# Geometry question posed in RMO 1999

Let $ABCD$ be a square and $M, N$ points on sides $AB, BC$ respectively, such that $\angle MDN = 45°$ . if $R$ is the midpoint of $MN$ show that $RP=RQ$ where $P,Q$ are the points of intersection of $AC$ with the lines $MD,ND$.
I have a feeling this problem can be solved by transformation geometry but since I'm new in this field, I have not proceeded a little. Any hint is greatly appreciated.

• There is a quite exceptional fact here: $P,Q,N,B,M$ all belong to the same circle with center $R$. – Jack D'Aurizio Mar 18 '15 at 16:51
• Jack D'Aurizio So you suggest to prove that MNQP is concyclic? – Aniket Bhattacharyea Mar 18 '15 at 16:59

Notice that $AMQD$ is cyclic with diameter $MD$ so $\angle QDM=\angle QMD=45^\circ$ thus $MQ\perp DN$.