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 equals \(\cfrac{1}{2}\).
Again, if \(x + y + z = 0\), then \(y + z = -x\)
\(\therefore \cfrac{x}{y+z} = \cfrac{x}{-x} = -1\)
Similarly, \(z + x = -y\)
\(\therefore \cfrac{y}{z+x} = \cfrac{y}{-y} = -1\)
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 \(\cfrac{1}{2}\) or \(-1\) (Proved).