Strong induction proof with polygon How can we show that if a simple polygon with at least four sides is triangu-lated, then at least two of the triangles in the triangulation have two sides that border the exterior of the polygon using strong induction
 A: Proposition. If $n\ge 3$ and $AB$ is an edge of a triangulated $n$gon then there exists a triangle with two sides bordering the exterior of the polygon and $AB$ is not among these two sides.
Proof:
For $n=3$ this is clear.
For $n>3$ note that removing the triangle $ABP$ having $AB$ as edge produces two smallerpolygons, a $k$gon and an $(n+1-k)$gon. At most one of these is degenerate (i.e. either $k\ge 3$ or $n+1-k\ge3$ because $k+(n+1-k)=n+1\ge 5$). By assumption, this smaller polygon has a triangle with two edges on the boundary of the smaller polygon and the edge $BP$ resp. $PA$ is not among these two edges. Then this triangle also works for the original $n$gon. $_\square$
Now the original problem is solved as follows: Pick any edge $AB$ to find a first triangle $XYZ$ having two boundary edges $XY$ and $YZ$. Now pick $XY$ instead of $AB$ to find a second triangle $UVW$ having two boundary edges different from $XY$. Since $n>3$, the triangles cannot be the same (as not all three edges can be boundary edges).
A: Consider a triangulation and its dual. The dual is a tree. Every tree with at least two nodes has at least two leaves (= vertices of degree 1). These leaves correspond to the triangles you seek.
