Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional

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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.

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:

$DB : BE = GB : BF$

$\Box$


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