The value of `(b-c)/r_1+ (c-a)/r_2+ (a-b)/r_3` is equal to

Options

A. ab
B. bc
C. ca
D. 0

Answer

To find the value of the expression \((b-c)/r_1 + (c-a)/r_2 + (a-b)/r_3\), we will use the formulas for the exradii of a triangle and some algebraic manipulation. Here’s a step-by-step solution:

### Step 1: Understand the Exradii
The exradii \(r_1\), \(r_2\), and \(r_3\) are defined as:
- \(r_1 = \frac{\Delta}{s - a}\)
- \(r_2 = \frac{\Delta}{s - b}\)
- \(r_3 = \frac{\Delta}{s - c}\)

where \(\Delta\) is the area of the triangle and \(s\) is the semi-perimeter given by \(s = \frac{a + b + c}{2}\).

### Step 2: Substitute Exradii into the Expression
Substituting the values of \(r_1\), \(r_2\), and \(r_3\) into the original expression, we have:

\[ \frac{b - c}{r_1} + \frac{c - a}{r_2} + \frac{a - b}{r_3} = \frac{(b - c)(s - a)}{\Delta} + \frac{(c - a)(s - b)}{\Delta} + \frac{(a - b)(s - c)}{\Delta} \]

### Step 3: Combine the Terms
Since \(\Delta\) is common in the denominator, we can combine the numerators:

\[ = \frac{1}{\Delta} \left( (b - c)(s - a) + (c - a)(s - b) + (a - b)(s - c) \right) \]

### Step 4: Expand the Numerators
Now, we will expand each term in the numerator:

1. \((b - c)(s - a) = (b - c)(\frac{a + b + c}{2} - a) = (b - c)(\frac{b + c - a}{2})\)
2. \((c - a)(s - b) = (c - a)(\frac{b + c + a}{2} - b) = (c - a)(\frac{c + a - b}{2})\)
3. \((a - b)(s - c) = (a - b)(\frac{a + b + c}{2} - c) = (a - b)(\frac{a + b - c}{2})\)

### Step 5: Combine and Simplify
Now, we will combine all these expanded terms:

\[ = \frac{1}{\Delta} \left( \frac{(b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)}{2} \right) \]

### Step 6: Analyze the Expression
Notice that each pair of terms will cancel out when we combine them:

- The terms \(b - c\), \(c - a\), and \(a - b\) will lead to cancellations due to symmetry.

Thus, when we simplify the entire expression, we find that:

\[ = \frac{0}{\Delta} = 0 \]

### Conclusion
The value of the expression \((b-c)/r_1 + (c-a)/r_2 + (a-b)/r_3\) is equal to **0**.

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