SHARYGIN 2018 Problem 5

The vertex $C$ of equilateral triangles $ABC$ and $CDE$ lies on the segment $AE$, and the vertices $B$ and $D$ lie on the same side with respect to this segment. The circumcircles of these triangles centered at $O_1$ and $O_2$ meet for the second time at point $F$. The lines $O_1O_2$ and $AD$ meet at point $K$. Prove that $AK = BF$.