AMC 10 B

Triangles $\triangle AFE$ and $\triangle CFB$ are similar by Angle-Angle. The ratio of corresponding sides is $1:2$, so the ratio of their areas is $1:4$. Furthermore, consider $\triangle AFE$ and $\triangle AFB$. Because $FE:FB = 1:2$ and the heights of the two triangles corresponding to bases $\overline{FE}$ and $\overline{FB}$ are the same, their areas are in a $1:2$ ratio. Finally, observe that the area of $\triangle ABE$ is $\frac{1}{4}$ the area of parallelogram $ABCD$. Therefore, using the area of $\triangle AFE$ as unit, the area of $\triangle CFB$ is $4$, the area of $\triangle AFB$ is $2$, the area of quadrilateral $ABCD$ is $4 \cdot (1+2) = 12$, and the area of quadrilateral $CDEF$ is $12 - (1+4+2) = 5$. The requested ratio of the areas of quadrilateral $CDEF$ and $\triangle CFB$ is $5:4$.