Points $A,B,C$are chosen inside the triangle $ A_{1}B_{1}C_{1},$ so that the quadrilaterals $B_{1}CBC_{1}, C_{1}ACA_{1}$ and $A_{1}BAB_{1}$ are inscribed in the circles $\Omega _{A}, \Omega _{B}$ and $\Omega _{C},$ respectively. The circle $Y_{A}$ internally touches the circles $\Omega _{B}, \Omega _{C}$ and externally touches the circle $\Omega _{A}.$ The common interior tangent to the circles $Y_{A}$ and $\Omega _{A}$ intersects the line $BC$ at point $A'.$ Points $B'$ and $C'$ are analogously defined. Prove that points $A',B'$ and $C'$ are lying on the same line.