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<PEQ = 2<APB , circles, tangents, cuba - romania

Source: 2017 Romania JBMO TST5 p3 - Cuban Olympiad 2003

May 16, 2020
geometrycirclescircumcircleangles

Problem Statement

Let AA be a point outside the circle ω\omega . The tangents from AA touch the circle at BB and CC. Let PP be an arbitrary point on extension of ACAC towards CC, QQ the projection of CC onto PBPB and EE the second intersection point of the circumcircle of ABPABP with the circle ω\omega . Prove that ∠PEQ=2∠APB\angle PEQ = 2\angle APB
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