The area of $E F G H$ is $E G \cdot F H \sin \theta / 2$, where $\theta$ is the angle between $E G$ and $F H$. This is at most 90. However, we claim the area of $A B C D$ is twice that of $E F G H$. To see this, notice that $E F=A C / 2=G H, F G=B D / 2=H E$, so $E F G H$ is a parallelogram. The half of this parallelogram lying inside triangle $D A B$ has area $(B D / 2)(h / 2)$, where $h$ is the height from $A$ to $B D$, and triangle $D A B$ itself has area $B D \cdot h / 2=2 \cdot(B D / 2)(h / 2)$. A similar computation holds in triangle $B C D$, proving the claim. Thus, the area of $A B C D$ is at most 180. And this maximum is attainable - just take a rectangle with $A B=C D=$ $15, B C=D A=12$.