8. As shown in the figure, let $E$ and $F$ be points on the sides $AB$ and $AC$ of $\triangle ABC$, respectively. The line segments $CE$ and $BF$ intersect at point $D$. If the areas of $\triangle CDF$, $\triangle BCD$, and $\triangle BDE$ are 3, 7, and 7, respectively, then the area of quadrilateral $AEDF$ is $\qquad$
8. As shown in the figure, let $E$ and $F$ be points on the sides $AB$ and $AC$ of $\triangle ABC$, respectively. The line segments $CE$ and $BF$ intersect at point $D$. If the areas of $\triangle CDF$, $\triangle BCD$, and $\triangle BDE$ are 3, 7, and 7, respectively, then the area of quadrilateral $AEDF$ is $\qquad$
【Solution】Solution: Connect $AD$. Since the heights of $\triangle CDF$ and $\triangle BCD$ are equal, we have $FD:DB = 3:7$, so the area ratio of $\triangle AFD$ to $\triangle ABD$ is also 3:7,

Thus, the area of $\triangle AFD$ can be considered as 3 parts, and the area of $\triangle ABD$ as 7 parts,
$$
S_{\triangle BCD} = 7, S_{\triangle BDE} = 7
$$

Therefore, $CD = DE$,
$$
S_{\triangle ACD} = S_{\triangle ADE}, S_{\triangle ACD} + S_{\triangle BDE} = S_{\triangle ABD},
$$
$S_{\triangle ACD} + S_{\triangle BDE} = 7$ parts,
$S_{\triangle AFD} + S_{\triangle CDF} + S_{\triangle BDE} = 7$ parts,
3 parts $+ 3 + 7 = 7$ parts, so 1 part $= 2.5$,
$S_{\text{quadrilateral } AEDF} = 10$ parts - 7
$= 10 \times 2.5 - 7$
$= 25 - 7$
$= 18$
Answer: The area of quadrilateral $AEDF$ is 18.
Therefore, the answer is: 18.