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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 5 of 6: Use the symmetric argument for P_1
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In plain words

Swapping the roles of A and B preserves the entire configuration and exchanges Q_1 with P_1.

A↔B,A1↔B1,A2↔B2,P↔Q,P1↔Q1⟹P,Q,B2,P1 are concyclicA\leftrightarrow B,\quad A_1\leftrightarrow B_1,\quad A_2\leftrightarrow B_2,\quad P\leftrightarrow Q,\quad P_1\leftrightarrow Q_1\Longrightarrow P,Q,B_2,P_1\text{ are concyclic}
Detailed analysis

Apply the preceding two angle-chasing steps after simultaneously swapping A with B, A_1 with B_1, A_2 with B_2, P with Q, and P_1 with Q_1. The hypothesis angle PP_1C = BAC becomes the corresponding angle condition in the swapped configuration. The same reasoning therefore proves that P,Q,B_2,P_1 are concyclic. Since P,Q,B_2 already lie on ω, this places P_1 on ω as well.