KOMATH

<Qua-2026-807>, hint

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

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

<Qua-2026-807>, cyclic, parallel

<Qua-2026-806>, hint

<Qua-2026-806>,cyclic, midpoint

<Tri-2026-2299 >,hint

Line BP cut AC at B' and line MP cut AB at M'.

Then APB and APB' are congruent and BP=PB'

since M is the midpoint of BC, line MP//AC and cut AB at midpoint M'

By Ceva's theorem (AM'/M'B)x(BD/DM)x(MQ/QA)=1

Since AM'=M'B, BD/DM=AQ/QM, Hence DQ//AB.