(2013•许昌二模)如图,已知PE切圆O于点E,割线PBA交圆O于A,B两点,∠APE的平分线和AE、BE分别交于点C

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  • 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]