if `(a)/(b + c) = (b)/(c + a) = (c)/(a + b)` and `a + b + c = 0`: show that each given ratio is equal to `-1`.

Answer

To solve the problem, we need to show that if \( \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} \) and \( a + b + c = 0 \), then each ratio is equal to \(-1\).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} \]
Let this common ratio be \( k \). Therefore, we can write:
\[ \frac{a}{b+c} = k, \quad \frac{b}{c+a} = k, \quad \frac{c}{a+b} = k \]

2. **Express \( a, b, c \) in terms of \( k \):**
From the first equation:
\[ a = k(b+c) \]
From the second equation:
\[ b = k(c+a) \]
From the third equation:
\[ c = k(a+b) \]

3. **Substitute \( a + b + c = 0 \):**
We know that:
\[ a + b + c = 0 \]
Substituting the expressions for \( a, b, c \):
\[ k(b+c) + k(c+a) + k(a+b) = 0 \]

4. **Simplify the equation:**
Expanding the left-hand side:
\[ k(b+c) + k(c+a) + k(a+b) = k[(b+c) + (c+a) + (a+b)] \]
This simplifies to:
\[ k[2(a+b+c)] = 0 \]
Since \( a + b + c = 0 \), we have:
\[ k[2 \cdot 0] = 0 \]
This means that \( k \) can be any value, but we will determine its specific value next.

5. **Substituting back to find \( k \):**
We can use any of the original equations. Let's use:
\[ \frac{a}{b+c} = k \]
Since \( b+c = -a \) (from \( a + b + c = 0 \)), we substitute:
\[ \frac{a}{-a} = k \Rightarrow k = -1 \]

6. **Conclusion:**
Since \( k = -1 \), we have:
\[ \frac{a}{b+c} = -1, \quad \frac{b}{c+a} = -1, \quad \frac{c}{a+b} = -1 \]
Thus, we have shown that each given ratio is equal to \(-1\).