Đáp án:

$\begin{array}{l}
Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {4^2} + {3^2} = 25\\
 \Rightarrow BC = 5\left( {cm} \right)\\
Theo\,t.c:\\
\dfrac{{AD}}{{AC}} = \dfrac{{BD}}{{BC}}\\
 \Rightarrow \dfrac{{AD}}{3} = \dfrac{{BD}}{5} = \dfrac{{AD + BD}}{{3 + 5}} = \dfrac{4}{8} = \dfrac{1}{2}\\
 \Rightarrow \left\{ \begin{array}{l}
AD = 1,5\left( {cm} \right)\\
BD = 2,5\left( {cm} \right)
\end{array} \right.\\
b)Xét\,\Delta HAC;\Delta ABC:\\
 + \widehat {AHC} = \widehat {BAC} = {90^0}\\
 + \widehat {ACB}\,chung\\
 \Rightarrow \Delta HAC \sim \Delta ABC\left( {g - g} \right)\\
c)Do:\Delta HAC \sim \Delta ABC\\
 \Rightarrow \dfrac{{CH}}{{AC}} = \dfrac{{AC}}{{BC}} = \dfrac{3}{5}\\
 \Rightarrow CH = \dfrac{3}{5}.3 = \dfrac{9}{5} = 1,8\left( {cm} \right)\\
d)Do:\\
\widehat {BCK} + \widehat {CBK} = {90^0}\\
\widehat {CBK} = \widehat {ADC}\\
\widehat {ADC} + \widehat {DAK} = {90^0}\\
 \Rightarrow \widehat {BCK} = \widehat {KAD}\\
Xét\,\Delta CDB;\Delta ADK\,:\\
 + \widehat {BCD} = \widehat {DAK}\\
 + \widehat {CDB} = \widehat {ADK}\left( {đối\,đỉnh} \right)\\
 \Rightarrow \Delta CDB \sim \Delta ADK\left( {g - g} \right)
\end{array}$