如图,∠DPB的顶点P在圆O外,其两条分别与圆O交与A,B,C,D四点,连接OA,OB,OC,OD.求证∠DPB=(∠B

连接BD
∴∠PDB=∠CDB=1/2(∠AOC+∠AOB) (利用圆周角=1/2圆心角）
∠PBD=∠ABD=1/2(∠AOC+∠COD)
∵∠COD+∠AOB=360°-∠AOC-∠BOD
∴∠PDB+∠PBD=1/2(2∠AOC+∠AOB+∠COD)
=1/2(2∠AOC+360°-∠AOC-∠BOD)
=180°+1/2(∠AOC-∠BOD)
∴∠DPB=180°-(∠PDB+∠PBD)
=180°-[180°+1/2(∠AOC-∠BOD)]
=-1/2(∠AOC-∠BOD)
=1/2(∠BOD-∠AOC)

没图啊！！！