Giải thích các bước giải:
3,
a,
Áp dụng công thức đường trung tuyến trong tam giác ta có:
\(\begin{array}{l}
A{M^2} = \frac{{A{B^2} + A{C^2}}}{2} - \frac{{B{C^2}}}{4}\\
\Leftrightarrow {c^2} = \frac{{{c^2} + {b^2}}}{2} - \frac{{{a^2}}}{4}\\
\Leftrightarrow 4{c^2} = 2{c^2} + 2{b^2} - {a^2}\\
\Leftrightarrow {a^2} = 2\left( {{b^2} - {c^2}} \right)
\end{array}\)
b,
Ta có:
\(\begin{array}{l}
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\\
{\sin ^2}A = {\left( {\frac{a}{{2R}}} \right)^2} = \frac{{{a^2}}}{{{{\left( {2R} \right)}^2}}} = \frac{{2\left( {{b^2} - {c^2}} \right)}}{{{{\left( {2R} \right)}^2}}}\\
= 2.\left[ {{{\left( {\frac{b}{{2R}}} \right)}^2} - {{\left( {\frac{c}{{2R}}} \right)}^2}} \right] = 2.\left( {{{\sin }^2}B - {{\sin }^2}C} \right)
\end{array}\)
4,
\(\begin{array}{l}
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\\
a,\\
\sin B.\cos C + \sin C.\cos B\\
= \frac{b}{{2R}}.\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} + \frac{c}{{2R}}.\frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\
= \frac{{{a^2} + {b^2} - {c^2}}}{{4R.a}} + \frac{{{a^2} + {c^2} - {b^2}}}{{4R.a}}\\
= \frac{{2{a^2}}}{{4R.a}} = \frac{a}{{2R}} = \sin A\\
b,\\
2R.\sin B.\sin C\\
= 2R.\frac{b}{{2R}}.\frac{c}{{2R}}\\
= \frac{{bc}}{{2R}}\\
\left. \begin{array}{l}
{S_{ABC}} = \frac{{abc}}{{4R}} = \frac{a}{2}.\frac{{bc}}{{2R}}\\
{S_{ABC}} = \frac{1}{2}a.{h_a}
\end{array} \right\} \Rightarrow \frac{{bc}}{{2R}} = {h_a} \Leftrightarrow {h_a} = 2R.\sin B.\sin C
\end{array}\)
