Imagine that the polygon's vertices are labeled ~0~ to ~N-1~ consecutively. Consider a diagonal from vertex ~0~ to some vertex ~i~. Vertices ~1~ through ~i-1~ are on one side of the diagonal, vertices ~i+1~ through ~N-1~ on the other. The diagonal intersects only those diagonals which have one vertex on one side and the other on the other side. Thus there are ~(i-1)(N-1-i)~ diagonals intersecting it.

We count all diagonals with one vertex being vertex ~0~ and multiply the number of diagonals by ~N~ (we could have labeled any vertex ~0~ and would get the same result). We counted each intersection on one diagonal twice (once from each vertex on a single diagonal) and each intersection a total of four times (because it lies on two diagonals). Hence we divide the result by ~4~.