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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?

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    Note that any proof will have to inherently use the simplicity of the polygon, because the theorem isn't true for self-intersecting polygons - imagine taking a heart-shaped (cardiod) loop that wraps twice around the origin and polygonalizing it.2012-12-29

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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.

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I will formulate my answer for general convex sets in vector spaces. Then convex closed polytopes (polygons) on the place accrue as a special case.

If $C$ is a convex set on a vector space $X=\mathbb{R}^n$ (One can easily generalise these facts to infinite-dimensional spaces) and $x\in\partial C$ (the boundary of $C$) then, there is a line $\mathcal{H}=\{x| Hx = K\}$ that supports $C$ at $x$, i.e. for all $x\in C$ it is $Hx\leq K$ and for $x\notin C$ it is $Hx>K$ (The relation $\leq$ is meant as the element-wise order of vectors of $X$.) This supporting hyperplane in the case of $\mathbb{R}^2$ is a line, i.e. its angle is $\pi$.

The angle you mentioned is actually the polar angle of the tangent cone $T_C(x)$ wheneven $x$ is a vertex of $C$. If $x$ is not a vertex $T_C(x)$ is a supporting hyper-plane; in general it is a cone that lies completely in a supporting plane of $C$ at $x$, so its angle is less than or equal to $\pi$. This proves your claim in a more general setting.

Back to your question again... you may also use purely Euclidean arguments (In Euclidean geometry by the way, $A_1A_2\ldots A_n$ is convex if it contains all line segments that can be drawn from its vertices. Assume that $\widehat{A_kA_{k+1} A_{k+2}}>180^\circ$. Then $A_kA_{k+2}$ has all its points outside the polygon.

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    I must say that I am much more interested in the Euclidean proof. The last paragraph you provided is one direction of my first claim. Do you have any idea how to prove the converse? If all interior angles are less than $\pi$ then the polygon is convex.2012-12-15