If S is a point on side PQ of a ` triangle PQR` such that PS = QS = RS, then `PR^(2) + QR^(2)` = ……………..

Answer

To solve the problem, we need to analyze the triangle \( PQR \) with point \( S \) on side \( PQ \) such that \( PS = QS = RS \). We want to find the value of \( PR^2 + QR^2 \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have a triangle \( PQR \).
- Point \( S \) lies on side \( PQ \) such that \( PS = QS = RS \).
- Let \( PS = QS = RS = x \).

2. **Identifying Angles**:
- Since \( PS = QS \), triangle \( PSQ \) is isosceles, which means \( \angle PSQ = \angle PQS \).
- Similarly, since \( QS = RS \), triangle \( QSR \) is also isosceles, which means \( \angle QSR = \angle QRS \).

3. **Using Angle Properties**:
- The sum of angles in triangle \( PQR \) is \( 180^\circ \).
- We can express the angles in terms of \( \angle R \):
\[ \angle P + \angle Q + \angle R = 180^\circ \]
- Since \( \angle PRS = \angle RPS \) and \( \angle QRS = \angle RQS \), we can denote these angles as equal.

4. **Applying the Pythagorean Theorem**:
- Since \( PS = QS = RS \) and the angles formed are equal, we can conclude that \( \triangle PRS \) and \( \triangle QRS \) are right triangles.
- By the Pythagorean theorem:
\[ PR^2 + QR^2 = PQ^2 \]
- Since \( PQ = PS + SQ = x + x = 2x \), we have:
\[ PQ^2 = (2x)^2 = 4x^2 \]

5. **Final Expression**:
- Thus, we can conclude:
\[ PR^2 + QR^2 = PQ^2 = 4x^2 \]

### Conclusion:
The value of \( PR^2 + QR^2 \) is \( 4x^2 \).

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