Question

如图,AB是⊙O的直径,点P在AB的延长线上,弦CE交AB于点D,连结OE、AC,已知∠POE=2∠CAB,∠P=∠E,

Answer 1

1.连接AE ∵OA=OE,∴∠OAE=∠OEA,∴∠POE=2∠OAE,∵∠POE=2∠CAB ,∴∠CAB=∠EAB

∵AB是直径,∴弧BC= 弧BE ,∴CE⊥AB

2.连接OC,∵OE=OC,∴∠OCE=∠OEC,∵∠P=∠CEO,∴∠OCE=∠P,∵ CE⊥AB,∴∠OCE+∠COD=90º,∴∠P+∠COB=90º,∴ OC⊥PC,∴PC是⊙O的切线