We define a 2D polygon to be a simple (ie. non-self-intersecting) closed path consisting of line segments. To be clear, only consecutive segments intersect and only at an endpoint. Moreover, we do not include polygons such that consecutive sides lie on the same line. A convex polygon is defined as a polygon that is a convex set (ie. if we define the interior of the polygon to include the boundary, the segment formed by joining any two points in the interior lies in the interior).
It is easy to prove that the following two conditions are necessary for convexity, but how can we prove that they are sufficient as well?
- Every diagonal (ie. segment created by joining two vertices) lies in the interior of the polygon.
- Every interior angle is strictly less than $180^{\circ}$.
It’s clear that 1 implies 2, so let’s prove that 2 implies the polygon is convex.
Given a polygon $P$ with every interior angle less than $180^\circ$, suppose to the contrary that part of a segment between interior points goes outside $P$. Let $AB$ be one of these outside parts with $A, B$ on the boundary of $P$. Then $AB$ divides the exterior of $P$ into two regions, one of which is a finite polygon $Q$.
Consider the vertex of $Q$ farthest from $AB$. It’s not $A$ or $B$, so its interior angle is an exterior angle of $P$ and must be greater than $180^\circ$. But that means we must be able travel in some direction interior to $Q$ that strictly increases the distance from $AB$. And for any interior point, there must be another vertex of $Q$ at least as far as that from $AB$. This contradicts our choice of the farthest vertex.
Therefore, there is no such segment and $P$ is convex.