Giải thích các bước giải:
Ta có:
$\begin{array}{l}
m_c^2 = \dfrac{{2\left( {{a^2} + {b^2}} \right) - {c^2}}}{4}\\
\Leftrightarrow {\left( {\dfrac{{c\sqrt 3 }}{2}} \right)^2} = \dfrac{{2\left( {{a^2} + {b^2}} \right) - {c^2}}}{4}\\
\Leftrightarrow \dfrac{{3{c^2}}}{4} = \dfrac{{2\left( {{a^2} + {b^2}} \right) - {c^2}}}{4}\\
\Leftrightarrow 2\left( {{a^2} + {b^2}} \right) - {c^2} = 3{c^2}\\
\Leftrightarrow {a^2} + {b^2} = 2{c^2}
\end{array}$
Khi đó:
$\begin{array}{l}
+) m_a^2 = \dfrac{{2\left( {{b^2} + {c^2}} \right) - {a^2}}}{4} = \dfrac{{2{b^2} + {a^2} + {b^2} - {a^2}}}{4} = \dfrac{{3{b^2}}}{4}\\
\Rightarrow {m_a} = \dfrac{{b\sqrt 3 }}{2}\\
+) m_b^2 = \dfrac{{2\left( {{a^2} + {c^2}} \right) - {b^2}}}{4} = \dfrac{{2{a^2} + {a^2} + {b^2} - {b^2}}}{4} = \dfrac{{3{a^2}}}{4}\\
\Rightarrow {m_b} = \dfrac{{a\sqrt 3 }}{2}\\
\Rightarrow {m_a} + {m_b} + {m_c} = \dfrac{{b\sqrt 3 }}{2} + \dfrac{{a\sqrt 3 }}{2} + \dfrac{{c\sqrt 3 }}{2} = \dfrac{{\sqrt 3 }}{2}\left( {a + b + c} \right)
\end{array}$
Ta có điều phải chứng minh.
