Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BAD} = \widehat {BHI} = {90^0}\\
\widehat {ABD} = \widehat {HBI} = \dfrac{1}{2}\widehat {ABC}
\end{array} \right.\\
\Rightarrow \Delta ABD \sim \Delta HBI\left( {g.g} \right)
\end{array}$
b) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {CHA} = {90^0}\\
\widehat {HAB} = \widehat {HCA}\left( { + \widehat {CAH} = {{90}^0}} \right)
\end{array} \right.\\
\Rightarrow \Delta AHB \sim \Delta CHA\left( {g.g} \right)\\
\Rightarrow \dfrac{{AH}}{{CH}} = \dfrac{{HB}}{{HA}}\\
\Rightarrow A{H^2} = HB.CH
\end{array}$
Lại có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {CAB} = {90^0}\\
\widehat Bchung
\end{array} \right.\\
\Rightarrow \Delta AHB \sim \Delta CAB\left( {g.g} \right)\\
\Rightarrow \dfrac{{AB}}{{CB}} = \dfrac{{HB}}{{AB}}\\
\Rightarrow A{B^2} = HB.CB\\
\Rightarrow AB = \sqrt {HB.\left( {HB + HC} \right)} = 15cm
\end{array}$
Vậy $AB=15cm$
