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.