Solution:
The area of $EFGH$ is $EG \cdot FH \sin \theta / 2$, where $\theta$ is the angle between $EG$ and $FH$. This is at most $90$. However, we claim the area of $ABCD$ is twice that of $EFGH$. To see this, notice that $EF = AC / 2 = GH$, $FG = BD / 2 = HE$, so $EFGH$ is a parallelogram. The half of this parallelogram lying inside triangle $DAB$ has area $(BD / 2)(h / 2)$, where $h$ is the height from $A$ to $BD$, and triangle $DAB$ itself has area $BD \cdot h / 2 = 2 \cdot (BD / 2)(h / 2)$. A similar computation holds in triangle $BCD$, proving the claim. Thus, the area of $ABCD$ is at most $180$. And this maximum is attainable—just take a rectangle with $AB = CD = 15$, $BC = DA = 12$.