Ireland

Whether or not $P$ lies between the parallel lines gives two cases to distinguish. Let $S$ and $T$ be points on the tangent to the circumcircle of $△ABP$ at $P$ such that $S$ and $A$ are on the same side of the line $BC$ and $T$ and $B$ are on the same side of $AD$.

First Solution: We have $\angle SPA=\angle PBA$ (chord tangent angle). Because $AB\parallel CD$, we also have $\angle PBA=\angle PCD$, hence $\angle SPA=\angle PCD$. If $P$ lies between the parallel lines, $\angle SPA=\angle TPD$ and so $\angle PCD=\angle TPD$. In both cases it follows that $ST$ is tangent to the circumcircle of $△DCP$.



Second Solution: If $P$ is between the parallel lines, reflect $A$ and $B$ at $P$ to get $A^{\prime }$ on $PD$ and $B^{\prime }$ on $PC$. If $P$ is not between the parallel lines we simply let $A^{\prime }=A$ and $B^{\prime }=B$.



Because $A^{\prime }{B}^{\prime }\parallel CD$ the triangle $PDC$ is obtained from $△P{A}^{\prime }{B}^{\prime }$ by a homothety with centre $P$. Hence, the circumcentres of the triangles $PDC$ and $P{A}^{\prime }{B}^{\prime }$ are on a line through $P$, on which we also find the circumcentre of $△PAB$. As the tangent at $P$ to these circles is perpendicular to the line through $P$ and the centres, it follows now that the circumcircles of $△PAB$ and $△PDC$ touch at $P$.