If in a triangle `(r)/(r_(1)) = (r_(2))/(r_(3))`, then

Options

A. `A = 90^(@)`
B. `B = 90^(@)`
C. `C = 90^(@)`
D. none of these

Explanation

To solve the problem, we need to prove that if \(\frac{r}{r_1} = \frac{r_2}{r_3}\), then angle \(C\) in triangle \(ABC\) is \(90^\circ\).

### Step-by-Step Solution:

1. **Understanding the Terms**:
- Let \(r\) be the inradius of triangle \(ABC\).
- Let \(r_1\), \(r_2\), and \(r_3\) be the exradii opposite to vertices \(A\), \(B\), and \(C\) respectively.
- The area of the triangle is denoted by \(\Delta\) and the semi-perimeter by \(s\).

2. **Expressing the Radii**:
- The inradius \(r\) is given by:
\[ r = \frac{\Delta}{s} \]
- The exradii are given by:
\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \]

3. **Substituting the Values**:
- From the problem statement, we have:
\[ \frac{r}{r_1} = \frac{r_2}{r_3} \]
- Substituting the values of \(r\), \(r_1\), \(r_2\), and \(r_3\):
\[ \frac{\frac{\Delta}{s}}{\frac{\Delta}{s - a}} = \frac{\frac{\Delta}{s - b}}{\frac{\Delta}{s - c}} \]

4. **Simplifying the Equation**:
- This simplifies to:
\[ \frac{s - a}{s} = \frac{s - b}{s - c} \]
- Cross-multiplying gives:
\[ (s - a)(s - c) = s(s - b) \]

5. **Expanding Both Sides**:
- Expanding both sides:
\[ s^2 - ac - as + cs = s^2 - bs \]
- Simplifying, we get:
\[ -ac + cs = -bs \]
- Rearranging gives:
\[ ac + bs = cs \]

6. **Using the Area Relation**:
- We know that:
\[ \tan^2\left(\frac{C}{2}\right) = \frac{(s-a)(s-b)}{s(s-c)} \]
- From our previous equation, we can relate this to the tangent:
\[ \tan^2\left(\frac{C}{2}\right) = 1 \]
- This implies:
\[ \frac{C}{2} = 45^\circ \quad \Rightarrow \quad C = 90^\circ \]

### Conclusion:
Thus, we conclude that if \(\frac{r}{r_1} = \frac{r_2}{r_3}\), then angle \(C\) is \(90^\circ\).