# Complements of Parallelograms are Equal

## Theorem

In the words of Euclid:

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

## Proof

Let $ABCD$ be a parallelogram, and let $AC$ be a diameter.

Let $EKHA$ and $FKGC$ be parallelograms about $AC$.

Let $BEKG$ and $DFKH$ be the complements of $EKHA$ and $FKGC$.

$\triangle ABC = \triangle ACD$
$\triangle AEK = \triangle AHK$
$\triangle KFC = \triangle KGC$

So from Common Notion 2:

$\triangle AEK + \triangle KGC = \triangle AHK + \triangle KFC$

But the whole of $\triangle ABC$ equals the whole of $\triangle ACD$.

So when $\triangle AEG + \triangle GIC$ and $\triangle AHG + \triangle GFC$ are subtracted from $\triangle ABC$ and $\triangle ACD$ respectively, the complement $BEKG$ which remains is equal in area to $DFKH$, from Common Notion 3.

$\blacksquare$

## Historical Note

This theorem is Proposition $43$ of Book $\text{I}$ of Euclid's The Elements.