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 $\frac{BZ}{ZC}$.
Solution 1: Construct the circumcircle $\odot O$ of $\triangle ABC$, and let the midpoint of arc $\overparen{BC}$ (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, so they have at most one intersection point, hence point $D$ coincides with point $P$.

Therefore, $DM \perp BC, DX \perp AB, DY \perp AC$, and $A, B, C, D$ are concyclic, so $X, M, Y$ are collinear (by Simson's Theorem), 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
$$
\frac{BZ}{ZC} = \frac{BM}{MC} = 1.
$$

Solution 2: As shown in the figure, connect $BP, CP$. Since $P$ lies on the perpendicular bisector of $BC$, we have $BP = CP$.

Since $P$ lies on the angle bisector of $\angle BAC$, we have $PX = PY, AX = AY, \triangle BPX \cong \triangle CPY$, so $BX = CY$.

Considering $XYZ$ as a transversal of $\triangle ABC$, by Menelaus' Theorem, we have
$$
\frac{AX}{XB} \cdot \frac{BZ}{ZC} \cdot \frac{CY}{YA} = 1,
$$

and since $AX = AY, BX = CY$, we have $\frac{BZ}{ZC} = 1$.