Ta có:[tex]AH^2=BH.CH\Rightarrow \frac{AH}{BH}=\frac{CH}{AH}[/tex]
Xét 2 tam giác [tex]\Delta AHB và \Delta CHA[/tex]:
[tex]\left\{\begin{matrix} \frac{AH}{BH}=\frac{CH}{AH}\\ \widehat{AHB}=\widehat{CHA}=90^o \end{matrix}\right.\Rightarrow \Delta AHB\sim \Delta CHA\Rightarrow \widehat{ABH}=\widehat{CAH}\Rightarrow \widehat{ABH}+\widehat{BAH}=\widehat{CAH}+\widehat{BAH}\Rightarrow \widehat{BAC}=90^o(đpcm)[/tex]