\(AH^2=HB.HC\)
\(↔\dfrac{AH}{HB}=\dfrac{HC}{AH}\)
Xét \(ΔAHB\) và \(ΔCHA\):
\(\dfrac{AH}{HB}=\dfrac{HC}{AH}\)
\(\widehat{AHB}=\widehat{CHA}(=90^\circ)\)
\(→ΔAHB\backsim ΔCHA\)
\(→\widehat{ABH}=\widehat{CAH}\) và \(\widehat{HAB}=\widehat{HCA}\)
mà \(\widehat{HCA}+\widehat{CAH}=90^\circ\)
\(→\widehat{HAB}+\widehat{CAH}=90^\circ\)
\(→\widehat A=90^\circ\)
\(→ΔABC\) vuông tại \(A\)