Polygon not Equal to Triangle has Chord

Theorem

Let $P$ be a polygon that is not a triangle.

Then $P$ has a chord that lies completely in the interior of $P$.

Let $A, B, C$ be vertices of $P$ such that $\angle BAC$ is a convex internal angle of the vertex $A$.

Let $BC$ be the subtend of $\angle BAC$.

Suppose $BC$ is equal to a side of $P$.

Then $P = \triangle ABC$, which would contradict our assumption that $P$ is not a triangle.

Suppose $BC$ is a chord of $P$ that lies completely in the interior of $P$.

Then the proof is done.

Suppose $BC$ does not lie completely in the interior of $P$.

Then either a side of $P$ intersects $BC$, or a vertex of $P$ lies on $BC$.

In either case, one or more vertices of $P$ lie inside $\triangle ABC$, or on $BC$.

Let $\mathcal V$ be the set of those vertices.

Define $D$ as the vertex in $\mathcal V$ that lies furthest away from $BC$.

If more than one vertex in $\mathcal V$ have the same maximal distance from $BC$, pick $D$ among them arbitrarily.

Now $AD$ is a chord of $P$ that lies completely in the interior.

As any side of $P$ that would intersect $AD$ would either intersect $AB$ or $AC$, or end in a vertex that lie further away from $BC$ than $D$ does.

In all of these cases, we have a contradiction.

$\blacksquare$