Example 4 (2003 China National Training Team) In $\triangle ABC$, $AC > AB$, $P$ is the intersection of the perpendicular bisector of $BC$ and the internal angle bisector of $\angle A$. Draw $PX \perp AB$, intersecting the extension of $AB$ at point $X$, and $PY \perp AC$ intersecting $AC$ at point $Y$, $Z$ is the intersection of $XY$ and $BC$. Find the value of $\frac{BZ}{ZC}$.
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
$$
\frac{BZ}{ZC} = \frac{BM}{MC} = 1
$$