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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 4 of 6: Place Q_1 on the main circle
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In plain words

A chain of equal angles travels from the auxiliary circle through Ω and ends at P, proving the same cyclic angle condition for ω.

∠QQ1A2=∠A1Q1A2=∠A1CA2=∠BCA2=∠BAA2=∠QPA2⟹P,Q,A2,Q1 are concyclic\angle QQ_1A_2=\angle A_1Q_1A_2=\angle A_1CA_2=\angle BCA_2=\angle BAA_2=\angle QPA_2\Longrightarrow P,Q,A_2,Q_1\text{ are concyclic}
Detailed analysis

Because Q,A_1,Q_1 are collinear, angle QQ_1A_2 equals angle A_1Q_1A_2. On the circle C,A_1,A_2,Q_1 this equals angle A_1CA_2. Since C,A_1,B are collinear, it equals angle BCA_2; since A,B,C,A_2 lie on Ω, it equals angle BAA_2. Finally AB is parallel to PQ and A,A_2,P are collinear, so angle BAA_2 equals angle QPA_2. Therefore P,Q,A_2,Q_1 are concyclic, and Q_1 lies on ω.