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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 5 of 6: Prove the two cyclicities
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In plain words

Equal directed angles put both target points on one circle.

∠ITP=∠IMP=∠MIP,∠MQP=∠MIP\angle ITP=\angle IMP=\angle MIP,\qquad \angle MQP=\angle MIP
Detailed analysis

Using the parallelogram and the parallelism MI∥AGMI\parallel AG, the right-angle relation at PP gives ∠ITP=∠IMP=∠MIP\angle ITP=\angle IMP=\angle MIP. Hence P,M,I,TP,M,I,T are cyclic. From B,Q,CB,Q,C collinear and the cyclic quadrilaterals PCEPCE and PBFPBF, angle chasing gives ∠MQP=∠CQP=∠CEP=∠DEP=∠DGP=∠GPI=∠MIP\angle MQP=\angle CQP=\angle CEP=\angle DEP=\angle DGP=\angle GPI=\angle MIP. Thus P,M,I,QP,M,I,Q are cyclic as well.