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Can we prove that for any simple polygon with more than 3 vertices there always exists a diagonal which:

  • is inside the polygon
  • doesn't intersect with any edges
  • splits the polygon in two polygons in such a way that the difference between their vertex counts is smaller than 2 (e.g. splits a polygon with 29 vertices into polygons with 15 and 16 vertices)?
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I think the following is a counterexample on six vertices: Start with an equilateral triangle $ABC$ and a larger equilateral triangle $A'B'C'$ with the same center and parallel edges. Now consider the hexagon $AA'BB'CC'$. The only candidates for bisecting diagonals (since we have an even number of vertices) are $AB'$, $BC'$, $CA'$, all of which intersect other edges.

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