BdMO Online Forum

The main idea of this solution is:If $\measuredangle APC=\dfrac {\pi} {2}$,then $CP$ must pass through the antipode of $A$,say,$A_{1}$ in $(O_{1})$.

Let $(O_{1}) \cap (O_{2})={P,Q}$.Then it is well known that $PQ$ passes through the midpoint of $AB$,say,$M$.

Now,let $O_{1}O_{2} \cap MQ=R$,then $\measuredangle O_{2}BM=\measuredangle MQO_{2}=\dfrac {\pi} {2}$.So,$O_{2}BMQ$ is cyclic.So,$\measuredangle BMP=\measuredangle BMQ=\measuredangle AO_{1}C$ and $\measuredangle O_AC=\dfrac {\pi} {2}-\measuredangle BAC=\measuredangle ABP$.So,$\triangle AO_{1}C \sim \triangle BMP$.So,$\triangle AA_{1}C \sim \triangle BAP$.So,$\measuredangle BAP=\measuredangle AA_{1}P=\measuredangle AA_{1}C$.Therefore, $A_{1},C,P$ are collinear.So,done.