Belarus — 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 $C_1B$ 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
$$\angle ALC = \angle ABC + \angle LAB = \beta + (90^\circ - (\gamma + \beta)/2) = 90^\circ - (\gamma - \beta)/2.$$
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$\text{Since}$ } $\angle$ BZX + $\angle$ ALC = $90^\circ, \text{}$ we have } AL $\perp$ XY.