Let $ABCD$ be a rectangle with $AB=10$ and $AD=5.$ Suppose points $P$ and $Q$ are on segments $CD$ and $BC,$ respectively, such that the following conditions hold:
[list]
[*] $BD \parallel PQ$
[*] $\angle APQ=90^{\circ}.$ 
 [/list]
 What is the area of $\triangle CPQ?$

[i]Proposed by Kyle Lee[/i]
1. **Identify the coordinates of the vertices of the rectangle:**
   - Let \( A = (0, 0) \), \( B = (10, 0) \), \( C = (10, 5) \), and \( D = (0, 5) \).

2. **Determine the slope of \( BD \):**
   - The coordinates of \( B \) and \( D \) are \( (10, 0) \) and \( (0, 5) \) respectively.
   - The slope of \( BD \) is given by:
     \[
     \text{slope of } BD = \frac{5 - 0}{0 - 10} = -\frac{1}{2}
     \]

3. **Since \( BD \parallel PQ \), the slope of \( PQ \) is also \( -\frac{1}{2} \).**

4. **Use the condition \( \angle APQ = 90^\circ \):**
   - This implies that the product of the slopes of \( AP \) and \( PQ \) is \( -1 \).
   - Let the slope of \( AP \) be \( m \). Then:
     \[
     m \cdot \left(-\frac{1}{2}\right) = -1 \implies m = 2
     \]

5. **Determine the coordinates of \( P \) and \( Q \):**
   - Let \( P = (x, 5) \) on segment \( CD \).
   - Let \( Q = (10, y) \) on segment \( BC \).

6. **Use the slope of \( AP \) to find \( x \):**
   - The slope of \( AP \) is \( 2 \), so:
     \[
     \frac{5 - 0}{x - 0} = 2 \implies x = \frac{5}{2}
     \]
   - Thus, \( P = \left(\frac{5}{2}, 5\right) \).

7. **Use the slope of \( PQ \) to find \( y \):**
   - The slope of \( PQ \) is \( -\frac{1}{2} \), so:
     \[
     \frac{y - 5}{10 - \frac{5}{2}} = -\frac{1}{2} \implies \frac{y - 5}{\frac{15}{2}} = -\frac{1}{2} \implies y - 5 = -\frac{15}{4} \implies y = 5 - \frac{15}{4} = \frac{5}{4}
     \]
   - Thus, \( Q = \left(10, \frac{5}{4}\right) \).

8. **Calculate the area of \( \triangle CPQ \):**
   - The coordinates of \( C \) are \( (10, 5) \), \( P = \left(\frac{5}{2}, 5\right) \), and \( Q = \left(10, \frac{5}{4}\right) \).
   - Use the formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
     \[
     \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
     \]
   - Substituting the coordinates:
     \[
     \text{Area} = \frac{1}{2} \left| 10 \left(5 - \frac{5}{4}\right) + \frac{5}{2} \left(\frac{5}{4} - 5\right) + 10 \left(5 - 5\right) \right|
     \]
     \[
     = \frac{1}{2} \left| 10 \left(\frac{20}{4} - \frac{5}{4}\right) + \frac{5}{2} \left(\frac{5}{4} - \frac{20}{4}\right) \right|
     \]
     \[
     = \frac{1}{2} \left| 10 \left(\frac{15}{4}\right) + \frac{5}{2} \left(-\frac{15}{4}\right) \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{150}{4} - \frac{75}{4} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{75}{4} \right|
     \]
     \[
     = \frac{1}{2} \cdot \frac{75}{4} = \frac{75}{8}
     \]

The final answer is \(\boxed{\frac{75}{8}}\)