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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 1 of 6: Introduce the auxiliary points
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In plain words

The second intersections with the circumcircle connect the two given cevians to the cyclic geometry of ABC.

Ω=(ABC),A2=(AA1∩Ω)∖{A},B2=(BB1∩Ω)∖{B}\Omega=(ABC),\qquad A_2=(AA_1\cap\Omega)\setminus\{A\},\qquad B_2=(BB_1\cap\Omega)\setminus\{B\}
Detailed analysis

Let Ω be the circumcircle of ABC. Let A_2 be the second point where line AA_1 meets Ω, and let B_2 be the second point where line BB_1 meets Ω. Thus A,P,A_1,A_2 are collinear and B,Q,B_1,B_2 are collinear, while A,B,A_2,B_2 all lie on Ω.