Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2}\\
= {6^2} + {8^2} = 100\\
\Leftrightarrow BC = 10\left( {cm} \right)\\
Vậy\,BC = 10cm\\
b)\\
Xet:\Delta ABC;\Delta HBA:\\
+ \widehat {BAC} = \widehat {BHA} = {90^0}\\
+ \widehat B\,chung\\
\Leftrightarrow \Delta ABC \sim \Delta HBA\left( {g - g} \right)\\
\Leftrightarrow \dfrac{{AB}}{{BH}} = \dfrac{{BC}}{{AB}}\\
\Leftrightarrow A{B^2} = BH.BC\\
c)\\
A{B^2} = BH.BC\\
\Leftrightarrow BH = \dfrac{{{6^2}}}{{10}} = 3,6\left( {cm} \right)\\
Theo\,t/c:\\
\dfrac{{BD}}{{AB}} = \dfrac{{CD}}{{AC}} = \dfrac{{BD + CD}}{{AB + AC}} = \dfrac{{BC}}{{6 + 8}} = \dfrac{5}{7}\\
\Leftrightarrow BD = \dfrac{5}{7}.AB = \dfrac{5}{7}.6 = \dfrac{{30}}{7}\left( {cm} \right)\\
Vậy\,BH = 3,6cm;BD = \dfrac{{30}}{7}cm\\
d)HD = BD - BH\\
= \dfrac{{30}}{7} - 3,6 = \dfrac{{24}}{{35}}\\
AH = \dfrac{{AB.AC}}{{BC}} = \dfrac{{6.8}}{{10}} = 4,8\left( {cm} \right)\\
Theo\,Pytago:\\
A{H^2} + H{D^2} = A{D^2}\\
\Leftrightarrow A{D^2} = 4,{8^2} + {\left( {\dfrac{{24}}{{35}}} \right)^2}\\
\Leftrightarrow AD = \dfrac{{24\sqrt 2 }}{7}\left( {cm} \right)\\
Vậy\,AD = \dfrac{{24\sqrt 2 }}{7}cm
\end{array}$
