jmc

We label the square, triangle, and intersections as above. Triangle $BCG$ and $FDG$ are congruent triangles. The area of the shaded region $DHF$ is the area of $FGD$ minus $DGH$.

Triangle $DGH$ is similar to triangle $BGC$. We can prove this because $\angle BGC=\angle DGH$. Also, $\overline{DE}$ has slope $2$ and $\overline{BF}$ has slope $-\frac{1}{2}$, which are negative reciprocals, so the two lines are perpendicular and create the right angle $\angle GHD$. Therefore, $\angle GHD=\angle BCG=90^{\circ }$. Since the two triangles have two of the same angle measures, they are similar. Therefore, we have the ratios $\frac{GD}{BG}=\frac{GH}{CG}=\frac{DH}{BC}$. We can find that $BG$ using the Pythagorean formula. $$\begin{array}{rl}B{C}^2+C{G}^2& =B{G}^2\\ 5^2+10^2=125& =B{G}^2\\ BG& =5\sqrt{5}.\end{array}$$ Therefore, we have $\frac{5}{5\sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{GH}{5}=\frac{DH}{10}$. We solve for the length of the two legs of triangle $DGH$ to find that $GH=\sqrt{5}$ and $DH=2\sqrt{5}$. Therefore, the area of triangle $DGH$ is $\frac{\sqrt{5}\cdot 2\sqrt{5}}{2}=5$.

The area of triangle $DGF$ is $\frac{5\cdot 10}{2}=25$. We subtract the area of $DGH$ from the area of $DGF$ to find the area of the shaded region to get $25-5=\boxed{20\text{ sq units}}$.