If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.
若一条直线任意分割，则整线上的正方形等于两段上正方形与两倍两段所成矩形之和。
正方形 ADEB 建于 AB 上，C 把 AB 分成两段。BD 为对角线，过 C 作 CF 平行于 AD，过 G 作 HK 平行于 AB，把正方形切成 HF（AC 上正方形）、CK（CB 上正方形）和两块面积相等的矩形 AG、GE。
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For let the straight line AB be cut at random at C; I say that the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB. For let the square ADEB be described on AB, [I. 46] let BD be joined; through C let CF be drawn parallel to either AD or EB, and through G let HK be drawn parallel to either AB or DE. [I. 31] Then, since CF is parallel to AD, and BD has fallen on them, the exterior angle CGB is equal to the interior and opposite angle ADB. [I. 29] But the angle ADB is equal to the angle ABD, since the side BA is also equal to AD; [I. 5] therefore the angle CGB is also equal to the angle GBC, so that the side BC is also equal to the side CG.
在整线上作正方形，并从分点向两边作平行线。
[I. 6] But CB is equal to GK, and CG to KB; [I. 34] therefore GK is also equal to KB; therefore CGKB is equilateral. I say next that it is also right-angled. For, since CG is parallel to BK, the angles KBC, GCB are equal to two right angles. [I. 29] But the angle KBC is right; therefore the angle BCG is also right, so that the opposite angles CGK, GKB are also right.
正方形被分成两段各自的正方形和两个相等的矩形。
[I. 34] Therefore CGKB is right-angled; and it was also proved equilateral; therefore it is a square; and it is described on CB. For the same reason HF is also a square; and it is described on HG, that is AC. [I. 34] Therefore the squares HF, KC are the squares on AC, CB. Now, since AG is equal to GE, and AG is the rectangle AC, CB, for GC is equal to CB, therefore GE is also equal to the rectangle AC, CB.
两个矩形都由两段所成。
Therefore AG, GE are equal to twice the rectangle AC, CB. But the squares HF, CK are also the squares on AC, CB; therefore the four areas HF, CK, AG, GE are equal to the squares on AC, CB and twice the rectangle contained by AC, CB. But HF, CK, AG, GE are the whole ADEB, which is the square on AB. Therefore the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB.
所以整线平方等于两段平方与两倍两段矩形之和。