∴∠ADB=90，

∴∠BAC+∠ABD=90，

∠∠DBC =∠BAC，

∴∠DBC+∠ABD=90，

∴AB⊥BC，

∫AB is the diameter,

∴bc⊙o tangency;

(2) Solution: Connect OD and cross O to make OM⊥BD in M,

∫∠BAC = 30，

∴∠BOD=2∠A=60，

OB = OD，

∴△OBD is an equilateral triangle,

∴OB=BD=OD=2，

∴BM=DM= 1，

From Pythagorean Theorem: OM=3,

∴ the area of the shaded part S=S sector DOB-sδDOB = 60π? 22360- 12×2×3=23π-3.