Make regular triangle ABE, regular triangle ADF,

Then there is △ Abd △ FBD △ AEF.

Take AG=AE on AC,

And △ AFG △ AFE.

∠ AEF =∠ Abd = 20，∠ EAF =∠ bad = 10。

So ∠ EFG = 2 ∠ 10+20) = 60.

FG=FE again. △ EFG is positive.

BE = AB = BF

∴∠FEB=80

∴∠GEC=40=∠GCE

∴GC=GE=GF

∴∠GCF=0.5∠FGA= 10=∠FAG

∴FC=FA=FD

∴∠DCA=0.5*60=30？

This is what I try to think, although it is very troublesome.