Question

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

• A. 0
• B. \(abc\)
• C. \(a + b + c\)
• D. \(ab + bc + ca\)

Hint:

\(r_1, r_2, r_3\) are exradii: \(r_1 = \frac{\Delta}{s - a}\), etc.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
\(r_1 = \frac{\Delta}{s - a}\), \(r_2 = \frac{\Delta}{s - b}\), \(r_3 = \frac{\Delta}{s - c}\).
Step 2: Detailed Explanation:
\(\frac{b - c}{r_1} = \frac{(b - c)(s - a)}{\Delta}\)
Similarly for other terms. Sum = \(\frac{1}{\Delta}[(b - c)(s - a) + (c - a)(s - b) + (a - b)(s - c)]\)
Expand: \((b - c)s - (b - c)a + (c - a)s - (c - a)b + (a - b)s - (a - b)c\)
= \(s[(b - c) + (c - a) + (a - b)] - [a(b - c) + b(c - a) + c(a - b)]\)
First bracket = 0, second bracket = \(ab - ac + bc - ab + ac - bc = 0\)
So sum = 0
Step 3: Final Answer:
0.