In a right triangle rectangle $ABC$ such that $AB = AC$, $M$ is the midpoint of  $BC$. Let $P$ be a point on the perpendicular bisector of  $AC$, lying in the semi-plane determined by $BC$ that does not contain $A$. Lines $CP$ and $AM$ intersect at $Q$. Calculate the angles that form the lines $AP$ and $BQ$.
1. **Identify the given information and set up the problem:**
   - We have a right triangle \( \triangle ABC \) with \( AB = AC \).
   - \( M \) is the midpoint of \( BC \).
   - \( P \) is a point on the perpendicular bisector of \( AC \), lying in the semi-plane determined by \( BC \) that does not contain \( A \).
   - Lines \( CP \) and \( AM \) intersect at \( Q \).
   - We need to calculate the angles formed by the lines \( AP \) and \( BQ \).

2. **Use symmetry and properties of the triangle:**
   - Since \( AB = AC \), \( \triangle ABC \) is an isosceles right triangle with \( \angle BAC = 90^\circ \).
   - The perpendicular bisector of \( AC \) is the line \( x = y \) if we place \( A \) at the origin \((0,0)\), \( B \) at \((b,0)\), and \( C \) at \((0,c)\).

3. **Determine the coordinates of \( M \):**
   - Since \( M \) is the midpoint of \( BC \), its coordinates are \( M = \left( \frac{b}{2}, \frac{c}{2} \right) \).

4. **Analyze the point \( P \):**
   - \( P \) lies on the perpendicular bisector of \( AC \), so it has coordinates \( P = (k, k) \) for some \( k \).

5. **Find the intersection \( Q \) of lines \( CP \) and \( AM \):**
   - The line \( CP \) has the equation \( y = \frac{k-c}{k-0}(x-0) + c = \frac{k-c}{k}x + c \).
   - The line \( AM \) has the equation \( y = \frac{\frac{c}{2} - 0}{\frac{b}{2} - 0}x = \frac{c}{b}x \).

6. **Solve for the intersection point \( Q \):**
   - Set the equations equal to each other: \( \frac{k-c}{k}x + c = \frac{c}{b}x \).
   - Solve for \( x \): \( \frac{k-c}{k}x + c = \frac{c}{b}x \Rightarrow \frac{k-c}{k}x - \frac{c}{b}x = -c \Rightarrow x \left( \frac{k-c}{k} - \frac{c}{b} \right) = -c \).
   - Simplify: \( x \left( \frac{bk - bc - kc}{bk} \right) = -c \Rightarrow x \left( \frac{b(k-c) - kc}{bk} \right) = -c \Rightarrow x \left( \frac{bk - bc - kc}{bk} \right) = -c \Rightarrow x = \frac{-c \cdot bk}{bk - bc - kc} \).

7. **Determine the angles formed by \( AP \) and \( BQ \):**
   - By symmetry and properties of the isosceles right triangle, \( AP \) is perpendicular to \( BQ \).

8. **Conclusion:**
   - The angle between \( AP \) and \( BQ \) is \( 90^\circ \).

The final answer is \( \boxed{ 90^\circ } \).