Question

If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{x}{a+b-c}$, then rove that $\frac{x+y+z}{a+b+c}=\frac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}$.

Answer 1

$\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{b+a-c}$ ....... $\left(i\right)$

$\frac{\text{sum of numerators}}{\text{sum of denominator}}=\frac{x+y+z}{b+c-a+c+a-b+b+a-c}$

$=\frac{x+y+z}{a+b+c}$ ........ $\left(ii\right)$

Multiply both numerator and denominator of $\left(i\right)$ by $y+z,z+x$ and $x+y$ respectively, we get

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

$\frac{\text{sum of numerators}}{\text{sum of denominator}}$

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

$=\frac{x(y+z)+y(z+x)+z(x+y)}{2ax+2by+2cz}$.....................(2)

As per (1) and (2) we get

$\frac{x+y+z}{a+b+c}=\frac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}$