a) Tam giác $BAE$ và tam giác $BHI$ có:
$\begin{array}{l} \left\{ \begin{array}{l} \widehat {BAE} = \widehat {BHI} = {90^o}\\ \widehat {IBH} = \widehat {EBA}\left( {gt} \right) \end{array} \right.\\ \Rightarrow \Delta BAE \sim \Delta BHI\left( {g - g} \right) \end{array}$
b) Tam giác $BHA$ và $BAC$ có:
$\begin{array}{l} \left\{ \begin{array}{l} \widehat {ABH}\,chung\\ \widehat {BHA} = \widehat {BAC} = {90^o} \end{array} \right.\\ \Rightarrow \Delta BHA \sim \Delta BAC\left( {g - g} \right)\\ \Rightarrow \dfrac{{BH}}{{BA}} = \dfrac{{BA}}{{BC}} \Rightarrow A{B^2} = BH.BC \end{array}$
c) $\Delta BAE \sim \Delta BHI$
$\begin{array}{l} \Delta BAE \sim \Delta BHI\left( {g - g} \right)\\ \Rightarrow \dfrac{{IH}}{{AE}} = \dfrac{{BH}}{{BA}} = \dfrac{{BA}}{{BC}} \end{array}$
Tam giác $BAI$ và tam giác $BCE$ có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BAI} = \widehat {BCE}\left( { + \widehat B = {{90}^o}} \right)\\
\widehat {ABI} = \widehat {CBE}\left( {gt} \right)
\end{array} \right.\\
\Rightarrow \Delta BAI \sim \Delta BCE\left( {g - g} \right)\\
\Rightarrow \dfrac{{BA}}{{IA}} = \dfrac{{BC}}{{EC}} \Rightarrow \dfrac{{BA}}{{BC}} = \dfrac{{IA}}{{EC}}\\
\Rightarrow \dfrac{{IH}}{{AE}} = \dfrac{{IA}}{{EC}}
\end{array}$
