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.

(The Elements: Book $\text{I}$: Proposition $43$)

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 $FKGC$.

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 $\triangle AEK + \triangle KGC$ and $\triangle AHK + \triangle KFC$ 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 proof is Proposition $43$ of Book $\text{I}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions