⊙O1与⊙O2内切于P,⊙O1的弦AB切⊙O2于C,PC交⊙O1于E.求证:PA:PE=PC:PB.
1个回答

延长AB交过点P的两圆公切线于Q.

∵QP、QC分别切⊙O2于P、C,∴由切线长定理,有:PQ=CQ,∴∠BCP=∠QPE.

∵QP切⊙O1于P,又∠EAP是⊙O1的圆周角,∴∠EAP=∠QPE.

由∠BCP=∠QPE、∠EAP=∠QPE,得:∠EAP=∠BCP.

∵A、P、B、E共圆,∴∠AEP=∠CBP.

由∠EAP=∠BCP、∠AEP=∠CBP,得:△AEP∽△CBP,∴PA∶PE=PC∶PB.