Book 2
4
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]
4
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]
4
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. [I. 6]
4
But CB is equal to GK, and CG to KB; [I. 34] therefore GK is also equal to KB; therefore CGKB is equilateral.
4
I say next that it is also right-angled.
4
For, since CG is parallel to BK, the angles KBC, GCB are equal to two right angles. [I. 29]
4
But the angle KBC is right; therefore the angle BCG is also right, so that the opposite angles CGK, GKB are also right. [I. 34]
4
Therefore CGKB is right-angled; and it was also proved equilateral; therefore it is a square; and it is described on CB.
4
For the same reason HF is also a square; and it is described on HG, that is AC. [I. 34]
4
Therefore the squares HF, KC are the squares on AC, CB.
4
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.
4
Therefore AG, GE are equal to twice the rectangle AC, CB.
4
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.
4
But HF, CK, AG, GE are the whole ADEB, which is the square on AB.
4
Therefore the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB.
