In the proof, a pentagon will be considered. However, the proof is valid for any polygon.

For any pentagon, two non-intersecting diagonals can be drawn to divide the pentagon into three triangles. In the case of an arbitrary polygon with $n$ vertices, $n-3$ non-intersecting diagonals can be drawn to divide the polygon into $n-2$ triangles.

Let $P$ be the sum of the measures of the angles of $ABCDE,$ and $S_1,$ $S_2,$ and $S_3$ be the sums of the measures of angles of $△BCD,$ $△BDE,$ and $△BEA,$ respectively. It can be noted that the sum of angle measures of the pentagon is equal to the total of the sums of angle measures of the three triangles. By the Triangle Angle Sum Theorem, the sum of angle measures of each triangle is equal to $180^{\circ }.$ Then, $180^{\circ }$ can be substituted for $S_1,$ $S_2,$ and $S_3.$ This means that $P$ is equal to $3\cdot 180^{\circ },$ or $(5-2)\cdot 180^{\circ }.$ In the case of an arbitrary polygon with $n$ vertices, there are $n-2$ triangles, so the sum of the angle measures of the polygon is $(n-2)\cdot 180^{\circ }.$ This proves the theorem.