Question

如图,P为⊙O外一点,PA,PB均为⊙O的切线,A和B是切点,BC是直径.求证：∠APB=2∠ABC;

Answer 1

证明：连接OA

∵PA、PB切圆O于A、B

∴PO垂直平分AB

∴∠APB＝2∠APO,∠PAB+∠APO＝90

∵PA切圆O于点A

∴∠OAP＝90

∴∠OAB+∠PAB＝90

∴∠APO＝∠OAB

∵OA＝OB

∴∠OAB＝∠ABC

∴∠APO＝∠ABC

∴∠APB＝2∠ABC