Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional

Theorem

 * 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.

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.


 * Euclid-VI-14.png

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:
 * $DB : BE = GB : BF$

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:
 * $\Box AB = \Box BC$