In the figure, it is given that angle $C = 90^{\circ}$, $\overline{AD} = \overline{DB}$, $DE \perp AB$, $\overline{AB} = 20$, and $\overline{AC} = 12$. The area of quadrilateral $ADEC$ is: 

$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 58\frac{1}{2}\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 37\frac{1}{2}\qquad\textbf{(E)}\ \text{none of these}$
$[ADEC]=[BCA]-[BDE]$
Note that, as right triangles sharing $\angle B$, $\triangle BDE \sim \triangle BCA$. 
$\overline{BD}=\frac{20}{2}=10$
Because the sides of $\triangle BCA$ are in the ratio $3:4:5$, $\overline{BC}=16$.
The sides of our triangles are in the ratio $\frac{10}{16}=\frac{5}{8}$, and the ratio of their areas is $\left(\frac{5}{8}\right)^2=\frac{25}{64}$.
$[BCA]=\frac{12 \cdot 16}{2}=96$, and $[BDE]=\frac{96 \cdot 25}{64}=\frac{75}{2}$.
$[ADEC]=\frac{192-75}{2}=\boxed{\textbf{(B)}\ 58\frac{1}{2}}$