Đáp án:
a) Xét ΔCEF và ΔHAB có:
$\begin{array}{l}
+ \widehat {FCE} = \widehat {AHB} = {90^0}\\
+ \widehat {CFE} = \widehat {HBA}\left( { + \widehat {{\rm{HAF}}} = {{90}^0}} \right)\\
\Rightarrow \Delta CEF \sim \Delta \Delta HAB\left( {g - g} \right)\\
b)Xét:\Delta ADE;\Delta DHE:\\
+ \widehat {ADE} = \widehat {DHE} = {90^0}\\
+ \widehat E\,chung\\
\Rightarrow \Delta ADE \sim \Delta DHE\left( {g - g} \right)\\
\Rightarrow \dfrac{{AD}}{{DH}} = \dfrac{{AE}}{{DE}}\\
\Rightarrow AD.DE = DH.AE\\
c)Xét:\Delta ADB;\Delta HAB:\\
+ \widehat {DAB} = \widehat {AHB} = {90^0}\\
+ \widehat {ABH}\,chung\\
\Rightarrow \Delta ADB \sim \Delta HAB\left( {g - g} \right)\\
\Rightarrow \dfrac{{AD}}{{AH}} = \dfrac{{BD}}{{AB}}\\
\Rightarrow AH = \dfrac{{AD.AB}}{{BD}} = \dfrac{{4.3}}{{\sqrt {{3^2} + {4^2}} }} = \dfrac{{12}}{5} = 2,4\left( {cm} \right)\\
Vay\,AH = 2,4cm
\end{array}$
