Best I can tell, it is not solvable, because I have been able to plug in any angle to cde and the resulting triangles on the diagram all equal 180. If I am wrong, message me why (don't spoil it).
This is what I found: Where X is the intersection between BE and CD, you can figure out all of other angles from several of the resulting triangles with no stress (Triangles ACD, AED, and XBC). But, plug in any number for angle CDE (some number between 1 and 109, since angle DXE is 70), and the resulting angles (BED, BEA, and ADE) are forced into place, and all angles work equally well (resulting in triangles with 180 degrees).
I've taken the tack that you were trying, Master, (bringing 90 angles down all over the place), but I can't make it such that any figure is missing only one angle; there are always two angles missing. I can narrow the range of Z(CDE) but that's all.
i think this is true.....opposite angles in a trapazoid are supplementary.....im not 100% sure bout this....but with this you can conclude that the answer is 60....im not taking random guesses...that may be wat it seems like