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.

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

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


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