Triangles with One Equal Angle and Two Other Sides Proportional are Similar

Theorem
Let two triangles be such that one of the angles of one triangle equals one of the angles of the other.

Let two corresponding sides which are adjacent to one of the other angles, be proportional.

Let the third angle in both triangles be either both acute or both not acute.

Then all of the corresponding angles of these triangles are equal.

Thus, by definition, such triangles are similar.


 * ''If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional.

Proof
Let $\triangle ABC, \triangle DEF$ be triangles such that:
 * $\angle BAC = \angle EDF$
 * two sides adjacent to $\angle ABC$ and $\angle DEF$ proportional, so that $AB : BC = DE : EF$
 * $\angle ACB$ and $\angle DFE$ either both acute or both not acute.

We need to show that $\angle ABC = \angle DEF$ and $\angle BCA = \angle EFD$.

First assume that $\angle ACB$ and $\angle DFE$ are both acute.


 * Euclid-VI-7a.png

Suppose $\angle ABC \ne \angle DEF$.

Then one of them is greater, WLOG $\angle ABC$.

On $AB$ and at point $B$, let $\angle ABG$ be constructed equal to $\angle DEF$.

Since $\angle BAG = \angle EDF$ and $\angle ABG = \angle DEF$, from Sum of Angles of Triangle Equals Two Right Angles:
 * $\angle AGB = \angle DFE$

So $\triangle ABG$ is equiangular with $\triangle DEF$.

So from Equiangular Triangles are Similar:
 * $AB : BG = DE : EF$

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

Thus from Equality of Ratios is Transitive:
 * $AB : BG = AB : BC$

So from Magnitudes with Same Ratios are Equal:
 * $BC = BG$

So from Isosceles Triangle has Two Equal Angles:
 * $\angle BCG = \angle BGC$

By hypothesis $\angle BCG$ is less than a right angle.

So $\angle BGC$ is less than a right angle.

So from Two Angles on Straight Line make Two Right Angles $\angle AGB$ is greater than a right angle.

But this was proved equal to $\angle DFE$.

But by hypothesis $\angle DFE$ is less than a right angle.

From this contradiction we conclude that:
 * $\angle ABC = \angle DEF$

We also have that $\angle BAC = \angle DEF$.

So from Sum of Angles of Triangle Equals Two Right Angles:
 * $\angle ACB = \angle DFE$

So, as we wanted to show, $\triangle ABC$ is equiangular with $\triangle DEF$.

Now, suppose $\angle ACB$ and $\angle DFE$ are both not acute:


 * Euclid-VI-7b.png

From the same construction we show similarly that $BC = BG$ and so from Isosceles Triangle has Two Equal Angles, $\angle BCG = \angle BGC$.

But $\angle BCG$ is not less than a right angle.

So neither is $\angle BGC$ less than a right angle.

So in $\triangle BGC$, $\angle BCG + \angle BGC$ are not less than two right angles.

From Two Angles of Triangle Less than Two Right Angles, this is impossible.

So from this contradiction we conclude that:
 * $\angle ABC = \angle DEF$

Hence, similarly to above, $\triangle ABC$ is equiangular with $\triangle DEF$.