given \frac{x}{b+c-a}= \frac{y}{c+a-b} = \frac{z}{a+b-c} prove that \frac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}= \frac{x+y+z}{a+b+c}
1let
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 ..
11please 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]
106satan why dont u keep ur goku pic.. it was luking nice rohan
11mere 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]
1it 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\\