Let $ABCD$ be a rectangle. Circles with diameters $AB$ and $CD$ meet at points $P$ and $Q$ inside the rectangle such that $P$ is closer to segment $BC$ than $Q$. Let $M$ and $N$ be the midpoints of segments $AB$ and $CD$. If $\angle MPN = 40^\circ$, find the degree measure of $\angle BPC$.

[i]Ray Li.[/i]
1. **Identify the given information and setup the problem:**
   - We have a rectangle \(ABCD\).
   - Circles with diameters \(AB\) and \(CD\) intersect at points \(P\) and \(Q\) inside the rectangle.
   - \(P\) is closer to segment \(BC\) than \(Q\).
   - \(M\) and \(N\) are the midpoints of segments \(AB\) and \(CD\) respectively.
   - \(\angle MPN = 40^\circ\).

2. **Understand the geometry of the problem:**
   - Since \(AB\) and \(CD\) are diameters of their respective circles, \(\angle APB = \angle CPD = 90^\circ\) because the angle subtended by a diameter in a circle is a right angle (Thales' theorem).
   - \(M\) and \(N\) being midpoints of \(AB\) and \(CD\) respectively, means \(M\) and \(N\) are the centers of the circles with diameters \(AB\) and \(CD\).

3. **Analyze the given angle \(\angle MPN = 40^\circ\):**
   - Since \(M\) and \(N\) are the centers of the circles, \(MP\) and \(NP\) are radii of the circles.
   - The angle \(\angle MPN = 40^\circ\) is given.

4. **Determine the relationship between \(\angle BPC\) and \(\angle MPN\):**
   - Since \(P\) and \(Q\) are the intersection points of the circles, and \(P\) is closer to \(BC\), we focus on \(\angle BPC\).
   - Note that \(\angle BPC\) is formed by the intersection of the two circles at \(P\).

5. **Use the properties of the circles and the given angle:**
   - Since \(\angle MPN = 40^\circ\), and \(M\) and \(N\) are the centers of the circles, the line segments \(MP\) and \(NP\) are radii.
   - The angle at the center of a circle subtended by a chord is twice the angle subtended at any point on the circumference by the same chord.

6. **Calculate \(\angle BPC\):**
   - Since \(\angle MPN = 40^\circ\), the angle subtended by the same arc at the circumference (i.e., \(\angle BPC\)) is half of this angle.
   - Therefore, \(\angle BPC = 2 \times \angle MPN = 2 \times 40^\circ = 80^\circ\).

The final answer is \(\boxed{80^\circ}\).