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Let's analyze and solve the problem step by step.

Given:
- \( \triangle ABC \) where \( BD \) is the altitude, and \( H \) is the orthocenter.
- The perpendicular bisector of \( HD \) meets line \( BCD \) at points \( P \) and \( Q \).
- We need to find \( \angle APB + \angle AQB \).

### Solution:
1. **Understanding the Geometry:**
   - Since \( H \) is the orthocenter, \( BD \) is perpendicular to \( AC \).
   - \( P \) and \( Q \) lie on the perpendicular bisector of \( HD \), which makes \( P \) and \( Q \) equidistant from \( H \) and \( D \).

2. **Key Concept:**
   - Since \( P \) and \( Q \) are on the perpendicular bisector, \( \triangle HPD \) and \( \triangle HQD \) are isosceles triangles, with \( HP = PD \) and \( HQ = QD \).
   - The configuration suggests that \( P \) and \( Q \) might be symmetric with respect to line \( AC \), indicating some symmetry in the angles we need to find.

3. **Utilizing Symmetry:**
   - By the properties of perpendicular bisectors and symmetry, \( \triangle APB \) and \( \triangle AQB \) should be congruent.
   - This implies that \( \angle APB = \angle AQB \).
   - Let \( \angle APB = \alpha \). Therefore, \( \angle AQB = \alpha \).

4. **Sum of Angles:**
   - \( \angle APB + \angle AQB = \alpha + \alpha = 2\alpha \).
   - Since \( APBQ \) is a cyclic quadrilateral (as \( P \) and \( Q \) lie on the perpendicular bisector and symmetry), \( \angle APB \) and \( \angle AQB \) add up to \( 180^\circ \).

Thus, the sum \( \angle APB + \angle AQB = 180^\circ \).

### Conclusion:
\[
\boxed{\angle APB + \angle AQB = 180^\circ}
\]

Would you like further details or clarifications on this solution? Here are some related questions:

1. Why does the perpendicular bisector ensure symmetry between points \( P \) and \( Q \)?
2. What properties of the orthocenter are crucial for this problem?
3. How can we prove that \( APBQ \) is a cyclic quadrilateral?
4. What role does symmetry play in determining the angles in this problem?
5. Can this method be generalized to other configurations involving perpendicular bisectors and orthocenters?

**Tip:** Remember that in many geometry problems involving symmetry, congruent triangles can reveal equal angles or distances, which can simplify the solution process.

In a triangle ABC, let BD be its altitude and let H be its orthocenter. The perpendicular bisector of HD meets (BCD) at P, Q. Find ∠APB + ∠AQB = ?

Solve this geometry problem involving triangle ABC with altitude BD, orthocenter H, and perpendicular bisector meeting at P and Q. Learn how to find ∠APB + ∠AQB using properties of the orthocenter and cyclic quadrilaterals. Suitable for Grades 10-12.

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