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.

In the words of Euclid:

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.

(The Elements: Book $\text{VI}$: Proposition $6$)

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 $\angle ACB = \angle DFE$.

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 Congruence:

$\triangle DEF = \triangle DGF$

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.

$\blacksquare$

Historical Note

This proof is Proposition $6$ of Book $\text{VI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VI}$. Propositions