15. (15 points) As shown in the figure, $E$ is a point on side $CD$ of parallelogram $ABCD$, and $BD$ intersects $AE$ at point $F$. It is known that the area of triangle $AFD$ is 6, and the area of triangle $DEF$ is 4. Find the area of quadrilateral $BCEF$.

保留源文本的换行和格式，直接输出翻译结果如下：

15. (15 points) As shown in the figure, $E$ is a point on side $CD$ of parallelogram $ABCD$, and $BD$ intersects $AE$ at point $F$. It is known that the area of triangle $AFD$ is 6, and the area of triangle $DEF$ is 4. Find the area of quadrilateral $BCEF$.
【Solution】Solution: According to the property that the area of a triangle is proportional to its base when the height is constant, we get: $A F: E F=6: 4=3$ : 2;

Since in parallelogram $A B C D$, triangle $A F B$ is similar to triangle $D E F$, we have $A B: D E=A F: E F$ $=3: 2$;
$A B=D C$, so $D C: D E=3: 2$,
Therefore, the area of triangle $D B C$: the area of triangle $A D E$ $=3: 2$,
The area of triangle $A D E$ is: $6+4=10$,
Thus, the area of triangle $D B C$ is: $3 \times 10 \div 2=15$,
Therefore, the area of quadrilateral $B C E F$ is: $15-4=11$;
Answer: The area of quadrilateral $B C E F$ is 11.