Xét tam giác ABC cân tại A, đường cao AH, BK, \(\widehat{BAH}=\widehat{HAC}=\alpha\)
Ta có \(\widehat{ABH}=\widehat{BCK};\widehat{AHB}=\widehat{BKC}\Rightarrow\Delta ABH\sim\Delta BCK\left(g-g\right)\)
\(\Rightarrow\frac{AH}{AB}=\frac{BK}{BC}\Rightarrow BK.AB=AH.BC=2AH.BH\)
Khi đó \(2sin\alpha.cos\alpha=2.\frac{AH}{AB}.\frac{BH}{AB}=\frac{BK.AB}{AB^2}=\frac{BK}{AB}=sinBAC=sin2\alpha\)