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


I have seen this question about proving this, but this deals with an inductive proof and not a geometric proof.

  • 1
    $\begingroup$ What kinds of polygon? Most polygons cannot be inscribed in a circle. $\endgroup$ Commented Jun 9, 2016 at 18:50
  • $\begingroup$ Any regular polygon. Will be edited. $\endgroup$ Commented Jun 9, 2016 at 18:53
  • $\begingroup$ That should be any regular convex polygon, to rule out regular stars. $\endgroup$
    – David K
    Commented Jun 9, 2016 at 18:55
  • $\begingroup$ It seems like your approach will end up being a variation of the Method 5 (spider theorem) at your link, or at least using many of the same ideas. $\endgroup$
    – Joffan
    Commented Jun 9, 2016 at 18:56
  • $\begingroup$ For a convex $n$-gon take an interior point $P$ and join it to each vertex. The $n$ triangles have total angles of $180n$ deg. $360$ deg is accounted for by the angles at $P$, leaving $180(n-2)$ for the angles of the $n$-gon. $\endgroup$
    – almagest
    Commented Jun 9, 2016 at 18:59

1 Answer 1


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.

  • $\begingroup$ I like your intution, but will circumscribing a circle around an $n$-gon help with my proof? $\endgroup$ Commented Jun 9, 2016 at 21:22
  • $\begingroup$ Are you requiring all vertices to lie on the circle? You can have a convex polygon for which you can't pass a single circle through all of the points. If you're not, then no, it doesn't directly help to circumscribe a circle, but that's not a shortcoming of the proof. It just means that it's not necessary to do so. But, if all of the vertices lie on the circle, then all of the triangles are inscribed as well, so it's "clean" to do it that way in a sense. $\endgroup$
    – John
    Commented Jun 9, 2016 at 21:45
  • $\begingroup$ I am circumscribing the circle to where the vertices do line on the circle, so that should answer my question. Thanks! $\endgroup$ Commented Jun 10, 2016 at 0:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .