Okay this is crazy long and complicated but I got it. I bet anything there's a simpler way, but this still works.
The answer is 20.
Try to follow along. Draw what I describe. Afterwards I'll create a few drawings to put into this at certain points and edit them in (if I have any time haha).
1) Create point F on AB such that BF = BC. Then angle BCE = angle BFC = 50.
2) Create point G on AC such that angle GBC = 20.
3) Since GBC = 20 and BCA = 80, we have CGB = 80. Thus triangle BCG is isosceles (BG = BC).
4) Since BF = BC and BC = BG, we have BF = BG.
5) Since CBG = 20 and ABC = 80, we have ABG = 60.
6) Draw line segment FG. Since BFG is isosceles and FBG = 60, we have BGF = GFB = 60. So triangle BFG is actually equilateral (BF = FG = BG).
7) Since FBG = 60 and ABE = 20, we have EBG = 40.
8) Since BGC = 80, we have BGA = 100.
9) Since BGA = 100 and EBG = 40, we have BEG = 40.
10) Since EBG = BEG = 40, triangle BEG is isosceles with BG = GE.
11) Since BG = FG (step 6) and BG = GE, we have FG = GE (triangle FGE is isosceles).
12) Since triangle FGE is isosceles and FGE is 40, we have GFE = FEG = 70.
13) Since BFG = 60 (step 6) and BFC = 50 (step 1), we have CFG = 10.
14) Since CFG = 10 and GFE = 70, we have CFE = 80.
15) Since BFC = 50 and CFE = 80, we have BFE = 130 (and also EFD = 50).
16) Since CFE = 80 and EFD = 50, we have CFD = 130.
17) Since BCD = 70 and ABC = 80, we have BDC = 30.
18) Since FEG = 70 (step 12) and BEG = 40 (step 9), we have FEB = 30.
19) Consider triangles BFE and CFD. In the former, angle B = 20, angle F = 130, and angle E = 30. In the latter, angle C = 20, angle F = 130, and angle D = 30. Since these are the same measurements, the two triangles are similar.
20) Since BFE and CFD are similar triangles, we note that DF/EF = CF/BF.
21) This can be arranged to read DF/CF = EF/BF.
22) Note triangles EFD and BFC. Note that in the former, angle F = 50 and in the latter, angle F = 50. Note that from step 21, DF/CF = EF/BF. So triangles EFD and BFC are also similar.
23) Since EFD and BFC are similar, their corresponding angles are equal in measurement. We already have angle F in both = 50. So in triangle EFD, we note that angle D has the same measurement as angle C does in triangle BFC. Since angle C in triangle BFC = 50, we have angle D in triangle EFD = 50.
24) Since angle EDF = 50 and angle CDF = 30, we have angle CDE = 20.
This is our desired angle.