Euclid's Foundations: The Exterior Angle Inequality Theorem

Theorem. The measure of an exterior angle of a triangle is greater than the measure of either opposite interior angle.

Proof.

Given $\triangle ABC$, extend side $\overline{BC}$ to ray $\overrightarrow{BC}$ and choose a point $D$ on this ray so that $C$ is between $B$ and $D$.

Then, $\angle AMB$ and $\angle CME$ are congruent vertical angles and $\triangle AMB \cong \triangle CME$ by SAS congruence. Consequently, $\angle MCE = \angle MAB = \angle A$, by the fact that corresponding parts of congruent triangles are congruent.

Now, $E$ lies in the half-plane of $A$ and $\overleftrightarrow{CD}$ since $A$ and $E$ are on the same side of $\overline{CD}$. Also, $E$ lies in the half-plane of $D$ and $\overline{AC}$ since $D$ and $E$ are on the same side of $\overline{AC}$. Therefore, $E$ lies in the intersection of these two half-planes, which is the interior of $\angle ACD$. Finally, $\angle ACD = \angle ACE + \angle ECD > \angle ACE = \angle A$.

The proof that $\angle ACD > \angle B$ is similar and left to the reader.

Proof attributed from Millersville.edu