A convex quadrilateral $ABCD$ is given. $\omega_1$ is a circle with diameter $BC$, $\omega_2$ is a circle with diameter $AD$. $AC$ meets $\omega_1$ and $\omega_2$ for the second time at $B_1$ and $D_1$. $BD$ meets $\omega_1$ and $\omega_2$ for the second time at $C_1$ and $A_1$. $AA_1$ meets $DD_1$ at $X$, $BB_1$ meets $CC_1$ at $Y$. $\omega_1$ intersects $\omega_2$ at $P$ and $Q$. $XY$ meets $PQ$ at $N$. Prove that $XN=NY$.