If a straight line is cut at random, then the square on the whole equals the sum of the squares on the segments plus twice the rectangle contained by the segments.

Let the straight line *AB* be cut at random at *C.*

I say that the square on *AB* equals the sum of the squares on *AC* and *CB* plus twice the rectangle *AC* by *CB.*

Describe the square *ADEB* on *AB.* Join *BD.* Draw *CF* through *C* parallel to either *AD* or *EB,* and draw *HK* through *G* parallel to either *AB* or *DE.*

Then, since *CF* is parallel to *AD,* and *BD* falls on them, the exterior angle *CGB* equals the interior and opposite angle *ADB.*

But the angle *ADB* equals the angle *ABD,* since the side *BA* also equals *AD.* Therefore the angle *CGB* also equals the angle *GBC,* so that the side *BC* also equals the side *CG.*

But *CB* equals *GK,* and *CG* to *KB.* Therefore *GK* also equals *KB.* Therefore *CGKB* is equilateral.

I say next that it is also right-angled.

Since *CG* is parallel to *BK,* the sum of the angles *KBC* and *GCB* equals two right angles.

But the angle *KBC* is right. Therefore the angle *BCG* is also right, so that the opposite angles *CGK* and *GKB* are also right.

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.* Therefore the squares *HF* and *KC* are the squares on *AC* and *CB.*

Now, since *AG* equals *GE,* and *AG* is the rectangle *AC* by *CB,* for *GC* equals *CB,* therefore *GE* also equals the rectangle *AC* by *CB.* Therefore the sum of *AG* and *GE* equals twice the rectangle *AC* by *CB.*

But the squares *HF* and *CK* are also the squares on *AC* and *CB,* therefore the sum of the four figures *HF, CK, AG,* and *GE* equals the sum of the squares on *AC* and *CB* plus twice the rectangle *AC* by *CB.*

But *HF, CK, AG,* and *GE* are the whole *ADEB,* which is the square on *AB.*

Therefore the square on *AB* equals the the sum of the squares on *AC* and *CB* plus twice the rectangle *AC* by *CB.*

Therefore *if a straight line is cut at random, the square on the whole equals the squares on the segments plus twice the rectangle contained by the segments.*

Q.E.D.