Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional
Theorem
In the words of Euclid:
- In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.
(The Elements: Book $\text{VI}$: Proposition $14$)
Note: in the above, equal is to be taken to mean of equal area.
Proof
Let $\Box AB$ and $\Box BC$ be two equiangular parallelograms of equal area such that the angles at $B$ are equal.
Let $DB, BE$ be placed in a straight line.
By Two Angles making Two Right Angles make Straight Line it follows that $FB, BG$ also make a straight line.
We need to show that $DB : BE = GB : BF$, that is, the sides about the equal angles are reciprocally proportional.
Let the parallelogram $\Box FE$ be completed.
We have that $\Box AB$ is of equal area with $\Box BC$, and $\Box FE$ is another area.
So from Ratios of Equal Magnitudes:
- $\Box AB : \Box FE = \Box BC : \Box FE$
But from Areas of Triangles and Parallelograms Proportional to Base:
- $\Box AB : \Box FE = DB : BE$
Also from Areas of Triangles and Parallelograms Proportional to Base:
- $\Box BC : \Box FE = GB : BF$
So from Equality of Ratios is Transitive:
- $DB : BE = GB : BF$
$\Box$
Next, suppose that $DB : BE = GB : BF$.
From Areas of Triangles and Parallelograms Proportional to Base:
- $DB : BE = \Box AB : \Box FE$
Also from Areas of Triangles and Parallelograms Proportional to Base:
- $GB : BF = \Box BC : \Box FE$
So from Equality of Ratios is Transitive:
- $\Box AB : \Box FE = \Box BC : \Box FE$
So from Magnitudes with Same Ratios are Equal:
- $\Box AB = \Box BC$
$\blacksquare$
Historical Note
This proof is Proposition $14$ 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