Acute triangle $\triangle ABC$ with $AB<AC$, circumcircle $\Gamma$ and circumcenter $O$ is given. Midpoint of side $AB$ is $D$. Point $E$ is chosen on side $AC$ so that $BE=CE$. Circumcircle of triangle $BDE$ intersects $\Gamma$ at point $F$ (different from point $B$). Point $K$ is chosen on line $AO$ satisfying $BK \perp AO$ (points $A$ and $K$ lie in different half-planes with respect to line $BE$). Prove that the intersection of lines $DF$ and $CK$ lies on $\Gamma$.
