Construct the convex hull CH(P) where P is a non-convex simple polygon. Since P is non-convex there must exist edges on the boundary bd(CH(P)) that are not edges of P. Each such edge forms the "lid" of a "pocket" of CH(P). We shall prove that in fact every such pocket yields a mouth. Let K_ij denote the pocket of CH(P) determined by vertices p_i and p_j of P. K_ij = {p_i, p_i+1, ..., p_j} U {p_j, p_i} forms itself a simple polygon. By the Two-Ears Theorem K_ij must have two ears and since they are non-overlapping they cannot both occur at p_i and p_j. Therefore, at least one ear must occur at p_k for i < k < j. Obviously such an ear for K_ij is a mouth for P.

Q.E.D.