SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote $S,T$ as the midpoints of $AM,AQ$ respectively. Hence, $KS$ is the perpendicular bisector of $AM$ and $TL$ is the perpendicular bisector of $AQ$. These imply that $AI,SK,TL$ are concurrent at the circumcenter $G$ of triangle $AMQ$. Note that $BI$ is the perpendicular bisector of $MN$ then $K\in BI,DI$ is the perpendicular bisector of $PQ$ then $L\in DI$. Consider two triangles IBD and GST with BS, DT, IG are concurrent at A then by applying the Desargues theorem, we have $S,T,R$ are collinear. In other word, $R$ belongs to $ST$ which is the perpendicular bisector of $AJ$ then $RA=RJ$.