IRL_ABooklet_2023

A point $X$ is on the perpendicular bisector of $AB$ iff $|XA|=|XB|$. Therefore, the centre $O$ of the circle is on the perpendicular bisectors of $AB$ and $CD$. As $AB\parallel CD$, this means that the two chords have the same perpendicular bisector. If $|PA|=|PB|$, $P$ must be on this common perpendicular bisector, hence we also have $|PC|=|PD|$.