Assume, \[ \frac{x}{y+z} = \frac{y}{z+x} = \frac{z}{x+y} = k \] where \(k\) is a non-zero constant. So, \[ x = k(y + z),\quad y = k(z + x),\quad z = k(x + y) \] Therefore, \[ x + y + z = k(y + z) + k(z + x) + k(x + y) \] Or, \[ x + y + z = k(y + z + z + x + x + y) \] Or, \[ x + y + z = 2k(x + y + z) \] Or, \[ (x + y + z) - 2k(x + y + z) = 0 \] Or, \[ (x + y + z)(1 - 2k) = 0 \] Therefore, either \(x + y + z = 0\) or \(1 - 2k = 0\), i.e., \(-2k = -1\) Hence, \(k = \frac{1}{2}\) So, the value of each ratio is either \(\frac{1}{2}\) Again, if \(x + y + z = 0\), then \(y + z = -x\) \[ \therefore \frac{x}{y + z} = \frac{x}{-x} = -1 \] Also, \(z + x = -y\) \[ \therefore \frac{y}{z + x} = \frac{y}{-y} = -1 \] And, \(x + y = -z\) \[ \therefore \frac{z}{x + y} = \frac{z}{-z} = -1 \] Therefore, if \[ \frac{x}{y+z} = \frac{y}{z+x} = \frac{z}{x+y} \] then each ratio is either \(\frac{1}{2}\) or \(-1\) (Proved)