Ratio of Areas of Equiangular Parallelograms

Theorem

As Euclid defined it:

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

(The Elements: Book VI: Proposition $23$)

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.

From Areas of Triangles and Parallelograms Proportional to Base, $BC : CG = \Box AC : \Box CH$.

From Equality of Ratios is Transitive, $K : L = \Box AC : \Box CH$.

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 rato compounded of the ratios of the sides.

Hence the result.

$\blacksquare$

Historical Note

This is Proposition 23 of Book VI of Euclid's The Elements.