A quadrilateral $ABCD$ is such that $AD$ is not parallel to $BC$. Let $O$ be the intersection of $AC$ and $BD$. Let $M$ and $N$ be points on $AD$ and $CB$ respectively so that $\frac {AM} {AD}= \frac {CN} {CB}$. Let $P$ and $Q$ be the intersections of $MN$ with $AC$ and $MN$ with $BD$ respectively. Prove that the centre of the circumcircle of triangle $OPQ$ belongs to a fixed line.
I do not know exactly what the centre of the circumcircle of triangle $OPQ$ belongs to a line or a circle. I do not have any ideas, anyone can help me?
I'll refer to Yufei-Zhao's article: http://yufeizhao.com/olympiad/cyclic_quad.pdf
Since map $M\mapsto N$ is spiral similarity $\mathcal{S}$, which takes $A$ to $C$ and $D$ to $B$, the center $S$ of $\mathcal{S}$ is intersection point of circles $(AOD)$ and $(BOD)$ which is fixed point since the mentioned circles are.
We use that property two more times. Since $\mathcal{S}$ takes
i) $A$ to $C$ and $M$ to $N$ and thus $S\in (APM)\cap (CPN)$ and
ii) $D$ to $B$ and $M$ to $N$ and thus $S\in (QNB)\cap (DQM)$.
Now usea Miquel theorem for the circles $(APM), (DQM)$ and $(OPQ)$ tell us that $S\in (OPQ)$. So a circle $(OPQ)$ has two fixed points and thus it center must lie on a fixed line.