6. (10 points) As shown in the figure, the area of square $A B C D$ is 840 square centimeters, $A E=E B, B F=2 F C, D F$ intersects $E C$ at $G$. Then the area of quadrilateral $A E G D$ is square centimeters.
510
【Analysis】This figure easily leads one to think about finding the areas of triangles $BCE$, $CDF$, and $CGF$. The most challenging part is finding the area of $\triangle CGF$. According to the given conditions, we should draw $GM \perp BC$ with the intersection point at $M$. This forms two sets of similar triangles corresponding to $\triangle BCE$ and $\triangle CDF$. By using the property that the ratio of the areas of similar triangles is equal to the square of the similarity ratio, we can find the area of $\triangle CGF$. Finally, based on the figure, we can calculate the area of the required shape.

【Solution】Draw $GM \perp BC$ with the intersection point at $M$,
$$
\begin{array}{l}
\therefore \triangle FMG \sim \triangle FCD \Rightarrow FM: FC = MG: CD, \\
\because BF = 2FC \Rightarrow BC = 3FC, \\
\therefore MG = 3FM, \\
\because \triangle CGM: \triangle CEB \Rightarrow CM: CB = GM: BE, \\
BC = 2BE, \\
\therefore GM = \frac{1}{2} CM = 3FM \Rightarrow CM = 6FM, \\
\therefore FM: FC = 1: 7, \\
CM: CB = 2: 7, \\
S \triangle BCE = \frac{1}{4} \times \square ABCD = 210, \\
S \triangle CGM = 4 \div 49 \times 210 = 840 \div 49 = 120 \div 7, \\
S \triangle CDF = S \square ABCD \div 6 = 140, \\
S \triangle MGF = 140 \times 1 \div 49 = 140 \div 49 = 20 \div 7, \\
S \triangle CGM + \triangle MGF = 120 \div 7 + 20 \div 7 = 20, \\
840 - 210 - 140 + 20 = 510 \text{ (square centimeters). }
\end{array}
$$

Therefore, the area of quadrilateral $AEGD$ is 510 square centimeters.