jmc

We will write the area of triangle $XYZ$ as $[XYZ].$ Since triangles $ADC$ and $BCD$ share a base and have the same altitude length to that base, they have the same area. Since $[BCD]=[ADC]$, we have $[BCE]+[CDE]=[ADE]+[CDE]$, so $[BCE]=[ADE]=20$.

To find the area of triangle $CDE$, we note that triangles $CDE$ and $ABE$ are similar, and the ratio of their sides is $DE/BE$. Triangles $ADE$ and $ABE$ share an altitude, so $DE/BE=[ADE]/[ABE]=20/50=2/5$. Since the ratio of the areas of two similar triangles is the square of the ratio of their sides, $[CDE]/[ABE]=(DE/BE{)}^2=4/25$, and $[CDE]=(4/25)[ABE]=(4/25)(50)=8$. Thus, the area of trapezoid $ABCD$ is $[ABE]+[ADE]+[BCE]+[CDE]=50+20+20+8=\boxed{98}$.