Let $AB$ intersect $CD$ at $E$. Let the circumcircles of $CEB$ and $DEA$ intersect again at $F$. Then $\angle BFC=\angle BEC=\angle AFD$ and $\angle BCF=\angle BEF=\angle ADF$, thus $\triangle ADF\sim\triangle BCF$ so by spiral symmetry, $\triangle ABF\sim\triangle DCF$. But since $\frac{AP}{PD}=\frac{BQ}{QC}$, we have $\triangle ABF\sim\triangle PQF\sim\triangle DCF$. Also note that $\frac{BF}{CF}=\frac{BA}{CD}=\frac{BQ}{CQ}$ thus $FQ$ is the angle bisector of $\angle BFC$. Therefore the angle formed by the lines $AB,PQ$ is the same as $DC$ with $PQ$, as they are equal to $\angle BFQ=\angle CFQ$.

Let $PQ$ meets $CD$ and $AB$ at $X$ and $Y$ respectively. Let $Z$ be the point on line $AQ$ such that $DZ\parallel PQ$. We have $\frac{AQ}{QZ}=\frac{AP}{PD}=\frac{BQ}{QC}$, so $\triangle AQB\sim\triangle ZQC$, therefore $CZ\parallel AB$. Also note that $\frac{AB}{CZ}=\frac{BQ}{QC}=\frac{AB}{CD}$, so $CZ=CD$, and hence $\angle CZD=\angle CDZ$. Since $CZ\parallel BY$ and $DZ\parallel PQ$, we get $\angle CZD=\angle QYB$ and $\angle CDZ=\angle CXQ$. Therefore we conclude that $\angle QYB=\angle CXQ$, as desired.

See also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=279955.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IbAMO1987Problem1 Vo Duc Dien