Let point $E$ be point on side $BN$ and point $F$ on side $AM$ such that $\angle FQM=\angle FQA=\angle BQE=\angle EQN=\angle QCB$. Since this is a rectangle, we can obviously see that triangle $EQC$ is right angle triangle, so $\angle EQC=\angle FQC=90^\circ$ Since we have perpendicular bisector of $MC$, triangle $NMC$ is isosceles. Let point $G$ be on side $MQ$ such that $CG \perp MQ$, since $CGQB$ is cyclic, $BC=CG$ Let point $H$ be on side $BC$ such that $MH \perp BC$ and $MH=AB$, since $NMC$ is isosceles, this means that $MH=CG=BC=AB$ So we proved that $AB=BC$ meaning that this rectangle is a square.

Clearly $\angle CMN=\angle MCN=\angle CMB\ (\ 1\ )$. Let $m(\widehat{BCQ})=\alpha$, then $\angle BQC=90^\circ-\alpha=\angle CQM\ (\ 2\ )$, thus $C$ is $A-$excenter of $\triangle AQM$ and $AC$ is the angle bisector of $\angle BAD$, that is, $ABCD$ is a square. Best regards, sunken rock