a) -△ABD và △HBI có: \(\widehat{ABD}=\widehat{HBI};\widehat{BAD}=\widehat{BHI}=90^0\).
\(\Rightarrow\)△ABD∼△HBI (g-g).
b) \(\widehat{BAH}=90^0-\widehat{ABC}=\widehat{ACH}\).
-△BAH và △ACH có: \(\widehat{BAH}=\widehat{ACH};\widehat{BHA}=\widehat{AHC}=90^0\).
\(\Rightarrow\)△BAH∼△ACH (g-g) \(\Rightarrow\dfrac{BH}{AH}=\dfrac{AH}{CH}\Rightarrow AH^2=BH.CH\)
\(\Rightarrow AH=\sqrt{BH.CH}=\sqrt{9.16}=12\left(cm\right)\)
c) △ABD∼△HBI \(\Rightarrow\widehat{ADI}=\widehat{HIB}\) mà \(\widehat{HIB}=\widehat{AID}\).
\(\Rightarrow\widehat{ADI}=\widehat{AID}\Rightarrow\)△AID cân tại A
△ABD∼△HBI \(\Rightarrow\dfrac{DA}{IH}=\dfrac{AB}{HB}\left(1\right)\)
△BAH và △BCA có: \(\widehat{BAC}=\widehat{BHA}=90^0;\widehat{ABC}\) là góc chung.
\(\Rightarrow\)△BAH∼△BCA (g-g) \(\Rightarrow\dfrac{BA}{BC}=\dfrac{BH}{BA}\Rightarrow\dfrac{BA}{BH}=\dfrac{BC}{BA}\left(2\right)\)
△ABC có: BD là phân giác \(\Rightarrow\dfrac{BC}{BA}=\dfrac{DC}{DA}\left(3\right)\)
-Từ (1) , (2) ,(3) suy ra: \(\dfrac{DA}{IH}=\dfrac{DC}{DA}\Rightarrow DA^2=DC.IH\)
