Equiangular Triangles are Similar

From ProofWiki
Jump to navigation Jump to search


Let two triangles have the same corresponding angles.

Then their corresponding sides are proportional.

Thus, by definition, such triangles are similar.

In the words of Euclid:

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

(The Elements: Book $\text{VI}$: Proposition $4$)


Let $\triangle ABC, \triangle DCE$ be equiangular triangles such that:

$\angle ABC = \angle DCE$
$\angle BAC = \angle CDE$
$\angle ACB = \angle CED$

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 Transversal Theorem 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 Transversal Theorem:

$BC : CE = FD : DE$

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 $BA : AC = CD : DE$.


Historical Note

This proof is Proposition $4$ of Book $\text{VI}$ of Euclid's The Elements.