9. (10 points) In triangle $ABC$, $BE=1, EF=6, FC=2, BD=2AD$, the area of triangle $AHG$ is 4.86, and the area of triangle $GFC$ is 2, then the area of quadrilateral $BEHD$ is $\qquad$ .
【Analysis】Connect $B H, F H$, analyze the area relationship between the triangles, and find out the size of each area.
【Solution】Solution:
Connect $B H, F H$. According to the problem, the area ratio of $\triangle B E H$, $\triangle E F H$, and $\triangle C F H$ is 2: 6: 1. From $B D=2 A D$, we can deduce that the area of $\triangle C B D$ is twice that of $\triangle C A D$, and the area of $\triangle D B H$ is twice that of $\triangle D A H$. From these two conditions, we can conclude that the area of $\triangle B H C$ is twice that of $\triangle A H C$.

If we consider the area of $\triangle B H C$ as 9 parts, then the area of $\triangle A H C$ is $9 \div 2=4.5$ parts. If we view $\triangle A H C$ as the sum of the areas of two triangles with base $A G$, and $\triangle C H F$ as the sum of the areas of two triangles with base $F G$, then the area ratio of $\triangle A H C$ to $\triangle C H F$ is equal to $A G: F G$, which is 4.5: 2 = 9: 4. Based on $A G: F G=9: 4$ and the area of $\triangle A H G$ being 4.86, we can find that the area of $\triangle H G F$ is $4.86 \div 9 \times 4=2.16$.
Since the area of $\triangle C F G$ is $2.16+2=4.16$, the area of $\triangle B E H$ is $4.16 \div 2=2.08$.
$A H: H E=4.5: 8=9: 16$
Therefore, the area of triangle $A B D$ is $2.08 \div 16 \times 9=1.17$.
Based on $A D: B D=1: 2$, the area of $B D H$ is $1.17 \div 3 \times 2=0.78$.
Thus, the area of quadrilateral $B E H D$ is $0.78+2.08=2.86$.