Ans: D
In $\Delta ABE$ and $\Delta ACD$,
In $\Delta ABE$ and $\Delta ACD$,
$\begin{array}{ll}
\angle BAE = \angle CAE & \text{(common angle)} \\
\angle ABE = \angle ACD & \text{(corr. $\angle$s, $BE\text{//}CD$)} \\
\angle AEB = \angle ADC & \text{(corr. $\angle$s, $BE\text{//}CD$)}
\end{array}$
$\therefore \Delta ABE \sim \Delta ACD$ (A.A.A.).
Hence, we have
$\begin{array}{rcl}
\dfrac{BE}{CD} & = & \dfrac{AB}{AC} ~\text{(corr. sides of $\sim\Delta$ in prop.)} \\
\dfrac{BE}{9} & = & \dfrac{8}{8+4} \\
BE & = & 6 \text{ cm}
\end{array}$
