Complements of Parallelograms are Equal

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

Proof

 * Euclid-I-43.png

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.