Question

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

Answer 1

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

$(y+z)=ak,(z+x)=bk,(x+y)=ck$

b+c-a = $\frac{1}{k}(z+x+x+y-y-z)$ = $\frac{2x}{k}$

$c+a-b=\frac{1}{k}(x+y+y+z-x)=\frac{2y}{k}$

$a+3-c=\frac{1}{k}(y+z+z+x-x-y)=\frac{2z}{k}$

$\therefore \frac{x}{b+c-a}=\frac{x}{2x/k}=k/2$

$\frac{y}{c+a-b}=k/2=\frac{z}{a+b-c}=\frac{x}{b+c-a}$