\(\cfrac{y+z-x}{b+c-a} = \cfrac{z+x-y}{c+a-b} = \cfrac{x+y-z}{a+b-c}\) \(= \cfrac{y+z-x + z+x-y + x+y-z}{b+c-a + c+a-b + a+b-c}\)
[Using the method of combination]
\(\therefore\) Each ratio \(= \cfrac{x+y+z}{a+b+c}\) So, \(\cfrac{y+z-x}{b+c-a} = \cfrac{x+y+z}{a+b+c}\) \(= \cfrac{(y+z-x) - (x+y+z)}{(b+c-a) - (a+b+c)} = \cfrac{-2x}{-2a} = \cfrac{x}{a}\)
Similarly, \(\cfrac{z+x-y}{c+a-b} = \cfrac{x+y+z}{a+b+c}\) \(= \cfrac{(z+x-y) - (x+y+z)}{(c+a-b) - (a+b+c)} = \cfrac{-2y}{-2b} = \cfrac{y}{b}\)
And, \(\cfrac{x+y-z}{a+b-c} = \cfrac{x+y+z}{a+b+c}\) \(= \cfrac{(x+y-z) - (x+y+z)}{(a+b-c) - (a+b+c)} = \cfrac{-2z}{-2c} = \cfrac{z}{c}\)
\(\therefore \cfrac{x}{a} = \cfrac{y}{b} = \cfrac{z}{c}\) [Proved]