Given triangle $ABC$. Point $B_1$ is marked on line $AC$ so that $AB = AB_1$, while $B_1$ and $C$ are on the same side of $A$. Through points $C$, $B_1$ and the foot of the bisector of angle $A$ of triangle $ABC$, a circle $\omega$ is drawn, intersecting for second time the circle circumscribed around triangle $ABC$, at point $Q$. Prove that the tangent drawn to $\omega$ at point $Q$ is parallel to $AC$.