By prolonging its sides, one can partition an arbitrary polygon into convex polygons

Question

It is straightforward to prove by induction on the number of sides that a convex n-sided polygon can be partitioned into n-2 triangles. My question here is whether there is an analogous proof that an arbitrary polygon can be partitioned into convex polygons by prolonging its sides (and if so, whether there is a known function of n that determines the number of convex polygons in the partition). Intuitively, this seems correct, and the result is immediate for n=3, but I am having some difficulty extending the argument to the case n>3. Any help would be much appreciated.