Đáp án:

a) 

$\begin{array}{l}
Xet:\Delta EHA;\Delta ECB:\\
 + \widehat {HEA} = \widehat {BEC} = {90^0}\\
 + \widehat {EAH} = \widehat {EBC}\left( { + \widehat C = {{90}^0}} \right)\\
 \Leftrightarrow \Delta EHA \sim \Delta ECB\left( {g - g} \right)\\
 \Leftrightarrow \dfrac{{EH}}{{EC}} = \dfrac{{EA}}{{EB}}\\
 \Leftrightarrow EH.EB = EA.EC\\
b)\\
Xet:\Delta EHA;\Delta DHB:\\
 + \widehat {EHA} = \widehat {BHD}\\
 + \widehat {HEA} = \widehat {HDB} = {90^0}\\
 \Leftrightarrow \Delta EHA \sim \Delta DHB\left( {g - g} \right)\\
 \Leftrightarrow \dfrac{{HE}}{{HD}} = \dfrac{{HA}}{{HB}}\\
 \Leftrightarrow \dfrac{3}{{HD}} = \dfrac{{HA}}{4}\\
 \Leftrightarrow HA.HD = 12\\
 \Leftrightarrow HA.\left( {AD - HA} \right) = 12\\
 \Leftrightarrow HA.\left( {8 - HA} \right) = 12\\
 \Leftrightarrow H{A^2} - 8HA + 12 = 0\\
 \Leftrightarrow \left( {HA - 6} \right)\left( {HA - 2} \right) = 0\\
 \Leftrightarrow HA = 2\left( {do:HA < HD} \right)\\
 \Leftrightarrow HD = 8 - 2 = 6\left( {cm} \right)\\
Vậy\,HA = 2cm;HD = 6cm
\end{array}$