FINAL ROUND

Let $\angle ABC=\beta$ and $\angle BCA=\gamma$. Then $\angle CAB=180^{\circ }-\beta -\gamma$. By $\xi$ denote the degree measure of the arc $C_{1}B$ that does not contain the vertex $C$, by $\eta$ denote the degree measure of the arc $\tilde{C}{B}_1$ that does not contain the vertex $B$. Then $\gamma =(\xi +\eta )/2$ and $\beta =(\eta +\zeta )/2$. Let $Z$ be the point of intersection of the lines $XY$ and $BC$. Let the figure shows the positions of $X$, $Y$, and $Z$. The angle $BZX$ is the external angle for the triangle $ZXC$, so $\angle BZX=\angle ZCX+\angle CXZ$. We have $$\angle ZCX=\frac{1}{2}\left(\xi +\eta -\frac{1}{2}(360^{\circ }-\zeta )\right)$$ and $$\angle CXZ=\frac{1}{2}\left(\frac{1}{2}(360^{\circ }-\xi )-(\eta +\zeta )\right).$$

Therefore, $$\angle BZX=\angle ZCX+\angle CXZ=(\xi -\zeta )/4=((\xi +\eta )-(\zeta +\eta ))/4=(\gamma -\beta )/2.$$ Let the bisector of the angle $CAB$ meet the side $BC$ at $L$. The angle $ALC$ is the external angle for the triangle $LBA$, so $$\angle ALC=\angle ABC+\angle LAB=\beta +(90^{\circ }-(\gamma +\beta )/2)=90^{\circ }-(\gamma -\beta )/2.$$ $$ \text{Since } \angle BZX + \angle ALC = 90^\circ, \text{ we have } AL \perp XY.