Suppose that we have a polygon $P$ whose vertices are $z_1, z_2,..., z_n$ inscribed in a circle $C$ in the Cartesian coordinate system. Furthermore, suppose that along the perimeter of the circle, $z_1$ is adjacent to $z_2$, $z_2$ is adjacent to $z_3$, ..., and $z_n$ is adjacent to $z_1$ (but it may happen that for instance the line segment determined by $z_1$ and $z_2$ is not an edge of $P$). We must prove that indeed every line segment determined by each pair of adjacent vertices is an edge which will prove the convexity of $P$ (which follows by using the cartesian equation of $P$). Toward a contradiction suppose that for example the line segment determined by $z_1$ and $z_2$ is not an edge of $P$. For some $i$ and $j$ (which may be the same) different from 1 and 2 we have that the line segment $l_1$ determined by $z_1$ and $z_i$, and the line segment $l_2$ determined by $z_2$ and $z_j$ are edges of $P$. $l_1$ and $l_2$ cannot “cross eachother” (this notion can be made more precise by using the polar coordinates of the vertices) since that would contradict the hypothesis that $P$ is a polygon. Accordingly, $the$ other edge $e_1$ incident on $z_1$, and the other edge $e_2$ incident on $z_2$ cannot cross eachother. Since $P$ is a polygon, there must be a chain of edges from $z_1$ to $z_2$. But any edge chain from $z_1$ to $z_2$ either leads to two edges crossing each other, or leads to one vertex having three incident edges, contradicting the fact that $P$ is a polygon.
Is this proof correct?
(I had posted this proof for Proof that every polygon with an inscribed circle is convex? by mistake as that question addresses a polygon with an $inscribed$ circle. If this coordinate proof is correct (or not), I think one can provide a similar coordinate proof for that question as well.)