Let us assume, \[ \frac{x}{y+z} = \frac{y}{z+x} = \frac{z}{x+y} = k \]
[where \(k\) is a non-zero constant]
Therefore, \[ x = k(y + z), \quad y = k(z + x), \quad z = k(x + y) \] \[ \therefore x + y + z = k(y + z) + k(z + x) + k(x + y) \] \[ \Rightarrow x + y + z = k[(y + z) + (z + x) + (x + y)] \] \[ \Rightarrow x + y + z = k[2(x + y + z)] \] \[ \Rightarrow (x + y + z) - 2k(x + y + z) = 0 \] \[ \Rightarrow (x + y + z)(1 - 2k) = 0 \] So either \[ x + y + z = 0 \] or \[ 1 - 2k = 0 \Rightarrow k = \frac{1}{2} \] Hence, in this case, the value of each ratio is \(\frac{1}{2}\). Now, if \(x + y + z = 0\), then \[ y + z = -x \Rightarrow \frac{x}{y + z} = \frac{x}{-x} = -1 \] Similarly, \[ z + x = -y \Rightarrow \frac{y}{z + x} = \frac{y}{-y} = -1 \] and \[ x + y = -z \Rightarrow \frac{z}{x + y} = \frac{z}{-z} = -1 \] \[ \therefore \frac{x}{y+z} = \frac{y}{z+x} = \frac{z}{x+y} \] implies that each ratio is either \(\frac{1}{2}\) or \(-1\). (Proved)