Proof that every polygon with an inscribed circle is convex? In many elementary (and not-so-elementary) Euclidean geometry texts, a (simple) polygon is said to be tangential&nbsp; if it is convex and has an inscribed circle (i.e., a circle that intersects and is tangent to each side of the polygon).  The assumption of convexity is not needed: I've come up with a rather laborious proof that every polygon with an inscribed circle is convex.  But I'd like to find either a simple elementary proof or a reference to a proof in the literature.  (By "elementary," I mean using only standard facts of axiomatic Euclidean geometry.)  

Does anyone know of a reference for a proof of this fact (elementary or not)?  Or can anyone think of a straightforward elementary proof?  You can use any definition of "convex polygon" that you like, but the easiest one to work with is that for each edge, the vertices not on that edge lie on one side of the line through that edge.

(Interestingly, the corresponding fact for circumscribed circles--i.e., that every polygon with a circumscribed circle is convex--is quite easy to prove: If P has a circumscribed circle, any two nonadjacent sides of P are non-intersecting chords of the circle; it is easy to show that both endpoints of each chord lie on the same side of the line through the other, and from there it is an easy matter to prove that P is convex.)
Thanks to everyone who suggested approaches to this problem. In the end, none of the suggested approaches fit into the axiomatic framework that I was working in, so I had to write up my own rather laborious proof. It's a bit long to post here, but in order to close this question, I just want to post the reference and a quick summary of the approach. 

You can find the complete proof in my textbook Axiomatic Geometry (Theorem 14.31). The basic idea is first to prove the following lemma:


  Lemma. Let $\mathscr P$ be a polygon circumscribed about a circle $\mathscr C$. Suppose $A$ is any vertex of $\mathscr P$, and $E$ and $F$ are the points of tangency of the two edges containing $A$. Then there are no points of $\mathscr P$ in the interior of $\triangle AEF$.


To prove that a tangential polygon $\mathscr P$ must be convex, the basic idea is to show that if $\ell$ is any edge line of $\mathscr P$, then there can't be any vertices of $\mathscr P$ on the "wrong" side of $\ell$ (the side not containing the inscribed circle), because that would violate the lemma.
Here is the sketch of a much shorter proof.

If a (simple) polygon is tangential, then it is convex (all its angles are each less than two right angles). Let the polygon be $n$-sided. Then draw radii from the centre of the circle to the $n$ points of tangency (by hypothesis). This splits the $n$-gon into $n$ quadrilaterals which are kites (that they are kites follows from the $RHS$ congruency test, after having connected each vertex of the $n$-gon to the centre of the circle). Each of these kites has a vertex of the $n$-gon as a vertex and (by a well-known result of Euclidean Geometry about tangent straight lines to circles and radii thereof) two right angles. The last angle of each kite is formed at the centre of the circle by two radii. Since the sum of angles in a quadrilateral is four right angles, it is clear that the two non-right angles in each quadrilateral must be supplementary, so that either one of them is always less than two right angles. The result follows.