On the side $AB$ of a triangle $ABC$, two points $X, Y$ are chosen so that $AX = BY$. Lines $CX$ and $CY$ meet the circumcircle of the triangle, for the second time, at points $U$ and $V$. Prove that all lines $UV$ (for all $X, Y$, given $A, B, C$) have a common point.
Problem
Source: 2007 Sharygin Geometry Olympiad Correspondence Round P13
Tags: geometry, Fixed point, circumcircle, equal segments