The problem
Two circles are inscribed in the corner. Points $A$ and $B$ are points of contact of the first circle with the sides of the angle, points $A_{1}$ and $B_{1}$ are points of contact of the second circle with the sides of the angle. The segment $AB_{1}$ intersects these circles at points $C$ and $C_{1}$. Prove that $AC = B_{1}C_{1}$
The figure
*Excuse me for the quality
So far of what might be helpful for the ultimate proof I've managed to prove:
- $ABB_{1}A_{1}$ is an isosceles trapezium.
- $ABB_{1} \sim B_{1}C_{1}A_{1} \sim BCB_{1}$.
I've tried to make a case for proving that $ACA_{1} = B_{1}C_{1}A_{1}$, but at first I need to prove somehow that two of their angles are equal. The problem itself is of the "First similarity property of triangles" matter and I've tried different ways to approach the problem, but I haven't proven what's initially required.