Question

If $a(y+z)=b(z+x)=c(x+y)$ and out of $a,b,c$ no two of them are equal then show that, $\frac{y-z}{a(b-c)}=\frac{z-x}{b(c-a)}=\frac{x-y}{c(a-b)}$.

Answer 1

$a(y+z)=b(z+x)=c(x+y)$

Assume $n=a(y+z)=b(z+x)=c(x+y)$

$n=a(y+z),n=b(z+x),n=c(x+y)$

$y+z=\frac{n}{a}---\left(1\right)$

$z+x=\frac{n}{b}-----\left(2\right)$

$x+y=\frac{n}{c}----\left(3\right)$

subtracting (2)-(1), (3)-(2), (1)-(3) we get

$x-y=n(a-b)/ab$----(4)

$y-z=n(b-c)/bc$----(5)

$z-x=n(c-a)/ac$----(6)

$x-y/c(a-b)=\frac{\left[n\frac{(a-b)}{ab}\right]}{c(a-b)}$

$=n/abc$

$y-z/a(b-c)=\frac{\left[n\frac{(b-c)}{bc}\right]}{a(b-c)}$

$=n/abc$

$z-x/b(c-a)=\frac{\left[n\frac{(c-a)}{ac}\right]}{b(c-a)}$

$=n/abc$

$\frac{y-z}{a(b-c)}=\frac{z-x}{b(c-a)}=\frac{x-y}{c(a-b)}$