Use similar triangles created by the crossing lines

The shaded kite area is twice one small triangle area

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.