Ta có:
$\quad 2AF = 3FC$
$\Leftrightarrow \dfrac{AF}{3} = \dfrac{FC}{2} = \dfrac{AF + FC}{3+2} =\dfrac{AC}{5}$
$\Rightarrow \dfrac{FC}{2} = \dfrac{AC}{5}$
$\Rightarrow \dfrac{FC}{AC} = \dfrac{2}{5}$
Ta lại có: $\dfrac{EB}{AB} = \dfrac{2}{5}$
Do đó: $EF//BC$ (Theo định lý $Thales$ đảo)