First, I know this question might have been asked by several times, see here, for an example. Before someone may want to mark it as dulplicate, I would like to calrify what I want to ask.

Mainly, I am asking for a rigorous proof, or why it is rigorous enough?

Notes: I will be considering simple polygon.

Consider the answer of this post by Misha Lavrov,

What makes things worse is that people often work with polygons on a somewhat intuitive level

I agree with this.

I definitely know that

$$Interior\space\angle\space sum\space of\space a\space N-sided\space polygon=(N-2)180^\circ$$ as every high school text shall states.

Most of the proofs which I have seen about the problem, has a similar idea as the accepted answer of this post.

Main idea of the proof:

For a polygon, we will just select points to join segments, and then we can devide the polygon into several pieces, and by $\angle$ sum of $\triangle$, we can find the $\angle$ sum of the polygon.

This proof is very intuitive, but I don't think it is rigorous enough, as I wonder, can we still connect every vertex to the point, even for a extremely ugly concave polygon, to seperate the polygon, into several $\triangle$s, such that each of the interior $\angle$ of each of the $\triangle$s is in an interior $\angle$ of the polygon and won't be counted twice. For an explaination and example for what I said right above, see below.


For a 'ugly' 23-sided polygon, which I drew 'randomly':enter image description here As in the image above, I discovered a way to divide the polygon into 21 pieces of $\triangle$, while it sounds to be eligible. However, I don't think this will certainly happen, if the polygon is even more ugly. How to explicitly consider the case, when different $\triangle$ share a same interior $\angle$ in the proof?

Also, as I asked above, can we always find a way to devide it properly? (I think it is important to prove it)

I also considered the way in this post.

For the induction part, asumming the $\angle$ sum formula for polygon is true for $N$-sided polygon. Then, consider any $N+1$-sided polygon, I used to think that we can select $2$ vertex which is saperate by one vertex in middle (can I present it more precisely?), join them together, and we form a $\triangle$ (Do we need to consider whether 'convex'enter image description here or 'concave'enter image description here?). Then we form a $N$-sided polygon, and by the induction hypothesis and $\angle$ sum of $\triangle$, we can prove the formula holds for $N+1$.

But, moreover, do we need to consider this caseenter image description here and/or this caseenter image description here, when the remaining polygon might not be $N$-sided?

TO SUM UP, How can we consider all possible cases and make a rigorous proof? Most importantly, I want a justification of the constructability(I mean whether the graph is constructable/valid, not for the Compass-and-straightedge construction)and generality of a graph, if there is a graph in the proof.

Remember, as I said above, I am looking for a rigorous proof, not an usual one.

Thank you so much for your answer :)

NOTES:Does the ways in this help?

  • $\begingroup$ Why not use induction. Then all you have to do is to draw a diagonal to reduce to a polygon with few sides? $\endgroup$
    – almagest
    Feb 19, 2018 at 13:26
  • $\begingroup$ @almagest Can you explain what you think with a rigorous proof as an answer? $\endgroup$
    – Tony Ma
    Feb 19, 2018 at 13:28
  • $\begingroup$ @almagest I had been considering induction also, but I cannot complete an rigorous proof, as stated above in the post. $\endgroup$
    – Tony Ma
    Feb 19, 2018 at 13:29

1 Answer 1


The key fact is that every simple polygon, not necessarily convex, can be decomposed into $n-2$ triangles by drawing $n-3$ diagonals. Then the sum of the interior angles of the polygon is equal to the sum of interior angles of all triangles, which is clearly $(n-2)\pi$.

The existence of triangulations for simple polygons follows by induction once we prove the existence of a diagonal. For a proof, see Chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke. This chapter is freely available. The picture below from that chapter that captures the gist of the proof:

enter image description here

See also Diagonals: Feature Column from the AMS by Malkevitch.

  • $\begingroup$ Adapted from math.stackexchange.com/questions/1103253/…. $\endgroup$
    – lhf
    Feb 19, 2018 at 13:30
  • $\begingroup$ Sorry, I just can't find a link of the freely available chapter... $\endgroup$
    – Tony Ma
    Feb 19, 2018 at 13:39
  • $\begingroup$ @TonyMa, see Resources in the book page. $\endgroup$
    – lhf
    Feb 19, 2018 at 13:40
  • $\begingroup$ oic, thanks, I will read it later. $\endgroup$
    – Tony Ma
    Feb 20, 2018 at 9:59

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