Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

But CB is equal to GK, and CG to KB; [I. 34]

I say next that it is also right-angled.

For, since CG is parallel to BK,

But the angle KBC is right;

Therefore CGKB is right-angled; and it was also proved equilateral;

For the same reason

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 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.

Therefore etc. Q. E. D. ^{1 2 3}

Proposition 5.

For let a straight line AB be cut into equal segments at C and into unequal segments at D; I say that the rectangle contained by AD, DB together with the square on CD is equal to the square on CB.

For let the square CEFB be described on CB, [I. 46] and let BE be joined; through D let DG be drawn parallel to either CE or BF, through H again let KM be drawn parallel to either AB or EF, and again through A let AK be drawn parallel to either CL or BM. [I. 31]

Then, since the complement CH is equal to the complement HF, [I. 43] let DM be added to each;

But CM is equal to AL,