Q.Given: \[ \frac{x}{b + c} = \frac{y}{c + a} = \frac{z}{a + b} \] Prove that: \[ \frac{a}{y + z - x} = \frac{b}{z + x - y} = \frac{c}{x + y - z} \]

Let \[ \frac{x}{b + c} = \frac{y}{c + a} = \frac{z}{a + b} = k \quad \text{(where } k \ne 0\text{)} \] Then, \[ x = k(b + c), \quad y = k(c + a), \quad z = k(a + b) \] Now, consider the first expression: \[ \frac{a}{y + z - x} = \frac{a}{k(c + a) + k(a + b) - k(b + c)} = \frac{a}{k(c + a + a + b - b - c)} = \frac{a}{2ak} = \frac{1}{2k} \] Second expression: \[ \frac{b}{z + x - y} = \frac{b}{k(a + b) + k(b + c) - k(c + a)} = \frac{b}{k(a + b + b + c - c - a)} = \frac{b}{2kb} = \frac{1}{2k} \] Third expression: \[ \frac{c}{x + y - z} = \frac{c}{k(b + c) + k(c + a) - k(a + b)} = \frac{c}{k(b + c + c + a - a - b)} = \frac{c}{2kc} = \frac{1}{2k} \] Therefore, \[ \frac{a}{y + z - x} = \frac{b}{z + x - y} = \frac{c}{x + y - z} \] (proved)
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