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:

$\begin{array}{l}
\dfrac{{BD}}{{AB}} = \dfrac{{CD}}{{AC}}\\
 \Rightarrow \dfrac{{BD}}{6} = \dfrac{{CD}}{8} = \dfrac{{BD + CD}}{{14}} = \dfrac{{10}}{{14}} = \dfrac{5}{7}\\
 \Rightarrow \left\{ \begin{array}{l}
BD = \dfrac{{30}}{7} = 4,3\left( {cm} \right)\\
CD = 5,7\left( {cm} \right)
\end{array} \right.
\end{array}$