Points $X$ , $Y$, $Z$ lie on a line $k$ in this order. Let $\omega_1$, $\omega_2$, $\omega_3$ be three circles of diameters $XZ$, $XY$ , $YZ$ , respectively. Line $\ell$ passing through point $Y$ intersects $\omega_1$ at points $A$ and $D$, $\omega_2$ at $B$ and $\omega_3$ at $C$ in such manner that points $A, B, Y, X, D$ lie on $\ell$ in this order. Prove that $AB =CD$.