Giải thích các bước giải:

Ta có:

$BD=3DC$

$\to \dfrac{BD}3=\dfrac{DC}1=\dfrac{BD+DC}{3+1}=\dfrac{BC}4$

$\to \dfrac{BD}{BC}=\dfrac34, \dfrac{DC}{BC}=\dfrac14$

$\to S_{ABD}=\dfrac34S_{ACB}=\dfrac34\cdot 140=105$

     $S_{ADC}=\dfrac14S_{ABC}=\dfrac14\cdot 140=35$

Vì $AE=ED\to E$ là trung điểm $AD$

$\to S_{ABE}=S_{BDE}=\dfrac12S_{ABD}=\dfrac12\cdot 105=52.5$

     $S_{ACE}=S_{DCE}=\dfrac12S_{ACD}=\dfrac12\cdot 35=17.5$

     $S_{BEC}=S_{ABC}-S_{ABE}-S_{ACE}=140-52.5-17.5=70$

$\to \dfrac{FA}{FC}=\dfrac{S_{BEA}}{S_{BEC}}=\dfrac34$

$\to \dfrac{FA}{FA+FC}=\dfrac3{3+4}$

$\to \dfrac{AF}{AC}=\dfrac37$

$\to S_{AEF}=\dfrac37S_{AEC}=\dfrac37\cdot 17.5=7.5$