Strange geo

Let us put the figure in $xy$ -coordinate system, with origin at $A$ . Then the coordinates of the points are $A=(0,0),E=\left(\frac{1}{2},0\right),B=(1,0),F=\left(1,\frac{1}{2}\right),C=(1,1),D=(0,1)$ Let $G=\overline{AF}\cap \overline{DE}$ and $H=\overline{AF}\cap \overline{DB}$ . The equations of segments $\overline{AF},\overline{DE},\overline{DB}$ are $\begin{aligned}\overline{AF} &: y=\frac{1}{2}x \\ \overline{DE} &: y = 1-2x \\ \overline{DB} &: y = 1-x \end{aligned}$ Therefore $\begin{aligned} G &= \overline{AF}\cap \overline{DE} =\left ( y=\frac{1}{2}x \right) \cap (y = 1-2x) = \left( \frac{2}{5},\frac{1}{5} \right) \\ H &= \overline{AF}\cap \overline{DB} =\left ( y=\frac{1}{2}x \right) \cap (y = 1-x) = \left( \frac{2}{3},\frac{1}{3} \right)\end{aligned}$ Hence, by the shoelace method , the area of the blue quadrilateral (i.e. the area $[EBHG]$ ) is given by $[EBHG] = \frac{1}{2}\left [ \begin{array}{cccc} \frac{1}{2} & 1 & \frac{2}{3} & \frac{2}{5} \\ 0 & 0 & \frac{1}{3} & \frac{1}{5} \end{array} \right] = \frac{1}{2} \left( \frac{1}{2}\times 0 + 1\times \frac{1}{3} + \frac{2}{3}\times \frac{1}{5} + \frac{2}{5}\times 0 - 0 \times 1 - 0\times \frac{2}{3} - \frac{1}{3} \times \frac{2}{5} - \frac{1}{5}\times\frac{1}{2}\right) = \frac{7}{60}$ Therefore, $p+q=7+60=\boxed{67}$