Taiwan IMO Selection Camp

Take $N'$ such that quadrilateral $NEN'F'$ forms a parallelogram, extend $AD$ to meet circle $O_1$ at $E'$, and extend $BC$ to meet circle $O_2$ at $F'$.

Since $M, C, F, E, D$ are five concyclic points, we get $\angle MFC = \angle MDC = \angle MBA = \angle MFB$,

therefore $B, C, F$ are three collinear points. That is, $F = F'$. Similarly, $E = E'$.

Connect $NM$ and extend it to meet $AB$ at point $L$. Since $LA^2 = LM \cdot LN$, $LB^2 = LM \cdot LN$,

we get $LA = LB$.

Also, $MB = ND$, so $LM \parallel AD$, that is, $MN \parallel AE$; similarly, $MN \parallel BF$. Thus $O_1O_2 \perp AE$, $O_1O_2 \perp BF$, therefore $A, M, B$ are respectively symmetric to $E, N, F$ with respect to $O_1O_2$.

Hence $\angle ENF = \angle AMB$, therefore
$$\begin{aligned} \angle EN'F + \angle EMF &= \angle ENF + \angle EMF \\ &= \angle EMF + \angle AMB \\ &= \angle EMN + \angle FMN + \angle AMB \\ &= \angle AEM + \angle BFM + \angle AMB \\ &= \angle BAM + \angle ABM + \angle AMB \\ &= \pi \end{aligned}$$
therefore $M, E, N', F'$ are four concyclic points.

Also, $ENFN'$ is a parallelogram, so $\triangle ENF$ and $\triangle FNE'$ are congruent. Therefore the circumradii of $\triangle ENF$ and $\triangle MEF$ are equal.