Show that, in every polygon with more than three edges, there must be two vertices A,B (not connected by any edge) such that the segment AB lies in the interior of the polygon and meets no edge of the plygon (except A and B).
The proof may be complex enough, but if we accept the triangulation polygon as a lemma or something, then the proof is done. Because select one triangle of the set of triangles, the triangle is inside the polygon, and its edges inside or belong to the polygon. Then there at least one side is not the edge of polygon, and this side of the triangle is what the proposition says. Read the wikipedia for the topic polygon triangulation to generate a proof.
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