<Qua-2026-809>, hint

Let <BAM=DAM=X, then <BMA=CNM=X

BF=BA=CD, OM=OC( radius), <OCD=<BCD+<OCB=2X+90-X=90+X

<OMB=<BMN+OMN=180-X+2X-90=90+X

Triangles OBM and ODC are congruent 

<Qua-2026-809>, parallelogram, angle

 

<Qa-2026-808>, hint

 

<Qua-2026-808>, square, cyclic

<Singapore Senior Math Olympiad 2023>

 

<Qua-2026-807>, hint

Since <PCQ=<PDQ, <PAQ=<PBQ

 PCDQ PBAQ are cyclic , Hence <PCD=<PQA, <PQA+<PBA=180, AB//CD