ἡ ΗΒ περιφέρεια τῆς ΒΜ ἐστι διπλῆ: ὥστε καὶ γωνία ἡ ὑπὸ ΗΖΒ γωνίας τῆς ὑπὸ ΒΖΜ ἐστι διπλῆ. ἔστι δὲ ἡ ὑπὸ ΗΖΒ καὶ τῆς ὑπὸ ΖΑΒ διπλῆ: ἴση γὰρ ἡ ὑπὸ ΖΑΒ τῇ ὑπὸ ΑΒΖ. καὶ ἡ ὑπὸ ΒΖΝ ἄρα τῇ ὑπὸ ΖΑΒ ἐστιν ἴση. κοινὴ δὲ τῶν δύο τριγώνων, τοῦ τε ΑΒΖ καὶ τοῦ ΒΖΝ, ἡ ὑπὸ ΑΒΖ γωνία: λοιπὴ ἄρα ἡ ὑπὸ ΑΖΒ λοιπῇ τῇ ὑπὸ ΒΝΖ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΖ τρίγωνον τῷ ΒΖΝ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΒ εὐθεῖα πρὸς τὴν ΒΖ, οὕτως ἡ ΖΒ πρὸς τὴν ΒΝ: τὸ ἄρα ὑπὸ τῶν ΑΒΝ ἴσον ἐστὶ τῷ ἀπὸ ΒΖ. πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΑΛ τῇ ΛΚ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΛΝ, βάσις ἄρα ἡ ΚΝ βάσει τῇ ΑΝ ἐστιν ἴση: καὶ γωνία ἄρα ἡ ὑπὸ ΛΚΝ γωνίᾳ τῇ ὑπὸ ΛΑΝ ἐστιν ἴση. ἀλλὰ ἡ ὑπὸ ΛΑΝ τῇ ὑπὸ ΚΒΝ ἐστιν ἴση: καὶ ἡ ὑπὸ ΛΚΝ ἄρα τῇ ὑπὸ ΚΒΝ ἐστιν ἴση. καὶ κοινὴ τῶν δύο τριγώνων τοῦ τε ΑΚΒ καὶ τοῦ ΑΚΝ ἡ πρὸς τῷ Α. λοιπὴ ἄρα ἡ ὑπὸ ΑΚΒ λοιπῇ τῇ ὑπὸ ΚΝΑ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΚΒΑ τρίγωνον τῷ ΚΝΑ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΑ εὐθεῖα πρὸς τὴν ΑΚ, οὕτως ἡ ΚΑ πρὸς τὴν ΑΝ: τὸ ἄρα ὑπὸ τῶν ΒΑΝ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΚ. ἐδείχθη δὲ καὶ τὸ ὑπὸ τῶν ΑΒΝ ἴσον τῷ ἀπὸ τῆς ΒΖ: τὸ ἄρα ὑπὸ τῶν ΑΒΝ μετὰ τοῦ ὑπὸ ΒΑΝ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΒΑ, ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΖ μετὰ τοῦ ἀπὸ τῆς ΑΚ. καί ἐστιν ἡ μὲν ΒΑ πενταγώνου πλευρά, ἡ δὲ ΒΖ ἑξαγώνου, ἡ δὲ ΑΚ δεκαγώνου. ἡ ἄρα τοῦ πενταγώνου πλευρὰ δύναται τήν τε τοῦ ἑξαγώνου καὶ τὴν τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",577]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",11]],["Association",["Rule","'VertexLabel'","'13.11'"],["Rule","'Text'","'If in a circle which has its diameter rational an equilateral pentagon be inscribed, the side of the pentagon is the irrational straight line called minor.'"],["Rule","'TextWordCount'",26],["Rule","'GreekText'","'ἐὰν εἰς κύκλον ῥητὴν ἔχοντα τὴν διάμετρον πεντάγωνον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",32]],["List",["Rule","'Book'",5],["Rule","'Theorem'",18]],["List",["Rule","'Book'",5],["Rule","'Theorem'",19]],["List",["Rule","'Book'",10],["Rule","'Definition'",3.4]],["List",["Rule","'Book'",10],["Rule","'Theorem'",9]],["List",["Rule","'Book'",10],["Rule","'Theorem'",12]],["List",["Rule","'Book'",10],["Rule","'Theorem'",15]],["List",["Rule","'Book'",10],["Rule","'Theorem'",73]],["List",["Rule","'Book'",10],["Rule","'Theorem'",94]],["List",["Rule","'Book'",13],["Rule","'Theorem'",1]],["List",["Rule","'Book'",13],["Rule","'Theorem'",8]]]],["Rule","'Proof'","'For in the circle ABCDE which has its diameter rational let the equilateral pentagon ABCDE be inscribed; I say that the side of the pentagon is the irrational straight line called minor. For let the centre of the circle, the point F, be taken, let AF, FB be joined and carried through to the points, G, H, let AC be joined, and let FK be made a fourth part of AF. Now AF is rational; therefore FK is also rational. But BF is also rational; therefore the whole BK is rational. And, since the circumference ACG is equal to the circumference ADG, and in them ABC is equal to AED, therefore the remainder CG is equal to the remainder GD. And, if we join AD, we conclude that the angles at L are right, and CD is double of CL. For the same reason the angles at M are also right, and AC is double of CM. Since then the angle ALC is equal to the angle AMF, and the angle LAC is common to the two triangles ACL and AMF, therefore the remaining angle ACL is equal to the remaining angle MFA; [I. 32] therefore the triangle ACL is equiangular with the triangle AMF; therefore, proportionally, as LC is to CA, so is MF to FA. And the doubles of the antecedents may be taken; therefore, as the double of LC is to CA, so is the double of MF to FA. But, as the double of MF is to FA, so is MF to the half of FA; therefore also, as the double of LC is to CA, so is MF to the half of FA. And the halves of the consequents may be taken; therefore, as the double of LC is to the half of CA, so is MF to the fourth of FA. And DC is double of LC, CM is half of CA, and FK a fourth part of FA; therefore, as DC is to CM, so is MF to FK. Componendo also, as the sum of DC, CM is to CM, so is MK to KF; [V. 18] therefore also, as the square on the sum of DC, CM is to the square on CM, so is the square on MK to the square on KF. And since, when the straight line subtending two sides of the pentagon, as AC, is cut in extreme and mean ratio, the greater segment is equal to the side of the pentagon, that is, to DC, [XIII. 8] while the square on the greater segment added to the half of the whole is five times the square on the half of the whole, [XIII. 1] and CM is half of the whole AC, therefore the square on DC, CM taken as one straight line is five times the square on CM. But it was proved that, as the square on DC, CM taken as one straight line is to the square on CM, so is the square on MK to the square on KF; therefore the square on MK is five times the square on KF. But the square on KF is rational, for the diameter is rational; therefore the square on MK is also rational; therefore MK is rational And, since BF is quadruple of FK, therefore BK is five times KF; therefore the square on BK is twenty-five times the square on KF. But the square on MK is five times the square on KF; therefore the square on BK is five times the square on KM; therefore the square on BK has not to the square on KM the ratio which a square number has to a square number; therefore BK is incommensurable in length with KM. [X. 9] And each of them is rational. Therefore BK, KM are rational straight lines commensurable in square only. But, if from a rational straight line there be subtracted a rational straight line which is commensurable with the whole in square only, the remainder is irrational, namely an apotome; therefore MB is an apotome and MK the annex to it. [X. 73] I say next that MB is also a fourth apotome. Let the square on N be equal to that by which the square on BK is greater than the square on KM; therefore the square on BK is greater than the square on KM by the square on N. And, since KF is commensurable with FB, componendo also, KB is commensurable with FB. [X. 15] But BF is commensurable with BH; therefore BK is also commensurable with BH. [X. 12] And, since the square on BK is five times the square on KM, therefore the square on BK has to the square on KM the ratio which 5 has to 1. Therefore, convertendo, the square on BK has to the square on N the ratio which 5 has to 4 [V. 19], and this is not the ratio which a square number has to a square number; therefore BK is incommensurable with N; [X. 9] therefore the square on BK is greater than the square on KM by the square on a straight line incommensurable with BK. Since then the square on the whole BK is greater than the square on the annex KM by the square on a straight line incommensurable with BK, and the whole BK is commensurable with the rational straight line, BH, set out, therefore MB is a fourth apotome. [X. Deff. III. 4] But the rectangle contained by a rational straight line and a fourth apotome is irrational, and its square root is irrational, and is called minor. [X. 94] But the square on AB is equal to the rectangle HB, BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and, as HB is to BA, so is AB to BM. Therefore the side AB of the pentagon is the irrational straight line called minor.'"],["Rule","'ProofWordCount'",990],["Rule","'GreekProof'","'εἰς γὰρ κύκλον τὸν ΑΒΓΔΕ ῥητὴν ἔχοντα τὴν διάμετρον πεντάγωνον ἰσόπλευρον ἐγγεγράφθω τὸ ΑΒΓΔΕ: λέγω, ὅτι ἡ τοῦ ΑΒΓΔΕ πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων. εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ζ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΖΒ καὶ διήχθωσαν ἐπὶ τὰ Η, Θ σημεῖα, καὶ ἐπεζεύχθω ἡ ΑΓ, καὶ κείσθω τῆς ΑΖ τέταρτον μέρος ἡ ΖΚ. ῥητὴ δὲ ἡ ΑΖ: ῥητὴ ἄρα καὶ ἡ ΖΚ. ἔστι δὲ καὶ ἡ ΒΖ ῥητή: ὅλη ἄρα ἡ ΒΚ ῥητή ἐστιν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓΗ περιφέρεια τῇ ΑΔΗ περιφερείᾳ, ὧν ἡ ΑΒΓ τῇ ΑΕΔ ἐστιν ἴση, λοιπὴ ἄρα ἡ ΓΗ λοιπῇ τῇ ΗΔ ἐστιν ἴση. καὶ ἐὰν ἐπιζεύξωμεν τὴν ΑΔ, συνάγονται ὀρθαὶ αἱ πρὸς τῷ Λ γωνίαι, καὶ διπλῆ ἡ ΓΔ τῆς ΓΛ. διὰ τὰ αὐτὰ δὴ καὶ αἱ πρὸς τῷ Μ ὀρθαί εἰσιν, καὶ διπλῆ ἡ ΑΓ τῆς ΓΜ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΑΛΓ γωνία τῇ ὑπὸ ΑΜΖ, κοινὴ δὲ τῶν δύο τριγώνων τοῦ τε ΑΓΛ καὶ τοῦ ΑΜΖ ἡ ὑπὸ ΛΑΓ, λοιπὴ ἄρα ἡ ὑπὸ ΑΓΛ λοιπῇ τῇ ὑπὸ ΜΖΑ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΓΛ τρίγωνον τῷ ΑΜΖ τριγώνῳ: ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΛΓ πρὸς ΓΑ, οὕτως ἡ ΜΖ πρὸς ΖΑ: καὶ τῶν ἡγουμένων τὰ διπλάσια: ὡς ἄρα ἡ τῆς ΛΓ διπλῆ πρὸς τὴν ΓΑ, οὕτως ἡ τῆς ΜΖ διπλῆ πρὸς τὴν ΖΑ. ὡς δὲ ἡ τῆς ΜΖ διπλῆ πρὸς τὴν ΖΑ, οὕτως ἡ ΜΖ πρὸς τὴν ἡμίσειαν τῆς ΖΑ: καὶ ὡς ἄρα ἡ τῆς ΛΓ διπλῆ πρὸς τὴν ΓΑ, οὕτως ἡ ΜΖ πρὸς τὴν ἡμίσειαν τῆς ΖΑ. καὶ τῶν ἑπομένων τὰ ἡμίσεα: ὡς ἄρα ἡ τῆς ΛΓ διπλῆ πρὸς τὴν ἡμίσειαν τῆς ΓΑ, οὕτως ἡ ΜΖ πρὸς τὸ τέταρτον τῆς ΖΑ. καί ἐστι τῆς μὲν ΛΓ διπλῆ ἡ ΔΓ, τῆς δὲ ΓΑ ἡμίσεια ἡ ΓΜ, τῆς δὲ ΖΑ τέταρτον μέρος ἡ ΖΚ: ἔστιν ἄρα ὡς ἡ ΔΓ πρὸς τὴν ΓΜ, οὕτως ἡ ΜΖ πρὸς τὴν ΖΚ. συνθέντι καὶ ὡς συναμφότερος ἡ ΔΓΜ πρὸς τὴν ΓΜ, οὕτως ἡ ΜΚ πρὸς ΚΖ: καὶ ὡς ἄρα τὸ ἀπὸ συναμφοτέρου τῆς ΔΓΜ πρὸς τὸ ἀπὸ ΓΜ, οὕτως τὸ ἀπὸ ΜΚ πρὸς τὸ ἀπὸ ΚΖ. καὶ ἐπεὶ τῆς ὑπὸ δύο πλευρὰς τοῦ πενταγώνου ὑποτεινούσης, οἷον τῆς ΑΓ, ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμα ἴσον ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ, τουτέστι τῇ ΔΓ, τὸ δὲ μεῖζον τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τῆς ὅλης, καί ἐστιν ὅλης τῆς ΑΓ ἡμίσεια ἡ ΓΜ, τὸ ἄρα ἀπὸ τῆς ΔΓΜ ὡς μιᾶς πενταπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΓΜ. ὡς δὲ τὸ ἀπὸ τῆς ΔΓΜ ὡς μιᾶς πρὸς τὸ ἀπὸ τῆς ΓΜ, οὕτως ἐδείχθη τὸ ἀπὸ τῆς ΜΚ πρὸς τὸ ἀπὸ τῆς ΚΖ: πενταπλάσιον ἄρα τὸ ἀπὸ τῆς ΜΚ τοῦ ἀπὸ τῆς ΚΖ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΚΖ: ῥητὴ γὰρ ἡ διάμετρος: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΜΚ: ῥητὴ ἄρα ἐστὶν ἡ ΜΚ δυνάμει μόνον. καὶ ἐπεὶ τετραπλασία ἐστὶν ἡ ΒΖ τῆς ΖΚ, πενταπλασία ἄρα ἐστὶν ἡ ΒΚ τῆς ΚΖ: εἰκοσιπενταπλάσιον ἄρα τὸ ἀπὸ τῆς ΒΚ τοῦ ἀπὸ τῆς ΚΖ. πενταπλάσιον δὲ τὸ ἀπὸ τῆς ΜΚ τοῦ ἀπὸ τῆς ΚΖ: πενταπλάσιον ἄρα τὸ ἀπὸ τῆς ΒΚ τοῦ ἀπὸ τῆς ΚΜ: τὸ ἄρα ἀπὸ τῆς ΒΚ πρὸς τὸ ἀπὸ ΚΜ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΚ τῇ ΚΜ μήκει. καί ἐστι ῥητὴ ἑκατέρα αὐτῶν. αἱ ΒΚ, ΚΜ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. ἐὰν δὲ ἀπὸ ῥητῆς ῥητὴ ἀφαιρεθῇ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, ἡ λοιπὴ ἄλογός ἐστιν ἀποτομή: ἀποτομὴ ἄρα ἐστὶν ἡ ΜΒ, προσαρμόζουσα δὲ αὐτῇ ἡ ΜΚ. λέγω δή, ὅτι καὶ τετάρτη. ᾧ δὴ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΚ τοῦ ἀπὸ τῆς ΚΜ, ἐκείνῳ ἴσον ἔστω τὸ ἀπὸ τῆς Ν: ἡ ΒΚ ἄρα τῆς ΚΜ μεῖζον δύναται τῇ Ν. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΚΖ τῇ ΖΒ, καὶ συνθέντι σύμμετρός ἐστιν ἡ ΚΒ τῇ ΖΒ. ἀλλὰ ἡ ΒΖ τῇ ΒΘ σύμμετρός ἐστιν: καὶ ἡ ΒΚ ἄρα τῇ ΒΘ σύμμετρός ἐστιν. καὶ ἐπεὶ πενταπλάσιόν ἐστι τὸ ἀπὸ τῆς ΒΚ τοῦ ἀπὸ τῆς ΚΜ, τὸ ἄρα ἀπὸ τῆς ΒΚ πρὸς τὸ ἀπὸ τῆς ΚΜ λόγον ἔχει, ὃν ε πρὸς ἕν. ἀναστρέψαντι ἄρα τὸ ἀπὸ τῆς ΒΚ πρὸς τὸ ἀπὸ τῆς Ν λόγον ἔχει, ὃν ε πρὸς δ, οὐχ ὃν τετράγωνος πρὸς τετράγωνον: ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΚ τῇ Ν: ἡ ΒΚ ἄρα τῆς ΚΜ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. ἐπεὶ οὖν ὅλη ἡ ΒΚ τῆς προσαρμοζούσης τῆς ΚΜ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ὅλη ἡ ΒΚ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΒΘ, ἀποτομὴ ἄρα τετάρτη ἐστὶν ἡ ΜΒ. τὸ δὲ ὑπὸ ῥητῆς καὶ ἀποτομῆς τετάρτης περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν, καλεῖται δὲ ἐλάττων. δύναται δὲ τὸ ὑπὸ τῶν ΘΒΜ ἡ ΑΒ διὰ τὸ ἐπιζευγνυμένης τῆς ΑΘ ἰσογώνιον γίνεσθαι τὸ ΑΒΘ τρίγωνον τῷ ΑΒΜ τριγώνῳ καὶ εἶναι ὡς τὴν ΘΒ πρὸς τὴν ΒΑ, οὕτως τὴν ΑΒ πρὸς τὴν ΒΜ. ἡ ἄρα ΑΒ τοῦ πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",819]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",12]],["Association",["Rule","'VertexLabel'","'13.12'"],["Rule","'Text'","'If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius of the circle.'"],["Rule","'TextWordCount'",28],["Rule","'GreekText'","'ἐὰν εἰς κύκλον τρίγωνον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ τριγώνου πλευρὰ δυνάμει τριπλασίων ἐστὶ τῆς ἐκ τοῦ κέντρου τοῦ κύκλου.'"],["Rule","'GreekTextWordCount'",19],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",4],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let ABC be a circle, and let the equilateral triangle ABC be inscribed in it; I say that the square on one side of the triangle ABC is triple of the square on the radius of the circle. For let the centre D of the circle ABC be taken, let AD be joined and carried through to E, and let BE be joined. Then, since the triangle ABC is equilateral, therefore the circumference BEC is a third part of the circumference of the circle ABC. Therefore the circumference BE is a sixth part of the circumference of the circle; therefore the straight line BE belongs to a hexagon; therefore it is equal to the radius DE. [IV. 15] And, since AE is double of DE, the square on AE is quadruple of the square on ED, that is, of the square on BE. But the square on AE is equal to the squares on AB, BE; [III. 31, I. 47] therefore the squares on AB, BE are quadruple of the square on BE. Therefore, separando, the square on AB is triple of the square on BE. But BE is equal to DE; therefore the square on AB is triple of the square on DE. Therefore the square on the side of the triangle is triple of the square on the radius.'"],["Rule","'ProofWordCount'",221],["Rule","'GreekProof'","'ἔστω κύκλος ὁ ΑΒΓ, καὶ εἰς αὐτὸν τρίγωνον ἰσόπλευρον ἐγγεγράφθω τὸ ΑΒΓ: λέγω, ὅτι τοῦ ΑΒΓ τριγώνου μία πλευρὰ δυνάμει τριπλασίων ἐστὶ τῆς ἐκ τοῦ κέντρου τοῦ ΑΒΓ κύκλου. εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου τὸ Δ, καὶ ἐπιζευχθεῖσα ἡ ΑΔ διήχθω ἐπὶ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΒΕ. καὶ ἐπεὶ ἰσόπλευρόν ἐστι τὸ ΑΒΓ τρίγωνον, ἡ ΒΕΓ ἄρα περιφέρεια τρίτον μέρος ἐστὶ τῆς τοῦ ΑΒΓ κύκλου περιφερείας. ἡ ἄρα ΒΕ περιφέρεια ἕκτον ἐστὶ μέρος τῆς τοῦ κύκλου περιφερείας: ἑξαγώνου ἄρα ἐστὶν ἡ ΒΕ εὐθεῖα: ἴση ἄρα ἐστὶ τῇ ἐκ τοῦ κέντρου τῇ ΔΕ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΕ τῆς ΔΕ, τετραπλάσιόν ἐστι τὸ ἀπὸ τῆς ΑΕ τοῦ ἀπὸ τῆς ΕΔ, τουτέστι τοῦ ἀπὸ τῆς ΒΕ. ἴσον δὲ τὸ ἀπὸ τῆς ΑΕ τοῖς ἀπὸ τῶν ΑΒ, ΒΕ: τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΕ τετραπλάσιά ἐστι τοῦ ἀπὸ τῆς ΒΕ. διελόντι ἄρα τὸ ἀπὸ τῆς ΑΒ τριπλάσιόν ἐστι τοῦ ἀπὸ ΒΕ. ἴση δὲ ἡ ΒΕ τῇ ΔΕ: τὸ ἄρα ἀπὸ τῆς ΑΒ τριπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΔΕ. ἡ ἄρα τοῦ τριγώνου πλευρὰ δυνάμει τριπλασία ἐστὶ τῆς ἐκ τοῦ κέντρου τοῦ κύκλου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",186]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",13]],["Association",["Rule","'VertexLabel'","'13.13'"],["Rule","'Text'","'To construct a pyramid, to comprehend it in a given sphere, and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.'"],["Rule","'TextWordCount'",37],["Rule","'GreekText'","'πυραμίδα συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει ἡμιολία ἐστὶ τῆς πλευρᾶς τῆς πυραμίδος.'"],["Rule","'GreekTextWordCount'",21],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",3],["Rule","'Theorem'",1]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",4],["Rule","'Theorem'",2]],["List",["Rule","'Book'",6],["Rule","'Theorem'",8]],["List",["Rule","'Book'",6],["Rule","'Theorem'",17]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]],["List",["Rule","'Book'",11],["Rule","'Theorem'",12]],["List",["Rule","'Book'",13],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'Let the diameter AB of the given sphere be set out, and let it be cut at the point C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from the point C at right angles to AB, and let DA be joined; let the circle EFG which has its radius equal to DC be set out, let the equilateral triangle EFG be inscribed in the circle EFG, [IV. 2] let the centre of the circle, the point H, be taken, [III. 1] let EH, HF, HG be joined; from the point H let HK be set up at right angles to the plane of the circle EFG, [XI. 12] let HK equal to the straight line AC be cut off from HK, and let KE, KF, KG be joined. Now, since KH is at right angles to the plane of the circle EFG, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle EFG. [XI. Def. 3] But each of the straight lines HE, HF, HG meets it: therefore HK is at right angles to each of the straight lines HE, HF, HG. And, since AC is equal to HK, and CD to HE, and they contain right angles, therefore the base DA is equal to the base KE. [I. 4] For the same reason each of the straight lines KF, KG is also equal to DA; therefore the three straight lines KE, KF, KG are equal to one another. And, since AC is double of CB, therefore AB is triple of BC. But, as AB is to BC, so is the square on AD to the square on DC, as will be proved afterwards. Therefore the square on AD is triple of the square on DC. But the square on FE is also triple of the square on EH, [XIII. 12] and DC is equal to EH; therefore DA is also equal to EF. But DA was proved equal to each of the straight lines KE, KF, KG; therefore each of the straight lines EF, FG, GE is also equal to each of the straight lines KE, KF, KG; therefore the four triangles EFG, KEF, KFG, KEG are equilateral. Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex. It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. For let the straight line HL be produced in a straight line with KH, and let HL be made equal to CB. Now, since, as AC is to CD, so is CD to CB, [VI. 8] while AC is equal to KH, CD to HE, and CB to HL, therefore, as KH is to HE, so is EH to HL; therefore the rectangle KH, HL is equal to the square on EH. [VI. 17] And each of the angles KHE. EHL is right; therefore the semicircle described on KL will pass through E also. [cf. VI. 8, III. 31.] If then, KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, since, if FL, LG be joined, the angles at F, G similarly become right angles; and the pyramid will be comprehended in the given sphere. For KL, the diameter of the sphere, is equal to the diameter AB of the given sphere, inasmuch as KH was made equal to AC, and HL to CB. I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid For, since AC is double of CB, therefore AB is triple of BC; and, convertendo, BA is one and a half times AC. But, as BA is to AC, so is the square on BA to the square on AD. Therefore the square on BA is also one and a half times the square on AD. And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid. Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.'"],["Rule","'ProofWordCount'",753],["Rule","'GreekProof'","'Ἐκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ, καὶ τετμήσθω κατὰ τὸ Γ σημεῖον, ὥστε διπλασίαν εἶναι τὴν ΑΓ τῆς ΓΒ: καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ ἤχθω ἀπὸ τοῦ Γ σημείου τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΑ: καὶ ἐκκείσθω κύκλος ὁ ΕΖΗ ἴσην ἔχων τὴν ἐκ τοῦ κέντρου τῇ ΔΓ, καὶ ἐγγεγράφθω εἰς τὸν ΕΖΗ κύκλον τρίγωνον ἰσόπλευρον τὸ ΕΖΗ: καὶ εἰλήφθω τὸ κέντρον τοῦ κύκλου τὸ Θ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΕΘ, ΘΖ, ΘΗ: καὶ ἀνεστάτω ἀπὸ τοῦ Θ σημείου τῷ τοῦ ΕΖΗ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΘΚ, καὶ ἀφῃρήσθω ἀπὸ τῆς ΘΚ τῇ ΑΓ εὐθείᾳ ἴση ἡ ΘΚ, καὶ ἐπεζεύχθωσαν αἱ ΚΕ, ΚΖ, ΚΗ. καὶ ἐπεὶ ἡ ΚΘ ὀρθή ἐστι πρὸς τὸ τοῦ ΕΖΗ κύκλου ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ τοῦ ΕΖΗ κύκλου ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. ἅπτεται δὲ αὐτῆς ἑκάστη τῶν ΘΕ, ΘΖ, ΘΗ: ἡ ΘΚ ἄρα πρὸς ἑκάστην τῶν ΘΕ, ΘΖ, ΘΗ ὀρθή ἐστιν. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΓ τῇ ΘΚ, ἡ δὲ ΓΔ τῇ ΘΕ, καὶ ὀρθὰς γωνίας περιέχουσιν, βάσις ἄρα ἡ ΔΑ βάσει τῇ ΚΕ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ΚΖ, ΚΗ τῇ ΔΑ ἐστιν ἴση: αἱ τρεῖς ἄρα αἱ ΚΕ, ΚΖ, ΚΗ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΓ τῆς ΓΒ, τριπλῆ ἄρα ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ, ὡς ἑξῆς δειχθήσεται. τριπλάσιον ἄρα τὸ ἀπὸ τῆς ΑΔ τοῦ ἀπὸ τῆς ΔΓ. ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΖΕ τοῦ ἀπὸ τῆς ΕΘ τριπλάσιον, καί ἐστιν ἴση ἡ ΔΓ τῇ ΕΘ: ἴση ἄρα καὶ ἡ ΔΑ τῇ ΕΖ. ἀλλὰ ἡ ΔΑ ἑκάστῃ τῶν ΚΕ, ΚΖ, ΚΗ ἐδείχθη ἴση: καὶ ἑκάστη ἄρα τῶν ΕΖ, ΖΗ, ΗΕ ἑκάστῃ τῶν ΚΕ, ΚΖ, ΚΗ ἐστιν ἴση: ἰσόπλευρα ἄρα ἐστὶ τὰ τέσσαρα τρίγωνα τὰ ΕΖΗ, ΚΕΖ, ΚΖΗ, ΚΕΗ. πυραμὶς ἄρα συνέσταται ἐκ τεσσάρων τριγώνων ἰσοπλεύρων, ἧς βάσις μέν ἐστι τὸ ΕΖΗ τρίγωνον, κορυφὴ δὲ τὸ Κ σημεῖον. δεῖ δὴ αὐτὴν καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ δυνάμει τῆς πλευρᾶς τῆς πυραμίδος. Ἐκβεβλήσθω γὰρ ἐπ᾽ εὐθείας τῇ ΚΘ εὐθεῖα ἡ ΘΛ, καὶ κείσθω τῇ ΓΒ ἴση ἡ ΘΛ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΓ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΔ πρὸς τὴν ΓΒ, ἴση δὲ ἡ μὲν ΑΓ τῇ ΚΘ, ἡ δὲ ΓΔ τῇ ΘΕ, ἡ δὲ ΓΒ τῇ ΘΛ, ἔστιν ἄρα ὡς ἡ ΚΘ πρὸς τὴν ΘΕ, οὕτως ἡ ΕΘ πρὸς τὴν ΘΛ: τὸ ἄρα ὑπὸ τῶν ΚΘ, ΘΛ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ. καί ἐστιν ὀρθὴ ἑκατέρα τῶν ὑπὸ ΚΘΕ, ΕΘΛ γωνιῶν: τὸ ἄρα ἐπὶ τῆς ΚΛ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Ε ἐπειδήπερ ἐὰν ἐπιζεύξωμεν τὴν ΕΛ, ὀρθὴ γίνεται ἡ ὑπὸ ΛΕΚ γωνία διὰ τὸ ἰσογώνιον γίνεσθαι τὸ ΕΛΚ τρίγωνον ἑκατέρῳ τῶν ΕΛΘ, ΕΘΚ τριγώνων. ἐὰν δὴ μενούσης τῆς ΚΛ περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, ἥξει καὶ διὰ τῶν Ζ, Η σημείων ἐπιζευγνυμένων τῶν ΖΛ, ΛΗ καὶ ὀρθῶν ὁμοίως γινομένων τῶν πρὸς τοῖς Ζ, Η γωνιῶν: καὶ ἔσται ἡ πυραμὶς σφαίρᾳ περιειλημμένη τῇ δοθείσῃ. ἡ γὰρ ΚΛ τῆς σφαίρας διάμετρος ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας διαμέτρῳ τῇ ΑΒ, ἐπειδήπερ τῇ μὲν ΑΓ ἴση κεῖται ἡ ΚΘ, τῇ δὲ ΓΒ ἡ ΘΛ. λέγω δή, ὅτι ἡ τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ δυνάμει τῆς πλευρᾶς τῆς πυραμίδος. ἐπεὶ γὰρ διπλῆ ἐστιν ἡ ΑΓ τῆς ΓΒ, τριπλῆ ἄρα ἐστὶν ἡ ΑΒ τῆς ΒΓ: ἀναστρέψαντι ἡμιολία ἄρα ἐστὶν ἡ ΒΑ τῆς ΑΓ. ὡς δὲ ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΔ ἐπειδήπερ ἐπιζευγνυμένης τῆς ΔΒ ἐστιν ὡς ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΔΑ πρὸς τὴν ΑΓ διὰ τὴν ὁμοιότητα τῶν ΔΑΒ, ΔΑΓ τριγώνων, καὶ εἶναι ὡς τὴν πρώτην πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ἀπὸ τῆς δευτέρας. ἡμιόλιον ἄρα καὶ τὸ ἀπὸ τῆς ΒΑ τοῦ ἀπὸ τῆς ΑΔ. καί ἐστιν ἡ μὲν ΒΑ ἡ τῆς δοθείσης σφαίρας διάμετρος, ἡ δὲ ΑΔ ἴση τῇ πλευρᾷ τῆς πυραμίδος. ἡ ἄρα τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ τῆς πλευρᾶς τῆς πυραμίδος: ὅπερ ἔδει δεῖξαι. λῆμμα δεικτέον, ὅτι ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ. Ἐκκείσθω γὰρ ἡ τοῦ ἡμικυκλίου καταγραφή, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΑΓ τετράγωνον τὸ ΕΓ, καὶ συμπεπληρώσθω τὸ ΖΒ παραλληλόγραμμον. ἐπεὶ οὖν διὰ τὸ ἰσογώνιον εἶναι τὸ ΔΑΒ τρίγωνον τῷ ΔΑΓ τριγώνῳ ἐστὶν ὡς ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΔΑ πρὸς τὴν ΑΓ, τὸ ἄρα ὑπὸ τῶν ΒΑ, ΑΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΔ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ΕΒ πρὸς τὸ ΒΖ, καί ἐστι τὸ μὲν ΕΒ τὸ ὑπὸ τῶν ΒΑ, ΑΓ: ἴση γὰρ ἡ ΕΑ τῇ ΑΓ: τὸ δὲ ΒΖ τὸ ὑπὸ τῶν ΑΓ, ΓΒ, ὡς ἄρα ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ὑπὸ τῶν ΒΑ, ΑΓ πρὸς τὸ ὑπὸ τῶν ΑΓ, ΓΒ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΒΑ, ΑΓ ἴσον τῷ ἀπὸ τῆς ΑΔ, τὸ δὲ ὑπὸ τῶν ΑΓΒ ἴσον τῷ ἀπὸ τῆς ΔΓ: ἡ γὰρ ΔΓ κάθετος τῶν τῆς βάσεως τμημάτων τῶν ΑΓ, ΓΒ μέση ἀνάλογόν ἐστι διὰ τὸ ὀρθὴν εἶναι τὴν ὑπὸ ΑΔΒ. ὡς ἄρα ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",894]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",14]],["Association",["Rule","'VertexLabel'","'13.14'"],["Rule","'Text'","'To construct an octahedron and comprehend it in a sphere, as in the preceding case; and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.'"],["Rule","'TextWordCount'",38],["Rule","'GreekText'","'ὀκτάεδρον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ τὰ πρότερα, καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει διπλασία ἐστὶ τῆς πλευρᾶς τοῦ ὀκταέδρου.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",4]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",11],["Rule","'Theorem'",12]]]],["Rule","'Proof'","'Let the diameter AB of the given sphere be set out, and let it be bisected at C; let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, let DB be joined; let the square EFGH, having each of its sides equal to DB, be set out, let HF, EG be joined, from the point K let the straight line KL be set up at right angles to the plane of the square EFGH [XI. 12], and let it be carried through to the other side of the plane, as KM; from the straight lines KL, KM let KL, KM be respectively cut off equal to one of the straight lines EK, FK, GK, HK, and let LE, LF, LG, LH, ME, MF, MG, MH be joined. Then, since KE is equal to KH, and the angle EKH is right, therefore the square on HE is double of the square on EK. [I. 47] Again, since LK is equal to KE, and the angle LKE is right, therefore the square on EL is double of the square on EK. [id.] But the square on HE was also proved double of the square on EK; therefore the square on LE is equal to the square on EH; therefore LE is equal to EH. For the same reason LH is also equal to HE; therefore the triangle LEH is equilateral. Similarly we can prove that each of the remaining triangles of which the sides of the square EFGH are the bases, and the points L, M the vertices, is equilateral; therefore an octahedron has been constructed which is contained by eight equilateral triangles. It is next required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron. For, since the three straight lines LK, KM, KE are equal to one another, therefore the semicircle described on LM will also pass through E. And for the same reason, if, LM remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, H, and the octahedron will have been comprehended in a sphere. I say next that it is also comprehended in the given sphere. For, since LK is equal to KM, while KE is common, and they contain right angles, therefore the base LE is equal to the base EM. [I. 4] And, since the angle LEM is right, for it is in a semicircle, [III. 31] therefore the square on LM is double of the square on LE. [I. 47] Again, since AC is equal to CB, AB is double of BC. But, as AB is to BC, so is the square on AB to the square on BD; therefore the square on AB is double of the square on BD. But the square on LM was also proved double of the square on LE. And the square on DB is equal to the square on LE, for EH was made equal to DB. Therefore the square on AB is also equal to the square on LM; therefore AB is equal to LM. And AB is the diameter of the given sphere; therefore LM is equal to the diameter of the given sphere. Therefore the octahedron has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is double of the square on the side of the octahedron.'"],["Rule","'ProofWordCount'",607],["Rule","'GreekProof'","'Ἐκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ, καὶ τετμήσθω δίχα κατὰ τὸ Γ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ ἤχθω ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ ἐκκείσθω τετράγωνον τὸ ΕΖΗΘ ἴσην ἔχον ἑκάστην τῶν πλευρῶν τῇ ΔΒ, καὶ ἐπεζεύχθωσαν αἱ ΘΖ, ΕΗ, καὶ ἀνεστάτω ἀπὸ τοῦ Κ σημείου τῷ τοῦ ΕΖΗΘ τετραγώνου ἐπιπέδῳ πρὸς ὀρθὰς εὐθεῖα ἡ ΚΛ καὶ διήχθω ἐπὶ τὰ ἕτερα μέρη τοῦ ἐπιπέδου ὡς ἡ ΚΜ, καὶ ἀφῃρήσθω ἀφ᾽ ἑκατέρας τῶν ΚΛ, ΚΜ μιᾷ τῶν ΕΚ, ΖΚ, ΗΚ, ΘΚ ἴση ἑκατέρα τῶν ΚΛ, ΚΜ, καὶ ἐπεζεύχθωσαν αἱ ΛΕ, ΛΖ, ΛΗ, ΛΘ, ΜΕ, ΜΖ, ΜΗ, ΜΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΚΕ τῇ ΚΘ, καί ἐστιν ὀρθὴ ἡ ὑπὸ ΕΚΘ γωνία, τὸ ἄρα ἀπὸ τῆς ΘΕ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΚ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΛΚ τῇ ΚΕ, καί ἐστιν ὀρθὴ ἡ ὑπὸ ΛΚΕ γωνία, τὸ ἄρα ἀπὸ τῆς ΕΛ διπλάσιόν ἐστι τοῦ ἀπὸ ΕΚ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΘΕ διπλάσιον τοῦ ἀπὸ τῆς ΕΚ: τὸ ἄρα ἀπὸ τῆς ΛΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ: ἴση ἄρα ἐστὶν ἡ ΛΕ τῇ ΕΘ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΛΘ τῇ ΘΕ ἐστιν ἴση: ἰσόπλευρον ἄρα ἐστὶ τὸ ΛΕΘ τρίγωνον. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἕκαστον τῶν λοιπῶν τριγώνων, ὧν βάσεις μέν εἰσιν αἱ τοῦ ΕΖΗΘ τετραγώνου πλευραί, κορυφαὶ δὲ τὰ Λ, Μ σημεῖα, ἰσόπλευρόν ἐστιν: ὀκτάεδρον ἄρα συνέσταται ὑπὸ ὀκτὼ τριγώνων ἰσοπλεύρων περιεχόμενον. δεῖ δὴ αὐτὸ καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει διπλασίων ἐστὶ τῆς τοῦ ὀκταέδρου πλευρᾶς. ἐπεὶ γὰρ αἱ τρεῖς αἱ ΛΚ, ΚΜ, ΚΕ ἴσαι ἀλλήλαις εἰσίν, τὸ ἄρα ἐπὶ τῆς ΛΜ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Ε. καὶ διὰ τὰ αὐτά, ἐὰν μενούσης τῆς ΛΜ περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, ἥξει καὶ διὰ τῶν Ζ, Η, Θ σημείων, καὶ ἔσται σφαίρᾳ περιειλημμένον τὸ ὀκτάεδρον. λέγω δή, ὅτι καὶ τῇ δοθείσῃ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΛΚ τῇ ΚΜ, κοινὴ δὲ ἡ ΚΕ, καὶ γωνίας ὀρθὰς περιέχουσιν, βάσις ἄρα ἡ ΛΕ βάσει τῇ ΕΜ ἐστιν ἴση. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΛΕΜ γωνία: ἐν ἡμικυκλίῳ γάρ: τὸ ἄρα ἀπὸ τῆς ΛΜ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΛΕ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, διπλασία ἐστὶν ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΔ: διπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΛΜ διπλάσιον τοῦ ἀπὸ τῆς ΛΕ. καί ἐστιν ἴσον τὸ ἀπὸ τῆς ΔΒ τῷ ἀπὸ τῆς ΛΕ: ἴση γὰρ κεῖται ἡ ΕΘ τῇ ΔΒ. ἴσον ἄρα καὶ τὸ ἀπὸ τῆς ΑΒ τῷ ἀπὸ τῆς ΛΜ: ἴση ἄρα ἡ ΑΒ τῇ ΛΜ. καί ἐστιν ἡ ΑΒ ἡ τῆς δοθείσης σφαίρας διάμετρος: ἡ ΛΜ ἄρα ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας διαμέτρῳ. περιείληπται ἄρα τὸ ὀκτάεδρον τῇ δοθείσῃ σφαίρᾳ. καὶ συναποδέδεικται, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει διπλασίων ἐστὶ τῆς τοῦ ὀκταέδρου πλευρᾶς: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",505]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",15]],["Association",["Rule","'VertexLabel'","'13.15'"],["Rule","'Text'","'To construct a cube and comprehend it in a sphere, like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.'"],["Rule","'TextWordCount'",36],["Rule","'GreekText'","'κύβον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ τὴν πυραμίδα, καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει τριπλασίων ἐστὶ τῆς τοῦ κύβου πλευρᾶς.'"],["Rule","'GreekTextWordCount'",23],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",11],["Rule","'Definition'",3]]]],["Rule","'Proof'","'Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, and let DB be joined; let the square EFGH having its side equal to DB be set out, from E, F, G, H let EK, FL, GM, HN be drawn at right angles to the plane of the square EFGH, from EK, FL, GM, HN let EK, FL, GM, HN respectively be cut off equal to one of the straight lines EF, FG, GH, HE, and let KL, LM, MN, NK be joined; therefore the cube FN has been constructed which is contained by six equal squares. It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. For let KG, EG be joined. Then, since the angle KEG is right, because KE is also at right angles to the plane EG and of course to the straight line EG also, [XI. Def. 3] therefore the semicircle described on KG will also pass through the point E. Again, since GF is at right angles to each of the straight lines FL, FE, GF is also at right angles to the plane FK; hence also, if we join FK, GF will be at right angles to FK; and for this reason again the semicircle described on GK will also pass through F. Similarly it will also pass through the remaining angular points of the cube. If then, KG remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, the cube will be comprehended in a sphere. I say next that it is also comprehended in the given sphere. For, since GF is equal to FE, and the angle at F is right, therefore the square on EG is double of the square on EF. But EF is equal to EK; therefore the square on EG is double of the square on EK; hence the squares on GE, EK, that is the square on GK [I. 47], is triple of the square on EK. And, since AB is triple of BC, while, as AB is to BC, so is the square on AB to the square on BD, therefore the square on AB is triple of the square on BD. But the square on GK was also proved triple of the square on KE. And KE was made equal to DB; therefore KG is also equal to AB. And AB is the diameter of the given sphere; therefore KG is also equal to the diameter of the given sphere. Therefore the cube has been comprehended in the given sphere; and it has been demonstrated at the same time that the square on the diameter of the sphere is triple of the square on the side of the cube.'"],["Rule","'ProofWordCount'",513],["Rule","'GreekProof'","'Ἐκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ καὶ τετμήσθω κατὰ τὸ Γ ὥστε διπλῆν εἶναι τὴν ΑΓ τῆς ΓΒ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ ἐκκείσθω τετράγωνον τὸ ΕΖΗΘ ἴσην ἔχον τὴν πλευρὰν τῇ ΔΒ, καὶ ἀπὸ τῶν Ε, Ζ, Η, Θ τῷ τοῦ ΕΖΗΘ τετραγώνου ἐπιπέδῳ πρὸς ὀρθὰς ἤχθωσαν αἱ ΕΚ, ΖΛ, ΗΜ, ΘΝ, καὶ ἀφῃρήσθω ἀπὸ ἑκάστης τῶν ΕΚ, ΖΛ, ΗΜ, ΘΝ μιᾷ τῶν ΕΖ, ΖΗ, ΗΘ, ΘΕ ἴση ἑκάστη τῶν ΕΚ, ΖΛ, ΗΜ, ΘΝ, καὶ ἐπεζεύχθωσαν αἱ ΚΛ, ΛΜ, ΜΝ, ΝΚ: κύβος ἄρα συνέσταται ὁ ΖΝ ὑπὸ ἓξ τετραγώνων ἴσων περιεχόμενος. δεῖ δὴ αὐτὸν καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει τριπλασία ἐστὶ τῆς πλευρᾶς τοῦ κύβου. ἐπεζεύχθωσαν γὰρ αἱ ΚΗ, ΕΗ. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΚΕΗ γωνία διὰ τὸ καὶ τὴν ΚΕ ὀρθὴν εἶναι πρὸς τὸ ΕΗ ἐπίπεδον δηλαδὴ καὶ πρὸς τὴν ΕΗ εὐθεῖαν, τὸ ἄρα ἐπὶ τῆς ΚΗ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Ε σημείου. πάλιν, ἐπεὶ ἡ ΗΖ ὀρθή ἐστι πρὸς ἑκατέραν τῶν ΖΛ, ΖΕ, καὶ πρὸς τὸ ΖΚ ἄρα ἐπίπεδον ὀρθή ἐστιν ἡ ΗΖ: ὥστε καὶ ἐὰν ἐπιζεύξωμεν τὴν ΖΚ, ἡ ΗΖ ὀρθὴ ἔσται καὶ πρὸς τὴν ΖΚ: καὶ διὰ τοῦτο πάλιν τὸ ἐπὶ τῆς ΗΚ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Ζ. ὁμοίως καὶ διὰ τῶν λοιπῶν τοῦ κύβου σημείων ἥξει. ἐὰν δὴ μενούσης τῆς ΚΗ περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, ἔσται σφαίρᾳ περιειλημμένος ὁ κύβος. λέγω δή, ὅτι καὶ τῇ δοθείσῃ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΗΖ τῇ ΖΕ, καί ἐστιν ὀρθὴ ἡ πρὸς τῷ Ζ γωνία, τὸ ἄρα ἀπὸ τῆς ΕΗ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΖ. ἴση δὲ ἡ ΕΖ τῇ ΕΚ: τὸ ἄρα ἀπὸ τῆς ΕΗ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΚ: ὥστε τὰ ἀπὸ τῶν ΗΕ, ΕΚ, τουτέστι τὸ ἀπὸ τῆς ΗΚ, τριπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΚ. καὶ ἐπεὶ τριπλασίων ἐστὶν ἡ ΑΒ τῆς ΒΓ, ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΔ, τριπλάσιον ἄρα τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΗΚ τοῦ ἀπὸ τῆς ΚΕ τριπλάσιον. καὶ κεῖται ἴση ἡ ΚΕ τῇ ΔΒ: ἴση ἄρα καὶ ἡ ΚΗ τῇ ΑΒ. καί ἐστιν ἡ ΑΒ τῆς δοθείσης σφαίρας διάμετρος: καὶ ἡ ΚΗ ἄρα ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας διαμέτρῳ. τῇ δοθείσῃ ἄρα σφαίρᾳ περιείληπται ὁ κύβος: καὶ συναποδέδεικται, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει τριπλασίων ἐστὶ τῆς τοῦ κύβου πλευρᾶς: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",428]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",16]],["Association",["Rule","'VertexLabel'","'13.16'"],["Rule","'Text'","'To construct an icosahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the side of the icosahedron is the irrational straight line called minor.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'εἰκοσάεδρον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ τὰ προειρημένα σχήματα, καὶ δεῖξαι, ὅτι ἡ τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",33]],["List",["Rule","'Book'",3],["Rule","'Theorem'",31]],["List",["Rule","'Book'",4],["Rule","'Theorem'",15]],["List",["Rule","'Book'",5],["Rule","'Definition'",9]],["List",["Rule","'Book'",6],["Rule","'Theorem'",8]],["List",["Rule","'Book'",11],["Rule","'Theorem'",6]],["List",["Rule","'Book'",13],["Rule","'Theorem'",3]],["List",["Rule","'Book'",13],["Rule","'Theorem'",9]],["List",["Rule","'Book'",13],["Rule","'Theorem'",10]],["List",["Rule","'Book'",13],["Rule","'Theorem'",11]]]],["Rule","'Proof'","'Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is quadruple of CB, let the semicircle ADB be described on AB, let the straight line CD be drawn from C at right angles to AB, and let DB be joined; let the circle EFGHK be set out and let its radius be equal to DB, let the equilateral and equiangular pentagon EFGHK be inscribed in the circle EFGHK, let the circumferences EF, FG, GH, HK, KE be bisected at the points L, M, N, O, P, and let LM, MN, NO, OP, PL, EP be joined. Therefore the pentagon LMNOP is also equilateral, and the straight line EP belongs to a decagon. Now from the points E, F, G, H, K let the straight lines EQ, FR, GS, HT, KU be set up at right angles to the plane of the circle, and let them be equal to the radius of the circle EFGHK, let QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, PQ be joined. Now, since each of the straight lines EQ, KU is at right angles to the same plane, therefore EQ is parallel to KU. [XI. 6] But it is also equal to it; and the straight lines joining those extremities of equal and parallel straight lines which are in the same direction are equal and parallel. [I. 33] Therefore QU is equal and parallel to EK. But EK belongs to an equilateral pentagon; therefore QU also belongs to the equilateral pentagon inscribed in the circle EFGHK. For the same reason each of the straight lines QR, RS, ST, TU also belongs to the equilateral pentagon inscribed in the circle EFGHK; therefore the pentagon QRSTU is equilateral. And, since QE belongs to a hexagon, and EP to a decagon, and the angle QEP is right, therefore QP belongs to a pentagon; for the square on the side of the pentagon is equal to the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. [XIII. 10] For the same reason PU is also a side of a pentagon. But QU also belongs to a pentagon; therefore the triangle QPU is equilateral. For the same reason each of the triangles QLR, RMS, SNT, TOU is also equilateral. And, since each of the straight lines QL, QP was proved to belong to a pentagon, and LP also belongs to a pentagon, therefore the triangle QLP is equilateral. For the same reason each of the triangles LRM, MSN, NTO, OUP is also equilateral. Let the centre of the circle EFGHK the point V, be taken; from V let VZ be set up at right angles to the plane of the circle, let it be produced in the other direction, as VX, let there be cut off VW, the side of a hexagon, and each of the straight lines VX, WZ, being sides of a decagon, and let QZ, QW, UZ, EV, LV, LX, XM be joined. Now, since each of the straight lines VW, QE is at right angles to the plane of the circle, therefore VW is parallel to QE. [XI. 6] But they are also equal; therefore EV, QW are also equal and parallel. [I. 33] But EV belongs to a hexagon; therefore QW also belongs to a hexagon. And, since QW belongs to a hexagon, and WZ to a decagon, and the angle QWZ is right, therefore QZ belongs to a pentagon. [XIII. 10] For the same reason UZ also belongs to a pentagon, inasmuch as, if we join VK, WU, they will be equal and opposite, and VK, being a radius, belongs to a hexagon; [IV. 15] therefore WU also belongs to a hexagon. But WZ belongs to a decagon, and the angle UWZ is right; therefore UZ belongs to a pentagon. [XIII. 10] But QU also belongs to a pentagon; therefore the triangle QUZ is equilateral. For the same reason each of the remaining triangles of which the straight lines QR, RS, ST, TU are the bases, and the point Z the vertex, is also equilateral. Again, since VL belongs to a hexagon, and VX to a decagon, and the angle LVX is right, therefore LX belongs to a pentagon. [XIII. 10] For the same reason, if we join MV, which belongs to a hexagon, MX is also inferred to belong to a pentagon. But LM also belongs to a pentagon; therefore the triangle LMX is equilateral. Similarly it can be proved that each of the remaining triangles of which MN, NO, OP, PL are the bases, and the point X the vertex, is also equilateral. Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles. It is next required to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor. For, since VW belongs to a hexagon, and WZ to a decagon, therefore VZ has been cut in extreme and mean ratio at W, and VW is its greater segment; [XIII. 9] therefore, as ZV is to VW, so is VW to WZ. But VW is equal to VE, and WZ to VX; therefore, as ZV is to VE, so is EV to VX. And the angles ZVE, EVX are right; therefore, if we join the straight line EZ, the angle XEZ will be right because of the similarity of the triangles XEZ, VEZ. For the same reason, since, as ZV is to VW, so is VW to WZ, and ZV is equal to XW, and VW to WQ, therefore, as XW is to WQ, so is QW to WZ. And for this reason again, if we join QX, the angle at Q will be right; [VI. 8] therefore the semicircle described on XZ will also pass through Q. [III. 31] And if, XZ remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through Q and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere. I say next that it is also comprehended in the given sphere. For let VW be bisected at A'. Then, since the straight line VZ has been cut in extreme and mean ratio at W, and ZW is its lesser segment, therefore the square on ZW added to the half of the greater segment, that is WA', is five times the square on the half of the greater segment; [XIII. 3] therefore the square on ZA' is five times the square on. And ZX is double of ZA', and VW double of; therefore the square on ZX is five times the square on WV. And, since AC is quadruple of CB, therefore AB is five times BC. But, as AB is to BC, so is the square on AB to the square on BD; [VI. 8, V. Def. 9] therefore the square on AB is five times the square on BD. But the square on ZX was also proved to be five times the square on VW. And DB is equal to VW, for each of them is equal to the radius of the circle EFGHK; therefore AB is also equal to XZ. And AB is the diameter of the given sphere; therefore XZ is also equal to the diameter of the given sphere. Therefore the icosahedron has been comprehended in the given sphere I say next that the side of the icosahedron is the irrational straight line called minor. For, since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle EFGHK, therefore the radius of the circle EFGHK is also rational; hence its diameter is also rational. But, if an equilateral pentagon be inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor. [XIII. 11] And the side of the pentagon EFGHK is the side of the icosahedron. Therefore the side of the icosahedron is the irrational straight line called minor.'"],["Rule","'ProofWordCount'",1376],["Rule","'GreekProof'","'Ἐκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ καὶ τετμήσθω κατὰ τὸ Γ ὥστε τετραπλῆν εἶναι τὴν ΑΓ τῆς ΓΒ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ ἤχθω ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ ἐκκείσθω κύκλος ὁ ΕΖΗΘΚ, οὗ ἡ ἐκ τοῦ κέντρου ἴση ἔστω τῇ ΔΒ, καὶ ἐγγεγράφθω εἰς τὸν ΕΖΗΘΚ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον τὸ ΕΖΗΘΚ, καὶ τετμήσθωσαν αἱ ΕΖ, ΖΗ, ΗΘ, ΘΚ, ΚΕ περιφέρειαι δίχα κατὰ τὰ Λ, Μ, Ν, Ξ, Ο σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΛΜ, ΜΝ, ΝΞ, ΞΟ, ΟΛ, ΕΟ. ἰσόπλευρον ἄρα ἐστὶ καὶ τὸ ΛΜΝΞΟ πεντάγωνον, καὶ δεκαγώνου ἡ ΕΟ εὐθεῖα. καὶ ἀνεστάτωσαν ἀπὸ τῶν Ε, Ζ, Η, Θ, Κ σημείων τῷ τοῦ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς γωνίας εὐθεῖαι αἱ ΕΠ, ΖΡ, ΗΣ, ΘΤ, ΚΥ ἴσαι οὖσαι τῇ ἐκ τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου, καὶ ἐπεζεύχθωσαν αἱ ΠΡ, ΡΣ, ΣΤ, ΤΥ, ΥΠ, ΠΛ, ΛΡ, ΡΜ, ΜΣ, ΣΝ, ΝΤ, ΤΞ, ΞΥ, ΥΟ, ΟΠ. καὶ ἐπεὶ ἑκατέρα τῶν ΕΠ, ΚΥ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, παράλληλος ἄρα ἐστὶν ἡ ΕΠ τῇ ΚΥ. ἔστι δὲ αὐτῇ καὶ ἴση: αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπιζευγνύουσαι ἐπὶ τὰ αὐτὰ μέρη εὐθεῖαι ἴσαι τε καὶ παράλληλοί εἰσιν. ἡ ΠΥ ἄρα τῇ ΕΚ ἴση τε καὶ παράλληλός ἐστιν. πενταγώνου δὲ ἰσοπλεύρου ἡ ΕΚ: πενταγώνου ἄρα ἰσοπλεύρου καὶ ἡ ΠΥ τοῦ εἰς τὸν ΕΖΗΘΚ κύκλον ἐγγραφομένου. διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ΠΡ, ΡΣ, ΣΤ, ΤΥ πενταγώνου ἐστὶν ἰσοπλεύρου τοῦ εἰς τὸν ΕΖΗΘΚ κύκλον ἐγγραφομένου: ἰσόπλευρον ἄρα τὸ ΠΡΣΤΥ πεντάγωνον. καὶ ἐπεὶ ἑξαγώνου μέν ἐστιν ἡ ΠΕ, δεκαγώνου δὲ ἡ ΕΟ, καί ἐστιν ὀρθὴ ἡ ὑπὸ ΠΕΟ, πενταγώνου ἄρα ἐστὶν ἡ ΠΟ: ἡ γὰρ τοῦ πενταγώνου πλευρὰ δύναται τήν τε τοῦ ἑξαγώνου καὶ τὴν τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΟΥ πενταγώνου ἐστὶ πλευρά. ἔστι δὲ καὶ ἡ ΠΥ πενταγώνου: ἰσόπλευρον ἄρα ἐστὶ τὸ ΠΟΥ τρίγωνον. διὰ τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν ΠΛΡ, ΡΜΣ, ΣΝΤ, ΤΞΥ ἰσόπλευρόν ἐστιν. καὶ ἐπεὶ πενταγώνου ἐδείχθη ἑκατέρα τῶν ΠΛ, ΠΟ, ἔστι δὲ καὶ ἡ ΛΟ πενταγώνου, ἰσόπλευρον ἄρα ἐστὶ τὸ ΠΛΟ τρίγωνον. διὰ τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν ΛΡΜ, ΜΣΝ, ΝΤΞ, ΞΥΟ τριγώνων ἰσόπλευρόν ἐστιν. εἰλήφθω τὸ κέντρον τοῦ ΕΖΗ ΘΚ κύκλου τὸ Φ σημεῖον: καὶ ἀπὸ τοῦ Φ τῷ τοῦ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἀνεστάτω ἡ ΦΩ, καὶ ἐκβεβλήσθω ἐπὶ τὰ ἕτερα μέρη ὡς ἡ ΦΨ, καὶ ἀφῃρήσθω ἑξαγώνου μὲν ἡ ΦΧ, δεκαγώνου δὲ ἑκατέρα τῶν ΦΨ, ΧΩ, καὶ ἐπεζεύχθωσαν αἱ ΠΩ, ΠΧ, ΥΩ, ΕΦ, ΛΦ, ΛΨ, ΨΜ. καὶ ἐπεὶ ἑκατέρα τῶν ΦΧ, ΠΕ τῷ τοῦ κύκλου ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, παράλληλος ἄρα ἐστὶν ἡ ΦΧ τῇ ΠΕ. εἰσὶ δὲ καὶ ἴσαι: καὶ αἱ ΕΦ, ΠΧ ἄρα ἴσαι τε καὶ παράλληλοί εἰσιν. ἑξαγώνου δὲ ἡ ΕΦ: ἑξαγώνου ἄρα καὶ ἡ ΠΧ. καὶ ἐπεὶ ἑξαγώνου μέν ἐστιν ἡ ΠΧ, δεκαγώνου δὲ ἡ ΧΩ, καὶ ὀρθή ἐστιν ἡ ὑπὸ ΠΧΩ γωνία, πενταγώνου ἄρα ἐστὶν ἡ ΠΩ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΥΩ πενταγώνου ἐστίν, ἐπειδήπερ, ἐὰν ἐπιζεύξωμεν τὰς ΦΚ, ΧΥ, ἴσαι καὶ ἀπεναντίον ἔσονται, καί ἐστιν ἡ ΦΚ ἐκ τοῦ κέντρου οὖσα ἑξαγώνου: ἑξαγώνου ἄρα καὶ ἡ ΧΥ. δεκαγώνου δὲ ἡ ΧΩ, καὶ ὀρθὴ ἡ ὑπὸ ΥΧΩ: πενταγώνου ἄρα ἡ ΥΩ. ἔστι δὲ καὶ ἡ ΠΥ πενταγώνου: ἰσόπλευρον ἄρα ἐστὶ τὸ ΠΥΩ τρίγωνον. διὰ τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν λοιπῶν τριγώνων, ὧν βάσεις μέν εἰσιν αἱ ΠΡ, ΡΣ, ΣΤ, ΤΥ εὐθεῖαι, κορυφὴ δὲ τὸ Ω σημεῖον, ἰσόπλευρόν ἐστιν. πάλιν, ἐπεὶ ἑξαγώνου μὲν ἡ ΦΛ, δεκαγώνου δὲ ἡ ΦΨ, καὶ ὀρθή ἐστιν ἡ ὑπὸ ΛΦΨ γωνία, πενταγώνου ἄρα ἐστὶν ἡ ΛΨ. διὰ τὰ αὐτὰ δὴ ἐὰν ἐπιζεύξωμεν τὴν ΜΦ οὖσαν ἑξαγώνου, συνάγεται καὶ ἡ ΜΨ πενταγώνου. ἔστι δὲ καὶ ἡ ΛΜ πενταγώνου: ἰσόπλευρον ἄρα ἐστὶ τὸ ΛΜΨ τρίγωνον. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἕκαστον τῶν λοιπῶν τριγώνων, ὧν βάσεις μέν εἰσιν αἱ ΜΝ, ΝΞ, ΞΟ, ΟΛ, κορυφὴ δὲ τὸ Ψ σημεῖον, ἰσόπλευρόν ἐστιν. συνέσταται ἄρα εἰκοσάεδρον ὑπὸ εἴκοσι τριγώνων ἰσοπλεύρων περιεχόμενον. δεῖ δὴ αὐτὸ καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων. ἐπεὶ γὰρ ἑξαγώνου ἐστὶν ἡ ΦΧ, δεκαγώνου δὲ ἡ ΧΩ, ἡ ΦΩ ἄρα ἄκρον καὶ μέσον λόγον τέμηται κατὰ τὸ Χ, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΦΧ: ἔστιν ἄρα ὡς ἡ ΩΦ πρὸς τὴν ΦΧ, οὕτως ἡ ΦΧ πρὸς τὴν ΧΩ. ἴση δὲ ἡ μὲν ΦΧ τῇ ΦΕ, ἡ δὲ ΧΩ τῇ ΦΨ: ἔστιν ἄρα ὡς ἡ ΩΦ πρὸς τὴν ΦΕ, οὕτως ἡ ΕΦ πρὸς τὴν ΦΨ. καί εἰσιν ὀρθαὶ αἱ ὑπὸ ΩΦΕ, ΕΦΨ γωνίαι: ἐὰν ἄρα ἐπιζεύξωμεν τὴν ΕΩ εὐθεῖαν, ὀρθὴ ἔσται ἡ ὑπὸ ΨΕΩ γωνία διὰ τὴν ὁμοιότητα τῶν ΨΕΩ, ΦΕΩ τριγώνων. διὰ τὰ αὐτὰ δὴ ἐπεί ἐστιν ὡς ἡ ΩΦ πρὸς τὴν ΦΧ, οὕτως ἡ ΦΧ πρὸς τὴν ΧΩ, ἴση δὲ ἡ μὲν ΩΦ τῇ ΨΧ, ἡ δὲ ΦΧ τῇ ΧΠ, ἔστιν ἄρα ὡς ἡ ΨΧ πρὸς τὴν ΧΠ, οὕτως ἡ ΠΧ πρὸς τὴν ΧΩ. καὶ διὰ τοῦτο πάλιν ἐὰν ἐπιζεύξωμεν τὴν ΠΨ, ὀρθὴ ἔσται ἡ πρὸς τῷ Π γωνία: τὸ ἄρα ἐπὶ τῆς ΨΩ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Π. καὶ ἐὰν μενούσης τῆς ΨΩ περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, ἥξει καὶ διὰ τοῦ Π καὶ τῶν λοιπῶν σημείων τοῦ εἰκοσαέδρου, καὶ ἔσται σφαίρᾳ περιειλημμένον τὸ εἰκοσάεδρον. λέγω δή, ὅτι καὶ τῇ δοθείσῃ. τετμήσθω γὰρ ἡ ΦΧ δίχα κατὰ τὸ Α #. καὶ ἐπεὶ εὐθεῖα γραμμὴ ἡ ΦΩ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Χ, καὶ τὸ ἔλασσον αὐτῆς τμῆμά ἐστιν ἡ ΩΧ, ἡ ἄρα ΩΧ προσλαβοῦσα τὴν ἡμίσειαν τοῦ μείζονος τμήματος τὴν ΧΑ # πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τοῦ μείζονος τμήματος: πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΩΑ # τοῦ ἀπὸ τῆς Α # Χ. καί ἐστι τῆς μὲν ΩΑ # διπλῆ ἡ ΩΨ, τῆς δὲ Α # Χ διπλῆ ἡ ΦΧ: πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΩΨ τοῦ ἀπὸ τῆς ΧΦ. καὶ ἐπεὶ τετραπλῆ ἐστιν ἡ ΑΓ τῆς ΓΒ, πενταπλῆ ἄρα ἐστὶν ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΔ: πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΩΨ πενταπλάσιον τοῦ ἀπὸ τῆς ΦΧ. καί ἐστιν ἴση ἡ ΔΒ τῇ ΦΧ: ἑκατέρα γὰρ αὐτῶν ἴση ἐστὶ τῇ ἐκ τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου: ἴση ἄρα καὶ ἡ ΑΒ τῇ ΨΩ. καί ἐστιν ἡ ΑΒ ἡ τῆς δοθείσης σφαίρας διάμετρος: καὶ ἡ ΨΩ ἄρα ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας διαμέτρῳ. τῇ ἄρα δοθείσῃ σφαίρᾳ περιείληπται τὸ εἰκοσάεδρον. λέγω δή, ὅτι ἡ τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων. ἐπεὶ γὰρ ῥητή ἐστιν ἡ τῆς σφαίρας διάμετρος, καί ἐστι δυνάμει πενταπλασίων τῆς ἐκ τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου, ῥητὴ ἄρα ἐστὶ καὶ ἡ ἐκ τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου: ὥστε καὶ ἡ διάμετρος αὐτοῦ ῥητή ἐστιν. ἐὰν δὲ εἰς κύκλον ῥητὴν ἔχοντα τὴν διάμετρον πεντάγωνον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων. ἡ δὲ τοῦ ΕΖΗΘΚ πενταγώνου πλευρὰ ἡ τοῦ εἰκοσαέδρου ἐστίν. ἡ ἄρα τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει πενταπλασίων ἐστὶ τῆς ἐκ τοῦ κέντρου τοῦ κύκλου, ἀφ᾽ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται, καὶ ὅτι ἡ τῆς σφαίρας διάμετρος σύγκειται ἔκ τε τῆς τοῦ ἑξαγώνου καὶ δύο τῶν τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",1216]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",17]],["Association",["Rule","'VertexLabel'","'13.17'"],["Rule","'Text'","'To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome.'"],["Rule","'TextWordCount'",30],["Rule","'GreekText'","'δωδεκάεδρον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ τὰ προειρημένα σχήματα, καὶ δεῖξαι, ὅτι ἡ τοῦ δωδεκαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἀποτομή.'"],["Rule","'GreekTextWordCount'",22],["Rule","'References'",["List",["List",["Rule","'Book'",1],["Rule","'Theorem'",8]],["List",["Rule","'Book'",1],["Rule","'Theorem'",47]],["List",["Rule","'Book'",5],["Rule","'Theorem'",15]],["List",["Rule","'Book'",6],["Rule","'Theorem'",32]],["List",["Rule","'Book'",11],["Rule","'Theorem'",1]],["List",["Rule","'Book'",11],["Rule","'Theorem'",6]],["List",["Rule","'Book'",11],["Rule","'Theorem'",38]],["List",["Rule","'Book'",13],["Rule","'Theorem'",4]],["List",["Rule","'Book'",13],["Rule","'Theorem'",5]],["List",["Rule","'Book'",13],["Rule","'Theorem'",6]],["List",["Rule","'Book'",13],["Rule","'Theorem'",7]],["List",["Rule","'Book'",13],["Rule","'Theorem'",15]]]],["Rule","'Proof'","'Let ABCD, CBEF, two planes of the aforesaid cube at right angles to one another, be set out, let the sides AB, BC, CD, DA, EF, EB, FC be bisected at G, H, K, L, M, N, O respectively, let GK, HL, MH, NO be joined, let the straight lines NP, PO, HQ be cut in extreme and mean ratio at the points R, S, T respectively, and let RP, PS, TQ be their greater segments; from the points R, S, T let RU, SV, TW be set up at right angles to the planes of the cube towards the outside of the cube, let them be made equal to RP, PS, TQ, and let UB, BW, WC, CV, VU be joined. I say that the pentagon UBWCV is equilateral, and in one plane, and is further equiangular. For let RB, SB, VB be joined. Then, since the straight line NP has been cut in extreme and mean ratio at R, and RP is the greater segment, therefore the squares on PN, NR are triple of the square on RP. [XIII. 4] But PN is equal to NB, and PR to RU; therefore the squares on BN, NR are triple of the square on RU. But the square on BR is equal to the squares on BN, NR; [I. 47] therefore the square on BR is triple of the square on RU; hence the squares on BR, RU are quadruple of the square on RU. But the square on BU is equal to the squares on BR, RU; therefore the square on BU is quadruple of the square on RU; therefore BU is double of RU. But VU is also double of UR, inasmuch as SR is also double of PR, that is, of RU; therefore BU is equal to UV. Similarly it can be proved that each of the straight lines BW, WC, CV is also equal to each of the straight lines BU, UV. Therefore the pentagon BUVCW is equilateral. I say next that it is also in one plane. For let PX be drawn from P parallel to each of the straight lines RU, SV and towards the outside of the cube, and let XH, HW be joined; I say that XHW is a straight line. For, since HQ has been cut in extreme and mean ratio at T, and QT is its greater segment, therefore, as HQ is to QT, so is QT to TH. But HQ is equal to HP, and QT to each of the straight lines TW, PX; therefore, as HP is to PX, so is WT to TH. And HP is parallel to TW, for each of them is at right angles to the plane BD; [XI. 6] and TH is parallel to PX, for each of them is at right angles to the plane BF. [id.] But if two triangles, as XPH, HTW, which have two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining straight lines will be in a straight line; [VI. 32] therefore XH is in a straight line with HW. But every straight line is in one plane; [XI. 1] therefore the pentagon UBWCV is in one plane. I say next that it is also equiangular. For, since the straight line NP has been cut in extreme and mean ratio at R, and PR is the greater segment, while PR is equal to PS, therefore NS has also been cut in extreme and mean ratio at P, and NP is the greater segment; [XIII. 5] therefore the squares on NS, SP are triple of the square on NP. [XIII. 4] But NP is equal to NB, and PS to SV; therefore the squares on NS, SV are triple of the square on NB; hence the squares on VS, SN, NB are quadruple of the square on NB. But the square on SB is equal to the squares on SN, NB; therefore the squares on BS, SV, that is, the square on BV —for the angle VSB is right—is quadruple of the square on NB; therefore VB is double of BN. But BC is also double of BN; therefore BV is equal to BC. And, since the two sides BU, UV are equal to the two sides BW, WC, and the base BV is equal to the base BC, therefore the angle BUV is equal to the angle BWC. [I. 8] Similarly we can prove that the angle UVC is also equal to the angle BWC; therefore the three angles BWC, BUV, UVC are equal to one another. But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular, [XIII. 7] therefore the pentagon BUVCW is equiangular. And it was also proved equilateral; therefore the pentagon BUVCW is equilateral and equiangular, and it is on one side BC of the cube. Therefore, if we make the same construction in the case of each of the twelve sides of the cube, a solid figure will have been constructed which is contained by twelve equilateral and equiangular pentagons, and which is called a dodecahedron. It is then required to comprehend it in the given sphere, and to prove that the side of the dodecahedron is the irrational straight line called apotome. For let XP be produced, and let the produced straight line be XZ; therefore PZ meets the diameter of the cube, and they bisect one another, for this has been proved in the last theorem but one of the eleventh book. [XI. 38] Let them cut at Z; therefore Z is the centre of the sphere which comprehends the cube, and ZP is half of the side of the cube. Let UZ be joined. Now, since the straight line NS has been cut in extreme and mean ratio at P, and NP is its greater segment, therefore the squares on NS, SP are triple of the square on NP. [XIII. 4] But NS is equal to XZ, inasmuch as NP is also equal to PZ, and XP to PS. But further PS is also equal to XU, since it is also equal to RP; therefore the squares on ZX, XU are triple of the square on NP. But the square on UZ is equal to the squares on ZX, XU; therefore the square on UZ is triple of the square on NP. But the square on the radius of the sphere which comprehends the cube is also triple of the square on the half of the side of the cube, for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. [XIII. 15] But, if whole is so related to whole, so is half to half also; and NP is half of the side of the cube; therefore UZ is equal to the radius of the sphere which comprehends the cube. And Z is the centre of the sphere which comprehends the cube; therefore the point U is on the surface of the sphere. Similarly we can prove that each of the remaining angles of the dodecahedron is also on the surface of the sphere; therefore the dodecahedron has been comprehended in the given sphere. I say next that the side of the dodecahedron is the irrational straight line called apotome. For since, when NP has been cut in extreme and mean ratio, RP is the greater segment, and, when PO has been cut in extreme and mean ratio, PS is the greater segment, therefore, when the whole NO is cut in extreme and mean ratio, RS is the greater segment. [Thus, since, as NP is to PR, so is PR to RN, the same is true of the doubles also, for parts have the same ratio as their equimultiples; [V. 15] therefore as NO is to RS, so is RS to the sum of NR, SO. But NO is greater than RS; therefore RS is also greater than the sum of NR, SO; therefore NO has been cut in extreme and mean ratio, and RS is its greater segment.] But RS is equal to UV; therefore, when NO is cut in extreme and mean ratio, UV is the greater segment. And, since the diameter of the sphere is rational, and the square on it is triple of the square on the side of the cube, therefore NO, being a side of the cube, is rational. [But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.] Therefore UV, being a side of the dodecahedron, is an irrational apotome. [XIII. 6]'"],["Rule","'ProofWordCount'",1470],["Rule","'GreekProof'","'Ἐκκείσθωσαν τοῦ προειρημένου κύβου δύο ἐπίπεδα πρὸς ὀρθὰς ἀλλήλοις τὰ ΑΒΓΔ, ΓΒΕΖ, καὶ τετμήσθω ἑκάστη τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ, ΕΖ, ΕΒ, ΖΓ πλευρῶν δίχα κατὰ τὰ Η, Θ, Κ, Λ, Μ, Ν, Ξ, καὶ ἐπεζεύχθωσαν αἱ ΗΚ, ΘΛ, ΜΘ, ΝΞ, καὶ τετμήσθω ἑκάστη τῶν ΝΟ, ΟΞ, ΘΠ ἄκρον καὶ μέσον λόγον κατὰ τὰ Ρ, Σ, Τ σημεῖα, καὶ ἔστω αὐτῶν μείζονα τμήματα τὰ ΡΟ, ΟΣ, ΤΠ, καὶ ἀνεστάτωσαν ἀπὸ τῶν Ρ, Σ, Τ σημείων τοῖς τοῦ κύβου ἐπιπέδοις πρὸς ὀρθὰς ἐπὶ τὰ ἐκτὸς μέρη τοῦ κύβου αἱ ΡΥ, ΣΦ, ΤΧ, καὶ κείσθωσαν ἴσαι ταῖς ΡΟ, ΟΣ, ΤΠ, καὶ ἐπεζεύχθωσαν αἱ ΥΒ, ΒΧ, ΧΓ, ΓΦ, ΦΥ. λέγω, ὅτι τὸ ΥΒΧΓΦ πεντάγωνον ἰσόπλευρόν τε καὶ ἐν ἑνὶ ἐπιπέδῳ καὶ ἔτι ἰσογώνιόν ἐστιν. ἐπεζεύχθωσαν γὰρ αἱ ΡΒ, ΣΒ, ΦΒ. καὶ ἐπεὶ εὐθεῖα ἡ ΝΟ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ρ, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΡΟ, τὰ ἄρα ἀπὸ τῶν ΟΝ, ΝΡ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΡΟ. ἴση δὲ ἡ μὲν ΟΝ τῇ ΝΒ, ἡ δὲ ΟΡ τῇ ΡΥ: τὰ ἄρα ἀπὸ τῶν ΒΝ, ΝΡ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΡΥ. τοῖς δὲ ἀπὸ τῶν ΒΝ, ΝΡ τὸ ἀπὸ τῆς ΒΡ ἐστιν ἴσον: τὸ ἄρα ἀπὸ τῆς ΒΡ τριπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΡΥ: ὥστε τὰ ἀπὸ τῶν ΒΡ, ΡΥ τετραπλάσιά ἐστι τοῦ ἀπὸ τῆς ΡΥ. τοῖς δὲ ἀπὸ τῶν ΒΡ, ΡΥ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΥ: τὸ ἄρα ἀπὸ τῆς ΒΥ τετραπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΥΡ: διπλῆ ἄρα ἐστὶν ἡ ΒΥ τῆς ΡΥ. ἔστι δὲ καὶ ἡ ΦΥ τῆς ΥΡ διπλῆ, ἐπειδήπερ καὶ ἡ ΣΡ τῆς ΟΡ, τουτέστι τῆς ΡΥ, ἐστι διπλῆ: ἴση ἄρα ἡ ΒΥ τῇ ΥΦ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΒΧ, ΧΓ, ΓΦ ἑκατέρᾳ τῶν ΒΥ, ΥΦ ἐστιν ἴση. ἰσόπλευρον ἄρα ἐστὶ τὸ ΒΥΦΓΧ πεντάγωνον. λέγω δή, ὅτι καὶ ἐν ἑνί ἐστιν ἐπιπέδῳ. ἤχθω γὰρ ἀπὸ τοῦ Ο ἑκατέρᾳ τῶν ΡΥ, ΣΦ παράλληλος ἐπὶ τὰ ἐκτὸς τοῦ κύβου μέρη ἡ ΟΨ, καὶ ἐπεζεύχθωσαν αἱ ΨΘ, ΘΧ: λέγω, ὅτι ἡ ΨΘΧ εὐθεῖά ἐστιν. ἐπεὶ γὰρ ἡ ΘΠ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Τ, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΠΤ, ἔστιν ἄρα ὡς ἡ ΘΠ πρὸς τὴν ΠΤ, οὕτως ἡ ΠΤ πρὸς τὴν ΤΘ. ἴση δὲ ἡ μὲν ΘΠ τῇ ΘΟ, ἡ δὲ ΠΤ ἑκατέρᾳ τῶν ΤΧ, ΟΨ: ἔστιν ἄρα ὡς ἡ ΘΟ πρὸς τὴν ΟΨ, οὕτως ἡ ΧΤ πρὸς τὴν ΤΘ. καί ἐστι παράλληλος ἡ μὲν ΘΟ τῇ ΤΧ: ἑκατέρα γὰρ αὐτῶν τῷ ΒΔ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν: ἡ δὲ ΤΘ τῇ ΟΨ: ἑκατέρα γὰρ αὐτῶν τῷ ΒΖ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. ἐὰν δὲ δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν, ὡς τὰ ΨΟΘ, ΘΤΧ, τὰς δύο πλευρὰς ταῖς δυσὶν ἀνάλογον ἔχοντα, ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ εὐθεῖαι ἐπ᾽ εὐθείας ἔσονται: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΨΘ τῇ ΘΧ. πᾶσα δὲ εὐθεῖα ἐν ἑνί ἐστιν ἐπιπέδῳ: ἐν ἑνὶ ἄρα ἐπιπέδῳ ἐστὶ τὸ ΥΒΧΓΦ πεντάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιόν ἐστιν. ἐπεὶ γὰρ εὐθεῖα γραμμὴ ἡ ΝΟ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ρ, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΟΡ ἔστιν ἄρα ὡς συναμφότερος ἡ ΝΟ, ΟΡ πρὸς τὴν ΟΝ, οὕτως ἡ ΝΟ πρὸς τὴν ΟΡ, ἴση δὲ ἡ ΟΡ τῇ ΟΣ ἔστιν ἄρα ὡς ἡ ΣΝ πρὸς τὴν ΝΟ, οὕτως ἡ ΝΟ πρὸς τὴν ΟΣ, ἡ ΝΣ ἄρα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ο, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΝΟ: τὰ ἄρα ἀπὸ τῶν ΝΣ, ΣΟ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΟ. ἴση δὲ ἡ μὲν ΝΟ τῇ ΝΒ, ἡ δὲ ΟΣ τῇ ΣΦ: τὰ ἄρα ἀπὸ τῶν ΝΣ, ΣΦ τετράγωνα τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΒ: ὥστε τὰ ἀπὸ τῶν ΦΣ, ΣΝ, ΝΒ τετραπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΒ. τοῖς δὲ ἀπὸ τῶν ΣΝ, ΝΒ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΣΒ: τὰ ἄρα ἀπὸ τῶν ΒΣ, ΣΦ, τουτέστι τὸ ἀπὸ τῆς ΒΦ ὀρθὴ γὰρ ἡ ὑπὸ ΦΣΒ γωνία 1, τετραπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΝΒ: διπλῆ ἄρα ἐστὶν ἡ ΦΒ τῆς ΒΝ. ἔστι δὲ καὶ ἡ ΒΓ τῆς ΒΝ διπλῆ: ἴση ἄρα ἐστὶν ἡ ΒΦ τῇ ΒΓ. καὶ ἐπεὶ δύο αἱ ΒΥ, ΥΦ δυσὶ ταῖς ΒΧ, ΧΓ ἴσαι εἰσίν, καὶ βάσις ἡ ΒΦ βάσει τῇ ΒΓ ἴση, γωνία ἄρα ἡ ὑπὸ ΒΥΦ γωνίᾳ τῇ ὑπὸ ΒΧΓ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ὑπὸ ΥΦΓ γωνία ἴση ἐστὶ τῇ ὑπὸ ΒΧΓ: αἱ ἄρα ὑπὸ ΒΧΓ, ΒΥΦ, ΥΦΓ τρεῖς γωνίαι ἴσαι ἀλλήλαις εἰσίν. ἐὰν δὲ πενταγώνου ἰσοπλεύρου αἱ τρεῖς γωνίαι ἴσαι ἀλλήλαις ὦσιν, ἰσογώνιον ἔσται τὸ πεντάγωνον: ἰσογώνιον ἄρα ἐστὶ τὸ ΒΥΦΓΧ πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον: τὸ ἄρα ΒΥΦΓΧ πεντάγωνον ἰσόπλευρόν ἐστι καὶ ἰσογώνιον, καί ἐστιν ἐπὶ μιᾶς τοῦ κύβου πλευρᾶς τῆς ΒΓ. ἐὰν ἄρα ἐφ᾽ ἑκάστης τῶν τοῦ κύβου δώδεκα πλευρῶν τὰ αὐτὰ κατασκευάσωμεν, συσταθήσεταί τι σχῆμα στερεὸν ὑπὸ δώδεκα πενταγώνων ἰσοπλεύρων τε καὶ ἰσογωνίων περιεχόμενον, ὃ καλεῖται δωδεκάεδρον. δεῖ δὴ αὐτὸ καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τοῦ δωδεκαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἀποτομή. Ἐκβεβλήσθω γὰρ ἡ ΨΟ, καὶ ἔστω ἡ ΨΩ: συμβάλλει ἄρα ἡ ΟΩ τῇ τοῦ κύβου διαμέτρῳ, καὶ δίχα τέμνουσιν ἀλλήλας: τοῦτο γὰρ δέδεικται ἐν τῷ παρατελεύτῳ θεωρήματι τοῦ ἑνδεκάτου βιβλίου. τεμνέτωσαν κατὰ τὸ Ω: τὸ Ω ἄρα κέντρον ἐστὶ τῆς σφαίρας τῆς περιλαμβανούσης τὸν κύβον, καὶ ἡ ΩΟ ἡμίσεια τῆς πλευρᾶς τοῦ κύβου. ἐπεζεύχθω δὴ ἡ ΥΩ. καὶ ἐπεὶ εὐθεῖα γραμμὴ ἡ ΝΣ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ο, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΝΟ, τὰ ἄρα ἀπὸ τῶν ΝΣ, ΣΟ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΟ. ἴση δὲ ἡ μὲν ΝΣ τῇ ΨΩ, ἐπειδήπερ καὶ ἡ μὲν ΝΟ τῇ ΟΩ ἐστιν ἴση, ἡ δὲ ΨΟ τῇ ΟΣ. ἀλλὰ μὴν καὶ ἡ ΟΣ τῇ ΨΥ, ἐπεὶ καὶ τῇ ΡΟ: τὰ ἄρα ἀπὸ τῶν ΩΨ, ΨΥ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΟ. τοῖς δὲ ἀπὸ τῶν ΩΨ, ΨΥ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΥΩ: τὸ ἄρα ἀπὸ τῆς ΥΩ τριπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΝΟ. ἔστι δὲ καὶ ἡ ἐκ τοῦ κέντρου τῆς σφαίρας τῆς περιλαμβανούσης τὸν κύβον δυνάμει τριπλασίων τῆς ἡμισείας τῆς τοῦ κύβου πλευρᾶς: προδέδεικται γὰρ κύβον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει τριπλασίων ἐστὶ τῆς πλευρᾶς τοῦ κύβου. εἰ δὲ ὅλη τῆς ὅλης, καὶ ἡ ἡμίσεια τῆς ἡμισείας: καί ἐστιν ἡ ΝΟ ἡμίσεια τῆς τοῦ κύβου πλευρᾶς: ἡ ἄρα ΥΩ ἴση ἐστὶ τῇ ἐκ τοῦ κέντρου τῆς σφαίρας τῆς περιλαμβανούσης τὸν κύβον. καί ἐστι τὸ Ω κέντρον τῆς σφαίρας τῆς περιλαμβανούσης τὸν κύβον: τὸ Υ ἄρα σημεῖον πρὸς τῇ ἐπιφανείᾳ ἐστὶ τῆς σφαίρας. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκάστη τῶν λοιπῶν γωνιῶν τοῦ δωδεκαέδρου πρὸς τῇ ἐπιφανείᾳ ἐστὶ τῆς σφαίρας: περιείληπται ἄρα τὸ δωδεκάεδρον τῇ δοθείσῃ σφαίρᾳ. λέγω δή, ὅτι ἡ τοῦ δωδεκαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἀποτομή. ἐπεὶ γὰρ τῆς ΝΟ ἄκρον καὶ μέσον λόγον τετμημένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΡΟ, τῆς δὲ ΟΞ ἄκρον καὶ μέσον λόγον τετμημένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΟΣ, ὅλης ἄρα τῆς ΝΞ ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΡΣ. οἷον ἐπεί ἐστιν ὡς ἡ ΝΟ πρὸς τὴν ΟΡ, ἡ ΟΡ πρὸς τὴν ΡΝ, καὶ τὰ διπλάσια: τὰ γὰρ μέρη τοῖς ἰσάκις πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον: ὡς ἄρα ἡ ΝΞ πρὸς τὴν ΡΣ, οὕτως ἡ ΡΣ πρὸς συναμφότερον τὴν ΝΡ, ΣΞ. μείζων δὲ ἡ ΝΞ τῆς ΡΣ: μείζων ἄρα καὶ ἡ ΡΣ συναμφοτέρου τῆς ΝΡ, ΣΞ: ἡ ΝΞ ἄρα ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΡΣ. ἴση δὲ ἡ ΡΣ τῇ ΥΦ: τῆς ἄρα ΝΞ ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΥΦ. καὶ ἐπεὶ ῥητή ἐστιν ἡ τῆς σφαίρας διάμετρος καί ἐστι δυνάμει τριπλασίων τῆς τοῦ κύβου πλευρᾶς, ῥητὴ ἄρα ἐστὶν ἡ ΝΞ πλευρὰ οὖσα τοῦ κύβου. ἐὰν δὲ ῥητὴ γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, ἑκάτερον τῶν τμημάτων ἄλογός ἐστιν ἀποτομή. ἡ ΥΦ ἄρα πλευρὰ οὖσα τοῦ δωδεκαέδρου ἄλογός ἐστιν ἀποτομή. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι τῆς τοῦ κύβου πλευρᾶς ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ τοῦ δωδεκαέδρου πλευρά. ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",1321]]],["Rule",["Association",["Rule","'Book'",13],["Rule","'Theorem'",18]],["Association",["Rule","'VertexLabel'","'13.18'"],["Rule","'Text'","'To set out the sides of the five figures and to compare them with one another.'"],["Rule","'TextWordCount'",16],["Rule","'GreekText'","'τὰς πλευρὰς τῶν πέντε σχημάτων ἐκθέσθαι καὶ συγκρῖναι πρὸς ἀλλήλας.'"],["Rule","'GreekTextWordCount'",10],["Rule","'References'",["List",["List",["Rule","'Book'",4],["Rule","'Theorem'",15]],["List",["Rule","'Book'",5],["Rule","'Definition'",9]],["List",["Rule","'Book'",6],["Rule","'Theorem'",4]],["List",["Rule","'Book'",6],["Rule","'Theorem'",8]],["List",["Rule","'Book'",6],["Rule","'Theorem'",20]],["List",["Rule","'Book'",13],["Rule","'Theorem'",9]],["List",["Rule","'Book'",13],["Rule","'Theorem'",10]],["List",["Rule","'Book'",13],["Rule","'Theorem'",13]],["List",["Rule","'Book'",13],["Rule","'Theorem'",14]],["List",["Rule","'Book'",13],["Rule","'Theorem'",15]],["List",["Rule","'Book'",13],["Rule","'Theorem'",16]],["List",["Rule","'Book'",13],["Rule","'Theorem'",17]]]],["Rule","'Proof'","'Let AB, the diameter of the given sphere, be set out, and let it be cut at C so that AC is equal to CB, and at D so that AD is double of DB; let the semicircle AEB be described on AB, from C, D let CE, DF be drawn at right angles to AB, and let AF, FB, EB be joined. Then, since AD is double of DB, therefore AB is triple of BD. Convertendo, therefore, BA is one and a half times AD. But, as BA is to AD, so is the square on BA to the square on AF, [V. Def. 9, VI. 8] for the triangle AFB is equiangular with the triangle AFD; therefore the square on BA is one and a half times the square on AF. But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [XIII. 13] And AB is the diameter of the sphere; therefore AF is equal to the side of the pyramid. Again, since AD is double of DB, therefore AB is triple of BD. But, as AB is to BD, so is the square on AB to the square on BF; [VI. 8, V. Def. 9] therefore the square on AB is triple of the square on BF. But the square on the diameter of the sphere is also triple of the square on the side of the cube. [XIII. 15] And AB is the diameter of the sphere; therefore BF is the side of the cube. And, since AC is equal to CB, therefore AB is double of BC. But, as AB is to BC, so is the square on AB to the square on BE; therefore the square on AB is double of the square on BE. But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [XIII. 14] And AB is the diameter of the given sphere; therefore BE is the side of the octahedron. Next, let AG be drawn from the point A at right angles to the straight line AB, let AG be made equal to AB, let GC be joined, and from H let HK be drawn perpendicular to AB. Then, since GA is double of AC, for GA is equal to AB, and, as GA is to AC, so is HK to KC, therefore HK is also double of KC. Therefore the square on HK is quadruple of the square on KC; therefore the squares on HK, KC, that is, the square on HC, is five times the square on KC. But HC is equal to CB; therefore the square on BC is five times the square on CK. And, since AB is double of CB, and, in them, AD is double of DB, therefore the remainder BD is double of the remainder DC. Therefore BC is triple of CD; therefore the square on BC is nine times the square on CD. But the square on BC is five times the square on CK; therefore the square on CK is greater than the square on CD; therefore CK is greater than CD. Let CL be made equal to CK, from L let LM be drawn at right angles to AB, and let MB be joined. Now, since the square on BC is five times the square on CK, and AB is double of BC, and KL double of CK, therefore the square on AB is five times the square on KL. But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [XIII. 16] And AB is the diameter of the sphere; therefore KL is the radius of the circle from which the icosahedron has been described; therefore KL is a side of the hexagon in the said circle. [IV. 15] And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [XIII. 16] and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK is equal to LB, therefore each of the straight lines AK, LB is a side of the decagon inscribed in the circle from which the icosahedron has been described. And, since LB belongs to a decagon, and ML to a hexagon, for ML is equal to KL, since it is also equal to HK, being the same distance from the centre, and each of the straight lines HK, KL is double of KC, therefore MB belongs to a pentagon. [XIII. 10] But the side of the pentagon is the side of the icosahedron; [XIII. 16] therefore MB belongs to the icosahedron. Now, since FB is a side of the cube, let it be cut in extreme and mean ratio at N, and let NB be the greater segment; therefore NB is a side of the dodecahedron. [XIII. 17] And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side AF of the pyramid, double of the square on the side BE of the octahedron and triple of the side FB of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two. Therefore the square on the side of the pyramid is fourthirds of the square on the side of the octahedron, and double of the square on the side of the cube; and the square on the side of the octahedron is one and a half times the square on the side of the cube. The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios. But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides; for they are irrational, the one being minor [XIII. 16] and the other an apotome [XIII. 17]. That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus. For, since the triangle FDB is equiangular with the triangle FAB, [VI. 8] proportionally, as DB is to BF, so is BF to BA. [VI. 4] And, since the three straight lines are proportional, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9, VI. 20] therefore, as DB is to BA, so is the square on DB to the square on BF; therefore, inversely, as AB is to BD, so is the square on FB to the square on BD. But AB is triple of BD; therefore the square on FB is triple of the square on BD. But the square on AD is also quadruple of the square on DB, for AD is double of DB; therefore the square on AD is greater than the square on FB; therefore AD is greater than FB; therefore AL is by far greater than FB. And, when AL is cut in extreme and mean ratio, KL is the greater segment, inasmuch as LK belongs to a hexagon, and KA to a decagon; [XIII. 9] and, when FB is cut in extreme and mean ratio, NB is the greater segment; therefore KL is greater than NB. But KL is equal to LM; therefore LM is greater than NB. Therefore MB, which is a side of the icosahedron, is by far greater than NB which is a side of the dodecahedron.'"],["Rule","'ProofWordCount'",1328],["Rule","'GreekProof'","'Ἐκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ, καὶ τετμήσθω κατὰ τὸ Γ ὥστε ἴσην εἶναι τὴν ΑΓ τῇ ΓΒ, κατὰ δὲ τὸ Δ ὥστε διπλασίονα εἶναι τὴν ΑΔ τῆς ΔΒ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΕΒ, καὶ ἀπὸ τῶν Γ, Δ τῇ ΑΒ πρὸς ὀρθὰς ἤχθωσαν αἱ ΓΕ, ΔΖ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΖΒ, ΕΒ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΔ τῆς ΔΒ, τριπλῆ ἄρα ἐστὶν ἡ ΑΒ τῆς ΒΔ. ἀναστρέψαντι ἡμιολία ἄρα ἐστὶν ἡ ΒΑ τῆς ΑΔ. ὡς δὲ ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ: ἰσογώνιον γάρ ἐστι τὸ ΑΖΒ τρίγωνον τῷ ΑΖΔ τριγώνῳ: ἡμιόλιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΑ τοῦ ἀπὸ τῆς ΑΖ. ἔστι δὲ καὶ ἡ τῆς σφαίρας διάμετρος δυνάμει ἡμιολία τῆς πλευρᾶς τῆς πυραμίδος. καί ἐστιν ἡ ΑΒ ἡ τῆς σφαίρας διάμετρος: ἡ ΑΖ ἄρα ἴση ἐστὶ τῇ πλευρᾷ τῆς πυραμίδος. πάλιν, ἐπεὶ διπλασίων ἐστὶν ἡ ΑΔ τῆς ΔΒ, τριπλῆ ἄρα ἐστὶν ἡ ΑΒ τῆς ΒΔ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΖ: τριπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΖ. ἔστι δὲ καὶ ἡ τῆς σφαίρας διάμετρος δυνάμει τριπλασίων τῆς τοῦ κύβου πλευρᾶς. καί ἐστιν ἡ ΑΒ ἡ τῆς σφαίρας διάμετρος: ἡ ΒΖ ἄρα τοῦ κύβου ἐστὶ πλευρά. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, διπλῆ ἄρα ἐστὶν ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΕ: διπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΕ. ἔστι δὲ καὶ ἡ τῆς σφαίρας διάμετρος δυνάμει διπλασίων τῆς τοῦ ὀκταέδρου πλευρᾶς. καί ἐστιν ἡ ΑΒ ἡ τῆς δοθείσης σφαίρας διάμετρος: ἡ ΒΕ ἄρα τοῦ ὀκταέδρου ἐστὶ πλευρά. ἤχθω δὴ ἀπὸ τοῦ Α σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς ἡ ΑΗ, καὶ κείσθω ἡ ΑΗ ἴση τῇ ΑΒ, καὶ ἐπεζεύχθω ἡ ΗΓ, καὶ ἀπὸ τοῦ Θ ἐπὶ τὴν ΑΒ κάθετος ἤχθω ἡ ΘΚ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΗΑ τῆς ΑΓ: ἴση γὰρ ἡ ΗΑ τῇ ΑΒ: ὡς δὲ ἡ ΗΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΘΚ πρὸς τὴν ΚΓ, διπλῆ ἄρα καὶ ἡ ΘΚ τῆς ΚΓ. τετραπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΘΚ τοῦ ἀπὸ τῆς ΚΓ: τὰ ἄρα ἀπὸ τῶν ΘΚ, ΚΓ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΘΓ, πενταπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΚΓ. ἴση δὲ ἡ ΘΓ τῇ ΓΒ: πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΚ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΒ τῆς ΓΒ, ὧν ἡ ΑΔ τῆς ΔΒ ἐστι διπλῆ, λοιπὴ ἄρα ἡ ΒΔ λοιπῆς τῆς ΔΓ ἐστι διπλῆ. τριπλῆ ἄρα ἡ ΒΓ τῆς ΓΔ: ἐνναπλάσιον ἄρα τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΔ. πενταπλάσιον δὲ τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΚ: μεῖζον ἄρα τὸ ἀπὸ τῆς ΓΚ τοῦ ἀπὸ τῆς ΓΔ. μείζων ἄρα ἐστὶν ἡ ΓΚ τῆς ΓΔ. κείσθω τῇ ΓΚ ἴση ἡ ΓΛ, καὶ ἀπὸ τοῦ Λ τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΛΜ, καὶ ἐπεζεύχθω ἡ ΜΒ. καὶ ἐπεὶ πενταπλάσιόν ἐστι τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΚ, καί ἐστι τῆς μὲν ΒΓ διπλῆ ἡ ΑΒ, τῆς δὲ ΓΚ διπλῆ ἡ ΚΛ, πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΚΛ. ἔστι δὲ καὶ ἡ τῆς σφαίρας διάμετρος δυνάμει πενταπλασίων τῆς ἐκ τοῦ κέντρου τοῦ κύκλου, ἀφ᾽ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται. καί ἐστιν ἡ ΑΒ ἡ τῆς σφαίρας διάμετρος: ἡ ΚΛ ἄρα ἐκ τοῦ κέντρου ἐστὶ τοῦ κύκλου, ἀφ᾽ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται: ἡ ΚΛ ἄρα ἑξαγώνου ἐστὶ πλευρὰ τοῦ εἰρημένου κύκλου. καὶ ἐπεὶ ἡ τῆς σφαίρας διάμετρος σύγκειται ἔκ τε τῆς τοῦ ἑξαγώνου καὶ δύο τῶν τοῦ δεκαγώνου τῶν εἰς τὸν εἰρημένον κύκλον ἐγγραφομένων, καί ἐστιν ἡ μὲν ΑΒ ἡ τῆς σφαίρας διάμετρος, ἡ δὲ ΚΛ ἑξαγώνου πλευρά, καὶ ἴση ἡ ΑΚ τῇ ΛΒ, ἑκατέρα ἄρα τῶν ΑΚ, ΛΒ δεκαγώνου ἐστὶ πλευρὰ τοῦ ἐγγραφομένου εἰς τὸν κύκλον, ἀφ᾽ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται. καὶ ἐπεὶ δεκαγώνου μὲν ἡ ΛΒ, ἑξαγώνου δὲ ἡ ΜΛ: ἴση γάρ ἐστι τῇ ΚΛ, ἐπεὶ καὶ τῇ ΘΚ: ἴσον γὰρ ἀπέχουσιν ἀπὸ τοῦ κέντρου: καί ἐστιν ἑκατέρα τῶν ΘΚ, ΚΛ διπλασίων τῆς ΚΓ: πενταγώνου ἄρα ἐστὶν ἡ ΜΒ. ἡ δὲ τοῦ πενταγώνου ἐστὶν ἡ τοῦ εἰκοσαέδρου: εἰκοσαέδρου ἄρα ἐστὶν ἡ ΜΒ. καὶ ἐπεὶ ἡ ΖΒ κύβου ἐστὶ πλευρά, τετμήσθω ἄκρον καὶ μέσον λόγον κατὰ τὸ Ν, καὶ ἔστω μεῖζον τμῆμα τὸ ΝΒ: ἡ ΝΒ ἄρα δωδεκαέδρου ἐστὶ πλευρά. καὶ ἐπεὶ ἡ τῆς σφαίρας διάμετρος ἐδείχθη τῆς μὲν ΑΖ πλευρᾶς τῆς πυραμίδος δυνάμει ἡμιολία, τῆς δὲ τοῦ ὀκταέδρου τῆς ΒΕ δυνάμει διπλασίων, τῆς δὲ τοῦ κύβου τῆς ΖΒ δυνάμει τριπλασίων, οἵων ἄρα ἡ τῆς σφαίρας διάμετρος δυνάμει ἕξ, τοιούτων ἡ μὲν τῆς πυραμίδος τεσσάρων, ἡ δὲ τοῦ ὀκταέδρου τριῶν, ἡ δὲ τοῦ κύβου δύο. ἡ μὲν ἄρα τῆς πυραμίδος πλευρὰ τῆς μὲν τοῦ ὀκταέδρου πλευρᾶς δυνάμει ἐστὶν ἐπίτριτος, τῆς δὲ τοῦ κύβου δυνάμει διπλῆ, ἡ δὲ τοῦ ὀκταέδρου τῆς τοῦ κύβου δυνάμει ἡμιολία. αἱ μὲν οὖν εἰρημέναι τῶν τριῶν σχημάτων πλευραί, λέγω δὴ πυραμίδος καὶ ὀκταέδρου καὶ κύβου, πρὸς ἀλλήλας εἰσὶν ἐν λόγοις ῥητοῖς. αἱ δὲ λοιπαὶ δύο, λέγω δὴ ἥ τε τοῦ εἰκοσαέδρου καὶ ἡ τοῦ δωδεκαέδρου, οὔτε πρὸς ἀλλήλας οὔτε πρὸς τὰς προειρημένας εἰσὶν ἐν λόγοις ῥητοῖς: ἄλογοι γάρ εἰσιν, ἡ μὲν ἐλάττων, ἡ δὲ ἀποτομή. ὅτι μείζων ἐστὶν ἡ τοῦ εἰκοσαέδρου πλευρὰ ἡ ΜΒ τῆς τοῦ δωδεκαέδρου τῆς ΝΒ, δείξομεν οὕτως. ἐπεὶ γὰρ ἰσογώνιόν ἐστι τὸ ΖΔΒ τρίγωνον τῷ ΖΑΒ τριγώνῳ, ἀνάλογόν ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΖ, οὕτως ἡ ΒΖ πρὸς τὴν ΒΑ. καὶ ἐπεὶ τρεῖς εὐθεῖαι ἀνάλογόν εἰσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ἀπὸ τῆς δευτέρας: ἔστιν ἄρα ὡς ἡ ΔΒ πρὸς τὴν ΒΑ, οὕτως τὸ ἀπὸ τῆς ΔΒ πρὸς τὸ ἀπὸ τῆς ΒΖ: ἀνάπαλιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΖΒ πρὸς τὸ ἀπὸ τῆς ΒΔ. τριπλῆ δὲ ἡ ΑΒ τῆς ΒΔ: τριπλάσιον ἄρα τὸ ἀπὸ τῆς ΖΒ τοῦ ἀπὸ τῆς ΒΔ. ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΑΔ τοῦ ἀπὸ τῆς ΔΒ τετραπλάσιον: διπλῆ γὰρ ἡ ΑΔ τῆς ΔΒ: μεῖζον ἄρα τὸ ἀπὸ τῆς ΑΔ τοῦ ἀπὸ τῆς ΖΒ: μείζων ἄρα ἡ ΑΔ τῆς ΖΒ: πολλῷ ἄρα ἡ ΑΛ τῆς ΖΒ μείζων ἐστίν. καὶ τῆς μὲν ΑΛ ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΚΛ, ἐπειδήπερ ἡ μὲν ΛΚ ἑξαγώνου ἐστίν, ἡ δὲ ΚΑ δεκαγώνου: τῆς δὲ ΖΒ ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΝΒ: μείζων ἄρα ἡ ΚΛ τῆς ΝΒ. ἴση δὲ ἡ ΚΛ τῇ ΛΜ: μείζων ἄρα ἡ ΛΜ τῆς ΝΒ τῆς δὲ ΛΜ μείζων ἐστὶν ἡ ΜΒ. πολλῷ ἄρα ἡ ΜΒ πλευρὰ οὖσα τοῦ εἰκοσαέδρου μείζων ἐστὶ τῆς ΝΒ πλευρᾶς οὔσης τοῦ δωδεκαέδρου: ὅπερ ἔδει δεῖξαι. λέγω δή, ὅτι παρὰ τὰ εἰρημένα πέντε σχήματα οὐ συσταθήσεται ἕτερον σχῆμα περιεχόμενον ὑπὸ ἰσοπλεύρων τε καὶ ἰσογωνίων ἴσων ἀλλήλοις. ὑπὸ μὲν γὰρ δύο τριγώνων ἢ ὅλως ἐπιπέδων στερεὰ γωνία οὐ συνίσταται. ὑπὸ δὲ τριῶν τριγώνων ἡ τῆς πυραμίδος, ὑπὸ δὲ τεσσάρων ἡ τοῦ ὀκταέδρου, ὑπὸ δὲ πέντε ἡ τοῦ εἰκοσαέδρου: ὑπὸ δὲ ἓξ τριγώνων ἰσοπλεύρων τε καὶ ἰσογωνίων πρὸς ἑνὶ σημείῳ συνισταμένων οὐκ ἔσται στερεὰ γωνία: οὔσης γὰρ τῆς τοῦ ἰσοπλεύρου τριγώνου γωνίας διμοίρου ὀρθῆς ἔσονται αἱ ἓξ τέσσαρσιν ὀρθαῖς ἴσαι: ὅπερ ἀδύνατον: ἅπασα γὰρ στερεὰ γωνία ὑπὸ ἐλασσόνων ἢ τεσσάρων ὀρθῶν περιέχεται. διὰ τὰ αὐτὰ δὴ οὐδὲ ὑπὸ πλειόνων ἢ ἓξ γωνιῶν ἐπιπέδων στερεὰ γωνία συνίσταται. ὑπὸ δὲ τετραγώνων τριῶν ἡ τοῦ κύβου γωνία περιέχεται: ὑπὸ δὲ τεσσάρων ἀδύνατον: ἔσονται γὰρ πάλιν τέσσαρες ὀρθαί. ὑπὸ δὲ πενταγώνων ἰσοπλεύρων καὶ ἰσογωνίων, ὑπὸ μὲν τριῶν ἡ τοῦ δωδεκαέδρου: ὑπὸ δὲ τεσσάρων ἀδύνατον: οὔσης γὰρ τῆς τοῦ πενταγώνου ἰσοπλεύρου γωνίας ὀρθῆς καὶ πέμπτου, ἔσονται αἱ τέσσαρες γωνίαι τεσσάρων ὀρθῶν μείζους: ὅπερ ἀδύνατον. οὐδὲ μὴν ὑπὸ πολυγώνων ἑτέρων σχημάτων περισχεθήσεται στερεὰ γωνία διὰ τὸ αὐτὸ ἄτοπον. οὐκ ἄρα παρὰ τὰ εἰρημένα πέντε σχήματα ἕτερον σχῆμα στερεὸν συσταθήσεται ὑπὸ ἰσοπλεύρων τε καὶ ἰσογωνίων περιεχόμενον: ὅπερ ἔδει δεῖξαι. λῆμμα ὅτι δὲ ἡ τοῦ ἰσοπλεύρου καὶ ἰσογωνίου πενταγώνου γωνία ὀρθή ἐστι καὶ πέμπτου, οὕτω δεικτέον. ἔστω γὰρ πεντάγωνον ἰσόπλευρον καὶ ἰσογώνιον τὸ ΑΒΓΔΕ, καὶ περιγεγράφθω περὶ αὐτὸ κύκλος ὁ ΑΒΓ ΔΕ, καὶ εἰλήφθω αὐτοῦ τὸ κέντρον τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ. δίχα ἄρα τέμνουσι τὰς πρὸς τοῖς Α, Β, Γ, Δ, Ε τοῦ πενταγώνου γωνίας. καὶ ἐπεὶ αἱ πρὸς τῷ Ζ πέντε γωνίαι τέσσαρσιν ὀρθαῖς ἴσαι εἰσὶ καί εἰσιν ἴσαι, μία ἄρα αὐτῶν, ὡς ἡ ὑπὸ ΑΖΒ, μιᾶς ὀρθῆς ἐστι παρὰ πέμπτον: λοιπαὶ ἄρα αἱ ὑπὸ ΖΑΒ, ΑΒΖ μιᾶς εἰσιν ὀρθῆς καὶ πέμπτου. ἴση δὲ ἡ ὑπὸ ΖΑΒ τῇ ὑπὸ ΖΒΓ: καὶ ὅλη ἄρα ἡ ὑπὸ ΑΒΓ τοῦ πενταγώνου γωνία μιᾶς ἐστιν ὀρθῆς καὶ πέμπτου: ὅπερ ἔδει δεῖξαι.'"],["Rule","'GreekProofWordCount'",1422]]]]