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.


 * ''If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

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.