Đáp án:
a)
Áp dụng t/c đường phân giác ta có:
$\begin{array}{l}
+ \frac{{EB}}{{AB}} = \frac{{EC}}{{AC}}\\
\Rightarrow \frac{{EB}}{{16}} = \frac{{EC}}{{32}} = \frac{{EC - EB}}{{32 - 16}} = \frac{{BC}}{{16}} = \frac{{21}}{{16}}\\
\Rightarrow \left\{ \begin{array}{l}
EB = 21\left( {cm} \right)\\
EC = 42\left( {cm} \right)
\end{array} \right.
\end{array}$
=> EB = BC = EC/2
=> B là trung điểm của EC
b)
$\begin{array}{l}
+ \frac{{BD}}{{AB}} = \frac{{CD}}{{AC}}\\
\Rightarrow \frac{{BD}}{{16}} = \frac{{CD}}{{32}} = \frac{{BD + CD}}{{16 + 32}} = \frac{{BC}}{{48}} = \frac{{21}}{{48}} = \frac{7}{{16}}\\
\Rightarrow \left\{ \begin{array}{l}
BD = 7\left( {cm} \right)\\
CD = 14\left( {cm} \right)
\end{array} \right.\\
\Rightarrow DE = EB + BD = 21 + 7 = 28\left( {cm} \right)
\end{array}$
