Example Mountain (2003 China National Training Team) 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 $\angle A$, construct $P X \perp A B$, 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 $\frac{B Z}{Z C}$.
Consider the circumcircle $\odot O$ of $\triangle ABC$, and let the midpoint of arc $BC$ (not containing $A$) be $D$, and the midpoint of chord $BC$ be $M$.
Since $\overparen{BD} = \overparen{CD}$, we have
$$
BD = DC, \angle BAD = \angle CAD,
$$

and $D$ lies on the angle bisector of $\angle BAC$ and the perpendicular bisector of $BC$.
Since $AC > AB$, the angle bisector of $\angle BAC$ and the perpendicular bisector of $BC$ do not coincide, thus they have at most one intersection point, so point $D$ coincides with point $P$.
Therefore, $DM \perp BC, DX \perp AB, DY \perp AC$, and $A, B, C, D$ are concyclic.
By the Simson line theorem, $X, M, Y$ are collinear, i.e., $M$ is the intersection of $BC$ and $XY$.
Since $BC$ and $XY$ do not coincide, they have at most one intersection point, so $M$ coincides with $Z$, thus
$$
\frac{BZ}{ZC} = \frac{BM}{MC} = 1
$$