Consider $Q$, isogonal conjugate of $A$ $wrt$ $BCD$. We have $QB=QD$; therefore, $M$ is a projection of $Q$ onto $BD$. It's well-known that then $E, F, N, M$ are concyclic.

@Rickyminer Are you Chinese？This problem was sent to Southeast race this year.

Note that $AF\perp CD$, $AE\perp BC$ and $AN\perp BD$. This means that $A,D,F,N$ and $A,B,E,N$ are cyclic. Therefore $$\angle ENF=360^{\circ}-\angle FNA-\angle ENA=360^{\circ}-(180^{\circ}-\angle FDA)-(180^{\circ}-\angle EBA)=\angle CDA+\angle CBA=2\angle CBA.$$ Let $P,Q$ be the midpoints of $AB,AD$ respectively. Note that $MQ=\frac{AB}{2}=AP=EP$ since $\triangle AEB$ is right-angled at $E$, while $MP=\frac{AD}{2}=DQ=FQ$ since $\triangle AFD$ is right-angled at $F$. At the same time, $$\angle FQM=180^{\circ}-\angle DQF-\angle AQM=180^{\circ}-\angle EPB-\angle APM=\angle MPE$$since $\triangle DFA$ is similar to $\triangle BEA$ and $AQMP$ is a parallelogram. Overall, we have $\triangle FQM$ being congruent to $\triangle MPE$. As such, $$\angle FMQ+\angle QMP+\angle PME=\angle FMQ+\angle QFM+\angle QAP=(180^{\circ}-\angle FQM)+(180^{\circ}-\angle AQM) =360^{\circ}-(\angle FQM+\angle MQA)=360^{\circ}-\angle FQA=360^{\circ}-2\angle FDA$$since $QF=QD$. Hence $\angle FME=2\angle CDA=2\angle CBA=\angle FNE$. Hence $E,F,M,N$ concyclic.

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