Let's analyze this problem using some properties of triangles and their orthocenters.

Step-by-Step Solution

1. Understanding the Setup:

   • $\triangle ABC$ has an altitude $BD$, meaning $BD \perp AC$.
   • $H$ is the orthocenter of $\triangle ABC$, so $H$ is the intersection of all the altitudes in $\triangle ABC$.
   • The perpendicular bisector of $HD$ intersects $\triangle BCD$ at points $P$ and $Q$.

2. Applying Properties of Orthocenters:

   • Since $H$ is the orthocenter, $BD$, $CE$, and $AF$ (where $E$ and $F$ are the feet of the altitudes from $C$ and $A$) are concurrent at $H$.
   • This means $H$ has certain symmetrical properties within $\triangle ABC$.

3. Angles of Interest:

   • To find $\angle APB + \angle AQB$, consider that $P$ and $Q$ are defined symmetrically with respect to $H$.
   • The points $P$ and $Q$, as intersections on the perpendicular bisector of $HD$, lie on the circumcircle of $\triangle ABC$.

4. Result from Cyclic Quadrilateral:

   • In cyclic quadrilateral $APQB$, the sum of the opposite angles $\angle APB + \angle AQB$ is equal to $180^\circ$ due to the property of cyclic quadrilaterals where opposite angles sum to $180^\circ$.

Final Answer

$\angle APB + \angle AQB = 180^\circ$

Would you like further details or have any questions on the properties used?

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Related Questions

1. What is the significance of an orthocenter in a triangle?
2. How does the perpendicular bisector help in finding angles in cyclic quadrilaterals?
3. What properties of cyclic quadrilaterals are useful for solving angle problems?
4. How do altitudes in a triangle contribute to symmetry?
5. What is the role of the circumcircle in angle calculations for cyclic figures?

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Tip: In geometry, using properties of cyclic quadrilaterals can often simplify angle calculations significantly.