Let \( \cfrac{x}{y+z} = \cfrac{y}{z+x} = \cfrac{z}{x+y} = k \)
[\(k\) is a nonzero constant]
Therefore, \(x = k(y+z)\), \(y = k(z+x)\), and
\(z = k(x+y)\)
\(\therefore x+y+z = k(y+z) + k(z+x)\)\(+ k(x+y)\)
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, \(1 - 2k = 0\) Or, \(-2k = -1\) \(\therefore k = \cfrac{1}{2}\)
\(\therefore\) Each ratio is \(\cfrac{1}{2}\)
Again, if \(x+y+z = 0\), then \(y+z = -x\)
\(\therefore \cfrac{x}{y+z} = \cfrac{x}{-x} = -1\)
Similarly, since \(z+x = -y\)
\(\therefore \cfrac{y}{z+x} = \cfrac{y}{-y} = -1\)
And \(x+y = -z\)
\(\therefore \cfrac{z}{x+y} = \cfrac{z}{-z} = -1\)
\(\therefore\) If \( \cfrac{x}{y+z} = \cfrac{y}{z+x} = \cfrac{z}{x+y} \), then each ratio is either \(\cfrac{1}{2}\) or \(-1\). (Proved)