An outerplanar graph is a graph that can be drawn as a planar graph where every vertex is incident to the outer region. A maximal outerplanar graph can be drawn such that every vertex is part of a simple cycle where every vertex is part of the outer region and every region inside the cycle is a triangle.
I have to show with proof by induction that for all $n\geq 2$ A graph with that drawing has exactly $2n-3$ edges.
The base case is easy: $n=2$: $v$---$u$ is the only graph with that drawing. $1=2\cdot2 -3 =1$ holds.
The induction hypothesis is stated above.
I can wrap my head around the induction step. I've seen that many induction proofs on graphs deconstruct a graph, use the induction hypothesis on the smaller graph and rebuild the graph to proof that the hypothesis holds. Here I don't see how I can do it. I thought about using rules such as that every new vertex needs to be on the circle and every inner region has to be a triangle which leads to the only possibility of adding 2 edges but I don't use the hypothesis there so it's no proof by induction right?