help!

This one was a little difficult...!!!

Note : angle AEB  = angle DEC

and angle CDB  = angle DBA

So...by AA congruency....ΔBEA ~ ΔDEC  →  DC /AB  = DE / BE

Area of  ΔDBA  = area of  ΔADE  + area of ΔABE  = 20 + 50  = 70

So  area of Δ DBA  = (1/2)sin DBA(AB * DB)  = 70

So...sin DBA = 140 / ( AB * DB)

And sin (DBA) = sin CDB

So area of  ΔCDB  = (1/2) sin (DBA)(DB * DC)  = (1/2) (140 / [(AB * DB] ) (DB * DC) =

70 (DC/AB)  = 70 ( DE /BE)

Now

Area ΔAEB  = (1/2)sin AEB ( AE * BE) = 50   (1)

Area Δ ADE  = (1/2) sin DEA (AE * DE)  = 20  (2)

But  angle AEB  is  supplemental to angle DEA  so thier sines are equal

So  sin AEB  = sinDEA

Rearranging (1), (2)  we have that

sin AEB  = 100 / [ AE * BE]      =     sin DEA = 40 / [ AE * DE ]

 So

100 / BE   = 40 / DE  →   DE / BE  = 40/100  = 2/5

But area of ΔCDB  was found to be  70 (DE / BE) = 70 (2/5)  =  28

So...area of trapezoid ABCD  = area of ΔDBA  + area of ΔCDB  =

                                                               70        +        28             =      98

Should be 98, I think...

Here's a pic  (not to scale) :