10. (10 points) As shown in the figure, in triangle $ABC$, $AD=2BD$, $AD=EC$, $BC=18$, the area of triangle $AFC$ is equal to the area of quadrilateral $DBEF$. What is the length of $AB$?
【Analysis】This problem can be approached by finding the lengths of $A D$ and $B D$ to determine the length of $A B$. Since $A D = E C$, we only need to find the length of $E C$ to know the length of $A D$. Given that the area of triangle $A F C$ is equal to the area of quadrilateral $D B E F$, we can find that $\frac{\mathrm{BE}}{\mathrm{EC}} = \frac{2}{1}$, which allows us to determine $E C = \frac{1}{3} B C = \frac{1}{3} \times 18 = 6$; then, using $A D = 2 B D$, we find that $2 B D = 6$, so $B D = 3$. Therefore, we can find the length of $A B$ and solve the problem.

【Solution】Solution: Let $S_{\triangle A F C} = S_{\text{quadrilateral } D B E F} = 1, S \triangle C E F = x, S \triangle A D F = y$,
Since $A D = 2 B D$, we have $\frac{S \triangle_{\mathrm{ADC}}}{S_{\mathrm{SCBD}}} = \frac{\mathrm{AD}}{\mathrm{BD}} = \frac{2}{1}$, which means $\frac{S_{\triangle \mathrm{ADF}} + S_{\triangle \mathrm{AFC}}}{S \text{ quadrilateralDBEF} + S_{\triangle C E F}} = \frac{y + 1}{x + 1} = \frac{2}{1}$,
Solving this, we get $y = 2x + 1$.
Then, $\frac{S_{\triangle A B E}}{S_{\triangle A E C}} = \frac{S \triangle A D F + S \text{ quadrilateral } \mathrm{DAEF}}{S \triangle A F C + S \triangle E F C} = \frac{(2x + 1) + 1}{x + 1} = \frac{2}{1}$,
So, $\frac{\mathrm{BE}}{\mathrm{EC}} = \frac{2}{1}$.
Therefore, $E C = \frac{1}{1+2} \times B C = \frac{1}{3} \times 18 = 6$, and $A D = E C = 6$.
Since $A D = 2 B D$, we have $B D = A D \div 2 = 6 \div 2 = 3$.
In summary, $A B = A D + B D = 6 + 3 = 9$.
Answer: The length of $A B$ is 9.