1. In $\triangle A B C$, $A C>A B, P$ is the intersection of the perpendicular bisector of $B C$ and the internal angle bisector of $\angle A$, draw $P X \perp A B$, intersecting the extension of $A B$ at point $X, P Y \perp A C$ intersecting $A C$ at point $Y, Z$ is the intersection of $X Y$ and $B C$. Find the value of $\frac{B Z}{Z C}$.
1. Connect $B P$ and $C P$. Since $P$ is on the perpendicular bisector of $B C$, we have $B P=C P$.
Since $P$ is on the bisector of $\angle B A C$,
we have $P X=P Y, A X=A Y, \triangle B P X \cong \triangle C P Y$
Therefore, $B X=C Y$
Consider $X Z Y$ as a transversal of $\triangle A B C$, by Menelaus' theorem we have
$$
\begin{array}{l}
\frac{A X}{X B} \cdot \frac{B Z}{Z C} \cdot \frac{C Y}{Y A}=1 \\
\text { Also, } A X=A Y \cdot B X=C Y
\end{array}
$$

Thus $\frac{B Z}{Z C}=1$
Figure for Question 1