Equiangular Triangles are Similar

Theorem
Let two triangles have the same corresponding angles.

Then their corresponding sides are proportional.

Thus, by definition, such triangles are similar.


 * In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

Proof
Let $\triangle ABC, \triangle DCE$ be equiangular triangles such that:
 * $\angle ABC = \angle DCE$
 * $\angle BAC = \angle CDE$
 * $\angle ACB = \angle CED$


 * Euclid-VI-4.png

Let $BC$ be placed in a straight line with $CE$.

From Two Angles of Triangle Less than Two Right Angles $\angle ABC + \angle ACB$ is less than two right angles.

As $\angle ACB = \angle DEC$, it follows that $\angle ABC + \angle DEC$ is also less than two right angles.

So from the Parallel Postulate, $BA$ and $ED$, when produced, will meet.

Let this happen at $F$.

We have that $\angle ABC = \angle DCE$.

So from Equal Corresponding Angles implies Parallel Lines:
 * $BF \parallel CD$

Again, we have that $\angle ACB = \angle CED$.

Again from Equal Corresponding Angles implies Parallel Lines:
 * $AC \parallel FE$

Therefore by definition $\Box FACD$ is a parallelogram.

Therefore from Opposite Sides and Angles of Parallelogram are Equal $FA = DC$ and $AC = FD$.

Since $AC \parallel FE$, it follows from Parallel Line in Triangle Cuts Sides Proportionally that $BA : AF = BC : CE$.

But $AF = CD$ so $BA : AF = BC : CE$.

From Proportional Magnitudes are Proportional Alternately, $AB : BC = DC : CE$.

Since $CD \parallel BF$, from Parallel Line in Triangle Cuts Sides Proportionally $BC : CE = FD : DE$.

But $FD = AC$ so $BC : CE = AC : DE$.

So from Proportional Magnitudes are Proportional Alternately, $BC : CA = CE : ED$.

It then follows from Equality of Ratios Ex Aequali that $BA : AC = CD : DE$.