Question 52
Let $\triangle ABC$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C$. The angle bisector of $\angle ABC$ intersects $CD$ at $E$ and meets the circumcircle $\omega$ of triangle $\triangle ADE$ again at $F$. If $\angle ADF = 45^\circ$, show that $CF$ is tangent to $\omega$.