证明:(Ⅰ)∵PE切圆O于E,∴∠PEB=∠A,
又∵PC平分∠APE,∴∠CPE=∠CPA,
∴∠PEB+∠CPE=∠A+∠CPA,
∴∠CDE=∠DCE,即CE=DE.
(Ⅱ)因为PC平分∠APE∴[CA/CE=
PA
PE],
又PE切圆O于点E,割线PBA交圆O于A,B两点,
∴PE2=PB•PA,
即[PA/PE=
PE
PB]
∴[CA/CE]=[PE/PB]
证明:(Ⅰ)∵PE切圆O于E,∴∠PEB=∠A,
又∵PC平分∠APE,∴∠CPE=∠CPA,
∴∠PEB+∠CPE=∠A+∠CPA,
∴∠CDE=∠DCE,即CE=DE.
(Ⅱ)因为PC平分∠APE∴[CA/CE=
PA
PE],
又PE切圆O于点E,割线PBA交圆O于A,B两点,
∴PE2=PB•PA,
即[PA/PE=
PE
PB]
∴[CA/CE]=[PE/PB]