Hai đường trung tuyến $BM,CN$ vuông góc với nhau thì
\(\begin{array}{l} {\left( {\dfrac{2}{3}{m_b}} \right)^2} + {\left( {\dfrac{2}{3}{m_c}} \right)^2} = {a^2}\\ \Leftrightarrow \dfrac{4}{9}\left( {\dfrac{{{a^2} + {b^2}}}{2} - \dfrac{{{c^2}}}{4}} \right) + \dfrac{4}{9}\left( {\dfrac{{{a^2} + {c^2}}}{2} - \dfrac{{{b^2}}}{4}} \right) = {a^2}\\ \Leftrightarrow 5{a^2} = {b^2} + {c^2} \end{array}\)
Mặt khác \(a^2=b^2+c^2-2bc.cosA\)
\(\begin{array}{l} \Leftrightarrow {a^2} = 5{a^2} - 2bc.\cos A \Rightarrow bc = \dfrac{{2{a^2}}}{{\cos A}} = \dfrac{{2{a^2}}}{{\cos \alpha }}\\ {S_{ABC}} = \dfrac{1}{2}bc.\sin A = {a^2}.\tan \alpha \end{array}\)
