a, Vì \(\left\{{}\begin{matrix}AB=AC\\BH=HC\\AH.chung\end{matrix}\right.\) nên \(\Delta AHB=\Delta AHC\left(c.c.c\right)\)
b, \(\Delta AHB=\Delta AHC\left(c.c.c\right)\Rightarrow\widehat{AHB}=\widehat{AHC}\)
Mà \(\widehat{AHB}+\widehat{AHC}=180^0\Rightarrow\widehat{AHB}=\widehat{AHC}=90^0\)
Vậy \(AH\perp BC\)

