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egmo 2018 p5

Source: EGMO 2018 P5

April 12, 2018
geometryTriangleanglesEGMOEGMO 2018geometry solvedpower of a point

Problem Statement

Let Γ\Gamma be the circumcircle of triangle ABCABC. A circle Ω\Omega is tangent to the line segment ABAB and is tangent to Γ\Gamma at a point lying on the same side of the line ABAB as CC. The angle bisector of ∠BCA\angle BCA intersects Ω\Omega at two different points PP and QQ. Prove that ∠ABP=∠QBC\angle ABP = \angle QBC.
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