∫∠A = 20，AB=AC，

∴∠ABC=∠ACB=80

∠ABE=∠ABC-∠CBE=60

Using BA=BE, we can know that △BAE is an equilateral triangle.

∴AE=AB=AC=BE and ∠ CAE = ∠ BAE-∠ BAC = 60-20 = 40.

AC = AE，∠CAE=40

∴∠ACE=∠AEC=70

∴∠BEC=∠AEC-∠AEB= 10

BC = AD，BE=AC，∠CBE=∠DAC=20

∴△BCE≌△ADC

∴∠ACD=∠BEC= 10

∴∠BDC=∠DAC+∠ACD=20+ 10=30

To sum up, ∠ BDC = 30

or

Let A be AF⊥BC, and take point E on AF to make BE=BC.

△ABC is an isosceles triangle, and it is easy to prove that △BCE is an equilateral triangle, ∠BCE=60.

△ABC is an isosceles triangle, ∠ A = 20, then ∠ BCA = 80.

∠ACE=80-60=20=∠DAC

EC=AD=BC

Communication * * *

According to the angular principle

△ACE?△ACD

∠EAC=∠ACD= 10

∠BDC=∠DAC+∠DCA=20+ 10=30

Hope to adopt, thank you.