Đáp án:
a) Áp dụng Pytago trong tam giác vuông ta có:
$\begin{array}{l}
B{C^2} = A{B^2} + A{C^2} = {8^2} + {6^2} = 100\\
\Rightarrow BC = 10\left( {cm} \right)\\
\Rightarrow {S_{ABC}} = \dfrac{1}{2}.AH.BC = \dfrac{1}{2}.AB.AC\\
\Rightarrow AH = \dfrac{{8.6}}{{10}} = 4,8\left( {cm} \right)\\
\Rightarrow C{H^2} = A{C^2} - A{H^2} = {6^2} - 4,{8^2} = 12,96\\
\Rightarrow CH = 3,6\left( {cm} \right)\\
2)CD//AB\\
\Rightarrow CD \bot AC\\
\Rightarrow \widehat {DCA} = {90^0}\\
\Rightarrow \widehat {HCD} + \widehat {HCA} = {90^0}\\
Do:\widehat {HAC} + \widehat {HCA} = {90^0}\\
\Rightarrow \widehat {HCD} = \widehat {HAC}\\
Xet:\Delta AHC;\Delta CHD:\\
+ \widehat {AHC} = \widehat {CHD} = {90^0}\\
+ \widehat {HAC} = \widehat {HCD}\\
\Rightarrow \Delta AHC \sim \Delta CHD\left( {g - g} \right)\\
3)\Delta ABC;\Delta CAD:\\
+ \widehat {BAC} = \widehat {ACD} = {90^0}\\
+ \widehat {ACB} = \widehat {CDA}\left( {cmt} \right)\\
\Rightarrow \Delta ABC \sim \Delta CAD\left( {g - g} \right)\\
\Rightarrow \dfrac{{AC}}{{DC}} = \dfrac{{AB}}{{AC}}\\
\Rightarrow A{C^2} = AB.DC
\end{array}$
