China Western Mathematical Olympiad

By the angle bisector theorem, we have
$$\frac{PF}{FC} \cdot \frac{CB}{BA} \cdot \frac{AD}{DP} = \frac{PB}{BC} \cdot \frac{BC}{BA} \cdot \frac{AB}{PB} = 1.$$
By the converse of Ceva theorem, the lines $AF$, $BP$ and $CD$ are concurrent.
Suppose there exists $\angle D'BF'$ satisfying the conditions: (a) $\angle D'BF' = 90^\circ$, (b) the lines $AF'$, $BP$ and $CD'$ are concurrent. We may assume that $F'$ lies on $PF$. Then, $D'$ is on $AD$.
So
$$\frac{PF'}{F'C} < \frac{PF}{FC}, \quad \frac{AD'}{PD'} < \frac{AD}{PD}.$$
Thus
$$\frac{PF'}{F'C} \cdot \frac{CB}{BA} \cdot \frac{AD'}{D'P} < \frac{PB}{BC} \cdot \frac{BC}{BA} \cdot \frac{AB}{PB} = 1,$$
which leads to a contradiction. This completes the proof.

(2) We may assume that the circle $O$ has radius 1. Let

$\angle PCB = \alpha$. By (1), $\angle PBE = \angle EBC = 30^\circ$. Therefore, $E$ is the midpoint of $\overline{PC}$.
Since $\angle MPE = \angle PBE = 30^\circ$, $\angle CPE = \angle CBE = 30^\circ$ and $PD = PE$, we obtain $\angle PDE = \angle PED = 15^\circ$, $PE = 2 \cdot 1 \cdot \sin 30^\circ$ and $DE = 2\cos 15^\circ$.

Since
$$BE = 2\sin \angle ECB = 2\sin(\alpha + 30^\circ)$$
and $\angle BED = \angle BEP - 15^\circ$, we have
$$\cos(\alpha - 15^\circ) = \frac{BE}{DE} = \frac{2\sin(\alpha + 30^\circ)}{2\cos 15^\circ},$$
$$\cos(\alpha - 15^\circ)\cos 15^\circ = \sin(\alpha + 30^\circ),$$
$$\cos \alpha + \cos(\alpha - 30^\circ) = 2\sin(\alpha + 30^\circ),$$
$$\cos \alpha + \cos \alpha \cos 30^\circ + \sin \alpha \sin 30^\circ = \sqrt{3} \sin \alpha + \cos \alpha,$$
$$1 + \frac{\sqrt{3}}{2} + \frac{1}{2} \tan \alpha = \sqrt{3} \tan \alpha + 1.$$
So
$$\tan \alpha = \frac{6 + \sqrt{3}}{11}.$$