jmc

First, we find out the coordinates of the vertices of quadrilateral $BEIH$, then use the Shoelace Theorem to solve for the area. Denote $B$ as $(0,0)$. Then $E(0,1)$. Since I is the intersection between lines $DE$ and $AF$, and since the equations of those lines are $y=\frac{1}{2}x+1$ and $y=-2x+2$, $I(\frac{2}{5},\frac{6}{5})$. Using the same method, the equation of line $BD$ is $y=x$, so $H(\frac{2}{3},\frac{2}{3})$. Using the Shoelace Theorem, the area of $BEIH$ is $\frac{1}{2}\cdot \frac{14}{15}=\boxed{\frac{7}{15}}$.