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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 2 of 6: Build the main circle
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In plain words

The parallel segment transfers an inscribed angle from Ω and gives a circle through P and Q.

PQ∥AB⟹P,Q,A2,B2 are concyclic;ω=(PQA2B2)PQ\parallel AB\Longrightarrow P,Q,A_2,B_2\text{ are concyclic};\qquad \omega=(PQA_2B_2)
Detailed analysis

Because P lies on AA_2, Q lies on BB_2, and PQ is parallel to AB, the directed-angle form of Reim's theorem applied to cyclic quadrilateral ABA_2B_2 gives that P,Q,A_2,B_2 are concyclic. Denote their circle by ω. Equivalently, one may check the cyclic criterion directly by replacing PQ with AB and the lines PA_2, QB_2 with AA_2, BB_2, then using the equal inscribed angles in Ω.