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Computable Euclid

Proposition 3

Theorem

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

Commentary

1. Let AB  be the given line segment, and let C be any point on AB .
2. On BC , construct a square CBED.
3. Extend ED  to F to construct rectangles ABEF and ACDF.
4. The area of the rectangle ABEF (AB CB ) is the sum of the areas of the two rectangles CBED (CB 2) and ACDF (AC CB ).
5. This geometric relationship can be expressed algebraically as follows: if x = y + z, then x y = y2 + y z (where AB  = x, CB  = y and AC  = z). This algebraic relationship is a special case of the distributive law.

Original statement

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.

English translation

If a straight line is 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.


Computable version


Additional instances


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