Let be given two concentric circles of radii $R$ and $R_1 > R$. Let quadrilateral $ABCD$ is inscribed in the smaller circle and let the rays $CD, DA, AB, BC$ meet the larger circle at $A_1, B_1, C_1, D_1$ respectively. Prove that $$ \frac{\sigma(A_1B_1C_1D_1)}{\sigma(ABCD)} \geq \frac{R_1^2}{R^2}$$where $\sigma(P)$ denotes the area of a polygon $P.$