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Easy angle equality

Source: European Mathematical Cup 2012, Junior Division, Problem 1

July 27, 2013
geometryangle bisectorgeometry proposed

Problem Statement

Let ABCABC be a triangle and QQ a point on the internal angle bisector of ∠BAC\angle BAC . Circle ω1\omega_1 is circumscribed to triangle BAQBAQ and intersects the segment ACAC in point P≠CP \neq C. Circle ω2\omega_2 is circumscribed to the triangle CQPCQP. Radius of the cirlce ω1\omega_1 is larger than the radius of ω2\omega_2. Circle centered at QQ with radius QAQA intersects the circle ω1\omega_1 in points AA and A1A_1. Circle centered at QQ with radius QCQC intersects ω1\omega_1 in points C1C_1 and C2C_2. Prove ∠A1BC1=∠C2PA\angle A_1BC_1 = \angle C_2PA .
Proposed by Matija Bucić.
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