**Proof: **The key idea of the proof goes by the method of
mathematical
induction. Let n = the number vertices = the number of sides in the
polygon.

The induction starts by considering n=3. Then we assume that we have justified the result for polygons with k vertices where k<n and justify the result for n based on this assumption. This allows us to use "induction" to prove the result for any value of n. Starting with n=3, we have the result is true for n= 4, and then n=5, and then n=6,.... and we can continue to any specific value of n.

The key idea of the proof goes by induction on the number n = the number vertices = the number of sides in the polygon, as follows:

Suppose n> 3 and that for any polygon with k vertices/ sides, where k<n, the polygon can be triangulated.

Now proceed to consider the vertices, v1, v2, ..., vn ordered so that vi is adjacent to vi+1 and vn is adjacent to v1.

Take a ray from v1 and rotate it from v1v2 so that it intersects the inside of the polygon. continue to rotate until it meets another vertex.

Case 1. This vertex is v3. Then consider the polygonal region Q1 = v1v3...vn which has n-1 vertices.

By induction Q1 can be triangulated, so the original polygon is triangulated using the triangulation of Q1 and the triangle v1v2v3.

By induction Q2 can be triangulated, so the original polygon is triangulated using the triangulation of Q2 and the triangle v1vnvn-1.

By induction Q3 and Q4 can be triangulated, so the original polygon is triangulated using the triangulations of Q3 and Q4.

For more discussion of proofs of this proposition see Triangulations and arrangements, Two lectures by Godfried Toussaint, transcribed by Laura Anderson and Peter Yamamoto.