In any parallelogram the complements of the parallelograms about the diameter are equal to one another.

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 $ACD$. [Prop. 1.34]Again, since $EH$ is a parallelogram, and $AK$ 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.