Let $D$ be foot of altitude from $A$ to $BC$ and let points $Y$ and $Z$ lie on rays $\overrightarrow{CA}$ and $\overrightarrow{BA}$ respectively s.t. $CY=BZ=\frac{BC}{2}$. Proof of $\text{only if}$ part. Lemma 1: $YMDZ$ is cyclic with centre $I$. Proof: Note that since $BM=\frac{BC}{2}=BZ$, $\angle MBI=\angle ZBI$ and $BI=BI$ we have that $\triangle BIM\cong\triangle BIZ$. Similarly $\triangle CIM\cong\triangle CIY$. Hence $IM=IZ=IY$. Now note that since $ADM$ is right angled at $D$ and $S$ is midpoint of $AM$ we have that $SD=SM$. Also $SI\perp DM$. Thus $SI$ is $\perp$ bisector of $DM$ which implies that $ID=IM$. Thus we have $ID=IM=IY=IZ$. Hence $YMDZ$ is cyclic with centre $I$. Lemma 2: $ALYIZ$ is cyclic with centre $T$. Proof: Since $\triangle BIM\cong\triangle BIZ$ and $\triangle CIM\cong\triangle CIY$ we have that $\angle AZI=180^{\circ}-\angle BZI=180^{\circ}-\angle BMI=\angle CMI=\angle CYI$ which means that $AYIZ$ is cyclic. Also note that $BZ=CY, LB=LC$ and $\angle LBZ=\angle LCY$. Hence $\triangle LBZ\cong\triangle LCY$. Thus there exists spiral similarity with centre $L$ taking $BZ$ to $CY$ which means there exists spiral similarity with centre $L$ taking $ZY$ to $BC$. Thus $\angle ZLY=\angle BLC=\angle BAC=\angle ZAY$ which means that $ALYZ$ is cyclic. Thus $ALIYZ$ is cyclic. Also $\angle IAL=90^{\circ}$ which implies that $LI$ is diameter of the circle. Thus midpoint of $LI$ ie $T$ is centre of cyclic quadrilateral $ALYIZ$ as desired. Lemma 3: $Q$ lies on $YZ$ Proof: Let midpoint of $YZ$ be $P'$. Note that $IY=IZ$. Also $\triangle LBZ\cong\triangle LCY$. Hence $LY=LZ$. Thus $LI$ is $\perp$ bisector of $YZ$. Thus $P'$ lies on $LI$. Now note that $\triangle LZY\sim\triangle LBC$ and $P'$ is midpoint of $YZ$ and $M$ is midpoint of $BC$. Hence $\triangle LZP'\sim\triangle LBM$. Hence there exists spiral similarity with centre $L$ taking $ZP'$ to $BM$ which means that there exists spiral similarity with centre $L$ taking $ZB$ to $P'M$. Thus $\triangle LZB\sim\triangle LP'M$. Hence $\angle LBZ=\angle LMP'$. Let $L'$ be midpoint of arc $BC$ of $\odot (ABC)$ not containing $A$. Then $\angle LBZ=\angle LL'A=\angle LL'I$. Thus angle that $L'I$, ie $AI$ makes with $\perp$ bisector of $BC$ and angle which $MP'$ makes with $\perp$ bisector of $BC$ is same. Thus $MP'\parallel AI$. And $P'$ lies on $LI$. Hence $P'\equiv P$. Also note that $YZ\perp LI$ since $LI$ is $\perp$ bisector of $YZ$. Hence we have that $Q$ lies on $YZ$ as desired. Now note that radical axis of $\odot (ADM)$ and $\odot (YMDZ)$ is $DM$. Also radical axis of $\odot (YMDZ)$ and $\odot (ALYIZ)$ is $YZ$. Hence by applying radical axes theorem on $\odot (ADM)$, $\odot (ALYIZ)$ and $\odot (YMDZ)$, we have that $Q=DM\cap YZ$ lies on radical axis of $\odot (ADM)$ and $\odot (ALYIZ)$. Also $A$ lies on radical axis of these 2 circles. Hence $AQ$ is the radical axis of $\odot (ADM)$ and $\odot (ALYIZ)$. Also the centres of circles $\odot (ADM)$ and $\odot (ALYIZ)$ are $S$ and $T$ respectively. Hence we have $AQ\perp ST$ as desired. Proof of $\text{if}$ part. Lemma 1: $I$ is centre of $\odot (YMZ)$. Proof: Note that since $BM=\frac{BC}{2}=BZ$, $\angle MBI=\angle ZBI$ and $BI=BI$ we have that $\triangle BIM\cong\triangle BIZ$. Similarly $\triangle CIM\cong\triangle CIY$. Hence $IM=IZ=IY$ which means that $I$ is centre of $\odot (YMZ)$ as desired. Lemma 2: $ALYIZ$ is cyclic with centre $T$. Proof: Since $\triangle BIM\cong\triangle BIZ$ and $\triangle CIM\cong\triangle CIY$ we have that $\angle AZI=180^{\circ}-\angle BZI=180^{\circ}-\angle BMI=\angle CMI=\angle CYI$ which means that $AYIZ$ is cyclic. Also note that $BZ=CY, LB=LC$ and $\angle LBZ=\angle LCY$. Hence $\triangle LBZ\cong\triangle LCY$. Thus there exists spiral similarity with centre $L$ taking $BZ$ to $CY$ which means there exists spiral similarity with centre $L$ taking $ZY$ to $BC$. Thus $\angle ZLY=\angle BLC=\angle BAC=\angle ZAY$ which means that $ALYZ$ is cyclic. Thus $ALIYZ$ is cyclic. Also $\angle IAL=90^{\circ}$ which implies that $LI$ is diameter of the circle. Thus midpoint of $LI$ ie $T$ is centre of cyclic quadrilateral $ALYIZ$ as desired. Lemma 3: $Q$ lies on $YZ$ Proof: Let midpoint of $YZ$ be $P'$. Note that $IY=IZ$. Also $\triangle LBZ\cong\triangle LCY$. Hence $LY=LZ$. Thus $LI$ is $\perp$ bisector of $YZ$. Thus $P'$ lies on $LI$. Now note that $\triangle LZY\sim\triangle LBC$ and $P'$ is midpoint of $YZ$ and $M$ is midpoint of $BC$. Hence $\triangle LZP'\sim\triangle LBM$. Hence there exists spiral similarity with centre $L$ taking $ZP'$ to $BM$ which means that there exists spiral similarity with centre $L$ taking $ZB$ to $P'M$. Thus $\triangle LZB\sim\triangle LP'M$. Hence $\angle LBZ=\angle LMP'$. Let $L'$ be midpoint of arc $BC$ of $\odot (ABC)$ not containing $A$. Then $\angle LBZ=\angle LL'A=\angle LL'I$. Thus angle that $L'I$, ie $AI$ makes with $\perp$ bisector of $BC$ and angle which $MP'$ makes with $\perp$ bisector of $BC$ is same. Thus $MP'\parallel AI$. And $P'$ lies on $LI$. Hence $P'\equiv P$. Also note that $YZ\perp LI$ since $LI$ is $\perp$ bisector of $YZ$. Hence we have that $Q$ lies on $YZ$ as desired. Note that radical axis of $\odot(YMZ)$ and $\odot (ALYIZ)$ is $YZ$. Also note that $A$ lies on radical axis of $\odot (ADM)$ and $\odot (ALYIZ)$. $T$ is centre of $\odot (ADM)$ and $S$ is centre of $\odot (ALYIZ)$. Also $AQ\perp ST$. Hence $AQ$ is radical axis of $\odot (ADM)$ and $\odot (ALYIZ)$. Hence by applying radical axes theorem on $\odot (ADM)$, $\odot (ALYIZ)$ and $\odot (YMZ)$, we have that $Q=AQ\cap YZ$ lies on radical axis of $\odot (ADM)$ and $\odot (YMZ)$. Also $M$ lies on radical axis of $\odot (ADM)$ and $\odot (YMZ)$. Thus $MQ$, ie $BC$ is radical axis of $\odot (ADM)$ and $\odot (YMZ)$. $I$ is centre of $\odot (YMZ)$ and $S$ is centre of $\odot (ADM)$. Hence $IS\perp BC$ as desired.

Very nice and magic problem! Solved with p_square, i3435. After dying for 2 hours, Rg230403 told us to add the magic points $Y,Z$ and we finished soon after that Ignore the given conditions for now. Let $Y,Z$ be points on $AC, AB$ such that $BZ = BM = CY$ (!). Since $BZ = CY$, we have $ALYZ$ cyclic and since $IZ = IM = IY$ and $AI$ is angle bisector, we have $I$ lies on this circle too. Let $M_A$ be the midpoint of minor arc $BC$ and $AI \cap EF = X$. Since $\frac{LP}{PI} = \frac{LM}{MM_A} = \frac{AX}{XI}$, we have $P \in EF$. Note that $\angle IPF = \angle ILA = \angle IYA$, so $IPEY$ and similarly, $IPFZ$ is cyclic. So $\angle IPY = \angle IEY = 90^\circ$ and $\angle IPZ = 90^\circ$ so $P,Y,Z$ are collinear and so $P$ is midpoint of $YZ$. This means $Y,Z,P,Q$ are collinear because $YZ \perp LI$ too. Now assume $SI \perp BC$, which means $S \in EF$ by a well known lemma. So if we let $R \in BC$ such that $AR \perp BC$, then $IM = IQ$ and so $MRZY$ cyclic because all four are equidistant from $I$. By radical axis theorem on $(ZYALI), (MRZY), (AMR)$, we obtain that $Q$ is the radical center, so $AQ$ is radical axis of $(AMR)$ and $(ZYALI)$, so the line connecting their centers is perpendicular to it, or $AQ \perp ST$, as desired. These last few steps can also be reversed, so the statement is indeed an if and only if. $\blacksquare$

Remark

@below - oops, thanks for noticing

Magical problem and solution. @above You used $Q$ twice.

Extremely beautiful problem! And the above solution is very nice! My proof doesn't involve the two magic points.

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Here's Part 3.

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Finally, the "if" part:

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Here's a solution for the only if direction (coordinate bash but clean). Let $\measuredangle$ denote directed angles modulo $180^\circ$, and let $d(P,\ell)$ denote the distance from a point $P$ to a line $\ell$. Let $DEF$ be the contact triangle of $ABC$. Claim 1: $P$ is unconditionally the intersection of the $I$-symmedian of $IBC$ with $\overline{EF}$. Proof: Notice that $P$ is the midpoint of $\overline{UV}$ from JMO 2014/6. $\square$ It is well known that $\overline{AM}$, $\overline{EF}$, and $\overline{DI}$ concur, so $S$ is their concurrence point. Let $r$ be the inradius of $ABC$, and let $M_A$ be the antipode of $L$ in the circumcircle of $ABC$. Claim 2: $d(A,\overline{BC})=3r$ and $d(M_A,\overline{BC})=r$. Since $\overline{IS} \parallel \overline{M_AM}$, we have $IA=IM$. If $K=\overline{AI} \cap \overline{BC}$, the incenter-excenter lemma gives $M_AI=M_B$ and the shooting lemma gives $MA \cdot MK=MB^2$, so $MK=\tfrac{MI}{2}$. Thus, we have $d(M_A,\overline{BC})=d(I,\overline{BC})=\tfrac{d(A,\overline{BC})}{3}$, as desired. $\square$ Let $D'$ be the reflection of $D$ over $I$. Since $d(S,\overline{BC})=\tfrac{3}{2}r$ and $d(D',\overline{BC})=2r$, we know that $S$ is the midpoint of $\overline{DD'}$. Thus, it suffices to show $\overline{AQ} \perp \overline{LD'}$. Claim 3: $LDM_AQ$ is cyclic. Proof: We have \[\measuredangle LQM=\measuredangle LPM=\measuredangle LIM_A=\measuredangle DIM=\measuredangle LM_AD,\]where the second-last equality is by the isogonality of $(\overline{IM_A},\overline{ID})$ and $(\overline{IP},\overline{IM})$ in $\angle BIC$ and the last equality is by parallelogram $IDM_AM$. $\square$ Let $x=MB$ and $y=MD$. Notice that \[r^2+x^2=BM^2+MM_A^2=M_AB^2=M_AI^2=4r^2+y^2 \implies x^2=3r^2+y^2.\]We have $MB \cdot MC=ML \cdot MM_A=MD \cdot MQ$ by power of a point, so $ML=\tfrac{x^2}{r}$ and $MQ=\tfrac{x^2}{y}$. Also, notice that since $S$ lies on $\overline{DI}$, we have $d(A,\overline{ML})=2d(S,\overline{ML})=2y$. If we set $M=(0,0)$, $B$ on the negative $x$-axis, and $L$ on the positive $y$-axis, we have $A=(-2y,3r)$, $L=(0,\tfrac{x^2}{r})$, $D'=(-y,2r)$, and $Q=(\tfrac{x^2}{y},0)$. Then, the slope of $\overline{AQ}$ is \[-\frac{3r}{2y+\frac{x^2}{y}}=-\frac{3ry}{2y^2+x^2}=-\frac{ry}{r^2+y^2}\]and the slope of $\overline{D'L}$ is \[\frac{\frac{x^2}{r}-2r}{y}=\frac{x^2-2r^2}{ry}=\frac{r^2+y^2}{ry}.\]These slopes multiply to $-1$, so $\overline{AQ} \perp \overline{D'L}$, as desired. $\square$