In `triangleABC`, if `a^(2)+b^(2)+c^(2)=ac+ab sqrt(3)`, then

Options

A. `r/r_(2) = r_(3)/r_(1)`
B. `r/r_(1) = r_(2)/r_(3)`
C. `r_(1)+r_(2)+r_(3)=2R+r`
D. `rr_(1)r_(2)r_(3)=4^(3)`

Answer

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

**Given:**
\[ a^2 + b^2 + c^2 = ac + ab\sqrt{3} \]

We need to determine the type of triangle based on this equation.

### Step 1: Rearranging the Equation
First, let's rearrange the equation to bring all terms to one side:
\[ a^2 + b^2 + c^2 - ac - ab\sqrt{3} = 0 \]

### Step 2: Grouping Terms
Next, we can group the terms:
\[ a^2 - ac + b^2 - ab\sqrt{3} + c^2 = 0 \]

### Step 3: Completing the Square
Now, we will complete the square for the terms involving \(a\) and \(b\):
1. For \(a^2 - ac\), we can write it as:
\[ a^2 - ac = \left(a - \frac{c}{2}\right)^2 - \frac{c^2}{4} \]

2. For \(b^2 - ab\sqrt{3}\), we can write it as:
\[ b^2 - ab\sqrt{3} = \left(b - \frac{a\sqrt{3}}{2}\right)^2 - \frac{3a^2}{4} \]

### Step 4: Substitute Back
Substituting these completed squares back into the equation gives:
\[ \left(a - \frac{c}{2}\right)^2 + \left(b - \frac{a\sqrt{3}}{2}\right)^2 + c^2 - \frac{c^2}{4} - \frac{3a^2}{4} = 0 \]

### Step 5: Analyzing the Equation
For the equation to hold true, each squared term must equal zero because squares are always non-negative. Thus, we set:
1. \( a - \frac{c}{2} = 0 \) → \( a = \frac{c}{2} \)
2. \( b - \frac{a\sqrt{3}}{2} = 0 \) → \( b = \frac{a\sqrt{3}}{2} \)

### Step 6: Relating the Sides
From the above relationships, we can express \(b\) and \(c\) in terms of \(a\):
- Since \(a = \frac{c}{2}\), we can write \(c = 2a\).
- Substituting \(c\) into the equation for \(b\):
\[ b = \frac{a\sqrt{3}}{2} \]

### Step 7: Finding Angles
Using the sine rule:
- The angles corresponding to sides \(a\), \(b\), and \(c\) can be derived from:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]

### Step 8: Conclusion
From the relationships derived:
- We can conclude that the triangle is a right triangle because the relationships between the sides correspond to a \(30^\circ-60^\circ-90^\circ\) triangle, which is a specific case of a right triangle.

Thus, the answer is:
**The triangle is a right triangle.**