Area of a Quadrilateral

Let the intersection be $O$

$\triangle DCO:\triangle ADO$

$=\overline{CO}:\overline{AO}=1:3$

$=\overline{DO}:\overline{BO}$

$=\triangle ADO:\triangle ABO=48:144$

$=\triangle DCO:\triangle BCO=16:48$

Hence, $16+48+48+144=256$

Let $G$ be the intersection of $AC$ and $BD$ . Let $x = EG$ be the altitude of $\triangle ADG$ and let $y = FG$ be the altitude of $\triangle CDG$ . Let $w = AD$ , $z = CD$ , and $v = AB$ .

By triangle areas of $\triangle ADG$ and $\triangle CDG$ , $yz = 32$ and $wx = 96$ .

Since $\triangle AEG$ is similar to $\triangle GFC$ , $\frac{w - y}{x} = \frac{w}{z}$ , which solves to $wz = yz + wx$ , and since $yz = 32$ and $wx = 96$ , $wz = 32 + 96 = 128$ .

Since $\triangle DEG$ is similar to $\triangle DAB$ , $\frac{v}{w} = \frac{x}{y}$ , which means $v = \frac{wx}{y}$ .

The area of $\triangle ABC$ is $A_{ABC} = \frac{1}{2}vw$ . Substituting $v = \frac{wx}{y}$ gives $A_{ABC} = \frac{w^2x}{2y} = \frac{wz \cdot wx}{2 \cdot yz}$ , and substituting $yz = 32$ , $wx = 96$ , and $wz = 128$ gives $A_{ABC} = \frac{128 \cdot 96}{2 \cdot 32} = 192$ .

Therefore, the area of $ABCD$ is $A_{ABCD} = A_{ADG} + A_{CDG} + A_{ABC} = 48 + 16 + 192 = \boxed{256}$ .