Complements of Parallelograms are Equal

Proof

 * Euclid-I-43.png

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 $FKGC$.

From Opposite Sides and Angles of Parallelogram are Equal:
 * $\triangle ABC = \triangle ACD$

Also from Opposite Sides and Angles of Parallelogram are Equal:
 * $\triangle AEK = \triangle AHK$
 * $\triangle KFC = \triangle KGC$

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.