The formula for the total degrees of the interior angles of a regular n-gon is (n-2)180.

To prove this, draw line segments from the center of the n-gon to each vertex, creating n isosceles triangles. What is the degree measure of the vertex angle of each isosceles triangle? 360/n, obviously, since there are 360 degrees divided up into n angles. And since the degrees of a triangle sum to 180 (by Euclid's 5th postulate), the other two angles of the triangle must sum to 180-(180/n). But remember that in an isosceles triangle, the base angles are congruent. So each base angle of each triangle must be (180-(360/n))/2. How many of these angles are there? 2n, since there are 2 angles per triangle and n triangles in the n-gon. So the sum of these angles is ((180-(360/n))/2)*2n, which multiplies out to 180n-360. Factor out the 180, and we get 180(n-2). Q.E.D.

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