Question

(13 * ) 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

$$\begin{gathered} \frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2}=\frac{ax+by}{a^2+b^2} \\ \Rightarrow \frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2}=\frac{ax+by}{a^2+b^2}=\frac{\left(by+cz\right)+\left(cz+ax\right)+\left(ax+by\right)}{\left(b^2+c^2\right)+\left(c^2+a^2\right)+\left(a^2+b^2\right)}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2}=\frac{ax+by}{a^2+b^2}=\frac{2ax+2by+2cz}{2a^2+2b^2+2c^2} \\ \Rightarrow \frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2}=\frac{ax+by}{a^2+b^2}=\frac{2\left(ax+by+cz\right)}{2\left(a^2+b^2+c^2\right)} \\ \Rightarrow \frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2}=\frac{ax+by}{a^2+b^2}=\frac{ax+by+cz}{a^2+b^2+c^2} \end{gathered}$$
Again applying theorem of equal ratios, we get

$\frac{\left(by+cz\right)-\left(ax+by+cz\right)}{\left(b^2+c^2\right)-\left(a^2+b^2+c^2\right)}=\frac{\left(cz+ax\right)-\left(ax+by+cz\right)}{\left(c^2+a^2\right)-\left(a^2+b^2+c^2\right)}=\frac{\left(ax+by\right)-\left(ax+by+cz\right)}{\left(a^2+b^2\right)-\left(a^2+b^2+c^2\right)}=\frac{ax+by+cz}{a^2+b^2+c^2}$
$$\begin{gathered} \Rightarrow \frac{-ax}{-a^2}=\frac{-by}{-b^2}=\frac{-cz}{-c^2} \\ \Rightarrow \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \end{gathered}$$