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Problem 2

In triangle ABCABC, point A1A_1 lies on side BCBC and point B1B_1 lies on side ACAC. Let PP and QQ be points on segments AA1AA_1 and BB1BB_1, respectively, such that PQPQ is parallel to ABAB. Let P1P_1 be a point on line PB1PB_1, such that B1B_1 lies strictly between PP and P1P_1, and ∠PP1C=∠BAC\angle PP_1C=\angle BAC. Similarly, let Q1Q_1 be the point on line QA1QA_1, such that A1A_1 lies strictly between QQ and Q1Q_1, and ∠CQ1Q=∠CBA\angle CQ_1Q=\angle CBA. Prove that points P,Q,P1,Q1P,Q,P_1,Q_1 are concyclic.
Step 3 of 6: Construct the first auxiliary circle
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In plain words

The defining angle at Q_1 matches an inscribed angle at A_2, so both points see the same chord C A_1.

∠CQ1A1=∠CQ1Q=∠CBA=∠CA2A=∠CA2A1⟹C,A1,A2,Q1 are concyclic\angle CQ_1A_1=\angle CQ_1Q=\angle CBA=\angle CA_2A=\angle CA_2A_1\Longrightarrow C,A_1,A_2,Q_1\text{ are concyclic}
Detailed analysis

Since A_1 lies strictly between Q and Q_1, the rays Q_1A_1 and Q_1Q coincide, so angle CQ_1A_1 equals angle CQ_1Q. By hypothesis this is angle CBA. Since A,B,C,A_2 lie on Ω, angle CBA equals angle CA_2A, and because A,A_1,A_2 are collinear on the same ray from A_2, this equals angle CA_2A_1. Thus C,A_1,A_2,Q_1 are concyclic.