Question:

In \( \triangle ABC \), \( \frac{b - c}{r_1} + \frac{c - a}{r_2} + \frac{a - b}{r_3} \) is equal to

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The sum \( \frac{b - c}{r_1} + \frac{c - a}{r_2} + \frac{a - b}{r_3} = 0 \) is a standard identity in triangle geometry that holds for all triangles.
Updated On: Apr 22, 2026
  • 0
  • \( abc \)
  • \( a + b + c \)
  • \( ab + bc + ca \)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the equation.
We are given the equation \( \frac{b - c}{r_1} + \frac{c - a}{r_2} + \frac{a - b}{r_3} \), where \( r_1, r_2, r_3 \) are the radii of the excircles opposite to the vertices \( A, B, C \) respectively.

Step 2: Use of known identity.

There is a well-known identity in triangle geometry which relates the terms involving the excircles and the sides of the triangle: \[ \frac{b - c}{r_1} + \frac{c - a}{r_2} + \frac{a - b}{r_3} = 0 \] This identity holds for any triangle.

Step 3: Simplification.

From the identity, it directly follows that: \[ \frac{b - c}{r_1} + \frac{c - a}{r_2} + \frac{a - b}{r_3} = 0 \]

Step 4: Conclusion.

Thus, the value of the given expression is 0, corresponding to option (A).
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