Circles $\omega_1$ and $\omega_2$ with centers at points $O_1$ and $O_2$ intersect at points $A$ and $B$. A point $C$ is constructed such that $AO_2CO_1$ is a parallelogram. An arbitrary line is drawn through point $A$, which intersects the circles $\omega_1$ and $\omega_2$ for the second time at points $X$ and $Y$, respectively. Prove that $CX = CY$. (Oleksii Masalitin)