In a triangle ABC, we have `acosB+bcosC+ccosA=(a+b+c)/2` where A, B, C are internal angles and a, b,c are length of sides opposite to them respectively.Then must be

Options

A. The triangle is isosceles
B. The triangle is equilateral
C. The triangle is isosceles right angled
D. The triangle is right angled

Answer

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

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

where \(A\), \(B\), and \(C\) are the internal angles of the triangle, and \(a\), \(b\), and \(c\) are the lengths of the sides opposite to these angles.

### Step 1: Understanding the Equation
We start with the left-hand side (LHS) of the equation:

\[ LHS = a \cos B + b \cos C + c \cos A \]

### Step 2: Assume Equal Angles
To explore the condition under which this equation holds, we can assume that the triangle is equilateral, meaning all angles are equal. Therefore, we set:

\[ A = B = C = \frac{\pi}{3} \]

### Step 3: Substitute the Angles
Now substituting \(A\), \(B\), and \(C\) into the LHS:

\[ LHS = a \cos \left(\frac{\pi}{3}\right) + b \cos \left(\frac{\pi}{3}\right) + c \cos \left(\frac{\pi}{3}\right) \]

### Step 4: Calculate Cosine Values
We know that:

\[ \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Substituting this value into the equation gives:

\[ LHS = a \cdot \frac{1}{2} + b \cdot \frac{1}{2} + c \cdot \frac{1}{2} = \frac{a + b + c}{2} \]

### Step 5: Compare with Right-Hand Side
Now, we compare LHS with the right-hand side (RHS):

\[ RHS = \frac{a + b + c}{2} \]

Since \(LHS = RHS\), we conclude that the equation holds true when the triangle is equilateral.

### Conclusion
Thus, the condition \(a \cos B + b \cos C + c \cos A = \frac{a + b + c}{2}\) implies that triangle ABC must be equilateral.

### Final Answer
The triangle is equilateral.

---