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Computable Euclid

Proposition 8

Theorem

If a line segment AB  is divided into any two parts at point C, then 4 AB CB  + AC 2 = AB  + CB 2.

Commentary

1. Let AB  be the given line segment, and let C be any point on AB .
2. Extend AB  to D so that B is the midpoint of line CD .
3. Construct square ADFE on AD , and find a point H on EF  such that CH  is parallel to AE  and intersects the diagonal DE  at a point I.
4. Find a point L on EF  such that BL  is parallel to AE  and intersects the diagonal DE  at a point K.
5. Through I, draw a line parallel to AD  and intersecting AE , BL  and DF  at points P, J and O, respectively.
6. Through K, draw a line parallel to AD  and intersecting AE , CH  and DF  at points M, G and N, respectively.
7. These constructed lines divide the square ADFE into nine polygons as follows: four rectangles ACGM, MGIP, IJLH, and JOFL with the same area; four squares CBKG, GKJI, BDNK, and KNOJ with the same area; and a square PIHE.
8. The sum of four times the area of rectangle ABKM (4 AB CB ) and the area of square PIHE (AC 2) is equal to the area of square ADFE (AB  + BC 2).
9. This geometric relationship is similar to the previous proposition, Book 2 Proposition 7, and can also be expressed algebraically as follows: if x = y + z, then 4 x y + z2 = x + y2 (where AB  = x, BC  = y and AC  = z). This algebraic relationship is a polynomial identity.

Original statement

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.

English translation

If a straight line is 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.


Computable version


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