Đáp án:
$\begin{array}{l}
1)\\
+ )Do:\frac{{\sin A}}{a} = \frac{{\sin B}}{b} = \frac{{{\mathop{\rm sinC}\nolimits} }}{c} = 2R\\
\Rightarrow \left\{ \begin{array}{l}
a = \frac{{\sin A}}{{2R}}\\
b = \frac{{\sin B}}{{2R}}\\
c = \frac{{\sin C}}{{2R}}
\end{array} \right.\\
Do:bc = {a^2} \Rightarrow \frac{{\sin B.\sin C}}{{4{R^2}}} = \frac{{{{\sin }^2}A}}{{4{R^2}}}\\
\Rightarrow sinB.\sin C = {\sin ^2}A\\
+ )S = \frac{1}{2}{h_a}.a = \frac{1}{2}{h_b}.b = \frac{1}{2}{h_{c.}}.c\\
\Rightarrow a = \frac{{2S}}{{{h_a}}};b = \frac{{2S}}{{{h_b}}};c = \frac{{2S}}{{{h_c}}}\\
b.c = {a^2}\\
\Rightarrow \frac{{2S}}{{{h_c}}}.\frac{{2S}}{{{h_b}}} = \frac{{4{S^2}}}{{h_a^2}}\\
\Rightarrow {h_b}.{h_c} = h_a^2\\
2)\\
\frac{{\sin A}}{a} = \frac{{\sin B}}{b} = \frac{{{\mathop{\rm sinC}\nolimits} }}{c} = \frac{{\sin B + \sin C}}{{b + c}}\\
\Rightarrow \left( {b + c} \right).\sin A = a.\left( {\sin B + \sin C} \right)
\end{array}$
