In triangle ABC, `R (b + c) = a sqrt(bc)`, 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 start with the given equation involving the circumradius \( R \) of triangle \( ABC \):

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

### Step 1: Express \( a \) in terms of \( R \)
We know that the side \( a \) can be expressed in terms of the circumradius \( R \) and the angle \( A \):

\[ a = 2R \sin A \]

### Step 2: Substitute \( a \) into the equation
Substituting \( a \) into the original equation gives:

\[ R(b + c) = 2R \sin A \sqrt{bc} \]

### Step 3: Cancel \( R \) from both sides
Assuming \( R \neq 0 \) (which is true for any triangle), we can cancel \( R \) from both sides:

\[ b + c = 2 \sin A \sqrt{bc} \]

### Step 4: Rearrange the equation
Rearranging the equation, we have:

\[ \sin A = \frac{b + c}{2 \sqrt{bc}} \]

### Step 5: Apply the sine inequality
We know that the sine of any angle is always less than or equal to 1:

\[ \frac{b + c}{2 \sqrt{bc}} \leq 1 \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:

\[ b + c \leq 2 \sqrt{bc} \]

### Step 7: Square both sides
Squaring both sides results in:

\[ (b + c)^2 \leq 4bc \]

### Step 8: Expand and rearrange
Expanding the left side:

\[ b^2 + 2bc + c^2 \leq 4bc \]

Rearranging gives:

\[ b^2 - 2bc + c^2 \leq 0 \]

### Step 9: Factor the expression
This can be factored as:

\[ (b - c)^2 \leq 0 \]

### Step 10: Conclude the equality
Since a square is always non-negative, the only solution is:

\[ (b - c)^2 = 0 \implies b = c \]

### Step 11: Determine the angle \( A \)
If \( b = c \), we can substitute back into the sine equation:

\[ \sin A = \frac{b + b}{2 \sqrt{b^2}} = \frac{2b}{2b} = 1 \]

This implies:

\[ A = 90^\circ \]

### Conclusion
Since angle \( A \) is \( 90^\circ \) and sides \( b \) and \( c \) are equal, triangle \( ABC \) is a right isosceles triangle.

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

---