(1)证明:连接OD.∵O为AB中点,D为BC中点,∴OD∥AC.∵DE为⊙O的切线,∴DE⊥OD.∴DE⊥AC.(2)过O作OF⊥BD,则BF=FD.在Rt△BFO中,∠B=30°,∴OF= 1 2 OB,BF= 3 2 OB.∵BD=DC,BF=FD,∴FC=3BF= 3 3 2 OB.在Rt△OFC中,tan∠BCO= OF FC = 1 2 OB 3 3 2 OB = 1 3 3 = 3 9 .
两图