a) Theo định lý sin ta có: \(\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\)
Ta suy ra: \(\dfrac{a}{{\sin A}} = \dfrac{{b + c}}{{\sin B + \sin C}} = \dfrac{{2a}}{{\sin B + \sin C}}\)
\( \Rightarrow 2\sin A = \sin B + \sin C\)
b) Đối với tam giác ABC ta có: \(S = \dfrac{1}{2}ab\sin C = \dfrac{1}{2}{h_c}.c = \dfrac{{abc}}{{4R}}\).
Ta suy ra \({h_c} = \dfrac{{ab}}{{2R}}\). Tương tự ta có \({h_b} = \dfrac{{ac}}{{2R}},{h_a} = \dfrac{{bc}}{{2R}}\). Do đó:
\(\dfrac{1}{{{h_b}}} + \dfrac{1}{{{h_c}}} = 2R\left( {\dfrac{1}{{ac}} + \dfrac{1}{{ab}}} \right) = 2R\dfrac{{b + c}}{{abc}}\)mà b + c = 2a
Nên \(\dfrac{1}{{{h_b}}} + \dfrac{1}{{{h_c}}} = \dfrac{{2R.2a}}{{abc}} = \dfrac{{2R.2}}{{bc}} = \dfrac{2}{{{h_a}}}\)
Vậy \(\dfrac{2}{{{h_a}}} = \dfrac{1}{{{h_b}}} + \dfrac{1}{{{h_c}}}\).
