ABC = 1 10, ∠ CBD = 40, so ∠ Abd = 70, ∠ ABC's external angle is also = 70; Because CE is the bisector of ∠ACB, draw the vertical lines to CA, CB and DB from point E, prove that the three vertical lines are equal, and get that DE is the bisector of ∠ADB; So DE is parallel to CB, ∠ CED = ∠ ECB = 20, so ∠ CDE = 140, ∠ CED = 20.

Solution 2: ∠ CBD = 40, ∠ DCB = 40, then △DBC is an isosceles triangle and ∠ BDC = 100.

In addition, < CEB = 50 = < BDC/2,

Then a circle with D as the center and CD as the radius passes through point E (because the central angle is equal to twice the peripheral angle).

So the triangle CDE is an isosceles triangle

Therefore, ∠ CDE = 140, ∠ CED = 20.