In ` Delta PQR ,PQ =PR . A ` is a point in PQ and B is a point in PR , so that
QR = RA = AB = BP
Find the vlaue of ` angle Q.`

Answer

To solve the problem regarding triangle \( \Delta PQR \) where \( PQ = PR \) and points \( A \) and \( B \) are placed such that \( QR = RA = AB = BP \), we need to find the value of \( \angle Q \).

### Step-by-Step Solution:

1. **Understanding the Triangle**:
Since \( PQ = PR \), triangle \( PQR \) is an isosceles triangle with \( PQ \) and \( PR \) as the equal sides.

**Hint**: Remember that in an isosceles triangle, the angles opposite the equal sides are also equal.

2. **Setting the Lengths**:
Let \( QR = x \). According to the problem, we have:
- \( RA = x \)
- \( AB = x \)
- \( BP = x \)

This means that \( A \) and \( B \) are points on \( PQ \) and \( PR \) respectively, and each segment \( QR, RA, AB, BP \) is equal to \( x \).

**Hint**: You can visualize this by drawing the triangle and marking the segments to see how they relate to each other.

3. **Finding the Lengths**:
Since \( A \) is on \( PQ \) and \( B \) is on \( PR \), we can express the lengths of \( PA \) and \( PB \):
- \( PA = PQ - QA = PQ - QR = PQ - x \)
- \( PB = PR - RB = PR - RA = PR - x \)

**Hint**: Use the property of the triangle to express the segments in terms of \( x \).

4. **Using the Isosceles Property**:
Since \( PQ = PR \), we can denote \( PQ = PR = k \) for some length \( k \). Thus:
- \( PA = k - x \)
- \( PB = k - x \)

**Hint**: This shows that both segments from point \( P \) to points \( A \) and \( B \) are equal, reinforcing the isosceles nature of the triangle.

5. **Finding Angles**:
In triangle \( PQR \), the angles opposite the equal sides are equal. Let \( \angle PQR = \angle PRQ = \theta \). Therefore, we have:
\[ \angle Q + 2\theta = 180^\circ \]
Rearranging gives:
\[ \angle Q = 180^\circ - 2\theta \]

**Hint**: Remember that the sum of angles in a triangle is always \( 180^\circ \).

6. **Using the Equal Lengths**:
Since \( QR = RA = AB = BP = x \), and all segments are equal, we can deduce that triangle \( QRA \) and triangle \( PAB \) are also isosceles. This means that the angles \( \angle QAR \) and \( \angle PAB \) can be expressed in terms of \( \theta \).

**Hint**: Look for relationships between the angles formed by the equal segments.

7. **Final Calculation**:
Since \( QR = RA = AB = BP \) and they are all equal, we can conclude that:
- \( \theta = 60^\circ \) (as all angles in an equilateral triangle are \( 60^\circ \)).
Therefore:
\[ \angle Q = 180^\circ - 2(60^\circ) = 60^\circ \]

**Hint**: Check if the angles make sense with the properties of the triangle.

### Conclusion:
Thus, the value of \( \angle Q \) is \( 60^\circ \).