Let $E$ and $F$ be the intersections of $r$ and $s$ with the circumcircle of $ABCD$, respectively. Since $ABCD$ is cyclic and $BA$ and $BF$ are symmetric with respect to the bisector of $\angle CBD$, $\angle ACD = \angle ABD = \angle CBF = \angle CAF$, hence $AF$ and $CD$ are parallel. Analogously, $BE$ and $CD$ are also parallel. Thus, since $CD$, $AF$ and $BE$ are parallel chords, its perpendicular bisectors coincide.

The intersection $P$ of $AE$ and $BF$ belong to this common perpendicular bisector because $\angle BEA = \angle EBF$. Thus this common perpendicular bisector is $OP$, which is perpendicular to $CD$ and, moreover, passes through its midpoint.