解:(1)由已知找出△ABE≌△ACD,
∵∠ABE=∠ACD,AB=AC,∠A=∠A,
∴△ABE≌△ACD.
(2)∵△ABE≌△ACD,
∴AE=AD,又∵AB=AC,∴AB-AD=AC-AE,即DB=EC.
又∵∠ABE=∠ACD,∠DOB=∠EOC,
∴△ODB≌△OEC.
∴OB=OC.
分析:(1)由已知可得到△ABE≌△ACD,因為∠ABE=∠ACD,AB=AC,∠A=∠A,所以△ABE≌△ACD;
(2)由(1)證得△ABE≌△ACD可得AE=AD,可推出DB=EC,則推出△ODB≌△OEC,所以OB=OC.
點評:此題考查的知識點是全等三角形的判定和性質(zhì),關鍵是先由已知證得△ABE≌△ACD,再證得△ODB≌△OEC.