Proposition 13
Theorem
In any triangle △ABC , with ∠ACB < 90° and AD ⊥ BC at point D , it follows that |BC|2 + |CA|2 = |AB|2 + 2·|BC|·|DC| .
Commentary
1. Let △ABC be the given triangle, with C being an acute angle.
2. FromA , draw a line that is perpendicular to BC at a point D .
3. Two right triangles are constructed:△ADB and △ADC .
4. Apply the Pythagorean Theorem to the right triangles so: in△ADB , |AB|2 = |BD|2 + |AD|2 , and in △ADC , |CA|2 = |DC|2 + |AD|2 . Using these equations, and the fact that |BD| + |DC| = |BC| , the following equation is obtained: |BC|2 + |CA|2 = |AB|2 + 2·|BC|·|DC| .
5. This algebraic relationship is similar to the previous proposition, Book 2 Proposition 12, and is a geometric version of the law of cosines in trigonometry.
6. This proposition is a variation of the Pythagorean Theorem (Book 1 Proposition 47) which applies to acute triangles.
2. From
3. Two right triangles are constructed:
4. Apply the Pythagorean Theorem to the right triangles so: in
5. This algebraic relationship is similar to the previous proposition, Book 2 Proposition 12, and is a geometric version of the law of cosines in trigonometry.
6. This proposition is a variation of the Pythagorean Theorem (Book 1 Proposition 47) which applies to acute triangles.
Original statement
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.
English translation
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.
