Xét \(\Delta ABCvà\Delta HBA\) có:
\(\widehat{B}\) chung
\(\widehat{BAC}=\widehat{AHB}=\left(90^o\right)\)
\(\Rightarrow\Delta ABC\) đồng dạng với \(\Delta HBA\)(1)
=> \(\dfrac{AB}{HB}=\dfrac{BC}{AB}\Leftrightarrow AB^2=HB.BC\)
b) Xét \(\Delta ABCvà\Delta HAC\) có:
\(\widehat{C}\) chung
\(\widehat{AHC}=\widehat{BAC}\left(=90^o\right)\)
\(\Rightarrow\Delta ABC\) đồng dạng với \(\Delta HAC\)(2)\(\Rightarrow\dfrac{AC}{HC}=\dfrac{BC}{AC}\Leftrightarrow AC^2=HC.BC\)
c)Từ (1)(2) \(\Rightarrow\Delta HBA\) đồng dạng với \(\Delta HAC\)
\(\Rightarrow\dfrac{AH}{CH}=\dfrac{HB}{AH}\Leftrightarrow AH^2=CH.HB\)