In the acute triangle $ABC$, $\angle BAC$ is less than $\angle ACB $. Let $AD$ be a diameter of $\omega$, the circle circumscribed to said triangle. Let $E$ be the point of intersection of the ray $AC$ and the tangent to $\omega$ passing through $B$. The perpendicular to $AD$ that passes through $E$ intersects the circle circumscribed to the triangle $BCE$, again, at the point $F$. Show that $CD$ is an angle bisector of $\angle BCF$.