Saudi Arabia Mathematical Competitions 2012

Let $P$ be on $BC$ such that $OP$ is parallel to $AD$. If $O$ is on $BC$, then $P=O$ and the result is clear. We claim that the circumcircles of $BO{O}_1$ and $CO{O}_2$ both pass through $P$. One of the angles $\widehat{ADB}$ and $\widehat{ADC}$ is not acute. Without loss of generality, assume that $\widehat{ADB}\ge 90^{\circ }$. Then $O_1$ does not lie in the interior of triangle $ADB$. Note that $$\widehat{O{O}_{1}B}=\frac{\widehat{A{O}_{1}B}}{2}=180^{\circ }-\widehat{ADB}=180^{\circ }-\widehat{OPB},$$ implying that $B{O}_{1}OP$ is cyclic.



Similarly, we can show that $C{O}_{2}OP$ is cyclic. Therefore, the circumcircles of $BO{O}_1$ and $CO{O}_2$ pass through $P$.