Polygon Interior Angles Theorem

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.