Answer: $\frac{8}{3}$
Solution: Refer to the figure below.
Observe that $\triangle G E C \cong \triangle B I G \cong \triangle F C H \cong \triangle H J D$.
Moreover, $B I=I J=J D$. This is because $I O=\frac{1}{2} B I$ and $O J=\frac{1}{2} J D$ (the points $I$ and $J$ are centroids)

Therefore:
$$
\begin{aligned}
{[A E F] } & =[A I J]+[E I J F] \\
& =[A I J]+[B C D] \\
& =\frac{1}{3}[B A D]+[B C D] \\
& =\frac{1}{3}(2)+2=\frac{8}{3}
\end{aligned}
$$
5. Given the square $A B C D$ with side length of 2 , triangle $A E F$ is constructed so that $A E$ bisects side $B C$, and $A F$ bisects side $C D$. Moreover, $E F$ is parallel to the diagonal $B D$ and passes through $C$. Find the area of triangle $A E F$.