Question

If $\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}$ then prove that. $\frac{a(b-c)}{y^2-z^2}=\frac{b(c-a)}{z^2-x^2}=\frac{c(a-b)}{x^2-y^2}$.

Answer 1

Let $\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=k$

$\Rightarrow a=k(y+z),b=k(z+x),c=k(x+y)$

(i) $\frac{a(b-c)}{y^{-}{z}^2}=\frac{k(y+z)\left[k\right(z+x)-k(x+y\left)\right]}{y^2-z^2}$

$=\frac{k^2(y+z)[z+x-x-y]}{(y+z)(y-z)}$

$=\frac{-k^2(y-z)}{(y-z)}=-k^2$

(ii) $\frac{b(c-a)}{z^2-x^2}=\frac{k(z+x)\left[k\right(x+y)-k(y+z\left)\right]}{z^2-x^2}$

$=\frac{k^2(z+x)[x+y-y-z]}{(z+x)(z-x)}$

$=\frac{-k^2(z-x)}{(z-x)}=-k^2$

(iii) $\frac{c(a-b)}{y^2-z^2}=\frac{k(x+y)\left[k\right(y+z)-k(z+x\left)\right]}{x^2-y^2}$

$=\frac{k^2(x+y)[y+z-z-x]}{(x+y)(x-y)}$

$=\frac{-k^2(x-y)}{(x-y)}=-k^2$

$\therefore \frac{a(b-c)}{y^2-z^2}=\frac{b(c-a)}{z^2-x^2}=\frac{c(a-b)}{x^2-y^2}$.