11. (16 points) As shown in Figure 1, the area of $\triangle ABC$ is 1. Through a point $O$ inside $\triangle ABC$, draw three lines parallel to the sides of the triangle: $DE$, $FG$, and $HI$. Points $D$, $E$, $F$, $G$, $H$, and $I$ are all on the sides of $\triangle ABC$. Find the minimum area of the hexagon $DGHFEI$.
11. From the given, we know that quadrilaterals $A G O H$, $B I O D$, and $C E O F$ are all parallelograms.

Also, $\triangle G D O$, $\triangle O I F$, and $\triangle H O E$ are all similar to $\triangle A B C$. Since the ratio of the areas of similar triangles is equal to the square of the similarity ratio, we have:
$$
\begin{array}{l}
\frac{S_{\triangle G D O}}{S_{\triangle A B C}}+\frac{S_{\triangle O I F}}{S_{\triangle A B C}}+\frac{S_{\triangle H O E}}{S_{\triangle A B C}} \\
=\left(\frac{D O}{B C}\right)^{2}+\left(\frac{I F}{B C}\right)^{2}+\left(\frac{O E}{B C}\right)^{2} \\
=\left(\frac{B I}{B C}\right)^{2}+\left(\frac{I F}{B C}\right)^{2}+\left(\frac{F C}{B C}\right)^{2} \\
\geqslant \frac{1}{3}\left(\frac{B I}{B C}+\frac{I F}{B C}+\frac{F C}{B C}\right)^{2}=\frac{1}{3} \\
\Rightarrow S_{\triangle G D O}+S_{\triangle O I F}+S_{\triangle H O E} \geqslant \frac{1}{3} \\
\Rightarrow S_{\triangle A G H}+S_{\triangle B I D}+S_{\triangle C E F} \\
=\frac{1}{2}\left(S_{\square A G O H}+S_{\square B I O D}+S_{\square C E O F}\right) \\
=\frac{1}{2}\left[S_{\triangle A B C}-\left(S_{\triangle G D O}+S_{\triangle O I F}+S_{\triangle H O B}\right)\right] \\
\leqslant \frac{1}{2}\left(1-\frac{1}{3}\right)=\frac{1}{3} \\
\Rightarrow S_{\text {hexagon } D G H E F I} \\
=S_{\triangle A B C}-\left(S_{\triangle A G H}+S_{\triangle B I D}+S_{\triangle C E F}\right) \\
\geqslant 1-\frac{1}{3}=\frac{2}{3} \text {. } \\
\end{array}
$$

When $O$ is the centroid of $\triangle A B C$, the equality holds.
Thus, the minimum area of hexagon $D G H E F I$ is $\frac{2}{3}$.