Let $ABC$ denote a triangle. The point $X$ lies on the extension of $AC$ beyond $A$, such that $AX = AB$. Similarly, the point $Y$ lies on the extension of $BC$ beyond $B$ such that $BY = AB$. Prove that the circumcircles of $ACY$ and $BCX$ intersect a second time in a point different from $C$ that lies on the bisector of the angle $\angle BCA$. (Theresia Eisenkölbl)