Lời giải:
Xét tam giác $ADC$ có \(F\in AC; E\in AD; B\in DC\) và $B,E,F$ thẳng hàng nên theo định lý Menelaus ta có:
\(\frac{AF}{CF}.\frac{DE}{AE}.\frac{CB}{DB}=1\)
\(\Leftrightarrow \frac{AF}{CF}.1.\frac{4}{3}=1\Leftrightarrow \frac{AF}{CF}=\frac{3}{4}\)
\(\Rightarrow \frac{AF}{AC}=\frac{3}{7}\)
\(\Rightarrow \frac{S_{ABF}}{S_{ABC}}=\frac{AF}{AC}=\frac{3}{7}\Leftrightarrow S_{ABF}=\frac{180}{7}\)
\(\Leftrightarrow S_{AEF}+S_{ABE}=\frac{180}{7}(1)\)
Mặt khác:
\(\frac{S_{ABD}}{S_{ABC}}=\frac{BD}{BC}=\frac{3}{4}\Rightarrow S_{ABD}=45\)
\(\frac{S_{ABE}}{S_{ABD}}=\frac{AE}{AD}=\frac{1}{2}\Rightarrow S_{ABE}=\frac{S_{ABD}}{2}=\frac{45}{2}(2)\)
Từ (1),(2)\(\Rightarrow S_{AEF}=\frac{180}{7}-S_{ABE}=\frac{180}{7}-\frac{45}{2}=\frac{45}{14}\) (cm)
