Triangles with One Equal Angle and Two Sides Proportional are Similar

Theorem
Let two triangles have two corresponding sides which are proportional.

Let the angles adjacent to both of these sides be equal.

Then all of their corresponding angles are equal.

Thus, by definition, such triangles are similar.


 * ''If two triangles have one angle equal to one angle and the sides about the equal angles 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 having $\angle BAC = \angle EDF$ and two sides proportional, so that $BA : AC = ED : DF$.

We need to show that $\triangle ABC$ is equiangular with $\triangle DEF$, such that $\angle ABC = \angle DEF$ and $\angle ACB = \angle DFE$.


 * Euclid-VI-6.png

Let $\angle FDG$ be constructed equal to either $\angle BAC$ or $\angle EDF$, and $\angle DFG = \angle ACB$.

From Sum of Angles of Triangle Equals Two Right Angles:
 * $\angle ABC = \angle DGF$

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

From Equiangular Triangles are Similar:
 * $BA : AC = GD : DF$

But by hypothesis:
 * $BA : AC = ED : DF$

So from Equality of Ratios is Transitive:
 * $ED : DF = GD : DF$

From Magnitudes with Same Ratios are Equal:
 * $ED = DG$

But $DF$ is common to $\triangle DEF$ and $\triangle DGF$, and $\angle EDF = \angle GDF$.

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

But $\angle DFG = \angle ACB$ and so:
 * $\angle ABC = \angle DFE$

But by hypothesis:
 * $\angle BAC = \angle EDF$

Therefore by Sum of Angles of Triangle Equals Two Right Angles:
 * $\angle ABC = \angle DEF$

Hence the result.