Consider a triangle with vertices $A,$ $B,$ and $C,$ and one of the exterior angles corresponding to $\angle C.$

The diagram shows that $\angle C$ and $\angle PCA$ form a linear pair, so the sum of their measures is $180^{\circ }.$ Additionally, by the Triangle Angle Sum Theorem, the sum of the angle measures of $△ABC$ is $180^{\circ }.$ Now $m\angle C$ can be isolated in Equation (I). Next, the expression of $m\angle C$ can be substituted into Equation (II). It has been proven that the measure of $\angle PCA$ is equal to the sum of the measures of $\angle A$ and $\angle B.$ Therefore, it can be said that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.

Consider $△ABC,$ where $D$ and $E$ are the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively. Let $\angle PCA$ be one exterior angle of $△ABC.$

Now, $△ABC$ can be rotated $180^{\circ }$ about $D.$ Since a rotation is a rigid motion, the image of $△ABC$ after the rotation is congruent to $△ABC.$ Corresponding parts of congruent figures are congruent, so the measures of the angles and the lengths of the sides remain unchanged. Since a $180^{\circ }\text{-}$rotation is equivalent to a reflection, $\overline{C^{\prime }A}$ is parallel to $\overline{BC}$ and $\overline{C^{\prime }B}$ is parallel to $\overline{AC}.$ Therefore, $C^{\prime }ACB$ is a parallelogram and $\angle C^{\prime }AC$ is congruent to $\angle CB{C}^{\prime }.$ Now the parallelogram $C^{\prime }ACB$ will be rotated $180^{\circ }$ about $E.$ By the Parallelogram Opposite Angles Theorem, $\angle PCA$ is congruent to $\angle A{B}^{\prime \prime }P.$ Congruent angles have the same measure by the definition. Since $m\angle A{B}^{\prime \prime }P$ is equal to the sum of $m\angle A$ and $m\angle B$ and because of the Transitive Property of Equality, $m\angle PCA$ is equal to the sum of $m\angle A$ and $m\angle B.$ This completes the proof.