All angles of a cyclic pentagon $ABCDE$ are obtuse. The sidelines $AB$ and $CD$ meet at point $E_1$, the sidelines $BC$ and $DE$ meet at point $A_1$. The tangent at $B$ to the circumcircle of the triangle $BE_1C$ meets the circumcircle $\omega$ of the pentagon for the second time at point $B_1$. The tangent at $D$ to the circumcircle of the triangle $DA_1C$ meets $\omega$ for the second time at point $D_1$. Prove that $B_1D_1 // AE$