Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
B{D^2} = A{B^2} + A{D^2}\\
= {8^2} + {6^2} = 100\\
\Rightarrow BD = 10\left( {cm} \right)\\
b)Xét:\Delta ADH;\Delta BDA:\\
+ \widehat {ADH}\,chung\\
+ \widehat {AHD} = \widehat {BAD} = {90^0}\\
\Rightarrow \Delta ADH \sim \Delta BDA\left( {g - g} \right)\\
c)Do:\Delta ADH \sim \Delta BDA\\
\Rightarrow \dfrac{{AD}}{{DB}} = \dfrac{{DH}}{{AD}}\\
\Rightarrow A{D^2} = DH.DB\\
d)Xét:\Delta AHB;\Delta BCD:\\
+ \widehat {AHB} = \widehat {BCD} = {90^0}\\
+ \widehat {ABH} = \widehat {BDC}\left( {so\,le\,trong} \right)\\
\Rightarrow \Delta AHB \sim \Delta BCD\left( {g - g} \right)\\
e)A{D^2} = DH.DB\\
\Rightarrow DH = \dfrac{{A{D^2}}}{{DB}} = \dfrac{{{6^2}}}{{10}} = 3,6\left( {cm} \right)\\
Do:\Delta AHB \sim \Delta BCD\\
\Rightarrow \dfrac{{AH}}{{BC}} = \dfrac{{AB}}{{BD}}\\
\Rightarrow AH = \dfrac{8}{{10}}.6 = 4,8\left( {cm} \right)
\end{array}$
