Tam giác \(ABC\) có ba đường trung tuyến \({m_a},{\rm{ }}{m_b},{\rm{ }}{m_c}\) thỏa mãn \(5m_a^2 = m_b^2 + m_c^2\). Khi đó tam giác này là tam giác gì?  

Ta có: \(\left\{ \begin{array}{l}m_a^2 = \dfrac{{{b^2} + {c^2}}}{2} - \dfrac{{{a^2}}}{4}\\m_b^2 = \dfrac{{{a^2} + {c^2}}}{2} - \dfrac{{{b^2}}}{4}\\m_c^2 = \dfrac{{{a^2} + {b^2}}}{2} - \dfrac{{{c^2}}}{4}\end{array} \right.\)

Mà: \(5m_a^2 = m_b^2 + m_c^2\)

\( \Rightarrow 5\left( {\dfrac{{{b^2} + {c^2}}}{2} - \dfrac{{{a^2}}}{4}} \right)\) \( = \dfrac{{{a^2} + {c^2}}}{2} - \dfrac{{{b^2}}}{4} + \dfrac{{{a^2} + {b^2}}}{2} - \dfrac{{{c^2}}}{4}\)

$\Leftrightarrow 5\left( {\dfrac{{{2b^2} + {2c^2}-a^2}}{4}} \right)$$=\dfrac{2{a^2} + 2{c^2} - {b^2} + 2{a^2} + 2{b^2} - {c^2}}{4}$

$\Leftrightarrow 5\left( {{2b^2} + {2c^2}-a^2} \right)$$=2{a^2} + 2{c^2} - {b^2} + 2{a^2} + 2{b^2} - {c^2}$

\( \Leftrightarrow 10{b^2} + 10{c^2} - 5{a^2} \)\(= 2{a^2} + 2{c^2} - {b^2} + 2{a^2} + 2{b^2} - {c^2}\)

\( \Leftrightarrow {b^2} + {c^2} = {a^2} \Rightarrow \) tam giác \(\Delta ABC\) vuông.