Proof: (Euclid)
(related to Proposition: 1.43: Complementary Segments of Parallelograms)
 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.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"