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,

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

Answer 1

Let ${\displaystyle a(y+z)=b(z+x)=c(x+y)=k}$

⇒ ${\displaystyle \frac{a(y+z)}{abc}=\frac{b(z+x)}{abc}=\frac{c(x+y)}{abc}}$

${\displaystyle \frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}}$ = k

${\displaystyle k=\frac{(x+y)-(z+x)}{ab-ac}=\frac{y-z}{a(b-c)}}$    ...(1)

${\displaystyle k=\frac{(y+z)-(x+y)}{bc-ab}=\frac{z-x}{b(c-a)}}$     ...(2)

${\displaystyle k=\frac{(z+x)-(y+z)}{ac-bc} =\frac{x-y}{c(a-b)}}$    ...(3)

${\displaystyle (\therefore \frac{a}{b}=\frac{c}{d}=k=\frac{a-c}{b-d})}$

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