Nice problem... Let $O$ be the center of the refered circuncircle and $O\widehat{A}M=\alpha \Rightarrow O\widehat{A}N=45-\alpha \Rightarrow O\widehat{M}A=\alpha$ and $O\widehat{N}A=45-\alpha$ once $O$ is circuncenter. Summing the external angles of $\Delta AMO$ and $\Delta ANO$ we get that $M\widehat{O}N=90$. $M\widehat{C}N=90$ so $ONCM$ is a cyclic quadrilateral. $\Delta ONM$ is a isosceles triangle so $O$ is the midpoint of the arc that pass throught $N,\ O,\ M$, so $OC$ bisects $B \widehat{C} D$, so it lies on $AC$ Just for remarks, what is the interval of the ratio that $O$ divides $AC$?

Attachments:

Let the perpendicular bisector of $AM$ be $PO$, where $P\in (AM)$ and $O\in (AC)$. Let $Q\in (AN)$ such that $OQ\perp AN$. We have $\angle BAM+\angle PAO=45^\circ=\angle MAN=\angle PAO+\angle OAN$, hence $\angle BAM=\angle QAO\Rightarrow\triangle ABM\sim\triangle AQO$. Therefore $AB\cdot AO=AM\cdot AQ\quad (1)$. Again, $\angle DAN+\angle QAO=45^\circ=\angle NAM=\angle QAO+\angle QAM$ hence $\angle DAN=\angle PAO\Rightarrow\triangle ADN\sim\triangle APO$. Therefore $AD\cdot AO=AN\cdot AP\quad (2)$. Since $AD=AB$, from (1) and (2) we get \[AM\cdot AQ=AN\cdot AP\implies\frac{AN}{AQ}=\frac{AM}{AP}=2.\] Hence $OQ$ is the perpendicular bisector of $AN$, as desired.

Dear Mathlinkers, an article concerning "Geometric miniatures on a square" with original proofs has been put on my website http://perso.orange.fr/jl.ayme vol. 7 Miniatures géométriques sur un carré p. 18 Sincerely Jean-Louis

From the problem http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=376897, it’s easy to see that $A$ is the excenter of $\Delta CMN$, so $\angle ANM=\angle ANB\ (\ 1\ )$, but $\angle ANB=90^\circ-\angle BAN\ (\ 2\ )$, while $\angle BAN=\angle CAM$, hence $\angle ANM=90^\circ-\angle MAC$, this relation showing that the circumcenter of $\Delta AMN$ lies on $AC$. Best regards, sunken rock