Belarus — Final Round

Let $\omega$ be the circle passing through $A$ and touching the side $BC$ at $X$. Since the angle $YAX$ subtends the arc $YX$, we have $\angle YAX = \frac{1}{2} \angle YX$.

Further, $\angle YXC = \frac{1}{2} \angle YX$ (as the angle between the tangent $BC$ and the chord $XY$). So, $\angle YAX = \angle YXC$. Let $W$ be the intersection point of the ray $AX$ and $\Omega$. Then $\angle YAX = \angle YAW$, so $\angle YAW = \angle YXC$. We have $\angle BXZ = \angle YXC$ as the vertical angles, so $\angle BXZ = \angle YAW$. Further, $\angle YAW = \angle YZW$ as inscribed angles of $\Omega$ subtending the same arc. Therefore, $\angle BXZ = \angle YZW$, so $BC \parallel WZ$. Hence $CW = ZB$, and then $\angle CAW = \angle ZAB$ as inscribed angles of $\Omega$ subtending the equal arcs.