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Problem 6

Let II be the incenter of acute triangle ABCABC with AB≠ACAB\ne AC. The incircle ω\omega of ABCABC is tangent to BCBC, CACA, and ABAB at DD, EE, and FF, respectively. The line through DD perpendicular to EFEF meets ω\omega again at RR (other than DD). Line ARAR meets ω\omega again at PP (other than RR). The circumcircles of triangles PCEPCE and PBFPBF meet again at QQ (other than PP). Prove that lines DIDI and PQPQ meet on the line through AA perpendicular to AIAI.
Step 3 of 6: Locate the second circle intersection
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In plain words

The two given circles force equal angles at QQ.

∠PQC=∠PEC=∠PED=∠PFD=∠PFB=∠PQB\angle PQC=\angle PEC=\angle PED=\angle PFD=\angle PFB=\angle PQB
Detailed analysis

Because P,C,E,QP,C,E,Q are cyclic, ∠PQC=∠PEC\angle PQC=\angle PEC. The tangency points and the fixed circle give ∠PEC=∠PED=∠PFD\angle PEC=\angle PED=\angle PFD. Since P,B,F,QP,B,F,Q are cyclic, ∠PFD=∠PFB=∠PQB\angle PFD=\angle PFB=\angle PQB. Hence ∠PQC=∠PQB\angle PQC=\angle PQB, so B,Q,CB,Q,C are collinear in the original figure; equivalently the inverse image Q′Q' lies on the corresponding line B′C′B'C'.