Answer: E

Solution:

We can find the area of the shaded region by finding the area of $\triangle DHC$ and subtracting out the two unshaded triangles.

Extend $\overline{DH}$ so that it hits $B.$ Let the intersection of $\overline{DB}$ and $\overline{AF}$ be $X.$

We have that $\triangle DXF\sim\triangle BXA.$ Since $AB=2\cdot DF$, corresponding sides give $BX=2\cdot DX$.

This means that $DX = \dfrac{1}{3} \cdot DB,$ which means that the altitude of $\triangle DXF$ is $\dfrac{1}{3}$ the height of the rectangle.

The area of $\triangle DXF$ is then $$\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{6}.$$

The area of both unshaded triangles is then $2 \cdot \dfrac{1}{6} = \dfrac{1}{3}.$ The area of $\triangle DHC$ is $$\dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{1}{2}.$$

The area of the shaded region is then $\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}.$

Thus, E is the correct answer.