If S is a point on side PQ of a `DeltaPQR` such that PS=QS=RS, then

Options

A. `PRcdotQR=RS^(2)`
B. `QS^(2)+RS^(2)=QR^(2)`
C. `PR^(2)+QR^(2)=PQ^(2)`
D. `PS^(2)+RS^(2)=PR^(2)`

Explanation

To solve the problem, we need to analyze the triangle \( \Delta PQR \) with point \( S \) on side \( PQ \) such that \( PS = QS = RS \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- We have triangle \( PQR \).
- Point \( S \) is on side \( PQ \).
- It is given that \( PS = QS = RS \).

2. **Label the Angles**:
- Let \( PS = QS = RS = x \).
- In triangle \( PSR \), since \( PS = RS \), by the property of triangles, the angles opposite to equal sides are equal. Let \( \angle P = \angle R = \alpha \).
- Therefore, \( \angle S = 180^\circ - 2\alpha \) (using the angle sum property of triangles).

3. **Consider Triangle \( RSQ \)**:
- In triangle \( RSQ \), since \( RS = SQ \), by the same property, let \( \angle R = \angle Q = \beta \).
- Thus, \( \angle S = 180^\circ - 2\beta \).

4. **Equate the Angles**:
- Since both triangles share angle \( S \), we can equate the two expressions for \( \angle S \):
\[ 180^\circ - 2\alpha = 180^\circ - 2\beta \]
- This simplifies to:
\[ 2\alpha = 2\beta \implies \alpha = \beta \]

5. **Conclusion about Angles**:
- Since \( \alpha = \beta \), we can conclude that \( \angle P = \angle Q \).
- Therefore, triangle \( PQR \) is isosceles with \( PQ = PR \).

6. **Using the Angle Sum Property**:
- The sum of angles in triangle \( PQR \) is:
\[ \angle P + \angle Q + \angle R = 180^\circ \]
- Since \( \angle P = \angle Q \), let \( \angle P = \angle Q = \theta \) and \( \angle R = 180^\circ - 2\theta \).

7. **Determine the Right Angle**:
- Since \( PS = QS = RS \), we can conclude that \( \angle S \) must be \( 90^\circ \) for the triangle to maintain the equality of sides.
- Thus, \( \angle P + \angle Q = 90^\circ \).

8. **Final Conclusion**:
- Therefore, triangle \( PQR \) is a right triangle at \( R \).
- By the Pythagorean theorem, we have:
\[ PQ^2 = PR^2 + QR^2 \]

### Final Result:
The correct conclusion is that \( PQ^2 = PR^2 + QR^2 \), confirming that triangle \( PQR \) is a right triangle.