KöMaL 2023 B5315

Consider a triangle $ABC$. Let $B'$ be a point on the extension of side $AB$ beyond $B$, and let $C'$ be the point on the extension of side $AC$ beyond $C$ such that $BB'=CC'$. Let $k$ and $k'$ denote the circumscribed circles of triangles $ABC'$ and $AB'C$, respectively. Prove that the common chord of $k$ and $k'$ lies on the angle bisector drawn from $A$.