In the square $ABCD$, the points $M$ and $N$ belong to the sides $BC$ and $CD$ respectively, such that $\angle MAN = 45^°$. Let $O$ be the point of intersection of the circle that goes through $C$, $M$ and $N$ with the segment $AC$. Is the point $O$ always the circumcenter of $\triangle MAN$?
The image shows $\angle NAD = \theta = 30^°$, but the question goes for all values of $\theta$ ranging from $0^°$ to $45^°$. The calculator suggests that it's true, but I can't find a way to prove it.
I've noticed that since $\angle MCN = 90^°$, the segment $MN$ must be a diameter of the small circle, and thus $\angle MON = 90^°$. If $O$ were to be the circumcenter of $\triangle MAN$, it would obey the rule of angles at the center being twice the angle at the circumference, however I don't know if that means it is necessarily the circumcenter.
First of all, if $N=D$, then $M=C$ and there are infinitely many different circles that can be drawn through $N$ and $M$. Similarily, if $N$=$C$. Therefore we are going to assume, that $N$ lies between points $D$ and $C$. WLOG $AD=1$. Put $\angle DAN=\theta$, $\angle NMC=\alpha$, $ND=q$, and let $r$ be the radius of the red circle.
$BM=tan(\frac{\pi}{4}-\theta)=\frac{1-tan(\theta)}{1+tan(\theta)}=\frac{1-q}{1+q}$, so $MC=1-\frac{1-q}{1+q}=\frac{2q}{1+q}$ and $CN=1-q$. Pythagoras theorem gives $MN^2=(1-q)^2+(\frac{2q}{1+q})^2=\frac{(1-q^2)^2+4q^2}{(1+q)^2}=\frac{(1+q^2)^2}{(1+q)^2}$, hence $r=\frac{MN}{2}=\frac{1+q^2}{2(1+q)}$. I'm going to switch to the complex plane now with the origin at $A$ and the real axis running along $AD$.
Note that $P=[\frac{1-q}{1+q}+\frac{q}{1+q}]+i[1-\frac{1-q}{2}]=\frac{1}{1+q}+i\frac{1+q}{2}$. Next, let $M'$ be the point obtained by rotating $M$ around $P$ by $\frac{\pi}{2}$ ccw. Since $M=-re^{-i\alpha}+P$, we get $M'=-re^{-i\alpha}e^{\frac{i\pi}{2}}+P=[-rcos(\alpha)+irsin(\alpha)]i+\frac{1}{1+q}+i\frac{1+q}{2}=[-ir\frac{2q}{2r(1+q)}-r\frac{1-q}{2r}]+\frac{1}{1+q}+i\frac{1+q}{2}=[\frac{1}{1+q}+\frac{q-1}{2}]+i[\frac{1+q}{2}-\frac{q}{1+q}]=\frac{1+q^2}{2(1+q)}+i\frac{1+q^2}{2(1+q)}$. Since $Re(M')=Im(M')$, $M'$ is a point that belongs to both the diagonal $AC$ and the red circle. There are precisely 2 such points, namely $O$ and $C$. Clearly, $C=1+i\neq \frac{1+q^2}{2(1+q)}+i\frac{1+q^2}{2(1+q)}$. Thus $M'=O$, so $\angle OPM=\frac{\pi}{2}$. Also, $AO=\sqrt{[Re(M')]^2+[Im(M')]^2}=\sqrt{2}(\frac{1+q^2}{2(1+q)})$. Aplying Pythagoras gives $MO=ON=\sqrt{2}r=\sqrt{2}(\frac{1+q^2}{2(1+q)})$. We therefore conclude that $O$ is the circumcenter of $\triangle MAN$ and the circumradius is $\sqrt{2}(\frac{1+q^2}{2(1+q)})$.