jmc

Let the polygon have $n$ sides. The number of diagonals then is $n(n-3)/2$, because each vertex is connected to $n-3$ other vertices by diagonals, but $n(n-3)$ counts each diagonal twice. We then have $$n=\frac{n(n-3)}{2}⟹1=\frac{n-3}{2}⟹n=\boxed{5}$$