In triangle `A B C ,R(b+c)=asqrt(b c)`,where `R` is the circumradius of the triangle. Then the triangle is

Options

A. isosceles but not right
B. right but not isosceles
C. right isosceles
D. equilateral

Explanation

To solve the problem, we need to analyze the given equation in the context of triangle properties and the circumradius. The equation provided is:

\[ R(b + c) = a \sqrt{bc} \]

where \( R \) is the circumradius of triangle \( ABC \), and \( a, b, c \) are the lengths of the sides opposite to angles \( A, B, C \) respectively.

### Step-by-Step Solution:

1. **Use the Law of Sines**:
From the Law of Sines, we know that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
Thus, we can express \( a \) in terms of \( R \) and \( \sin A \):
\[ a = 2R \sin A \]

2. **Substitute \( a \) in the given equation**:
Substitute \( a \) into the equation \( R(b + c) = a \sqrt{bc} \):
\[ R(b + c) = (2R \sin A) \sqrt{bc} \]

3. **Cancel \( R \) (assuming \( R \neq 0 \))**:
Dividing both sides by \( R \) gives:
\[ b + c = 2 \sin A \sqrt{bc} \]

4. **Rearranging the equation**:
Rearranging gives:
\[ \frac{b + c}{\sqrt{bc}} = 2 \sin A \]

5. **Using the Cauchy-Schwarz Inequality**:
By the Cauchy-Schwarz inequality, we know:
\[ (b + c)^2 \leq (1^2 + 1^2)(b + c) = 2(b + c) \]
Thus:
\[ \frac{(b + c)^2}{bc} \geq 4 \]
This implies:
\[ \frac{b + c}{\sqrt{bc}} \geq 2 \]

6. **Combining results**:
From the previous steps, we have:
\[ 2 \sin A \leq 2 \]
This implies:
\[ \sin A \leq 1 \]
The equality holds when \( b = c \) (which means the triangle is isosceles).

7. **Conclusion about the angles**:
Since \( \sin A = 1 \), we have:
\[ A = 90^\circ \]
This means triangle \( ABC \) is a right triangle.

8. **Final classification**:
Since \( b = c \) and \( A = 90^\circ \), triangle \( ABC \) is a right isosceles triangle.

### Final Answer:
The triangle is a **right isosceles triangle**.