Let the straight line AB be divided into any two parts in C;

the square of AB is equal to

the squares of AC, CB,

and to twice the rectangle contained by AC, CB.

Upon AB describe the square ADEB,

and join BD,

and through C draw CGF parallel to AD or BE,

and through G draw HK parallel to AB or DE.

And because CF is parallel to AD, and BD falls upon them,

the exterior angle BGC is equal to

the interior and opposite angle ADB;

but ADB is equal to the angle ABD;

because BA is equal to AD,

being sides of a square;

wherefore the angle CGB is equal to the angle GBC;

and therefore the side BC is equal to the side CG;

But CB is equal also to GK,

and CG to BK;

wherefore the figure CGKB is equilateral :

It is likewise rectangular;

for CG is parallel to BK, and CB meets them

the angles KBC, GCB are therefore equal to two right angles,

and KBC is a right angle;

wherefore GCB is a right angle;

and therefore also the angles CGK, GKB opposite to these are right angles,

and CGKB is rectangular;

but it is also equilateral, as was demonstrated;

wherefore it is a square, and it is upon the side CB:

For the same reason HF also is a square,

and it is upon the side HG, which is equal to AC:

Therefore HF, CK are the squares of AC, CB;

and because the complement AG is equal to the complement GE,

and that AG is the rectangle contained by AC, CB,

for GC is equal to CB;

therefore GE is also equal to the rectangle AC, CB;

wherefore AG, GE are equal to twice the rectangle AC, CB:

And HF, CK are the squares of AC, CB;

wherefore the four figures HF, CK, AG, GE are

equal to the squares of AC, CB, and to twice the rectangle AC, CB:

But HF, CK, AG, GE, make up the whole figure ADEB,

which is the square of AB:

Therefore the square of AB is equal to

the squares of AC, CB and twice the rectangle AC, CB.

Wherefore if a straight line, &c. QED.