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Computable Euclid

Proposition 10

Theorem

If a line AB  is bisected at point C, and divided externally at point D, then AD 2 + BD 2 = 2 AC 2 + 2 CD 2.

Commentary

1. Let AB  be the given line segment. Let C be the midpoint of AB , and let D be any point on the extension of AB .
2. Construct a line CE  that is perpendicular to AB  and has the same length as AC . Connect AE  and EB .
3. Extend EB  to a point F such that CDFG is a rectangle with G on the extension of EC . Connect AF .
4. Five isosceles right triangles are constructed: ACE, BCE, AEB, EGF and FDB. Two right triangles that are not isosceles are constructed: ADF and AEF.
5. Apply the Pythagorean Theorem to the right triangles so: in ADF, AF 2 = AD 2 + DF 2, in AEF, AF 2 = AE 2 + EF 2, in ACE, AE 2 = AC 2 + CE 2 and in EGF, EF 2 = EG 2 + GF 2. With substitution of lines with the same length, the following equation is obtained: AD 2 + BD 2 = 2 AC 2 + 2 CD 2.
6. This geometric relationship is similar to the previous proposition, Book 2 Proposition 9, and can also be expressed algebraically as x + y2 + x - y2 = 2 x2 + 2 y2 (where AC  = CE  = x and CD  = y). This algebraic relationship is an interpretation of a commonly used polynomial identity.

Original statement

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.

English translation

If a straight line is bisected, and a straight line is 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.


Computable version


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