For any parallelogram, the complements of the parallelograms about the diagonal are equal to one another. Let $ABCD$ be a parallelogram, and $AC$ its diagonal. And let $EH$ and $FG$ be the parallelograms about  $AC$, and $BK$ and $KD$ the so-called complements (about $AC$). I say that the complement $BK$ is equal to the complement $KD$.  For since $ABCD$ is a parallelogram, and $AC$ its diagonal,  triangle $ABC$ is equal to triangle $ACD$ [Prop.~1.34]. Again, since $EH$ is a parallelogram, and $AK$ is its diagonal, triangle $AEK$ is equal to triangle $AHK$ [Prop.~1.34]. So, for the same (reasons), triangle $KFC$ is also equal to (triangle) $KGC$. Therefore, since triangle $AEK$ is equal to triangle $AHK$, and $KFC$ to $KGC$, triangle $AEK$ plus $KGC$ is equal to triangle $AHK$ plus $KFC$. And the whole triangle $ABC$ is also equal to the whole (triangle) $ADC$. Thus, the remaining complement $BK$ is equal to the remaining complement $KD$.  Thus, for any parallelogramic figure, the complements of the parallelograms about the diagonal are equal to one another. (Which is) the very thing it was required to show.