nice one

let

x/(b+c-a)=y/(c+a-b)=z/(a+b-c)=k ---- (i)
then

x+y/(b+c-a+c+a-b)=k
hence

x+y=2kc
similarly

x+z=2kb and y+z=2ka

so the expression in L.H.S reduces to

x(2ka)+y(2kb)+z(2kc)/2(ax+by+cz)=k

further from (i)

k=x+y+z/(b+c-a+c+a-b+a+b-c)=x+y+z/(a+b+c)

hence L.H.S=R.H.S proved ..

please can u justify this step

x+y/(b+c-a+c+a-b)=k
hence

x+y=2kc
similarly........................

if u will add both then u will have to multiply the denominator
how u can add them in denominator[7][7][7]

mere bhai tu keh raha hai

well do you remember that if

a/b=c/d=k

then k=a+b/c+d as a=bk and c=dk then it cancels out to k ...

to k = a+c/b+d nahin hona chahiya??[7][7][7]

thanx 4 the concept[1]

it could also be solved using the property
\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k ,\ k=\frac{ma+lb+ne}{mb+ld+nf}\ where\ l,m,n\ are\ positive\ integers\\