Xét hai tam giác AHB và CHA có:
\(\left\{{}\begin{matrix}\widehat{AHB}=\widehat{CHA}=90^0\\\widehat{HAB}=\widehat{HCA}\left(\text{cùng phụ }\widehat{B}\right)\end{matrix}\right.\)
\(\Rightarrow\Delta AHB\sim\Delta CHA\left(g.g\right)\)
\(\Rightarrow\dfrac{AH}{HC}=\dfrac{BH}{AH}\Rightarrow AH^2=BH.HC\)
\(\Rightarrow AH=\sqrt{BH.HC}=30\)