We have $\angle FBD = \angle DFE = \angle FDB$, so $BF=FD$. Similarly $DE=EC$. $\angle DFE = \angle FBD = \angle IFE$, similarly $\angle FED = \angle IEF$. Hence $\bigtriangleup IEF \cong \bigtriangleup DEF$. This implies that $D$ is the reflection of $I$ about $EF$, and hence $EF$ is the midline of $\bigtriangleup IBC$. $A$ is a Miquel point of $IBC$, thus $IFAE$ cyclic. We have $\angle FEQ = \angle QDE = \angle FDR = \angle RFE$, hence $EM=MF$. Also $\angle FEM = \angle FDE = \angle FIE$, hence $M$ is the $I$-symmedian point of $\bigtriangleup IFE$. It suffices to show that $\angle FIA = \angle EIN$, where $N$ is the midpoint of $FE$. Now it is well-known that $D,A,N$ are collinear. Thus $\angle NIE = \angle NDE = \angle ADE = \angle AEF = \angle AIF$. Hence, $IA$ is the $I$-symmedian of $\bigtriangleup IFE$, and the conclusion of the problem follows.

Hello.A slightly different approach. From $\hat{IFA}=\hat{ADB}=\hat{AEC}$,it follows that $IFAE$ is cyclic.Observe that $\hat{IFE}=\hat{IBD}=\hat{EFD}=\hat{FDB}$ and $\hat{IEF}=\hat{ICD}=\hat{FED}=\hat{EDC}$.Hence,the triangles $\triangle{DEF},\triangle{IFE}$ are symmetric with respect to $EF$,thus equal.Also $FI=FB=FD$ and $EI=ED=EC$.We have $\hat{MFE}=\hat{RFE}=\hat{FDR}=\hat{FDE}=\hat{FIE}$,whence we get that $MF$ is tangent to the circumcircle of $IFAE$.Similarly we can prove that so is $ME$.Finally,observe that $\hat{AFM}=\hat{AFR}=\hat{ADR}=\hat{ADE}=\hat{AQE}=\hat{AQM}$.We deduce that $AMQF$ is cyclic.It follows that $\hat{MAF}=\hat{MQD}=\hat{EQD}=180^{\circ}-\hat{DCE}=180^{\circ}-\hat{FEI}=180^{\circ}-\hat{FAI}$,thus $\hat{MAI}=180^{\circ}$,q.e.d. [asy][asy]import graph; size(7.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.5, xmax = 7., ymin = -3., ymax = 6.; /* image dimensions */ /* draw figures */ draw(circle((1.862334388882474,2.4150740754174103), 1.548049567874546), linewidth(0.8)); draw(circle((4.27306917491673,3.04448150207501), 2.0364954027835105), linewidth(0.8)); draw((1.1519619747384653,3.7905116413649824)--(2.7523810260589,-2.0512447616183915), linewidth(0.8)); draw((2.7523810260589,-2.0512447616183915)--(6.030122965287318,4.074081183166025), linewidth(0.8)); draw((1.1519619747384653,3.7905116413649824)--(6.030122965287318,4.074081183166025), linewidth(0.8)); draw((1.9521715003986821,0.8696334398732969)--(4.391251995673109,1.0114182107738174), linewidth(0.8)); draw((2.408551081738495,3.863557692529458)--(1.9521715003986821,0.8696334398732969), linewidth(0.8)); draw((2.408551081738495,3.863557692529458)--(4.391251995673109,1.0114182107738174), linewidth(0.8)); draw((1.9521715003986821,0.8696334398732969)--(3.4103664607771735,2.422434033431555), linewidth(0.8)); draw((2.2491916417528834,2.8181338572606793)--(4.391251995673109,1.0114182107738174), linewidth(0.8)); draw((1.9521715003986821,0.8696334398732969)--(3.0469252312762256,1.4184788713784637), linewidth(0.8)); draw((3.0469252312762256,1.4184788713784637)--(4.391251995673109,1.0114182107738174), linewidth(0.8)); draw(circle((3.2464879002019615,-0.3458256855018819), 1.7755550760980399), linewidth(0.8)); draw((3.0469252312762256,1.4184788713784637)--(6.030122965287318,4.074081183166025), linewidth(0.8)); draw((3.0469252312762256,1.4184788713784637)--(3.4103664607771735,2.422434033431555), linewidth(0.8)); draw((2.2491916417528834,2.8181338572606793)--(3.0469252312762256,1.4184788713784637), linewidth(0.8)); draw((3.1045014119963117,2.096724891512446)--(3.0469252312762256,1.4184788713784637), linewidth(0.8) + linetype("2 2")); draw((2.408551081738495,3.863557692529458)--(3.0469252312762256,1.4184788713784637), linewidth(0.8)); /* dots and labels */ dot((1.1519619747384653,3.7905116413649824),linewidth(3.pt) + dotstyle); label("$B$", (0.9481303655861019,3.9638885884981674), NE * labelscalefactor); dot((6.030122965287318,4.074081183166025),linewidth(3.pt) + dotstyle); label("$C$", (6.100912952639475,4.170689227443455), NE * labelscalefactor); dot((2.408551081738495,3.863557692529458),linewidth(3.pt) + dotstyle); label("$D$", (2.2,4.1), NE * labelscalefactor); dot((1.9521715003986821,0.8696334398732969),linewidth(3.pt) + dotstyle); label("$F$", (1.4,0.5), NE * labelscalefactor); dot((4.391251995673109,1.0114182107738174),linewidth(3.pt) + dotstyle); label("$E$", (4.3,0.5), NE * labelscalefactor); dot((2.7523810260589,-2.0512447616183915),linewidth(3.pt) + dotstyle); label("$I$", (2.45,-2.4), NE * labelscalefactor); dot((3.0469252312762256,1.4184788713784637),linewidth(3.pt) + dotstyle); label("$A$", (2.9,1), NE * labelscalefactor); dot((2.2491916417528834,2.8181338572606793),linewidth(3.pt) + dotstyle); label("$Q$", (1.9,2.8), NE * labelscalefactor); dot((3.4103664607771735,2.422434033431555),linewidth(3.pt) + dotstyle); label("$R$", (3.4814381926658537,2.5335175024599326), NE * labelscalefactor); dot((3.1045014119963117,2.096724891512446),linewidth(3.pt) + dotstyle); label("$M$", (2.9,2.3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

And yet another slightly different approach First of all , with just simple angle chasing we see that $\bigtriangleup IEF \cong \bigtriangleup DEF$ and that $IFAE$ is cyclic We know that $FE^2=FQ \cdot FD$ so $\angle{FDE}=\angle{FEM}=\angle{FIE}$ so $ME$ is tangent to the circumcircle of $IFAE$ and in the same way $MF$ is also tangent $\implies MF=ME$ so we have to prove that $IA$ is the symmedian of triangle $IFE$ and to prove this it is well-known that we have to prove that : $\frac{AF}{FI}=\frac{AE}{EI}$ it is also easy to see that $ \bigtriangleup IAF \sim \bigtriangleup AEF$ and since $EC$=$IE$ we get the desired result .

Invert with center $D$ , then $I$ will be the incenter of triangle $ABC$, while $DEF$ is the contact triangle, $Q = AE \cap FD$, $R = AF \cap ED$, we see that M is the Miquel point of complete quadrilateral $ADEF$. We want to show $M,A,I,D$ concyclic, which is straight-forward angle-chasing.

Barycentrics wrt. $\triangle DEF$. Let $a=EF$, $b=FD$, $c=DE$. We easily get: $M=(a^2:c^2-a^2:b^2-a^2)$ $A=(a^2:b^2+c^2-a^2:b^2+c^2-a^2)$ $I=(-a^2:a^2+c^2-b^2:a^2+b^2-c^2)$ The area matrix becomes zero after 3 row-column operations.