Complements of Parallelograms are Equal

Theorem

In any parallelogram, the complements of the parallelograms about either diameter are equal.

Proof

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

Let $EGHA$ and $FGIC$ be parallelograms about $AC$.

Let $BEGI$ and $DFGH$ be the complements of $EGHA$ and $FGIC$.

From Opposite Sides and Angles of Parallelogram are Equal, $\triangle ABC = \triangle ACD$.

Similarly, $\triangle AEG = \triangle AHG$, and $\triangle GIC = \triangle GFC$.

So $\triangle AEG + \triangle GIC = \triangle AHG + \triangle GFC$ from Common Notion 2.

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 $BEGI$ which remains is equal in area to $DFGH$, from Common Notion 3.

$\blacksquare$

Historical Note

This is Proposition 43 of Book I of Euclid's The Elements.