SHARYGIN 2011 Problem 97

Circles $\omega$ and $\Omega$ are inscribed into the same angle. Line $\ell$ meets the sides of angles, $\omega$ and $\Omega$ in points $A$ and $F, B$ and $C, D$ and $E$ respectively (the order of points on the line is $A,B,C,D,E, F$ ). It is known that $BC = DE$ . Prove that $AB = EF$ .