CWMO 2006 Problem 3

In $\triangle PBC$ , $\angle PBC=60^{o}$ , the tangent at point $P$ to the circumcircle $g$ of $\triangle PBC$ intersects with line $CB$ at $A$ . Points $D$ and $E$ lie on the line segment $PA$ and $g$ respectively, satisfying $\angle DBE=90^{o}$ , $PD=PE$ . $BE$ and $PC$ meet at $F$ . It is known that lines $AF,BP,CD$ are concurrent. a) Prove that $BF$ bisect $\angle PBC$ b) Find $\tan \angle PCB$