Can anyone post your solution?

Any help???

Let $R=AP\cap DQ,S=BP\cap CQ$. If $AD\parallel BC$ and $AD=BC$,then the quadrilateral $ABCD$ is a parallelogram,so $AB\parallel CD$,which is a contradiction,therefore $AD\neq BC.$ Without loss of generality,we may assume that $AD<BC$ and $T=AB\cap CD$. $AD\parallel BC\implies \frac{TP^2}{TQ^2}=\frac{TA\cdot TB}{TD\cdot TC}=\frac{TA^2}{TD^2}=\frac{TB^2}{TC^2}$,and hence $\frac{TP}{TQ}=\frac{TA}{TD}=\frac{TB}{TC}$,so $A,D,P,Q$ are concyclic and $B,C,P,Q$ are concyclic.Therefore,$AR\cdot RP=DR\cdot RQ,BS\cdot SP=CS\cdot SQ$,so $R,S$ are on the radical axis of $\omega_1,\omega_2$,namely $R,S\in EF$. Moreobver,from $\angle BAP=\angle BPC=\angle BQC$,we obtain that $AP\parallel CQ$.Similarly,$BP\parallel DQ$,and hence the quadrilateral $PRQS$ is a parallelogram,which follows that $PQ$ bisects $EF$. If $EF$ bisects $PQ$, we have $IP=IQ$,where $I= EF\cap PQ$.Let $G,H$ be the second points of intersection of $PQ$ with $\omega_1,\omega_2$ respectively,then $(GQ+QI)\cdot IP=GI\cdot IP=EI\cdot IF=HI\cdot IQ=(HP+PI)\cdot IQ$,so we obtain that $GQ\cdot IP=HP\cdot IQ\implies GQ\cdot QP=HP\cdot PQ\implies AQ\cdot QB=DP\cdot PC$. Moreover,since $\angle AQR=\angle PCS,\angle RAQ=\angle SPC$,we get that $\triangle AQR\sim \triangle PCS$.Similarly,$\triangle DPR\sim \triangle QBS$.And hence $\frac{QR}{CS}=\frac{AQ}{PC}=\frac{DP}{QB}=\frac{PR}{BS}$.Also,$\angle PRQ=\angle AQR+\angle RAQ=\angle PCS+\angle SPC=\angle BSC$,then $\triangle PRQ\sim \triangle BSC$.Similarly,$\triangle DPR\sim \triangle QBS$.And hence $\frac{QR}{CS}=\frac{AQ}{PC}=\frac{DP}{QB}=\frac{PR}{BS}$.Also,$\angle PRQ=\angle AQR+\angle RAQ=\angle PCS+\angle SPC=\angle BSC$,then $\triangle PRQ\sim \triangle BSC$.Analogously, $\triangle ARD\sim \triangle PSQ$. Consequently,$\angle ABC+\angle BAD=\angle QBS+\angle SBC+\angle RAQ+\angle RAD=\angle DPR+\angle RPQ+$ $\angle SPC+\angle SPQ=180^\circ$,and hence $AD\parallel BC$,as desired. $Q.E.D.$