# Ratio of Areas of Equiangular Parallelograms

## Theorem

In the words of Euclid:

Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides.

## Proof

Let $\Box AC, \Box CF$ be equiangular parallelograms having $\angle BCD = \angle ECG$.

Let them be arranged so that $BC$ is in a straight line with $CG$.

Then $DC$ is in a straight line with $CE$.

Let the parallelogram $DG$ be completed.

Let a straight line $K$ be set out.

Using Construction of Fourth Proportional Straight Line, let $L$ and $M$ be constructed such that:

$K : L = BC : CG$
$L : M = DC : CE$

Then $K : L$ and $L : M$ are the same as the ratios of the sides.

But $K : M$ is compounded of $K : L$ and $L : M$.

So $K : M$ is the ratio compounded of the ratios of the sides.

$BC : CG = \Box AC : \Box CH$
$K : L = \Box AC : \Box CH$

Similarly, from Areas of Triangles and Parallelograms Proportional to Base:

$DC : CE = \Box CH : \Box CF$

Since we have $DC : CE = L : M$, it follows from Equality of Ratios is Transitive that:

$L : M = \Box CH : \Box CF$

Since we have:

$K : L = \Box AC : \Box CH$
$L : M = \Box CH : \Box CF$

it follows that:

$K : M = \Box AC : \Box CF$

But $K : M$ is the ratio compounded of the ratios of the sides.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $23$ of Book $\text{VI}$ of Euclid's The Elements.