Obviously $AD \cap (BIC)=I_A$ is the $A$-excenter. We want $\angle AFC=90^{o}$ and we have that $\angle MFC=90^{o}-\frac{\alpha}{2}$, so we want $\angle AFM =\angle BAM \iff MD^2=MA^2=MF \cdot MB=MI \cdot MI_A$, but the last one easily follows since $(AD; II_A)=-1$.

Let $L$ be the midpoint of the arc $BC$ not containing $A$ in $(ABC)$ and let $N$ be the midpoint of $AC$. It is easy to check that $\triangle ALC\sim \triangle ABD$, as $LN$ and $BM$ are medians we have $\angle NLC=\angle MBD$. On the other hand $\angle FLC = 2\angle FBD$ so $LN$ bisects $\angle FLC$, this implies that $\triangle NLC = \triangle NLF$. Finally we have $NA = NC = NF$ which implies the result.

Let the centre of the circumcirle of triangle $BIC$ be $T$,then because $BT$=$IT$=$CT$,$T$ is on the circumcirle of triangle $BAC$,easily we have $MF \cdot MB$=$MN^2-BT^2$=$MN^2-DT \cdot TA$=$MN^2-(MN-MT) \cdot(MN+AM)$=$MN^2$ then $\triangle ABM $ is similar to $\triangle AFM$ just angle chasing we can know that $AF\bot FC$

$J$ is the excentre wrt $A$ of $\Delta ABC$ its welknow $(ADIJ)=-1$ so $MA^2 =MF.MB \rightarrow \angle MFA = \angle BAM$ $\rightarrow \angle AFC = \angle CFM + \frac{ \angle A}{2} = 180^{\circ} - \angle BIC + \frac{ \angle A}{2} =90^{\circ} $

Let $AD\cap \odot(BIC)=J$,suppose$T\in BC,AT\bot BC$ Note that $\dfrac{JD}{DI}=\dfrac{JA}{AI}$ So $MI^{2}=MI\cdot MJ=MF\cdot MB,MT^{2}=MF\cdot MB=MA^{2}$ Which means that $\angle AFM=\dfrac{1}{2}\angle A,\angle MTF=\angle MDF$ So $M,F,T,D$ are cyclic Hence $\angle MFT=\angle ADC=180^{\circ}-\dfrac{1}{2}\angle A-\angle C$ So $\angle AFT=180^{\circ}-C$ Which means that $AFTC$ are cyclic So $\angle AFC=\angle ATC=90^{\circ}\Rightarrow AF\bot FC$

Note that the condition that $AF \perp FC$ simply says that $F$ lies on the circle with diameter $AC$. Thus, we wish to show that points $A$ , $F$ , $H$ and $C$ are concyclic, where $H$ is the foot of the $A-$altitude. Let $N_A$ denote the $BC$ arc midpoint which is well known to be the center of $(BIC)$. We perform a $\sqrt{AB \cdot AC}$ inversion followed by a reflection across the $A$-angle bisector. Then, the problem rewrites to Inverted Problem wrote: Let $\triangle ABC$ be a triangle with minor arc midpoint of $BC$, $N_A$. Let $P$ be the reflection of $A$ across $N_A$. Now, let $R$ be the second intersection of circles $(ABP)$ and $(BIC)$. Show that the points $R$ , $A'$ (the $A$-antipodal point) and $C$ are collinear. Let $(BIC) = \omega$. We instead let $R = \overline{AC'} \cap \omega$ and show that points $A$ , $B$ , $R$ and $P$ are concyclic. We first prove the following claim. Claim : Lines $\overline{BR}$ and $\overline{AI}$ are parallel. Proof : Let $R'$ be the intersection of the line through $B$ parallel to the $A$-angle bisector and $\overline{CA'}$. Clearly, \[\measuredangle BR'C = \measuredangle IAC + \measuredangle ACA' = 90 + \measuredangle IAC = \measuredangle BIC \]from which it follows that $R'$ indeed lies on $\omega$ which proves the claim. Now, we can observe the following pair of congruent triangles. Note that, \[AN_A = N_AP \text{ and }BN_A=N_AR\]due to reflection and radii respectively. Further, \[\measuredangle PN_AR = \measuredangle BRN_A = \measuredangle N_ABR = \measuredangle BN_AA\]from which it follows that $\triangle ABN_A \cong \triangle N_ARP$. Now, we already know that $ABRP$ is a trapezoid due to the previous claim. Due to the proved pair of congruent triangles, we further have $RP=AB$ so this trapezoid is in fact isosceles, and in turn cyclic. Thus, points $A$ , $B$ , $R$ and $P$ are concyclic which finishes the proof.

We do inversion with center $M$ with radius $AM$. If $J$ is $A$-excenter and $E= MC \cap (IBC)$, then as $(A,D;I,J)=-1$ we have from $AM^2 = MI.MJ \Rightarrow I \leftrightarrow J$ and hence so $F \leftrightarrow B$ and $E \leftrightarrow C$. Now proving $\angle CFA= 90 \Leftrightarrow \angle BAM + \angle MEB = 90$. But as $\angle MEB = \angle CJB = 90 - \angle BAI$, our statement is true.

Let $N$ be midpoint of $CA;$ $AD$ intersects $(ABC)$ again at $S,$ $B'$ is reflection of $B$ in $AD$. We have $\angle{MQB'} = \angle{ACB} = 180^{\circ} - \angle{MNC}$ and $\angle{MSB'} = \angle{MSB} = \angle{ACB} = 180^{\circ} - \angle{MNC}$. Then $Q, S \in (MNB')$. From this, we have $\angle{NCQ} = \angle{QBB'}$. But $\angle{QMB'} = \angle{QNB'}$ then $\triangle NQC \sim \triangle MBB'$. Combine with $\triangle MBB'$ is isosceles at $M,$ we have $\triangle NQC$ is isosceles at $N$ or $NC = NQ$. Hence $\triangle AQC$ is $Q$ - right triangle