The circle $\gamma$ passing through the vertex $A$ of triangle $ABC$ intersects its sides $AB$ and $AC$ for the second time at points $X$ and $Y$, respectively. Also, the circle $\gamma$ intersects side $BC$ at points $D$ and $E$ so that $AD = AE$. Prove that the points $B, X, Y, C$ lie on the same circle. Proposed by Mykhailo Shtandenko