Kẻ các đường cao AH, BK, CP
\(=>\left\{{}\begin{matrix}AH=AB.sin\widehat{B}\\CP=AC.sin\widehat{A}\\BK=CB.sin\widehat{C}\end{matrix}\right.\left(1\right)\)
\(S_{ABC}=\dfrac{1}{2}AB.CP=\dfrac{1}{2}BC.AH=\dfrac{1}{2}CA.BK\left(2\right)\)
Thay (1) vào (2) ta đc
\(S_{ABC}=\dfrac{1}{2}AB.AC.sin\widehat{A}=\dfrac{1}{2}BC.BA.sin\widehat{B}=\dfrac{1}{2}CA.CB.sin\widehat{C}\)