Ratio of Areas of Equiangular Parallelograms

Theorem

 * 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 $CE$.


 * Euclid-VI-23.png

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 ratio compounded of the ratios of the sides.

Hence the result.