Xét hai tam giác CHA và CAB có:
\(\left\{{}\begin{matrix}\widehat{CHA}=\widehat{CAB}=90^0\\\widehat{C}-chung\end{matrix}\right.\)
\(\Rightarrow\Delta CHA\sim\Delta CAB\left(g.g\right)\)
\(\Rightarrow\dfrac{CH}{CA}=\dfrac{CA}{CB}\Rightarrow CA^2=CB.CH\)









