Consider a closed simple polygon in the plane. It is intuitively obvious that the polygon is convex if and only if all the interior angles measure less than or equal to $\pi$ radians.
I have never seen a rigorous proof of this fact and I was wondering if anyone could provide such a proof.
A related question:
Given a concave polygon (or more generally a higher dimensional polytope), how can we prove that there will always be two vertices of the polygon which cannot be joined by a line lying entirely inside the polygon?
Best Answer
Suppose a concave polyhedron exists so that all vertex connections produce lines entirely inside the polyhedron.
Since the object is concave, there are points on two faces with a connecting segment outside of the polyhedron. Now consider the hull of the points for just these two faces. The connecting segment must be inside this hull. Contradiction.