in a triangle ABC let BD be its altitude and let H be its orthocentre. The perpendicular bisector of HD meets BCD at P Q find angle APB + angle AQB

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.

In a triangle ABC, let BD be its altitude and let H be its orthocentre. The perpendicular bisector of HD meets BCD at P and Q. Find angle APB + angle AQB.

Explore a step-by-step solution to finding angle APB + angle AQB in a triangle with orthocenter H. This problem involves geometric properties of cyclic quadrilaterals and orthocenters, perfect for Grades 10-12.

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