From the lengths of the sides, we get that $M$ is on the extension of segment $CB$ close to $B$ and $N$ is on the extension of $ED$ close to $D$. Now $\angle BAC= \alpha$ and $\angle ABC = \beta$. Let $H$ be the orthocenter of triangle $ABC$. Then quadrilaterals $ AFHE, BDHF, CEHD, ABDE, BCEF$ and $CAFD$ are cyclic. As, $ABDE$ is cyclic $$\angle CED = 180^{\circ} - \angle AED = \angle ABD = \beta$$. Analogically $\angle CDE = \angle BDF = \alpha$. We have $$\angle CED = \beta = \angle AEF$$then, $$\angle DEH = 90^{\circ} - \beta =\angle FEH$$. Thus, $EH$ is the angle bisector of $\angle DEF$. From here we get $DE \parallel FM$ $$\angle MFE = 180^{\circ} - \angle FED = 180^{\circ} - 2(90^{\circ} - \beta) = 2\beta$$Because $FN$ is bisector of $\angle MFE$, $\angle EFN = \frac {1}{2} \cdot 2\beta = \beta$. As we kMow that $\angle FEH = 90^{\circ} - \beta$ and that $FN \perp EH$ we get that $EH$ is not only a bisector but also an altitude in $\triangle FEN$. Then $FN$ is a side bisector. $B$ is a midpoint, $|BF| = |BN|$. We have earlier stated that $\angle CDE = \alpha = \angle BDF$. As $DE \parallel FM$ we get $$\angle BMF = \angle CDE = \alpha$$Thus, $$\angle DMF = \angle BMF = \angle BDF = \angle MDF, |FM| = |FD|$$Now call $S$ the intersection of $AC$ and $MF$. Then $\angle BFM = \angle AFS$ and from $DE \parallel FM$ we have $\angle CED = \angle CSF$. Due to outer angle theorem in $\triangle AFS$ $$\angle CSF = \angle SAF + \angle AFS = \alpha + \angle AFS$$Combining all the facts above, $$\angle CED = \alpha + \angle BFM$$Applying $\angle CED = \beta$, thus $\angle BFM = \beta - \alpha$. We also know that $\angle BMF = \alpha$. We conclude all these with $|BF| =|BM|$ if and only if $\beta - \alpha = \alpha$ if and only if $\beta = 2\alpha$. We already have $|BF| =|BN|$ which gives us $B$ is the center of circumcircle of $\triangle FMN$ if and only if $\beta = 2\alpha$. We have earlier stated that $EH$ is both bisector and altitude in $\triangle EFN$, thus it is an isosceles triangle with vertex $E$. $$\angle DNF = \angle ENF = \angle EFN = \beta$$. We also know that $$\angle CDE = \alpha = \angle BDF$$which leads to $$\angle NDF = \angle NDB + \angle BDF = \angle CDE + \angle BDF = 2\alpha$$Thus $|FD| =|FM|$ which holds $F$ is the circumcenter of $\triangle DMN$ if and only if $\beta = 2\alpha$. We conclude all this and get that $F$ is the circumcenter of $\triangle DMN$ if and only if $B$ is the circumcenter of $\triangle FMN$ because both properties hold if and only if $\beta = 2\alpha$. I really got tired writing this