67th Czech and Slovak Mathematical Olympiad

Reflect $k$ and $C$ about $PQ$ to get $l$ and $C^{\prime }$, respectively (Fig. 1). Then $C^{\prime }$ lies on $l$ and since $\angle QA{C}^{\prime }=\angle QAC=\angle PAB$ it also lies on $BA$. Triangle $C^{\prime }CA$ is isosceles, hence $$\angle BAC=\angle A{C}^{\prime }C+\angle AC{C}^{\prime }=2\cdot \angle B{C}^{\prime }C$$ The chord $BC$ of circle $k\cup l$ has a fixed length, hence the corresponding inscribed angle $B{C}^{\prime }C$ has fixed size and we may conclude.

Fig. 1

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Alternative solution.

Let $O$ be the midpoint of $PQ$. We will show that $O$ lies on the circumcircle of triangle $ABC$ (Fig. 2). This will imply that $\angle BAC=\angle BOC$ which is clearly fixed. Observe that $O$ lies on the perpendicular bisector of $BC$. Moreover, if $O\ne A$ then $AO$ is the external $A$-angle bisector with respect to triangle $ABC$. Therefore $O$ is the midpoint of arc $BAC$.

Fig. 2