Let \(\cfrac{x}{y+z}=\cfrac{y}{z+x}=\cfrac{z}{x+y}=k\).
[\(k\) is a nonzero constant]
So, \(x=k(y+z)\), \(y=k(z+x)\), 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, \(1-2k=0\) → \(-2k=-1\) → \(k=\cfrac{1}{2}\)
\(\therefore\) 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 equals \(\cfrac{1}{2}\) or \(-1\). (Proved)