In the triangle ABC, take A as the origin, and write b = overrightarrowAB and c = overrightarrowAC…

Step 1. The point $D$. With $BD : DC = 2 : 3$, the section formula gives $\overrightarrow{AD} = \dfrac{3\vec b + 2\vec c}{5}$.

Step 2. The point $E$. $\overrightarrow{AE} = \dfrac{1}{2}\overrightarrow{AD} = \dfrac{3\vec b + 2\vec c}{10}$.

Step 3. The line $BE$. A point on it is $\vec b + s(\overrightarrow{AE} - \vec b) = \left(1 - \dfrac{7s}{10}\right)\vec b + \dfrac{s}{5}\vec c$.

Step 4. The point $F$. It lies on $AC$, so the $\vec b$ part is zero. Then $s = \dfrac{10}{7}$, and $\overrightarrow{AF} = \dfrac{10}{7} \times \dfrac{1}{5}\vec c = \dfrac{2}{7}\vec c$.

Step 5. The ratio. $AF = \dfrac{2}{7}AC$ and $FC = \dfrac{5}{7}AC$, so $AF : FC = 2 : 5$.