In parallelogram $ABCD$, angle $A$ is acute. Let $X$ be a point, symmetrical to point $C$ wrt to straight line $AD$, $Y$ is a point symmetrical to the point $C$ wrt point $D$, and $M$ is the intersection point of $AC$ and $BD$. It turned out, that the circumcircles of triangles $BMC$ and $AXY$ are tangent internally. Prove that $AM = AB$.