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