${S_{ABC}} = \frac{1}{2}AB.CF = \frac{1}{2}.AB.BC\sin B(\alpha )$
${S_{ABC}} = \frac{1}{2}AC.BE = \frac{1}{2}.AC.AB\operatorname{sinA} (\beta )$
${S_{ABC}} = \frac{1}{2}BC.AD = \frac{1}{2}.BC.AC\operatorname{sinC} (\gamma )$
${S_{ABC}} = \frac{{AB.AC.BC}}{{4R}}(\delta )$
$\because (\alpha ) \wedge (\beta ) \wedge (\gamma ) \wedge (\delta )$
$\therefore \frac{{AB}}{{\sin C}} = \frac{{AC}}{{\sin B}} = \frac{{BC}}{{\sin A}} = 2R(Q.E.D.)$
Bài viết đã được chỉnh sửa nội dung bởi Hoang Huynh: 21-09-2021 - 21:32