smc

$[ADEC]=[BCA]-[BDE]$ Note that, as right triangles sharing $\angle B$, $△BDE\sim △BCA$. $\overline{BD}=\frac{20}{2}=10$ Because the sides of $△BCA$ 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 $[BDE]=\frac{96\cdot 25}{64}=\frac{75}{2}$. $[ADEC]=\frac{192-75}{2}=\boxed{58\frac{1}{2}}$