Proof: (Euclid)

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 parallelogrammic figure, the complements of the parallelograms about the diagonal are equal to one another.
(Which is) the very thing it was required to show.
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