Can circumscribing a circle around a polygon prove that the sum of the interior angles of an n-sided polygon is $180(n - 2)$? I am trying to create my own proof that the sum of the interior angles in a regular polygon is $180(n - 2)$, where $n$ is the number of sides in the polygon. I have seen these proofs for this formula, and I have also seen this inductive proof for the formula. I'm trying to prove this by circumscribing a circle around a polygon. My question about this is, is this a possible way to prove that the sum of the interior angles of an $n$-gon is $180(n - 2)$? If it is, I have the "first" step completed, which is to circumscribe a circle around a polygon, but I don't know where to go from here. Can anyone help me with the proof if it's possible? Here is an illustration of what I have got so far: ![Proof of sum of interior angles]( ## Notes I have seen this question about proving this, but this deals with an inductive proof and not a geometric proof.

Show that the sum of the angles in a triangle is $180^{\circ}$ (use the fact that an angle inscribed in a circle subtends twice the arc length).

Then, break up an inscribed $n$-gon ($n > 3$) into $n-2$ triangles by picking a vertex and drawing a line segment to each non-adjacent vertex.

The angles of each triangle add up to $180^{\circ}$, and also sum precisely to the angles in the $n-$gon.