Let $PQ \cap AB=G$, $AQ \cap PC=E$ and $BQ \cap PD=F$ $\implies$ $PEQF$ and $AEFBCD$ is cyclic $\implies$ $\angle EPG$ $=$ $\angle EFQ$ $=$ $\angle EAG$ $\implies$ $EAGP$ is cyclic $\implies$ $PQ$ $\perp$ $AB$

Clearly $AQ\parallel DP,BQ\parallel CP,AB\parallel CD$ thus the triangles $AQB$ and $DPC$ are perspective with perspectrix as the line of $\infty$.Thus by Desargue's Theorm $AD\cap BC \in PQ$ but Since $AD\cap BC=\infty_{AD}$.Thus $PQ\parallel AD \implies PQ\perp AB$ as desired.$\blacksquare$

AlastorMoody wrote: Let $PQ \cap AB=G$, $AQ \cap PC=E$ and $BQ \cap PD=F$ $\implies$ $PEQF$ and $AEFBCD$ is cyclic $\implies$ $\angle EPG$ $=$ $\angle EFQ$ $=$ $\angle EAG$ $\implies$ $EAGP$ is cyclic $\implies$ $PQ$ $\perp$ $AB$ Hi Your solution is wrong because the in this statement "AEFBCD is cyclic" you have to use consequent of the problem to solve it.

Points $A,AQ\cap PC,C,B,BQ\cap PD,D$ lie on one circle, so by Pascal theorem $PQ\parallel BC\perp AB$ $\blacksquare$

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