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 $KD$.
Let \(\boxdot{ABCD}\) be a parallelogram with a diagonal \(\overline{AC}\) and let \(K\) be any point on that diagonal. Then segments, which are parallel to the sides of the parallelogram and pass through \(K\), (specifically \(\overline{EF}\), \(\overline{GH}\)) divide \(\boxdot{ABCD}\) into four smaller parallelograms: the two segments through which the diagonal does not pass (\(\boxdot{EBGK}\), \(\boxdot{HKFD}\)) are called the complements of the other two (\(\boxdot{AEKH}\), \(\boxdot{KGCF}\)). Moreover, the complements are equal in area: \(\boxdot{EBGK}=\boxdot{HKFD}\).
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