Question Number 169101 by mnjuly1970 last updated on 24/Apr/22
$$ \\ $$$$\:\:\:\:\:\:{In}\:\:{A}\overset{\Delta} {{B}C}\::\:\:\:\:{m}_{{b}} ^{\:\mathrm{2}} \:+\:{m}_{{c}} ^{\:\mathrm{2}} =\:\mathrm{5}\:{m}_{{a}} ^{\:\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\::\:\:\:\overset{\:\:\wedge} {{A}}\:=\:\mathrm{90}^{\:°} \\ $$$$\:\:\:\:\:\:\:\:\:\:{m}_{{a}} :\:\:\left(\:{median}\:\right) \\ $$
Answered by mr W last updated on 24/Apr/22
$${m}_{{a}} =\frac{\sqrt{\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{b}} =\frac{\sqrt{\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{c}} =\frac{\sqrt{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} =\frac{\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} =\frac{\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{5}{m}_{{a}} ^{\mathrm{2}} =\frac{\mathrm{5}\left(\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{\mathrm{4}}=\frac{\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{10}{b}^{\mathrm{2}} +\mathrm{10}{c}^{\mathrm{2}} −\mathrm{5}{a}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \: \\ $$$$\Rightarrow{right}\:{angled}\:{triangle}\:{with}\:\angle{A}=\mathrm{90}° \\ $$
Commented by mnjuly1970 last updated on 24/Apr/22
$$\:\:\:{thanks}\:{alot}\:\:{sir}\:\:{W}\:…. \\ $$