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To solve this question, we need to use angle properties in geometry, particularly focusing on the isosceles triangle, parallel lines, and straight line properties.

### Steps:

1. **Identify key properties:**
   - \( PQR \) is an isosceles triangle, meaning \( PQ = PR \), and therefore angles \( \angle PQR \) and \( \angle PRQ \) are equal.
   - \( APR \parallel CQD \) means that corresponding and alternate angles can be used with these parallel lines.
   - \( BPQ \) is a straight line, meaning angles along that line will sum to \( 180^\circ \).
   - \( \angle APB = 56^\circ \) is given.

2. **Use straight line property:**
   Along the straight line \( BPQ \), the angle \( \angle APB + \angle BPR = 180^\circ \) because they form a straight angle.
   \[
   \angle BPR = 180^\circ - 56^\circ = 124^\circ
   \]

3. **Apply isosceles triangle properties:**
   In the isosceles triangle \( PQR \), the two base angles \( \angle PQR \) and \( \angle PRQ \) are equal. The sum of angles in any triangle is \( 180^\circ \), so:
   \[
   \angle PQR + \angle PRQ + \angle QPR = 180^\circ
   \]
   Since \( \angle PQR = \angle PRQ \) and \( \angle QPR = 124^\circ \) (calculated earlier), we can solve for the base angles:
   \[
   2 \times \angle PQR + 124^\circ = 180^\circ
   \]
   \[
   2 \times \angle PQR = 56^\circ
   \]
   \[
   \angle PQR = \frac{56^\circ}{2} = 28^\circ
   \]

4. **Use parallel line properties:**
   Since \( APR \parallel CQD \), \( \angle CQR \) is an alternate angle to \( \angle PQR \). Hence, \( \angle CQR = 28^\circ \).

### Conclusion:
The size of angle \( CQR \) is \( 28^\circ \).

Would you like further explanation on any step?

Here are 5 related questions for practice:
1. How do you calculate base angles in an isosceles triangle?
2. What is the sum of angles in any triangle?
3. How are alternate angles used in parallel line problems?
4. How can straight lines help in finding unknown angles?
5. What is the relationship between corresponding angles in parallel lines?

**Tip:** When working with isosceles triangles, always remember that the base angles are equal!

In the diagram, PQR is an isosceles triangle with PQ = PR. APR and CQD are parallel lines. BPQ is a straight line. Angle APB = 56°. Work out the size of angle CQR.

Learn how to find the size of angle CQR in an isosceles triangle with parallel lines. Using key concepts like base angles, alternate angles, and angles on a straight line, this problem is designed for Grades 9-10 geometry students.

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