# Proof of $\angle$ sum of polygon.

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.

e.g.)

For a 'ugly' 23-sided polygon, which I drew 'randomly': 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' or 'concave'?). 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 case and/or this case, 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.

NOTES:Does the ways in this help?

• Why not use induction. Then all you have to do is to draw a diagonal to reduce to a polygon with few sides? – almagest Feb 19 '18 at 13:26
• @almagest Can you explain what you think with a rigorous proof as an answer? – Tony Ma Feb 19 '18 at 13:28
• @almagest I had been considering induction also, but I cannot complete an rigorous proof, as stated above in the post. – Tony Ma Feb 19 '18 at 13:29

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 $(n-2)\pi$.
• @TonyMa, see Resources in the book page. – lhf Feb 19 '18 at 13:40