As shown in the figure: extend CE⊥BA, drop your feet to point E, make the middle line of the bottom of the isosceles triangle af ABC, pass the CD at point G, and make DH⊥AF.

Connecting BG, △BGC is also an isosceles triangle, ∠ GBF = 20, and BG is the bisector of ∠DBC.

(BD/BC)=(DG/GC)=(DH/FC)=(DH/BC/2)？ BD=2DH

BD=2DE？ DH=DE？ ∠ bud =∠DAF= 10

∠DAC=20 + 10 =30