Đáp án:
$\begin{array}{l}
Do:2AF = 3FC\\
\Rightarrow \dfrac{{AF}}{{FC}} = \dfrac{3}{2}\\
\Rightarrow \dfrac{{AF}}{{AC}} = \dfrac{3}{5}\left( {AF + FC = AC} \right)\\
\dfrac{{EB}}{{AB}} = \dfrac{2}{5} \Rightarrow \dfrac{{AE}}{{AB}} = \dfrac{3}{5}\left( {AE + EB = AB} \right)\\
\Rightarrow \dfrac{{AF}}{{AC}} = \dfrac{{AE}}{{AB}} = \dfrac{3}{5}\\
\Rightarrow EF//BC\left( {Theo\,Talet\,đảo} \right)
\end{array}$