Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC = 20$, and $CD = 30$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E$, and $AE = 5$. What is the area of quadrilateral $ABCD$ ?
Because the area of $\triangle{ACD}$ is simply $\frac{1}{2}\times 20 \times 30 = 300$, it is sufficient to compute the area of $\triangle{ABC}$ in order to get the final answer.
In order to compute the area of $\triangle{ABC}$, let's draw the attitude $AF$ and let $\overline{EF}=x$. Now because $\triangle{BFE}\sim\triangle{DCE}$, we have $$\frac{\overline{BF}}{\overline{EF}}=\frac{\overline{DC}}{\overline{CE}}\implies \frac{BF}{x}=\frac{30}{20-5}\implies \overline{BF}=2x$$
Meanwhile, in the right $\triangle{ABC}$, we have $$\overline{BF}^2 = \overline{AF}\cdot\overline{FC}\implies (2x)^2=(5-x)(15+x)\implies x=3\quad(\because x > 0)$$
It follows that the area of $\triangle{ABC}$ equals $\frac{1}{2}\times 20\times 6=60$. This means the area of $ABCD=\boxed{360}$.