Question

If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c},$ prove that

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

Answer 1

Each of the given fractions $=\frac{sumofnumerators}{sumofdenominators}$

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

Again, if we multiply both numerator and denominator of the three given fractions by $y+z,z+x,x+y$ respectively,

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

$=\frac{sumofnumerators}{sumofdenominators}$

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

$\therefore$ From $\left(i\right)$ and $\left(ii\right)$,

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

Hence proved.