Proving that a closed simple polygon in the plane is convex if and only if all the interior angles measure less than or equal to $\pi$ radians 
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?

 A: 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.
A: 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.
