## Proposition 2

 If a straight line is cut at random, then the sum of the rectangles contained by the whole and each of the segments equals the square on the whole. Let the straight line AB be cut at random at the point C. I say that the sum of the rectangle BA by AC and the rectangle AB by BC equals the square on AB Describe the square ADEB on AB, and draw CF through C parallel to either AD or BE. Then AE equals AF plus CE. I.46 I.31 Now AE is the square on AB; AF is the rectangle BA by AC, for it is contained by DA and AC, and AD equals AB; and CE is the rectangle AB by BC, for BE equals AB. II.Def.1 Therefore the sum of the rectangle BA by AC and the rectangle AB by BC equals the square on AB. Therefore if a straight line is cut at random, then the sum of the rectangles contained by the whole and each of the segments equals the square on the whole. Q.E.D.
This proposition is actually a special case of II.1. In II.1 Euclid shows that the product of one line by a sum of any number of lines is the sum of the products of that line by each of the lines. In this proposition, there are just two of those lines and their sum equals the one line. Rather than using II.1 to prove II.2, Euclid proves II.2 directly. This suggests that II.1 may have been inserted into the Elements after II.2 was included, either by Euclid or someone else.

In modern algebraic notation this proposition says that if y = y1 + y2, then xy = x y1 + x y2. This can also be stated in a single equation as

x (y1 + y2) = x y1 + x y2.

#### Use of this proposition

This proposition is used in the proof of proposition XIII.10 which shows that a certain relationship holds for the sides of a regular pentagon, regular hexagon, and regular decagon that are all inscribed in the same circle, namely, the square on the side of the pentagon equals the sum of the squares on the side of a hexagon and on the side of a decagon.

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