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Point in a triangle, with angle constraints

Source: Tuymaada 2012, Problem 3, Day 1, Seniors

July 21, 2012
geometrycircumcirclesymmetrytrapezoidincentergeometric transformationreflection

Problem Statement

Point PP is taken in the interior of the triangle ABCABC, so that ∠PAB=∠PCB=14(∠A+∠C).\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C). Let LL be the foot of the angle bisector of ∠B\angle B. The line PLPL meets the circumcircle of △APC\triangle APC at point QQ. Prove that QBQB is the angle bisector of ∠AQC\angle AQC.
Proposed by S. Berlov
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