Đáp án:
a) Do AD là phân giác góc A nên ta có:
$\begin{array}{l}
\frac{{BD}}{{AB}} = \frac{{CD}}{{AC}}\\
\Rightarrow \frac{{BD}}{{12}} = \frac{{CD}}{{20}} = \frac{{BD + CD}}{{12 + 20}} = \frac{{BC}}{{32}} = \frac{{28}}{{32}} = \frac{7}{8}\\
\Rightarrow \left\{ \begin{array}{l}
BD = 12.\frac{7}{8} = 10,5\left( {cm} \right)\\
CD = 20.\frac{7}{8} = 17,5\left( {cm} \right)
\end{array} \right.\\
Theo\,talet:do\,DE//AB\\
\Rightarrow \frac{{DE}}{{AB}} = \frac{{CD}}{{BC}} = \frac{{17,5}}{{28}} = \frac{5}{8}\\
\Rightarrow DE = 12.\frac{5}{8} = 7,5\left( {cm} \right)
\end{array}$
b) Ta có:
$\begin{array}{l}
\frac{{BD}}{{BC}} = \frac{{10,5}}{{28}} = \frac{3}{8}\\
\Rightarrow \frac{{{S_{ABD}}}}{{{S_{ABC}}}} = \frac{{\frac{1}{2}.h.BD}}{{\frac{1}{2}.h.BC}} = \frac{3}{8}\\
\Rightarrow {S_{ABD}} = \frac{3}{8}S\\
\Rightarrow {S_{DEC}} = \frac{{CD}}{{BC}}.\frac{{DE}}{{AB}} = \frac{5}{8}.\frac{5}{8} = \frac{{25}}{{64}}S\\
\Rightarrow {S_{ADE}} = S - \frac{3}{8}S - \frac{{25}}{{64}}S = \frac{{15}}{{64}}S
\end{array}$
