Given a convex quadrilateral $ABCD$. Let $R_a, R_b, R_c$ and $R_d$ be the circumradii of triangles $DAB, ABC, BCD, CDA$. Prove that inequality $R_a < R_b < R_c < R_d$ is equivalent to $180^o - \angle CDB < \angle CAB < \angle CDB$ . (O.Musin)