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Question Number 169101 by mnjuly1970 last updated on 24/Apr/22

In AB^Δ C : m_b ^( 2) + m_c ^( 2) = 5 m_a ^( 2) prove that : A^( ∧) = 90^( °) m_a : ( median )

$$ \\ $$$$\:\:\:\:\:\:{In}\:\:{A}\overset{\Delta} {{B}C}\::\:\:\:\:{m}_{{b}} ^{\:\mathrm{2}} \:+\:{m}_{{c}} ^{\:\mathrm{2}} =\:\mathrm{5}\:{m}_{{a}} ^{\:\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\::\:\:\:\overset{\:\:\wedge} {{A}}\:=\:\mathrm{90}^{\:°} \\ $$$$\:\:\:\:\:\:\:\:\:\:{m}_{{a}} :\:\:\left(\:{median}\:\right) \\ $$

Answered by mr W last updated on 24/Apr/22

m_a =((√(2b^2 +2c^2 −a^2 ))/2) m_b =((√(2c^2 +2a^2 −b^2 ))/2) m_c =((√(2a^2 +2b^2 −c^2 ))/2) m_b ^2 +m_c ^2 =((2c^2 +2a^2 −b^2 +2a^2 +2b^2 −c^2 )/4) m_b ^2 +m_c ^2 =((4a^2 +b^2 +c^2 )/4) 5m_a ^2 =((5(2b^2 +2c^2 −a^2 ))/4)=((4a^2 +b^2 +c^2 )/4) 10b^2 +10c^2 −5a^2 =4a^2 +b^2 +c^2 b^2 +c^2 =a^2 ⇒right angled triangle with ∠A=90°

$${m}_{{a}} =\frac{\sqrt{\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{b}} =\frac{\sqrt{\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{c}} =\frac{\sqrt{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} =\frac{\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} =\frac{\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{5}{m}_{{a}} ^{\mathrm{2}} =\frac{\mathrm{5}\left(\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{\mathrm{4}}=\frac{\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{10}{b}^{\mathrm{2}} +\mathrm{10}{c}^{\mathrm{2}} −\mathrm{5}{a}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \: \\ $$$$\Rightarrow{right}\:{angled}\:{triangle}\:{with}\:\angle{A}=\mathrm{90}° \\ $$

Commented by mnjuly1970 last updated on 24/Apr/22

thanks alot sir W ....

$$\:\:\:{thanks}\:{alot}\:\:{sir}\:\:{W}\:.... \\ $$

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