Question Number 225932 by Rojarani last updated on 16/Nov/25
$$\:{If},\:\frac{{by}+{cz}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\frac{{cz}+{ax}}{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{ax}+{by}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\:{then}\:{prove}\:{that},\:\frac{{x}}{{a}}=\frac{{y}}{{b}}=\frac{{z}}{{c}} \\ $$
Answered by som(math1967) last updated on 16/Nov/25
![((by+cz)/(b^2 +c^2 ))=((cz+ax)/(c^2 +a^2 ))=((ax+by)/(a^2 +b^2 )) =((2(by+cz+ax))/(2(a^2 +b^2 +c^2 ))) [∵(a/b)=(c/d)=(e/f)=((a+c+e)/(b+d+f))] (((by+cz+ax)−(by+cz))/((a^2 +b^2 +c^2 )−(b^2 +c^2 ))) =(((by+cz+ax)−(cz+ax))/((a^2 +b^2 +c^2 )−(c^2 +a^2 ))) =(((by+cz+ax)−(ax+by))/((a^2 +b^2 +c^2 )−(a^2 +b^2 ))) ⇒((ax)/a^2 )=((by)/b^2 )=((cz)/c^2 ) ∴(x/a)=(y/b)=(z/c)](https://www.tinkutara.com/question/Q225936.png)
$$\:\frac{{by}+{cz}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\frac{{cz}+{ax}}{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{ax}+{by}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left({by}+{cz}+{ax}\right)}{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)} \\ $$$$\left[\because\frac{{a}}{{b}}=\frac{{c}}{{d}}=\frac{{e}}{{f}}=\frac{{a}+{c}+{e}}{{b}+{d}+{f}}\right] \\ $$$$\:\frac{\left({by}+{cz}+{ax}\right)−\left({by}+{cz}\right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)} \\ $$$$=\frac{\left({by}+{cz}+{ax}\right)−\left({cz}+{ax}\right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\: \\ $$$$=\frac{\left({by}+{cz}+{ax}\right)−\left({ax}+{by}\right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\frac{{ax}}{{a}^{\mathrm{2}} }=\frac{{by}}{{b}^{\mathrm{2}} }=\frac{{cz}}{{c}^{\mathrm{2}} } \\ $$$$\therefore\frac{{x}}{{a}}=\frac{{y}}{{b}}=\frac{{z}}{{c}} \\ $$