Question Number 173369 by AgniMath last updated on 10/Jul/22
Answered by mr W last updated on 10/Jul/22
$$\frac{{by}+{cz}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\frac{{cz}+{ax}}{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{ax}+{by}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={k},\:{say} \\ $$$${ax}+{by}={k}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\:\:\:\:…\left({i}\right) \\ $$$${by}+{cz}={k}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:\:\:\:\:…\left({ii}\right) \\ $$$${cz}+{ax}={k}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\:\:\:…\left({iii}\right) \\ $$$$\Sigma: \\ $$$${ax}+{by}+{cz}={k}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:\:\:…\left({iv}\right) \\ $$$$\left({iv}\right)−\left({ii}\right): \\ $$$$\Rightarrow{ax}={ka}^{\mathrm{2}} \:\Rightarrow\frac{{x}}{{a}}={k} \\ $$$${similarly} \\ $$$$\Rightarrow\frac{{y}}{{b}}={k},\:\frac{{z}}{{c}}={k} \\ $$$$\Rightarrow\frac{{a}}{{x}}=\frac{{b}}{{b}}=\frac{{c}}{{z}}={k} \\ $$