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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 6 of 6: Conclude the proof
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In plain words

Both newly constructed points have now been placed on the circle already containing P and Q.

P,Q,P1,Q1∈ω⟹P,Q,P1,Q1 are concyclicP,Q,P_1,Q_1\in\omega\Longrightarrow P,Q,P_1,Q_1\text{ are concyclic}
Detailed analysis

The circle ω contains P and Q by construction, contains Q_1 by the first angle chase, and contains P_1 by the symmetric angle chase. Hence all four points P,Q,P_1,Q_1 lie on one circle, exactly as required.