∠∠aed = 60, ∠∠a = 60, ∴△AED is an equilateral triangle, AE=AD=EB,∴DE+DB=AB.

Make an auxiliary line to extend AB to point F, so that CF=CA, ∴∠ CAB = ∠ CFA = 60, ∴△CAF is an equilateral triangle, ∴CF=AF.

∵CA=AB+AD,∵AF=AB+BF,∴BF=AD

In △CAD and △CFB, CA=CF, ∠CAF=∠CFA, AD=AF,∴△CAD are all equal to △CFB, ∴CD=CB.

∴∠CDB=∠CBD，in △CED，∠ A =∠ AED =∠ DCE+∠ CDE = 60。

In △CDA, ∠CDB=∠A+∠DCE=2∠DCE+∠CDE, ∠CDB = 2∠CDE = 2∞.

∴3∠dce=60 =∠acf=2∠dce+∠dcb,∴∠dcb=∠dce=60÷3 = 20