<Tri-2025-2242>, hint 

Suppose line AQ cuts  BM and BC at P', P" respectively and line through M, parallel to AQ

cuts BC at M'. In triangle ABM, by Ceva's  AP', MQ, BN concurr at O, then BP'/P'M=AN/NM=(1-t)/t.  Since NP//AB, BP/PM=BP'/P'M, Hence P=P'

NS=(1+t)/2 AP", AP=AP"-PP"=(1-(1-t)/2)AP=(1+t)/2 AP", PP"=MM'(1-t)=AP" (1-t)/2

Hence  NS//=AP, APSN is a parallelogram.