Đáp án:
$\begin{array}{l}
a)Theo\,t.c:\\
\dfrac{{DB}}{{DC}} = \dfrac{{AB}}{{AC}} = \dfrac{8}{6} = \dfrac{4}{3}\\
b)Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {8^2} + {6^2} = 100\\
\Rightarrow BC = 10\left( {cm} \right)\\
\Rightarrow DB + DC = 10\left( {cm} \right)\\
Do:\dfrac{{DB}}{{DC}} = \dfrac{4}{3}\\
\Rightarrow \dfrac{{DB}}{4} = \dfrac{{DC}}{3} = \dfrac{{DB + DC}}{{4 + 3}} = \dfrac{{10}}{7}\\
\Rightarrow \left\{ \begin{array}{l}
DB = \dfrac{{40}}{7} = 5,71\left( {cm} \right)\\
DC = \dfrac{{30}}{7} = 4,29\left( {cm} \right)
\end{array} \right.\\
c)\\
Xet:\Delta AHB;\Delta CHA:\\
+ \widehat {AHB} = \widehat {CHA} = {90^0}\\
+ \widehat {HAB} = \widehat {HCA}\left( {cùng\,phụ\,với\,\widehat {ABC}} \right)\\
\Rightarrow \Delta AHB \sim \Delta CHA\left( {g - g} \right)\\
\Rightarrow \dfrac{{{S_{AHB}}}}{{{S_{CHA}}}} = {\left( {\dfrac{{AB}}{{AC}}} \right)^2} = \dfrac{{16}}{9}\\
d){S_{ABC}} = \dfrac{1}{2}.AH.BC = \dfrac{1}{2}.AB.AC\\
\Rightarrow AH = \dfrac{{AB.AC}}{{BC}} = \dfrac{{6.8}}{{10}} = 4,8\left( {cm} \right)
\end{array}$
