<Qua-2025-794>, hint

Suppose P on AD, Q on BC.

Let x be the angle BAQ, then BA=BQ means <BAQ=<BQA=x.. Let M be the intersection

of AQ with BP. Since AMP and AQP are congruent, <PAQ=<PQA=A-x

Let T be the intersection of lines AQ with DC.

Since <BCD=<BAD=A, <QTD=A-x(=<BQA) <QTC=<QCD-<TQC(=x)=A-x

Since <AQP=<ATD=A-x PQ//DC.