Hint1

hint 2

This problem is just a simple angle chasing. At first let's call $\angle BAC = 2\theta, \angle CBA = 2\alpha, \angle BCA = 2\beta. => \angle DBI = \angle EBI = \alpha$ and $\angle DCI = \angle ICF = \beta$. As we can see $BDIE, DCFI, NFEM$ and $EFDM$ are all cyclic => $\angle DBI = \angle EDI = \angle IED = \alpha, \angle DCI = \angle IFD = \angle IDF = \beta$, but this implies that $IE = ID = IF$, i.e., $I$ is the circuncenter of $\triangle DEF$ (not usefull at all), and $\angle EDF = \angle EMF = \alpha + \beta$. But $NFME$ is cyclic, thus $\angle ENF + \angle EMF = 180^\circ$ $=>$ $\alpha + \beta + \angle ENF = 180^\circ$ => $\angle ENF = 2\theta + \alpha + \beta$ and $\angle ANF = \alpha + \beta$, thus $\angle ANF = \angle AFN = \angle AME = \alpha + \beta$ if and only if $EM \parallel FN $.

Easy angle chasing

$\angle BAC=2\alpha$ , $\angle ABC=2\beta$ , $\angle ACB=2\gamma$ $I$ is the incenter of traingle $ABC$ $=>$ $\angle ABI$=$\angle CBI$=$\frac{\angle ABC}{2}$=$\frac{2\beta}{2}$=$\beta$ $\angle BCI$=$\angle ACI$=$\frac{\angle ACB }{2}$=$\frac{2\gamma}{2}$=$\gamma$ $\angle BAC+\angle ABC+\angle ACB=180$ $2\alpha+2\beta+2\gamma=180$ $\alpha+\beta+\gamma=90$ $...(1)$ From $CFID$ cyclic we get: $\angle BCI$=$\angle DCI$=$\angle DFI$=$\gamma$ $\angle ACI$=$\angle FCI$=$\angle FDI$=$\gamma$ $=>$ $\angle DFI$=$\angle FDI$ $<=>$ The traingle $IDF$ is isoscleles $<=>$ $IF$=$ID$ From $BEID$ cyclic we get: $\angle ABI$=$\angle EBI$=$\angle EDI$=$\beta$ $\angle CBI$=$\angle DBI$=$\angle DEI$=$\beta$ $=>$ $\angle EDI$=$\angle DEI$ $<=>$ The traingle $IDE$ is isosceles $<=>$ $ID$=$IE$ From $IF$=$ID$ and $ID$=$IE$ we get that $ID$=$IE$=$IF$ Since $DEMF$-cyclic we have: $\angle EDF+\angle EMF=180$ $\angle EDI+\angle FDI+\angle EMF=180$ $\beta+\gamma+\angle EMF=180$. Combining with $(1)$ $90-\alpha+\angle EMF$=$180$ $\angle EMF=\alpha+90$ $\angle AME=180-\angle EMF$ $\angle AME=180-(\alpha+90)$ $\angle AME=90-\alpha$ From traingle $AME$ we have: $\angle AME+\angle MAE+\angle AEM=180$ $90-\alpha+2\alpha+\angle AEM=180$ $90+\alpha+\angle AEM=180$ $\angle AEM=90-\alpha$ $=>$ $\angle AME=\angle AEM$ $<=>$ $AEM$ is an isoscele traingle $\angle NEM=180-\angle AEM$ $\angle NEM=180-(90-\alpha)$ $\angle NEM=90+\alpha$ $NEMA- cyclic$ $<=>$ $\angle NEM+\angle NFM=180$ , $\angle FME+\angle FNE=180$ $90+\alpha+\angle NFM=180$ $\angle NFM=90-\alpha$ $=>$ $\angle AME=\angle NFM$ $\angle AME=\angle AFN$ $...(2)$ $\angle FME+\angle FNE=180$ $\alpha+90+\angle FNE=180$ $\angle FNE=90-\alpha$ $=>$ $\angle AEM=\angle FNE$ $\angle AEM=\angle ANF$ $...(3)$ From $(2)$ we had: $\angle AME=\angle AFN$ And from $(3)$ we had: $\angle AEM=\angle ANF$ $<=>$ $EM$ $\parallel$ $NF$

Attachments:

It took me a longer time to draw a good diagram than it did for me to solve this . Let $\angle BAC=2x$ and $\angle ACB=2z$, and $\angle CBA=2y$, now since I is the incenter, $\angle DCI=\angle ICF=z$ and $\angle IBE=\angle DBI=y$, now it is easy to see that $\angle DBI = \angle EDI = \angle IED = y , \angle DCI = \angle IFD = \angle IDF = z$ , so I is also the cirmumcenter of $\triangle DEF$ and $\angle EDF=\angle FME=y+z$, but since $EMFN$ is cyclic $\angle ANF=x+y$, which is only true if and only if $EM \parallel FN.$ Hence, proved.

Notation: (XY) stands for small arc XY <AEM=<EMF (because EMFN is cyclic) = (EM)/2 + (EN)/2 (1) <AEM= <EDF (because EMFD is cyclic) = (EM)/2 + (MF)/2 (2) From (1), (2) => (EN )= (MF) => <EMN = < MNF => EM //AF Solution is valid for random I, not only the incenter, as I didn’t use any property of the incenter