First notice that $\angle BFC=\angle BIC=90-a$ (we denote $\angle BAC=2a$) Let $G$ be the A-excenter of $\triangle ABC$ and $AD$ intersects $(ABC)$ at $E$ It is well-known that $(A, D; I, G)=-1$. We have $M$ and $E$ the midpoints of $AD$ and $IG$, so $MD^2=MI * MG=MF*MB$, so $AD$ is tangent to $(AFB)$ and $(AFD)$. Thus $\angle FAM= \angle ABF=x$ In $\triangle AFB$ we have $ \angle AFB=180- \angle BAF- \angle ABF=180-( \angle BAM- \angle FAM)-x=180-(a-x)-x= 180-a+x-x=180-a$ But $\angle AFC= 360- \angle AFB- \angle BFC=360-(180-a)-(90+a)=90$ So $\angle AFC=90$ and we are done

Let $AD$ intersect circle $(ABC)$ at point $T$. It is well known that $T$ is the midpoint of the minor arc $BC$ and is center of circle $(BIC)$. The following Claim is the pith of the problem. Claim: $MA$ is tangent to circle $(AFB)$. Proof: It suffices to prove that $MA^2=MF \cdot MB$. Note that $MF \cdot MB={\rm pow} (M,(BIC))=MT^2-IT^2$. Therefore, $MF \cdot MB-MA^2=(MT^2-IT^2)-MA^2=(MT^2-MA^2)-IT^2=$ $=(MT-MA)(MT+MA)-IT^2=TD \cdot TA-TB^2=0,$ with the last equality being true since $\angle TBD=\angle BAD$ $\blacksquare$ To the problem, note that from the Claim $\angle MAF=\angle ABM,$ and so $\angle AFC=\angle AFM+\angle MFC=\angle BAF+\angle ABF+\angle MFC=$ $=\angle BAF+\angle FAM+\angle MFC=\angle BAM+180^\circ-\angle BFC=$ $=\dfrac{\angle A}{2}+180^\circ-(90^\circ+\dfrac{\angle A}{2})=90^\circ,$ and so $AF \perp FC,$ as desired.

Let $\overline{AM}$ intersect the circumcircle of $\triangle ABC$ at $S\neq A$. Let $N$ be the midpoint of side $AC$. Angle chasing shows that $\triangle BAD\sim\triangle SAC$ and therefore $\triangle MBD\sim\triangle NSC$. Also, by the Incenter-Excenter Lemma, $S$ is the circumcenter of the cyclic $BFIC$, so \[\angle FSC = 2\angle FBC = 2\angle NSC\Longrightarrow \angle CSN = \angle FSN\]Hence $NS$ is the perpendicular bisector of $FC$, implying that $NF = NC = NA$ as $N$ is the midpoint of $AC$, so $\angle AFC = 90^{\circ}$ as desired.

Redefine $F$ to be the $B$-humpty point in $\triangle BAD$. Then clearly $F$ lies on $BM$. Since $\frac{DC}{AC} = \frac{DB}{AB}$, $C$ lies on the $B$-apollonian circle of $\triangle BAD$. But also, $F,I$ lie on the apollonian circle, so points $B,F,I,C$ are concyclic, implying this is the same as the $F$ in the problem. To finish, note that $\angle AFC = \angle AFM + \angle MFC = \angle BAD + (180 - \angle BFC) = 180 + \angle BAD - \angle BIC = 90^\circ$, so we're done. $\blacksquare$

This problem used in Azerbaijan 2023 BMO TST. And this is my solution during exam: We use complex numbers.Let $(BIC)$ be unit circle and $I_a , K$ be circumcenter of $(BIC)$ and antipode of $I_a$ wrt $(BIC)$ , respectively. Then $$i=1 , k=-1 , i_a=0 , a=\bar{a}$$$$D=BC \cap IK \implies d=\frac{bc(i+k)-ik(b+c)}{bc-ik} \implies d=\frac{b+c}{bc+1} \wedge \ \ \bar{d}=d$$$$ABOC \text{ is cyclic} \implies \frac{c(b-a)}{a(b-c)}=\frac{b(\bar{b}-a)}{a(c-b)} \implies a=\frac{bc+1}{b+c}$$$$m=\frac{a+d}{2} \implies m=\frac{(bc+1)^2+(b+c)^2}{2(bc+1)(b+c)} \wedge m=\bar{m}$$Since $M$ lies on $BF$ we have $$m+\bar{m}bf=b+f \implies f=\frac{b-m}{bm-1}$$After simplifying the expression we get $$f=\frac{2b(bc+1)(b+c)-(bc+1)^2-(b+c)^2}{b(bc+1)^2+b(b+c)^2-2(bc+1)(b+c)}$$So we need to show $$\frac{a-f}{c-f} \in \mathbb{R} \implies \frac{b(bc+1)^3+(b+c)(bc+1)^2+b(bc+1)(b+c)^2+(b+c)^3-2(bc+1)(b+c)(b^2+2bc+1)}{(bc+1)^3(b+c)+(bc+1)(b+c)^3-2(b+c)^3(bc+1)} \in \mathbb{R} $$ And the rest is simple algebra.$\blacksquare$

This problem was used in Uzbekistan MO 2023. I want to give a bit different solution. Let $AD$ meets $(ABC)$ at $N$, and $I_a$ be the excenter of $\triangle ABC$. Our aim is to define the point $F$ in a different way. Let $\triangle ABC$ and $\triangle AB'C'$ be symmetrical w.r.t. $AD$, i.e. $B'\in AC$, $C'\in AB$ and $BB'\perp AD$, $CC'\perp AD$. Assume that $(ABD)$ and $(AB'D)$ meet $(BIC)$ at points $X$ and $X'$, respectively. It is clear that they are symmetrical w.r.t. $AD$, and $\angle ABX = \angle AB'X' = \angle CBX' = \angle DBX'$. We know that the points $\{A,I,D,I_a\}$ are harmonic (forms harmonic bundle), and the circle $(BIC)$ (which is centred at $N$) be the Apollonian circle for $\triangle ABD$. That means $NB$ is tangent to $(ABD)$ and $\{A,B,D,X'\}$ are harmonic as well. In other words, $BX'$ should be the symmedian line for the triangle $ABD$. Therefore, $\angle DBX' = \angle ABM$. Putting together the above results, we obtain that $\angle ABX = \angle ABM$, that is $B,X,M$ are collinear. So the point $X$ should be our $F$, $X\equiv F$. Hence, we have a harmonic quadrilateral $AFDB'$, which gives $\angle AFM = \angle DFB' = \angle DAB'$. Now, it is clear that \[ \angle AFC = \angle AFM + \angle CFM = \angle DAB' + \angle BB'A=90^{\circ} \]as $BB'\perp AD$.

Let $AD$ intersect $(A B C)$ at point $K$. From $\angle DBK=\angle KAB$ we have $KB^2=KD*KA$ $(1)$. Power of $M$ respect to $(B F I C)$ is equal to (we use that $K$ is center of $(B F I C)$, from thrillium and $(1)$): $MF*MB=MK^2-KB^2=(MD+DK)^2-DK*(2*MD+DK) = MD^2+DK^2+2*MD*DK-DK^2-2*MD*DK=MD^2$. Then $MA^2=MD^2=MF*MB$, then $AM$ is tangent to $(A B F)$, Then $\angle AFB=180-\angle AFM=180-(\angle BAC/2)$ $(2)$. And we know that $90+(\angle BAC/2)=\angle BIC=\angle BFC$ $(3)$. Then from $(2)$ and $(3)$, $\angle AFC=360-(\angle AFB+\angle BFC)=90$, as desired.

$(A,D;I,I_a)=-1 \implies MA^2=MI.MI_a=MF.MB \implies MA$ is tangent to circle $(AFB)$. $\angle CFA=\angle CFM + \angle MFA = \angle BIC + \angle FBA + \angle BAF = \angle BIC + \angle FAM + \angle BAF = 90^\circ$

Solution by Barycentric Coordinates: Let our reference triangle be $ABC$. $ I = (a:b:c) \implies D=(0:b:c) \implies M = D/2 + A/2= ( \frac{b+c}{2}:b/2:c/2)=(b+c:b:c)$ Let us calculate the circumcircle of $BIC$. Let the circle equation be \[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]By substituting $B$ and $C$ we get $v=w=0$ Lastly we need to substitute $I=(a:b:c) \implies abc(a+b+c)=ua(a+b+c)\implies u=bc$ We than get the circle of $BIC$ to be\[-a^2yz-b^2xz-c^2xy+bcx(x+y+z)=0\]Let $F=MB\cap(BIC)$ Than $F=(b+c:t:c)$ If we put it in the equation we get $t=\frac {bc(b+c)}{S_b} \implies F=(1:\frac{bc}{S_b}:\frac{c}{b+c})$ Let the perpendicular foot from $A$ to $BC$ be $K$ $H=(S_{BC}:S_{AC}:S_{AB}\implies K=(0:S_{AC}:S_{AB})$ Let the circle equation be \[-a^2yz-b^2xz-c^2xy+(px+qy+rz)(x+y+z)=0\]By $A$ and $C$ we get $p=r=0$ and by substituting $K$ we get $a^2S_{AC}S_{AB}=qS_{AC}(0+S_{AC}+S_{AB})\implies q=S_B$ Then the circle equation is \[-a^2yz-b^2xz-c^2xy+S_By(x+y+z)=0\]Then the equation of Radical Axis of the two circles is $S_By=bcx$ $F$ is easily seen to satisfy the condition meaning $F$ is in the radical axis of the two circles. Then $F\in (ADC)\implies AF\perp FC$ $ \blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -1.232482102569296, xmax = 11.908982883586937, ymin = -12.071873610784134, ymax = 2.7778272510863076; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((0.23098333107937805,1.6619537759168523)--(0.26,-4.29)--(10.38,-2.73)--cycle, linewidth(0.4) + zzttqq); /* draw figures */ draw((0.23098333107937805,1.6619537759168523)--(0.26,-4.29), linewidth(0.4) + zzttqq); draw((0.26,-4.29)--(10.38,-2.73), linewidth(0.4) + zzttqq); draw((10.38,-2.73)--(0.23098333107937805,1.6619537759168523), linewidth(0.4) + zzttqq); draw(circle((4.977933273433024,-1.2909517481680761), 5.5904547927143255), linewidth(0.4)); draw((xmin, -1.5143088594153271*xmin + 2.011733880547618)--(xmax, -1.5143088594153271*xmax + 2.011733880547618), linewidth(0.4)); /* line */ draw(circle((5.8296431098352155,-6.816146327905367), 6.115745223676725), linewidth(0.4)); draw((0.26,-4.29)--(2.015991682353883,-1.0411001845484766), linewidth(0.4)); draw((0.23098333107937805,1.6619537759168523)--(0.6650579712660512,-3.540570044654556), linewidth(0.4)); draw((0.6650579712660512,-3.540570044654556)--(10.38,-2.73), linewidth(0.4)); draw((xmin, -0.06491959873961704*xmin-3.49739474802138)--(xmax, -0.06491959873961704*xmax-3.49739474802138), linewidth(0.4)); /* line */ draw(circle((6.242551700591986,-0.45583913603740017), 6.3736960297184035), linewidth(0.4)); draw((9.199759190359845,-11.91954296590187)--(11.374391377142317,-4.235815672132819), linewidth(0.4)); draw((5.8296431098352155,-6.816146327905367)--(11.374391377142317,-4.235815672132819), linewidth(0.4)); draw((2.459527029310588,-1.7127496899088663)--(11.374391377142317,-4.235815672132819), linewidth(0.4)); draw((5.8296431098352155,-6.816146327905367)--(10.38,-2.73), linewidth(0.4)); draw((10.38,-2.73)--(11.374391377142317,-4.235815672132819), linewidth(0.4)); /* dots and labels */ dot((0.23098333107937805,1.6619537759168523),dotstyle); label("$A$", (-0.24111698533846962,1.4473627229996495), NE * labelscalefactor); dot((0.26,-4.29),dotstyle); label("$B$", (-0.04798503771298647,-4.625564074557192), NE * labelscalefactor); dot((10.38,-2.73),dotstyle); label("$C$", (10.466976555229984,-2.522571755968604), NE * labelscalefactor); dot((3.801000033628388,-3.744154145013806),linewidth(4pt) + dotstyle); label("$D$", (3.879031230671837,-3.5740679152628982), NE * labelscalefactor); dot((2.015991682353883,-1.0411001845484766),linewidth(4pt) + dotstyle); label("$M$", (2.097925491459048,-0.8702206485061418), NE * labelscalefactor); dot((2.459527029310588,-1.7127496899088663),linewidth(4pt) + dotstyle); label("$I$", (2.5485667025851755,-1.5354529125494707), NE * labelscalefactor); dot((5.8296431098352155,-6.816146327905367),linewidth(4pt) + dotstyle); label("$E$", (5.917646233385271,-6.6427199719789), NE * labelscalefactor); dot((0.6650579712660512,-3.540570044654556),linewidth(4pt) + dotstyle); label("$F$", (0.7460018580806664,-3.359476862345695), NE * labelscalefactor); dot((9.199759190359845,-11.91954296590187),linewidth(4pt) + dotstyle); label("$I_A$", (9.286725764185364,-11.74998703140833), NE * labelscalefactor); dot((11.374391377142317,-4.235815672132819),linewidth(4pt) + dotstyle); label("$G$", (11.45409539864912,-4.067627336972465), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Since $(AD,II_A)=-1$ $MA^2 = MD^2=MI \cdot MI_A=MF \cdot MD $ Let $G=FD \cap (BIC)$, $\measuredangle BCG = \measuredangle BFD = \measuredangle ADC$ and hence $CG \parallel AD$ $\measuredangle AFC = \measuredangle AFG - \measuredangle GFC = 180^\circ - \measuredangle B - (90^\circ-\measuredangle B)=90^\circ$