\[ \frac{b + c - a}{y + z - x} = \frac{c + a - b}{z + x - y} = \frac{a + b - c}{x + y - z} \] \[ = \frac{(b + c - a) + (c + a - b) + (a + b - c)}{(y + z - x) + (z + x - y) + (x + y - z)} \quad \text{[By addition]} \] \[ = \frac{a + b + c}{x + y + z} \] Now, \[ \frac{b + c - a}{y + z - x} = \frac{a + b + c}{x + y + z} = \frac{(a + b + c) - (b + c - a)}{(x + y + z) - (y + z - x)} = \frac{2a}{2x} = \frac{a}{x} \] ∴ Each ratio equals \(\frac{a}{x}\) Similarly, \[ \frac{c + a - b}{z + x - y} = \frac{a + b + c}{x + y + z} = \frac{(a + b + c) - (c + a - b)}{(x + y + z) - (z + x - y)} = \frac{2b}{2y} = \frac{b}{y} \] ∴ Each ratio equals \(\frac{b}{y}\) And, \[ \frac{a + b - c}{x + y - z} = \frac{a + b + c}{x + y + z} = \frac{(a + b + c) - (a + b - c)}{(x + y + z) - (x + y - z)} = \frac{2c}{2z} = \frac{c}{z} \] ∴ Each ratio equals \(\frac{c}{z}\) Therefore, \[ \frac{a}{x} = \frac{b}{y} = \frac{c}{z} \quad \text{[Proved]} \]