19. In the triangle $A B C, A B=A C, \angle A=90^{\circ}, D$ is the midpoint of $B C, E$ is the midpoint of $A C$ and $F$ is a point on $A B$ such that $B E$ intersects $C F$ at $P$ and $B, D, P, F$ lie on a circle. Let $A D$ intersect $C P$ at $H$. Given $A P=\sqrt{5}+2$, find the length of $P H$.
19. Answer: 1
Solution.

Join $P D$. Then $\angle D P C=\angle F B D=45^{\circ}=\angle D A C$ so that $D, P, A, C$ are concyclic. Thus $\angle A P C=\angle A D C=90^{\circ}$. It follows that $E A=E P=E D=E C$. Let $\angle P A H=\theta$. Then $\angle P C D=\theta$. Thus $\angle E P C=\angle E C P=45^{\circ}-\theta$ so that $\angle A E B=90^{\circ}-2 \theta$. That is $\angle A B E=2 \theta$. Thus $\tan 2 \theta=A E / A B=1 / 2$. From this, we get $\tan \theta=\sqrt{5}-2$. Therefore, $P H=A P \tan \theta=(\sqrt{5}+2)(\sqrt{5}-2)=1$