Let ABCD be a parallelogram, and AC its diameter;
and about AC let EH, FG be parallelograms, and BK, KD the so-called complements;
I say that the complement BK is equal to the complement KD.

For since ABCD is a parallelogram, and AC its diameter,
the triangle ABC is equal to the triangle ACD. [I. 34]

Again, since EH is a parallelogram, and AK is its diameter,
the triangle AEK is equal to the triangle AHK.

For the same reason
the triangle KFC is also equal to KGC.

Now, since the triangle AEK is equal to the triangle AHK,
and KFC to KGC,
the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C.N. 2]

And the whole triangle ABC is also equal to the whole ADC;
therefore the complement BK which remains is equal to the complement KD which remains. [C.N. 3]

Therefore etc. Q.E.D.