I had this question asked a few weeks ago and gave an argument that involved finding an ear and clipping it, professor said it was not quite the correct answer and that I lacked some insights.
How would you prove this? I'm not allowed to use the dual graph of a triangulation, because given that any polygon has at least 2 ears, then we can triangulate the polygon by ear clipping.
So how can I prove that every polygon has at least 2 ears?
We can prove that any $P$ polygon with $n \geq 4$ vertices has a diagonal in its interior.
Let $x$ be a convex vertex, so the inner angle given by its two neighbours $y$ and $z$ is less than $\pi$.
If $[y,z] \in int(P)$ than we have found such a diagonal, if there are any edges intersects $[y,z]$ than we have a vertex $v$ in the interior of the $xyz$ triangle that is furthest from the line given by $yz$, (since there are finite vertices) means that $xv$ is a diagonal in $P$'s interior.
We have used that every simple polygon has a convex point; this can be proved by induction. For $n = 3$ we have a convex point.
Assuming we have one for $n=k$
Let's see the $n=k+1$ case.
If we replace an edge with a vertex and two edges it is possible we have messed up the convex point $x$ in our polygon with $k$ vertices.
So our original convex point $x$ now not convex. Assume its original neighbour $y$ is also not convex.
If our new point $z$ is not convex, the $xyz$ triangle lies in the interior of the polygon and has at least one nonconvex point and that is a contradiction.
So from here we can find a diagonal in $P$'s interior and it divides our polygon into two other polygons with at least $3$ vertices. From here it is easy to show as we continue the search for the inner diagonals in the new polygons.
Note that a polygon with $4$ verices has at least $2$ ears.