A fixed circle $\omega$ is inscribed into an angle with vertex $C$. An arbitrary circle passing through $C$, touches $\omega$ externally and meets the sides of the angle at points $A$ and $B$. Prove that the perimeters of all triangles $ABC$ are equal.
Let $\Gamma$ be the $C$-excircle in $\triangle ABC$. It is sufficient to prove $\Gamma$ stays fixed. Consider inversion at $C$ fixing $\omega$. Then $\omega$ is the incircle of $\triangle CA'B'$ while $\Gamma'$ is the $C$-mixtilinear incircle. Now observe that the touch-chord of $\Gamma'$ passes through the center of $\omega$ so $\Gamma'$ does not depend on the choice of line $\overline{A'B'}$ tangent to $\omega$.