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.