Let $ABC$ be a triangle with $AC > AB$.  Let $P$ be the intersection point of the perpendicular bisector of $BC$ and the internal angle bisector of $\angle{A}$.  Construct points $X$ on $AB$ (extended) and $Y$ on $AC$ such that $PX$ is perpendicular to $AB$ and $PY$ is perpendicular to $AC$.  Let $Z$ be the intersection point of $XY$ and $BC$.
Determine the value of $\frac{BZ}{ZC}$.
1. **Identify the key points and properties:**
   - Let $ABC$ be a triangle with $AC > AB$.
   - $P$ is the intersection point of the perpendicular bisector of $BC$ and the internal angle bisector of $\angle A$.
   - Construct points $X$ on $AB$ (extended) and $Y$ on $AC$ such that $PX \perp AB$ and $PY \perp AC$.
   - Let $Z$ be the intersection point of $XY$ and $BC$.

2. **Analyze the properties of point $P$:**
   - Since $P$ lies on the perpendicular bisector of $BC$, it is equidistant from $B$ and $C$. Therefore, $PB = PC$.
   - Since $P$ also lies on the internal angle bisector of $\angle A$, it is on the arc midpoint of $\overarc{BC}$ in the circumcircle of $\triangle ABC$.

3. **Simpson Line and Perpendicularity:**
   - The line $XY$ is the Simpson line of $P$ with respect to $\triangle ABC$. By definition, the Simpson line of a point on the circumcircle of a triangle is perpendicular to the sides of the triangle at the feet of the perpendiculars from the point to the sides.
   - Therefore, $PZ \perp BC$ and $Z \in BC$.

4. **Determine the position of $Z$:**
   - Since $P$ lies on the perpendicular bisector of $BC$, the perpendicular from $P$ to $BC$ will intersect $BC$ at its midpoint.
   - Thus, $Z$ is the midpoint of $BC$.

5. **Calculate the ratio $\frac{BZ}{ZC}$:**
   - Since $Z$ is the midpoint of $BC$, we have $BZ = ZC$.
   - Therefore, $\frac{BZ}{ZC} = 1$.

\[
\boxed{1}
\]