Question

by + cz/b^2 + c^2 = cz + ax/c^2 + a^2 = ax + by/a^2 + b^2 then prove that x/a = y/b = z/c

please help me​

Answer 1

Answer:

$\frac{by+cz}{b^{2}+c^{2} } =\frac{cz+ax}{c^{2}+a^{2} }=\frac{ax+by}{a^{2}+b^{2} }\\=\frac{2(ax+by+cz)}{2(a^{2}+b^{2}+c^{2}) }\\=\frac{ax+by+cz}{a^{2}+b^{2}+c^{2} }...Adding\\\\ \frac{by+cz}{b^{2}+c^{2} }=\frac{ax+by+cz}{a^{2}+b^{2}+c^{2} }\\=>\frac{by+cz}{b^{2}+c^{2} }=\frac{b^{2}+c^{2}}{a^{2}+b^{2}+c^{2} }...(by alternendo)\\=>\frac{by+ cz - ax- by- cz }{ax+by+cz }=\frac{b^2 + c^2 - a^2 - b^2 - c^2}{a^2 + b^2 + c^2 }$

$=>\frac{- ax }{ax+by+cz }=\frac{-a}{a^2 + b^2 + c^2 }\\=>\frac{x }{ax+by+cz }=\frac{a}{a^2 + b^2 + c^2 }\\=>\frac{x }{a }=\frac{ax+by+cz }{a^2 + b^2 + c^2 }\\\\Similarly, \frac{y }{b }=\frac{ax+by+cz }{a^2 + b^2 + c^2 }\\Similarly, \frac{z }{c }=\frac{ax+by+cz }{a^2 + b^2 + c^2 }\\\\Hence, \frac{x }{a }=\frac{y }{b }=\frac{z }{c }\\Hence proved.$