Question

If $\frac{bc}{y+z}=\frac{ac}{z+x}=\frac{ab}{x+y}$. then show that $\frac{a(b-c)}{y-z}=\frac{b(c-a)}{z-x}=\frac{c(a-b)}{x-y}$

Answer 1

let $\frac{bc}{y+z}$ ----Eq$1$

$\frac{ac}{z+x}$ ----Eq$2$

$\frac{ab}{x+y}$ ----Eq$3$

now,

subtract Eq$3$ from Eq$1$

$\frac{bc-ab}{y+z-x-y}$ [as they are equal}

subtract Eq$1$ from Eq$2$

$\frac{ac-bc}{z+x-y-z}$ [as they are equal}

subtract Eq$2$ from Eq$2$

$\frac{ab-ac}{x+y-z-x}$ [as they are equal}

We Get,

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