Triangles with Proportional Sides are Similar

Theorem
Let two triangles have corresponding sides which are proportional.

Then their corresponding angles are equal.

Thus, by definition, such triangles are similar.

Proof
Let $\triangle ABC, \triangle DEF$ be triangles whose sides are proportional, so that:
 * $ AB : BC = DE : EF$
 * $ BC : CA = EF : FD$
 * $ BA : AC = ED : DF$

We need to show that
 * $\angle ABC = \angle DEF$
 * $\angle BCA = \angle EFD$
 * $\angle BAC = \angle EDF$


 * Euclid-VI-5.png

On the straight line $EF$, and at the points $E, F$ on it, construct $\angle FEG = \angle ABC$ and $\angle EFG = \angle ACB$.

From Sum of Angles of Triangle Equals Two Right Angles, the remaining angle at $A$ equals the remaining angle at $G$.

Therefore $\triangle ABC$ is equiangular with $\triangle GEF$.

From Equiangular Triangles are Similar, the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

So:
 * $AB : BD = GE : EF$

But by hypothesis:
 * $AB : BC = DE : EF$

So from Equality of Ratios is Transitive
 * $DE : EF = GE : EF$

So each of $DE, GE$ has the same ratio to $EF$.

So from Magnitudes with Same Ratios are Equal:
 * $DE = GE$

For the same reason:
 * $DF = GF$

So we have that $DE = EG$, $EF$ is common and $DF = FG$.

So from Triangle Side-Side-Side Equality:
 * $\triangle DEF = \triangle GEF$

That is:
 * $\angle DEF = \angle GEF, \angle DFE = \angle GFE, \angle EDF = \angle EGF$

As $\angle GEF = \angle ABC$ it follows that:
 * $\angle ABC = \angle DEF$

For the same reason $\angle ACB = \angle DFE$ and $\angle BAC = \angle EDF$.

Hence the result.