Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {6^2} + {8^2} = 100\\
\Rightarrow BC = 10\\
\Rightarrow AH = \dfrac{{AB.AC}}{{BC}} = \dfrac{{6.8}}{{10}} = 4,8\\
+ )B{H^2} = A{B^2} - A{H^2}\\
= {6^2} - 4,{8^2} = 12,96\\
\Rightarrow BH = 3,6\\
\Rightarrow CH = 10 - 3,6 = 6,4\\
b)Theo\,t/c:\\
\dfrac{{BD}}{{AB}} = \dfrac{{DC}}{{AC}}\\
\Rightarrow \dfrac{{BD}}{6} = \dfrac{{DC}}{8} = \dfrac{{BD + DC}}{{6 + 8}} = \dfrac{{10}}{{14}} = \dfrac{5}{7}\\
\Rightarrow DC = \dfrac{{40}}{7}\left( {cm} \right)\\
\Rightarrow DH = CH - DC = 6,4 - \dfrac{{40}}{7} = \dfrac{{24}}{{35}}\\
\Rightarrow AD = \sqrt {A{H^2} + D{H^2}} \\
= \sqrt {4,{8^2} + {{\left( {\dfrac{{24}}{{35}}} \right)}^2}} = \dfrac{{24\sqrt 2 }}{7}
\end{array}$
$\begin{array}{l}
c)Xet:\Delta ABE;\Delta HBI:\\
+ \widehat {ABE} = \widehat {HBI}\left( {gt} \right)\\
+ \widehat {BAE} = \widehat {BHI} = {90^0}\\
\Rightarrow \Delta ABE \sim \Delta HBI\left( {g - g} \right)\\
\Rightarrow \dfrac{{AB}}{{BH}} = \dfrac{{BE}}{{BI}}\\
\Rightarrow AB.BI = BE.BH\\
d){S_{ABH}} = \dfrac{1}{2}.BH.AH = \dfrac{1}{2}.3,6.4,8 = 8,64
\end{array}$
