Sides of Equiangular Triangles are Reciprocally Proportional

Theorem

 * In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and 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 $\triangle ABC, \triangle ADE$ be triangles of equal area which have one angle equal to one angle, namely $\angle BAC = \angle DAE$.

We need to show that $CA : AD = EA : AB$, that is, the sides about the equal angles are reciprocally proportional.


 * Euclid-VI-15.png

Place them so $CA$ is in a straight line with $AD$.

From Two Angles making Two Right Angles make Straight Line $EA$ is also in a straight line with $AB$.

Join $BD$.

It follows from Ratios of Equal Magnitudes that:
 * $\triangle CAB : \triangle BAD = \triangle EAD : \triangle BAD$

But from Areas of Triangles and Parallelograms Proportional to Base:
 * $\triangle CAB : \triangle BAD = CA : AD$

Also from Areas of Triangles and Parallelograms Proportional to Base:
 * $\triangle EAD : \triangle BAD = EA : AB$

So from Equality of Ratios is Transitive:
 * $CA : AD = EA : AB$

Now let the sides in $\triangle ABC, \triangle ADE$ be reciprocally proportional.

That is, $CA : AD = EA : AB$.

Join $BD$.

From Areas of Triangles and Parallelograms Proportional to Base:
 * $\triangle CAB : \triangle BAD = CA : AD$

Also from Areas of Triangles and Parallelograms Proportional to Base:
 * $\triangle EAD : \triangle BAD = EA : AB$

It follows from Equality of Ratios is Transitive that:
 * $\triangle CAB : \triangle BAD = \triangle EAD : \triangle BAD$

So from Magnitudes with Same Ratios are Equal:
 * $\triangle ABC = \triangle ADE$