Cho tam giác ABC có h_b + h_c = 2h_a. Chứng minh rằng:

$\frac{1}{{\sin B}} + \frac{1}{{\sin C}} = \frac{2}{{\sin {\rm{A}}}}$.

Ta có ${S_{ABC}} = \frac{1}{2}{h_a}a = \frac{1}{2}{h_b}b = \frac{1}{2}{h_c}c$

Suy ra ${h_b} = \frac{{2{{\rm{S}}_{ABC}}}}{b},{h_c} = \frac{{2{{\rm{S}}_{ABC}}}}{c},{h_a} = \frac{{2{{\rm{S}}_{ABC}}}}{a}$

Ta có h_b + h_c = 2h_a

$\Leftrightarrow \frac{{2{{\rm{S}}_{ABC}}}}{b} + \frac{{2{{\rm{S}}_{ABC}}}}{c} = \frac{{{\rm{4}}{{\rm{S}}_{ABC}}}}{a}$

$\Leftrightarrow \frac{1}{b} + \frac{1}{c} = \frac{2}{a}$

Áp dụng định lí sin ta có

$\frac{1}{{\sin {\rm{A}}}} + \frac{1}{{\sin B}} = \frac{{2R}}{b} + \frac{{2{\rm{R}}}}{c} = 2{\rm{R}}\left( {\frac{1}{b} + \frac{1}{c}} \right) = 2{\rm{R}}.\frac{2}{a} = \frac{2}{{\sin {\rm{A}}}}$

Vậy $\frac{1}{{\sin B}} + \frac{1}{{\sin C}} = \frac{2}{{\sin {\rm{A}}}}$.