如图所示,已知PA与⊙O相切,A为切点,过点P的割线交圆于B、C两点,弦CD∥AP,AD、BC相交于点E,F为CE上一点,且DE2=EF•EC.
(1)求证:CE•EB=EF•EP;
(2)若CE:BE=3:2,DE=3,EF=2,求PA的长.
证明:∵DE2=EF•EC,∠DEF公用,
∴△DEF∽△CED,
∴∠EDF=∠C.
又∵弦CD∥AP,∴∠P=∠C,
∴∠EDF=∠P,∠DEF=∠PEA
∴△EDF∽△EPA.
∴EAEF
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