3. Let $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, E^{\prime}, F^{\prime}$ be the midpoints of the six sides $AB, BC, CD, DE, EF, FA$ of hexagon $ABCDEF$. Try to express the area of hexagon $ABCDEF$ using the areas of $\triangle ABC^{\prime}, \triangle BCD^{\prime}, \triangle CDE^{\prime}, \triangle DEF^{\prime}, \triangle EFA^{\prime}, \triangle FAB^{\prime}$.
3. $A^{\prime}$ is the midpoint of $A B$, so $2 S_{\triangle E F A^{\prime}}=S_{\triangle E F A}+S_{\triangle E F B}$. Similarly, we have:
$$
\begin{array}{l}
2 S_{\triangle F A B^{\prime}}=S_{\triangle F A B}+S_{\triangle F A C}, \\
2 S_{\triangle A B C}=S_{\triangle A B C}+S_{\triangle A B D}, \\
2 S_{\triangle B C D^{\prime}}=S_{\triangle B C D}+S_{\triangle B C E}, \\
2 S_{\triangle C D E}=S_{\angle, C D E}-S_{\triangle A C D F}, \\
2 S_{\triangle \triangle E N^{T}}=S_{\triangle, D E F}-S_{\triangle D E A} .
\end{array}
$$

On the other hand,
$$
\begin{aligned}
\dot{S}_{A B C D E F} & =S_{\triangle E F A}+S_{\triangle D E A}+S_{\triangle A B D}+S_{\triangle B C D} \\
& =S_{\triangle F A B}+S_{\triangle E F B}+S_{\triangle B C E}+S_{\triangle C D E} \\
& =S_{\triangle A B C}+S_{\triangle F A C}+S_{\triangle C D F}+S_{\triangle D E F}
\end{aligned}
$$

Therefore, the area of hexagon $A B C D E F$ is exactly $\frac{2}{3}$ of the sum of the areas of the six given triangles.