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PE is bisector of BPC

Source: Iran TST 2012 -first day- problem 2

April 23, 2012
geometrycircumcircleincentertrigonometryangle bisectorpower of a point

Problem Statement

Consider ω\omega is circumcircle of an acute triangle ABCABC. DD is midpoint of arc BACBAC and II is incenter of triangle ABCABC. Let DIDI intersect BCBC in EE and ω\omega for second time in FF. Let PP be a point on line AFAF such that PEPE is parallel to AIAI. Prove that PEPE is bisector of angle BPCBPC.
Proposed by Mr.Etesami
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