Problem 6. Given is a scalene triangle $ABC$. Outside the triangle, isosceles triangles $AB'C$ and $CA'B$ are constructed with bases $AC$ and $BC$ respectively, and angles at the bases equal to $\varphi$. The line through vertex $C$, perpendicular to $A'B'$, intersects the perpendicular bisector of segment $AB$ at point $C_{1}$. Find $\angle AC_{1}B$.
Solution: Let $A_{1}$ and $B_{1}$ be the midpoints of $B C$ and $A C$ respectively, and let the intersection point of $C C_{1}$ and $A^{\prime} B^{\prime}$ be $C^{\prime}$. Then $B^{\prime} C C^{\prime} B_{1}$ and $A^{\prime} C C^{\prime} A_{1}$ are cyclic quadrilaterals, and therefore $\angle B^{\prime} C^{\prime} B_{1} = \angle A^{\prime} C^{\prime} A_{1} = \varphi$. From this, we obtain that $C^{\prime} C_{1}$ is the angle bisector of $\angle A_{1} C^{\prime} B_{1} = 180^{\circ} - 2 \varphi$.

Under a homothety with center $C$ and coefficient $\frac{1}{2}$, the segment $A B$ is mapped to $B_{1} A_{1}$, and the perpendicular bisector of $A B$ is mapped to the perpendicular bisector of $B_{1} A_{1}$. Since the line $C C_{1}$ is mapped to itself, the image of point $C_{1}$ under the homothety is the intersection point $X$ of the perpendicular bisector of $B_{1} A_{1}$ and the line $C C_{1}$. Then the desired angle $\angle A C_{1} B$ is equal to its image, $\angle B_{1} X A_{1}$.

On the other hand, in $\triangle A_{1} C^{\prime} B_{1}$, the intersection point $X$ of the angle bisector $C^{\prime} C_{1}$ and the perpendicular bisector of $A_{1} B_{1}$ lies on the circumcircle of the triangle. Then $\angle B_{1} X A_{1} = 180^{\circ} - \angle B_{1} C^{\prime} A_{1} = 2 \varphi$.