Question

If $\frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2}=\frac{ax+by}{a^2+b^2}$ then prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

Answer 1

Let $\frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2}=\frac{ax+by}{a^2+b^2}=k$ …………$\left(1\right)$

$by+cz=k(b^2+c^2)$, $cz+ax=k(c^2+a^2)$,

$ax+by=k(a^2+b^2)$

Adding these we get $2(ax+by+cz)=2k(a^2+b^2+c^2)$

$\Rightarrow ax+by+cz=k(a^2+b^2+c^2)$

$\Rightarrow ax=k(a^2+b^2+c^2)-(by+cz)$

$=k(a^2+b^2+c^2)-k(b^2+c^2)$ from above

$=k(a^2+b^2+c^2-b^2-c^2)=k{a}^2$

$\therefore x=ka$

Again $by=k(a^2+b^2+c^2)-(ax+cz)$

$by=k(a^2+b^2+c^2)-k(c^2+a^2)$ from above

$\Rightarrow by=k(a^2+b^2+c^2-a^2-c^2)=k{b}^2$

$\Rightarrow y=kb$

Similarly, $cz=k(a^2+b^2+c^2)-(ax+by)$

$\Rightarrow cz=k(a^2+b^2+c^2)-k(a^2+b^2)$ from above

$\Rightarrow cz=k(a^2+b^2+c^2-a^2-b^2)=k{c}^2$

$\therefore z=kc$

$\therefore \frac{x}{a}=k,\frac{y}{b}=k$ and $\frac{z}{c}=k$

Hence $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$ (or)

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$ hence proved.