连接OB,BC,AO,OD//AB,得到∠ACO=∠CAB=∠OAC(等腰),∠ABO=∠BOD,∠OAB=∠OBA=∠BOD,∠OAC=∠OCA,所以∠OAB=∠OBA=∠BOD=2∠OCA,∠D=∠ACB,∠ABC=∠BCD,且∠ACB=∠D,∠OBC=∠OCB(等腰),所以∠CBD=∠OAC=∠ACO,∠BOD+∠ODB+∠OBD=∠BOC+∠ACB+∠OBC+∠CBD=∠BOD+∠D+∠CBD+∠CBD+∠D=∠BOD+2∠D+2∠CBD=2∠BOD+2∠D=180,所以∠BOD+∠D=90,所以∠OBD=90,所以OB垂直于BD,所以BD为圆O的切线。... |