In a `triangle ABC`, acosB + b cosC + c cosA `=(a+b+c)/2` then

Options

A. Triangle in isosceles
B. Triangle may be equilateral
C. `sin(A-B) + sin(B-C) + sin(C-A)=3/2`
D. `4sin( (A-B)/2) sin((B-C)/2) sin((C-A)/2)=1`

Answer

To solve the problem, we need to prove that in triangle \( ABC \), if \( a \cos B + b \cos C + c \cos A = \frac{a + b + c}{2} \), then the triangle is equilateral.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ a \cos B + b \cos C + c \cos A = \frac{a + b + c}{2} \]
This is our equation (1).

2. **Use the cosine rule to express \( \cos A \), \( \cos B \), and \( \cos C \):**
- From the cosine rule, we have:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

3. **Substitute these values into equation (1):**
\[ a \left( \frac{a^2 + c^2 - b^2}{2ac} \right) + b \left( \frac{a^2 + b^2 - c^2}{2ab} \right) + c \left( \frac{b^2 + c^2 - a^2}{2bc} \right) = \frac{a + b + c}{2} \]

4. **Simplify the left-hand side:**
\[ \frac{a^2 + c^2 - b^2}{2c} + \frac{a^2 + b^2 - c^2}{2a} + \frac{b^2 + c^2 - a^2}{2b} \]

5. **Combine the fractions:**
To combine these fractions, we can find a common denominator, which is \( 2abc \):
\[ \frac{b(a^2 + c^2 - b^2) + c(a^2 + b^2 - c^2) + a(b^2 + c^2 - a^2)}{2abc} \]

6. **Expand and simplify the numerator:**
After expanding and combining like terms, we will have:
\[ A^3B + B^3C + C^3A - (A^2B^2 + B^2C^2 + C^2A^2) = 0 \]
This means that the terms must balance out.

7. **Analyze the resulting equation:**
The equation \( A^3B - C + B^3C - A + C^3A - B = 0 \) implies that \( A = B = C \) (the triangle is equilateral).

8. **Conclusion:**
Therefore, if \( a \cos B + b \cos C + c \cos A = \frac{a + b + c}{2} \), then it must be that \( a = b = c \), meaning triangle \( ABC \) is equilateral.