Q.If \[ \frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} \] then prove that each of these ratios is either \(\frac{1}{2}\) or \(-1\).

Let \[ \frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = k \quad \text{[where \(k\) is a non-zero constant]} \] Then: \[ a = k(b + c), \quad b = k(c + a), \quad c = k(a + b) \] Adding all three: \[ a + b + c = k(b + c) + k(c + a) + k(a + b) = k[(b + c) + (c + a) + (a + b)] = k[2(a + b + c)] \] So: \[ a + b + c = 2k(a + b + c) \Rightarrow (a + b + c)(1 - 2k) = 0 \] Therefore, either: - \(a + b + c = 0\), or - \(1 - 2k = 0 \Rightarrow k = \frac{1}{2}\) So, each ratio is either \(\frac{1}{2}\) Now, if \(a + b + c = 0\), then: - \(b + c = -a \Rightarrow \frac{a}{b + c} = \frac{a}{-a} = -1\) - \(c + a = -b \Rightarrow \frac{b}{c + a} = \frac{b}{-b} = -1\) - \(a + b = -c \Rightarrow \frac{c}{a + b} = \frac{c}{-c} = -1\) Hence, if \(\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b}\), then each ratio is either \(\frac{1}{2}\) or \(-1\) — proven.
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