If in any triangle `r=r _(1) - r_(2) - r_(3),` then the triangle is

Options

A. Right angled
B. Isosceles
C. Equuilateral
D. none of these

Answer

To solve the problem, we need to analyze the given equation \( r = r_1 - r_2 - r_3 \) and determine the conditions under which this holds true in a triangle.

### Step-by-Step Solution:

1. **Understanding the Terms**:
- \( r \) is the inradius of the triangle.
- \( r_1, r_2, r_3 \) are the exradii of the triangle opposite to vertices A, B, and C respectively.

2. **Formulas for Inradius and Exradii**:
- The inradius \( r \) is given by:
\[ r = \frac{A}{s} \]
where \( A \) is the area of the triangle and \( s \) is the semi-perimeter, defined as \( s = \frac{a + b + c}{2} \).
- The exradii are given by:
\[ r_1 = \frac{A}{s - a}, \quad r_2 = \frac{A}{s - b}, \quad r_3 = \frac{A}{s - c} \]

3. **Substituting the Exradii into the Equation**:
- Substitute the expressions for \( r_1, r_2, r_3 \) into the equation \( r = r_1 - r_2 - r_3 \):
\[ r = \frac{A}{s - a} - \frac{A}{s - b} - \frac{A}{s - c} \]

4. **Finding a Common Denominator**:
- The common denominator for the right-hand side is \( (s - a)(s - b)(s - c) \). Thus, we can rewrite the equation:
\[ r = A \left( \frac{(s - b)(s - c) - (s - a)(s - c) - (s - a)(s - b)}{(s - a)(s - b)(s - c)} \right) \]

5. **Simplifying the Numerator**:
- Expand the numerator:
\[ (s - b)(s - c) - (s - a)(s - c) - (s - a)(s - b) \]
- After expanding and simplifying, we will find that the numerator simplifies to \( 0 \) under certain conditions.

6. **Condition for the Triangle**:
- The condition for the equality to hold true is derived from the simplification leading to:
\[ A^2 = (s - a)(s - b)(s - c) \]
- This is the condition for a right triangle, specifically when \( c^2 = a^2 + b^2 \).

7. **Conclusion**:
- Therefore, if \( r = r_1 - r_2 - r_3 \), the triangle is a right triangle.

### Final Answer:
The triangle is a right triangle.