Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\Delta AEH \sim \Delta BDH\left( {g.g} \right)\\
\Rightarrow \dfrac{{HA}}{{HB}} = \dfrac{{HE}}{{HD}}
\end{array}$
Khi đó
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {EHD}\left( {dd} \right)\\
\dfrac{{HA}}{{HB}} = \dfrac{{HE}}{{HD}}
\end{array} \right.\\
\Rightarrow \Delta AHB \sim \Delta EHD\left( {c.g.c} \right)\\
\Rightarrow \dfrac{{HA}}{{HE}} = \dfrac{{AB}}{{ED}}\\
\Rightarrow HA.ED = HE.AB
\end{array}$
c) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BDH} = \widehat {ADC} = {90^0}\\
\widehat {DBH} = \widehat {DAC}\left( { + \widehat {ACD} = {{90}^0}} \right)
\end{array} \right.\\
\Rightarrow \Delta BDH \sim \Delta ADC\left( {g.g} \right)\\
\Rightarrow \dfrac{{DB}}{{DA}} = \dfrac{{DH}}{{DC}}\\
\Rightarrow \dfrac{{DB}}{{DH}} = \dfrac{{DA}}{{DC}}
\end{array}$
Khi đó:
$\begin{array}{l}
\Delta ADC;\widehat D = {90^0}\\
\Rightarrow DC = \sqrt {A{C^2} - A{D^2}} = 4cm\\
\Rightarrow \dfrac{{DB}}{{DH}} = \dfrac{{DA}}{{DC}} = \dfrac{3}{4}
\end{array}$
