Đáp án:
a) Theo Pytago và tính chất đường phân giác ta có:
$\begin{array}{l}
+ B{C^2} = A{B^2} + A{C^2}\\
= {12^2} + {16^2}\\
= 400\\
\Rightarrow BC = 20\left( {cm} \right)\\
+ \dfrac{{BD}}{{AB}} = \dfrac{{CD}}{{AC}} = \dfrac{{BD + CD}}{{AB + AC}} = \dfrac{{BC}}{{12 + 16}} = \dfrac{{20}}{{28}} = \dfrac{5}{7}\\
\Rightarrow \left\{ \begin{array}{l}
BD = \dfrac{5}{7}.12 = \dfrac{{60}}{7}\\
CD = \dfrac{5}{7}.16 = \dfrac{{80}}{7}
\end{array} \right.\\
Vậy:BC = 20cm;BD = \dfrac{{60}}{7}cm;CD = \dfrac{{80}}{7}cm\\
b){S_{ABC}} = \dfrac{1}{2}.AH.BC = \dfrac{1}{2}.AB.AC\\
\Rightarrow AH = \dfrac{{12.16}}{{20}} = 9,6\left( {cm} \right)\\
B{H^2} = A{B^2} - A{H^2}\\
\Rightarrow B{H^2} = {12^2} - 9,{6^2}\\
\Rightarrow BH = 7,2\left( {cm} \right)\\
\Rightarrow HD = BD - BH = \dfrac{{60}}{7} - 7,2 = \dfrac{{48}}{{35}}\left( {cm} \right)\\
Theo\,Pytago:\\
A{D^2} = A{H^2} + H{D^2}\\
\Rightarrow AD = \sqrt {9,{6^2} + {{\left( {\dfrac{{48}}{{35}}} \right)}^2}} = \dfrac{{48\sqrt 2 }}{7}\left( {cm} \right)
\end{array}$
