Sides of Equiangular Parallelograms are Reciprocally Proportional

Theorem

As Euclid defined it:

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 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 a 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, we have that $\Box AB : \Box FE = \Box BC : \Box FE$.

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 is Proposition 14 of Book VI of Euclid's The Elements.