Question

Answer 1

Vẽ đường cao AD,BE

Ta có: \(\dfrac{a}{sinA}=\dfrac{BC}{\dfrac{BE}{AB}}=\dfrac{AB.BC}{BE}=\dfrac{AB.BC.AC}{BE.AC}=\dfrac{AB.AC.BC}{2S_{ABC}}\)

Lại có; \(\dfrac{b}{sinB}=\dfrac{AC}{\dfrac{AD}{AB}}=\dfrac{AB.AC}{AD}=\dfrac{AB.AC.BC}{AD.BC}=\dfrac{AB.AC.BC}{2S_{ABC}}\)

Mặt khác: \(\dfrac{c}{sinC}=\dfrac{AB}{\dfrac{AD}{AC}}=\dfrac{AB.AC}{AD}=\dfrac{AB.AC.BC}{AD.BC}=\dfrac{AB.AC.BC}{2S_{ABC}}\)

Từ đó \(\Rightarrow\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\)