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$(n−2)180_{∘}$

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.$P=S_{1}+S_{2}+S_{3} $

By the Triangle Angle Sum Theorem, the sum of angle measures of each triangle is equal to $180_{∘}.$ Then, $180_{∘}$ can be substituted for $S_{1},$ $S_{2},$ and $S_{3}.$
$P=S_{1}+S_{2}+S_{3}⇕P=180_{∘}+180_{∘}+180_{∘} $

This means that $P$ is equal to $3⋅180_{∘},$ or $(5−2)⋅180_{∘}.$ 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 $(n−2)⋅180_{∘}.$ This proves the theorem. $P=(n−2)180_{∘}$