If in `Delta ABC, c(a+b) cos ""B/2=b (a+c) cos ""C/2,` the triangle is

Options

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

Answer

To solve the problem, we need to analyze the equation given in the triangle \( ABC \):

\[ c(a + b) \cos \frac{B}{2} = b(a + c) \cos \frac{C}{2} \]

### Step 1: Use the Sine Rule
We will apply the sine rule in triangle \( ABC \):
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
where \( R \) is the circumradius of the triangle.

From the sine rule, we can express \( a, b, c \) in terms of \( R \) and the angles:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

### Step 2: Substitute Values into the Given Equation
Now substitute \( a, b, c \) into the equation:
\[ (2R \sin C)(2R \sin A + 2R \sin B) \cos \frac{B}{2} = (2R \sin B)(2R \sin A + 2R \sin C) \cos \frac{C}{2} \]

### Step 3: Simplify the Equation
Factor out \( 2R \):
\[ 2R \sin C (2R (\sin A + \sin B) \cos \frac{B}{2}) = 2R \sin B (2R (\sin A + \sin C) \cos \frac{C}{2}) \]
Cancelling \( 2R \) from both sides (assuming \( R \neq 0 \)):
\[ \sin C (\sin A + \sin B) \cos \frac{B}{2} = \sin B (\sin A + \sin C) \cos \frac{C}{2} \]

### Step 4: Analyze the Resulting Equation
The equation we derived can be analyzed to check for specific conditions. Notably, if we set \( b = c \), we can check if the triangle is isosceles:
- If \( b = c \), then \( \angle B = \angle C \).

### Step 5: Conclusion
From the analysis, we find that the condition \( c(a + b) \cos \frac{B}{2} = b(a + c) \cos \frac{C}{2} \) leads us to conclude that the triangle must be isosceles, specifically where \( b = c \).

Thus, the triangle \( ABC \) is **isosceles**.