Question

⊙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.

Answer 1

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.