Question Number 40960 by behi83417@gmail.com last updated on 30/Jul/18
Commented by MrW3 last updated on 30/Jul/18
$${a}×\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}+{b}×\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}={c}×\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$×\mathrm{2}{abc}: \\ $$$${a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)={c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{4}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} ={a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{c}^{\mathrm{4}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{4}} −{b}^{\mathrm{4}} =−{c}^{\mathrm{4}} \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} ={c}^{\mathrm{4}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\pm{c}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:{or}\:{b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\Rightarrow{right}\:{angled}\:{triangle}\:{in}\:{both}\:{cases} \\ $$