Đáp án:
$\begin{array}{l}
1)\\
Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {6^2} + {8^2} = 100\\
\Rightarrow BC = 10\left( {cm} \right)\\
\Rightarrow AH = \dfrac{{AB.AC}}{{BC}} = 4,8\left( {cm} \right)\\
B{H^2} + A{H^2} = A{B^2}\\
\Rightarrow B{H^2} = {6^2} - 4,{8^2} = 12,96\\
\Rightarrow BH = 3,6\left( {cm} \right)\\
\Rightarrow CH = 10 - 3,6 = 6,4\left( {cm} \right)\\
2)\Delta AEH;\Delta AHB:\\
+ \widehat {AEH} = \widehat {AHB} = {90^0}\\
+ \widehat {EAH}\,chung\\
\Rightarrow \Delta AEH \sim \Delta AHB\left( {g - g} \right)\\
\Rightarrow \dfrac{{AE}}{{AH}} = \dfrac{{AH}}{{AB}}\\
\Rightarrow AE.AB = A{H^2}\\
TT:\Delta AFH \sim \Delta AHC\left( {g - g} \right)\\
\Rightarrow AF.AC = A{H^2}\\
\Rightarrow AE.AB = AF.AC
\end{array}$
3) Tứ giác AEHF có 3 góc vuông
=> AEHF là hình chữ nhật
=> AH = EF = 4,8cm
4) Theo t.c đường phân giác:
$\begin{array}{l}
\dfrac{{BD}}{{AB}} = \dfrac{{CD}}{{AC}}\\
\Rightarrow \dfrac{{BD}}{6} = \dfrac{{CD}}{8} = \dfrac{{BD + CD}}{{14}} = \dfrac{{10}}{{14}} = \dfrac{5}{7}\\
\Rightarrow \left\{ \begin{array}{l}
BD = \dfrac{{30}}{7} = 4,3\left( {cm} \right)\\
CD = 5,7\left( {cm} \right)
\end{array} \right.
\end{array}$
