Just a mix of well known lemmas . Here's my solution: Suppose that the perpendicular from $D$ to line $EF$ meets line $EF$ and $\omega$ at $Z,S$ respectively. Let $N=AS \cap \omega$ and $X=DN \cap EF$. Note that $N$ is the $D$-Queue point of $\triangle DEF$ (since $SENF$ is harmonic), and so $X$ must be the $D$-Ex point of $\triangle DEF$. Since $X$ lies on the orthic axis of $\triangle DEF$, so using the well known fact that the orthic axis is the radical axis of the circumcircle and the tangential circle, we get $$XP \cdot XQ=XE \cdot XF=XD \cdot XN \Rightarrow N \in \odot (DPQ)$$Taking $K=EF \cap BC$, we have $(K,D;E,F)=-1$, which means that $D,K$ are inverses in the circle with $BC$ as diameter. This means that $$KD \cdot KM=KB \cdot KC=KP \cdot KQ \Rightarrow M \in \odot (DPQ)$$Combining the above two observations, we get that $N=\omega \cap \Gamma$, and so it suffices to prove that $R,N,S$ are collinear (since we have $N \in AS$). Since $MR \parallel DS$, so this follows from $$\measuredangle MRN=\measuredangle MDN=\measuredangle BDN=\measuredangle DSN \Rightarrow S \in NR \quad \blacksquare$$ NOTE: This configuration is pretty well known, and has appeared numerous times on various contests. Off the top of my head, this example is the foremost one.

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Surprised that all solutions here involve harmonics or powers. Angle-chase ftw JustPostTaiwanTST wrote: Given a triangle $ \triangle{ABC} $. Denote its incircle and circumcircle by $ \omega, \Omega $, respectively. Assume that $ \omega $ tangents the sides $ AB, AC $ at $ F, E $, respectively. Then, let the intersections of line $ EF $ and $ \Omega $ to be $ P,Q $. Let $ M $ to be the mid-point of $ BC $. Take a point $ R $ on the circumcircle of $ \triangle{MPQ} $, say $ \Gamma $, such that $ MR \perp EF $. Prove that the line $ AR $, $ \omega $ and $ \Gamma $ intersect at one point. WLOG $AB<AC$. Let $I, O$ be the centres of $\omega, \Omega$ and $M_A$ be the midpoint of arc $\widehat{BAC}$ of $\Omega$. Clearly, $D \in \odot(MPQ)$. The centre $N$ of $\odot(MPQ)$ lies on the perpendicular bisector of $\overline{DM}$ and the line through $O$ parallel to $\overline{AI}$, hence it is the midpoint of $\overline{IM_A}$. Suppose $\odot(MPQ)$ meets $\omega$ at $S \ne D$. Since $\overline{IN} \perp \overline{DS}$, we see that $\overline{AS}$ is the $A$-mixtilinear cevian (follows from an angle-chase after noticing $\overline{IM_A}$ is a symmedian in $\triangle IBC$ and $\overline{IM}$ is parallel to the $A$-Nagell cevian). Let $K \ne S$ be the common point of $\overline{AS}$ and $\odot(MPQ)$. It is enough to prove $\overline{MK} \parallel \overline{AI}$. Let $\alpha=\frac{1}{2}\angle MND$ and $\beta=\frac{1}{2}(\angle B-\angle C)$. Now $\angle KSM=\angle KSD-\angle MSD=90^{\circ}+\beta-\alpha$ so $\angle MNK=2(180^{\circ}-\angle KSM)=180^{\circ}-2\beta+2\alpha$ hence $\angle KMN =\beta-\alpha$ since $N$ lies on the perpendicular bisector of $\overline{MK}$. Since $\angle NMD=90^{\circ}-\alpha$ we conclude that $\angle KMD=90^{\circ}-\beta=\angle (\overline{AI}, \overline{BC})$ proving the claim.

Let $\odot(AEF)$ intersects $\odot(ABC)$ at $\mathcal {H}\implies\triangle HBF\sim\triangle HCE.$ Also, $\frac {\mathcal {H}B}{\mathcal {H}C}=\frac{BF}{CE}=\frac{BD}{DC}.$ Hence, $\mathcal {H},K,D $ are collinear$\implies KD\cdot K\mathcal{H}=K\mathcal{J}\cdot KA=LB^2. $ $A\mathcal {J}D\mathcal {H}$ is concyclic $(GD;BC)=-1.$ By applying Maclaurin's theorem $GP\cdot GQ=GD\cdot GM. $ Thus, $MDPQ $ is concylic. Applying Radical axis theorem on $\odot(ABC),\odot(MDPQ),\odot(AEF).$ $A\mathcal {H}$ cuts $PQ$ on the radical axis of $\odot(AEF) $ and $\odot(MDPQ). $ If $\odot(MDPQ) $ intersects $(I) $ at $\ell.$ Then $D\ell,A\mathcal {H},PQ $ concur. Let $\odot(A\mathcal {H}D)$ intersects $(I) $ at $\ell'. $ Then $D\ell',A\mathcal {H},PQ $ concur cause applying Radical axis theorem on $\odot(AEF),(I),\odot (AHD).$ So $\ell=\ell'. $ So note by Reim's theorem $MR\parallel A\mathcal {J}. $

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Taiwan TST 2019 Round 2 wrote: Given a triangle $ \triangle{ABC} $. Denote its incircle and circumcircle by $ \omega, \Omega $, respectively. Assume that $ \omega $ tangents the sides $ AB, AC $ at $ F, E $, respectively. Then, let the intersections of line $ EF $ and $ \Omega $ to be $ P,Q $. Let $ M $ to be the mid-point of $ BC $. Take a point $ R $ on the circumcircle of $ \triangle{MPQ} $, say $ \Gamma $, such that $ MR \perp EF $. Prove that the line $ AR $, $ \omega $ and $ \Gamma $ intersect at one point. Let $X$ be the midpoint of $\widehat{BAC}$ and $X^*$ denote it's antipode WRT $\Delta ABC$. Let $\odot(DPQ)\cap\odot(I)=A'$. Let $\overline{PQ}\cap\overline{BC}=\{V\}$. Then $VP\cdot VQ=VB\cdot VC=VD\cdot VM$ from Ma'laurin. So, $\{D\}\in\odot(MPQ)$. Then from OGM- (#109) we get that $\overline{AA'}$ is the $A-$ Mixtillinear Cevian of $\Delta ABC$ $(\bigstar)$. Let $\{I\}$ denote the Incenter of $\Delta ABC$. So from here (#1,#2) we get that $\overline{X-R-I-T}$ where $\{T\}$ is the $A-$ Mixtillinear Incircle touchpoint with $\odot(ABC)$ and from $(\bigstar)$ we get $\overline{A-A'-T}$. Let $\overline{AT}\cap\overline{MR}=K$. Then by Reims $KTXX^*$ is cyclic. If $\overline{MR}\cap\odot(MPQ)=K^*$. Then $MR\cdot MK^*=RP\cdot RQ=RX\cdot RT\implies K^*\in\odot(XX^*T)\implies K^*\equiv K$. $\blacksquare$

Solution from Twitch Solves ISL: Let $D$ be the foot from $I$ to $\overline{BC}$. Claim: Quadrilateral $MDPQ$ is cyclic. Proof. Let $X = PQ \cap BC$. Then $(XD;BC) = -1$ by Ceva/Menelaus with the Gergonne point. Hence, $XD \cdot XM = XB \cdot XC = XP \cdot XQ$ as needed. $\blacksquare$ We let $G$ be the intersection of the $D$-altitude to $\overline{EF}$ with the incircle. Let line $AG$ meet the incircle again at $H$. [asy][asy] size(12cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair M = midpoint(B--C); pair Z = orthocenter(D, E, F); pair K = foot(D, E, F); pair N = midpoint(E--F); pair G = -Z+2*K; pair H = -G+2*foot(I, A, G); filldraw(A--B--C--cycle, invisible, blue); filldraw(incircle(A, B, C), invisible, blue); pair X = extension(E, F, B, C); pair S = foot(A, I, -A); pair Y = extension(H, D, E, F); filldraw(unitcircle, invisible, blue); pair P = IP(circumcircle(H, D, M), unitcircle); pair Q = OP(circumcircle(H, D, M), unitcircle); filldraw(circumcircle(H, D, M), invisible, orange); filldraw(circumcircle(A, E, F), invisible, deepgreen); draw(E--X--P, blue); draw(X--B, blue); pair Rp = extension(M, foot(M, P, Q), A, H); draw(H--Rp--M, red); draw(S--I--A, deepgreen); draw(A--Y--D, lightred); draw(D--G, lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(240)); dot("$C$", C, dir(300)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); dot("$E$", E, dir(350)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); dot("$K$", K, dir(K)); dot("$N$", N, dir(N)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$X$", X, dir(X)); dot("$S$", S, dir(S)); dot("$Y$", Y, dir(140)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(160)); dot("$R'$", Rp, dir(Rp)); /* TSQ Source: !size(12cm); A = dir 130 B = dir 210 R240 C = dir 330 R300 I = incenter A B C D = foot I B C E = foot I C A R350 F = foot I A B M = midpoint B--C Z := orthocenter D E F K = foot D E F N = midpoint E--F G = -Z+2*K H = -G+2*foot I A G A--B--C--cycle 0.1 lightcyan / blue incircle A B C 0.1 lightcyan / blue X = extension E F B C S = foot A I -A Y = extension H D E F R140 unitcircle 0.1 palecyan / blue P = IP circumcircle H D M unitcircle Q = OP circumcircle H D M unitcircle R160 circumcircle H D M 0.1 yellow / orange circumcircle A E F 0.1 lightgreen / deepgreen E--X--P blue X--B blue R' = extension M foot M P Q A H H--Rp--M red S--I--A deepgreen A--Y--D lightred D--G lightblue */ [/asy][/asy] Claim: Quadrilateral $HDPQ$ is cyclic. Proof. [Proof $HDPQ$ cyclic] For this proof, we need to introduce several points: Let $K$ be the foot from $D$ to $\overline{EF}$. Let $N$ be the midpoint of $\overline{EF}$. Let $S$ be the inverse of $K$ with respect to the incircle, which is known to satisfy $\angle AIS = 90^{\circ}$. Let $Y = \overline{HD} \cap \overline{EF}$. We have $-1 = (EF;GH) \overset{D}{=} (EF;KY)$. Since $\overline{SK}$ bisects $\angle FSE$, this implies $Y$ also lies on line $AS$. We can then calculate \[ YH \cdot YD = YF \cdot YE = YK \cdot YN = YS \cdot YA = YP \cdot YQ \]which implies the concyclic condition. $\blacksquare$ We let $R'$ be a point on line $\overline{AGH}$ such that $\overline{R'M} \perp \overline{PQ}$. Then \[ \measuredangle HR'M = \measuredangle HGD = \measuredangle HDB = \measuredangle HDM \]so $R' = R$ and the problem is solved (the concurrence point in the problem is $H$).

why are there 2 threads lol [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(130), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), M = (B+C)/2, P = intersectionpoint(unitcircle,E--E+dir(E--F)*100), Q = intersectionpoint(unitcircle,F--F+dir(F--E)*100), R = intersectionpoint(foot(M,E,F)--foot(M,E,F)+dir(M--foot(M,E,F))*100,circumcircle(P,Q,M)), S = intersectionpoints(incircle(A,B,C),circumcircle(P,Q,M))[0], T = extension(I,foot(I,D,S),A,S), K = extension(A,S,D,foot(D,E,F)), L = dir(90), N = (I+L)/2; draw(A--B--C--A); draw(unitcircle); draw(incircle(A,B,C)); draw(circumcircle(M,P,Q), blue); draw(P--Q^^M--R^^D--K^^I--A--L, heavyred); draw(R--T--L^^T--D, orange); draw(I--D^^M--L, magenta); dot("$A$", A, dir(140)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$D$", D, dir(270)); dot("$E$", E, dir(80)); dot("$F$", F, dir(135)); dot("$P$", P, dir(200)); dot("$Q$", Q, dir(20)); dot("$R$", R, dir(110)); dot("$S$", S, dir(210)); dot("$M$", M, dir(270)); dot("$T$", T, dir(225)); dot("$K$", K, dir(140)); dot("$L$", L, dir(90)); dot("$I$", I, dir(330)); dot("$N$", N, dir(140)); [/asy][/asy] Claim: $D \in \Gamma$, where $D$ is the tangency point of $\omega$ on $\overline{BC}$. Proof. Define a function $f$ such that $f(X) = XB/XC$ for any point $X$. By Ratio Lemma $$f(P)f(Q) = f(\overline{PQ} \, \cap \, \overline{BC}) = f(\overline{EF} \, \cap \, \overline{BC}) = f(D) = f(D)f(M)$$and we may conclude. Claim: Let $T$ be the $A$-mixtilinear intouch point. If $\overline{AT}$ meet $\omega$ at $K$ and $S$ where $AKST$ are in that order, then $S \in \Gamma$ as well. Proof. Let $L$ be the midpoint of arc $BAC$ so that $T \in \overline{LI}$. From the fact that $\angle DTI = \angle ATI$ (well-known), we know that $\overline{IL}$ is the perpendicular bisector of $\overline{DS}$. It follows that the midpoint $N$ of $\overline{IL}$ is the circumcenter of $\triangle DMS$. But since $\overline{AL} \parallel \overline{PQ}$ and $\angle LAI = 90^\circ$, we have $NP = NQ$, so $N$ is also the center of $\Gamma$ which implies the conclusion. Claim: Redefine $R = \overline{AS} \cap \Gamma$; then $\overline{MR} \perp \overline{PQ}$. Proof. By Reim's $\overline{DK} \parallel \overline{RM}$, but it is well-known that $\overline{DK} \perp \overline{EF}$, so we are done. $\blacksquare$

Let $L$ midpoint of arc $BC$, $N$ midpoint of arc $BAC$ in $\Omega$. Let $M$ midpoint of $BC$, and $D=\omega\cap BC$. Let $G=AI\cap EF$. Let $H$ be the projection of $D$ on $EF$. Let $K=NI \cap EF$, and $S=ND\cap \Omega$. Let $T=NI\cap \Omega$ and $X=AT\cap \omega$. Let $V=NI\cap BC$. Let $Y\in \omega$ such that $DY\perp EF$. Let $Z=AY\cap BC$. $$ -1=(B,C;N,L)\stackrel{(S)}=(B,C;D,LS\cap BC), $$so $LS, EF, BC$ concur at a certain point $W$. $$ WD\cdot WM=WS\cdot WL=WP\cdot WQ \Longrightarrow D\in \Gamma $$Also, $$ \frac{NK}{NI}=\frac{AG}{AI}=\frac{NM}{NL}, \Longrightarrow MK\parallel LI \Longrightarrow M-K-R \thickspace \text{collinear.} $$Now, $$ TK\cdot KN=PK\cdot KQ=MK\cdot KR \Longrightarrow TMNR \thickspace \text{cyclic.} $$Now, $$ \measuredangle NTR=\measuredangle NMR=\measuredangle NLA=\measuredangle NTA \Longrightarrow A-T-R \thickspace \text{collinear.} $$Also, by $\triangle YFHE\sim IBDC$ we get $$ \frac{FH}{HE}=\frac{BD}{DC} \Longrightarrow AFHE\sim NBDC \Longrightarrow \measuredangle FAH=\measuredangle BND \Longrightarrow A-H-S \thickspace \text{collinear.} $$By radical axis on $(LTI), (BIC), \Gamma$ we get that $U=LT\cap BC$ satisfies $UI\perp LI$, so $(\overline{UI})=(UTDI)$ tangent to $LI$, $$ \measuredangle DSV=\measuredangle DTV=\measuredangle DTI=\measuredangle DIL=\measuredangle NLA=\measuredangle NSA=\measuredangle DSA \Longrightarrow A-S-V \thickspace \text{collinear.} $$where we have used $DVTS$ cyclic, which is trivial using inversion $NB=NC$. Therefore $A-H-V-S$ collinear. Claim. $A-Y-T$ collinear. Proof. Let $X'=AY\cap \omega$. Taking polars wrt $\omega$, Brocard on $DEFY$ gives $DE\cap FY\in \mathcal{P}(H)$. Pascal on $DX'YFEE$ gives $DX'\cap FE - A - DE\cap FY$ collinear, and $A\in \mathcal{P}(H)$ by La Hire. Hence $DX'\cap FE\in \mathcal{P}(H)\Longrightarrow H\in \mathcal{P}(DX'\cap FE)$. And clearly by La Hire, $A\in \mathcal{P}(DX'\cap FE)$. Since $A-H-V$ collinear, then $$ V\in \mathcal{P}(DX'\cap FE) \Longrightarrow DX'\cap FE\in\mathcal{P}(V) \Longrightarrow VX'=VD \Longrightarrow VI \thickspace \text{is the perpendicular bisector line of} \thickspace \overline{XD'}. $$And since $T,V\in IN$, then $T\in VI$, so $$ \measuredangle VTX'=\measuredangle DTV=\measuredangle DSV=\measuredangle NSA=\measuredangle NTA=\measuredangle VTA $$Therefore $A-X'-T$ collinear, and so $X\equiv X'$, which proves the Claim. Finally, $\triangle ZDY$ and $\triangle ZMR$ are clearly homothetic from $Z$ because $DY\parallel MR$, so $$ \frac{ZM}{ZR}=\frac{ZD}{ZY}=\frac{ZD}{\frac{ZD^2}{ZX}}=\frac{ZX}{ZD} \Longrightarrow ZM\cdot ZD=ZX\cdot ZR, $$which means $Z\in \Gamma$, as desired.

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Cute problem, took like 20 mins (yay, strict oly training back). Ok so let $I$ the incenter of $\triangle ABC$, let $D$ the intouch point of $\omega$ with $BC$, let the perpendicular from $D$ to $EF$ meet $\omega$ again at $G$ and hit $EF$ at $K$, let $EF$ meet $BC$ at $S$, let $J$ the $D$-queque point on $\triangle DEF$, let $DJ$ meet $EF$ at $U$ and let $(AEF)$ meet $\Omega$ again at $V$ Claim 1: $PJDMQ$ is cyclic Proof: Clearly $-1=(S, D; E, F)$ so by McLaurins theorem and PoP $$SD \cdot SM=SB \cdot SC=SP \cdot SQ \implies PDMQ \; \text{cyclic}$$Now since its well known that $VK$ bisects $\angle EVF$ and $-1=(U, K; E, F)$ projecting $$-1=(A, I; E, F) \overset{V}{=} (AV \cap EF, K; E, F) \implies AV \cap EF \cap DJ=U$$And now by PoP $$UP \cdot UQ=UA \cdot UV=UE \cdot UF=UD \cdot UJ \implies PJDQ \; \text{cyclic}$$Hence $PJMDQ$ is cyclic. Finishing: Using Claim 1 let $JA \cap \Gamma =R'$, since its known that $J,G,A$ are colinear by Reim's theorem on $\omega, (PJMDQ)$ we get $GD \parallel MR'$ which means that $R=R'$ hence $AR,\omega, \Gamma$ share $J$ as a common point. Thus, we are done

In the diagram attached below I have swapped $R$ and $N$ JustPostTaiwanTST wrote: Given a triangle $ \triangle{ABC} $. Denote its incircle and circumcircle by $ \omega, \Omega $, respectively. Assume that $ \omega $ tangents the sides $ AB, AC $ at $ F, E $, respectively. Then, let the intersections of line $ EF $ and $ \Omega $ to be $ P,Q $. Let $ M $ to be the mid-point of $ BC $. Take a point $ R $ on the circumcircle of $ \triangle{MPQ} $, say $ \Gamma $, such that $ MR \perp EF $. Prove that the line $ AR $, $ \omega $ and $ \Gamma $ intersect at one point. Say $G \in \omega$ such that $GENF$ is harmonic where $N \in \omega,DN \perp EF$.By the harmonic condition we have $A-G-N$.Say $A_{SD}$ be the $\text{A-Sharkydevil}$ point.So by Incircle Inversion $I,A_{SD},DN\cap EF=J$ are collinear.Now say $EF \cap BC=X$ Claim 1: $D \in \Omega$ Proof : Since $AD,BE,CF$ concurr,by ceva-menelaus we have,$(XD;BC)=-1$.So by midpoint lengths, $$XD.XM=XB.XC=XP.XQ \implies D \in \Omega \text{ }\blacksquare $$Claim 2: $G \in \Omega$ Proof : Say $AA_{SD}\cap EF=H$ and $DG\cap EF=H'$,we will show that $H\equiv H'$ and then prove the claim.Consider the following Cross-ratio chase $$(H,J;E,F)\stackrel{A_{SD}}=(A,I;E,F)=-1=(G,N;E,F)\stackrel{D}=(H',J;E,F)$$This proves $H\equiv H'$.Now by Power of Point, $$HD.HG=HE.HF=HA_{SD}.HA=HP.HQ \implies G\in \Omega \text{ } \blacksquare$$Finally say $GR \cap MR=R'$ Claim 3: $N' \equiv N$ and thus $A-G-R$ Proof : Since $DR || MN$,$\angle GND=\angle GR'M$.Thus, \begin{align*}\angle GR'M+\angle GDM &=\angle GND+\angle GDN+\angle MDN \\ &=\angle GND+\angle GDN+\angle DGN \\ &=180 \end{align*}Which means $R'\in \Omega \implies R'\equiv R,R-A-G$,we get $AR,\Gamma,\omega$ meet at the same point $G$ $\blacksquare $

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We first prove that $D$ and $H$ lie on $(MPQ)$ where the incircle touches $BC$ at $D$ and $H$ is the $D$ queue point of $\Delta DEF$. Let $EF \cap BC= J$ then it is well known that $(JD;BC)=-1$ so by the harmonic midpoint lemma, $JB.JC=JD.JM=JP.JQ$ so $D$ lies on $(MPQ)$. Now let $S$ be the $A$ sharky devil point and $AS \cap EF=K$. It is well known that $K$ is the $D$ expoint in $\Delta DEF$ so it is collinear with $D$ and $H$. Now by power of point- $KP.KQ=KS.KA=KF.KE=KH.KD$ so $D,H \in (MPQ)$. Now let $D$ altitude cuts $EF$ and $\omega$ at $G$ and $T$ respectively. It is well known that $\overline{I-G-S}$ and $\overline{A-T-H}$ so it suffices to prove that $R,T,H$ are collinear which follows by some angle chasing.

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$EF\cap BC=T$ Let $L,N$ be the midpoints of arcs $BC,BAC$ respectively and let $S$ be $A-$sharky devil point. Let $D$ be the tangency point of incircle with $BC$. Since $T,S,N$ and $S,D,L$ are collinear, we have $TP.TQ=TS.TN=TD.TM$ thus, $D,M,P,Q$ are concyclic. Invert around the incircle. New Problem Statement: $ABC$ is a triangle with circumcenter $O$. Nine point circle and $(BOC)$ meet at $P,Q$ and $M$ is the intersection of $(APQ)$ and the circle with diameter $AO$. $(APQ)$ intersects $(ABC)$ at $V$ and $(VON)$ meets $(APQ)$ at $W$, prove that $(MOW)$ is tangent to the perpendicular bisector of $BC$. Perform $\sqrt{\frac{bc}{2}}$ inversion and reflect over the angle bisector of $\measuredangle CAB$. New Problem Statement: $ABC$ is a triangle with circumcenter $O$ and its nine point circle meets $(BOC)$ at $P,Q$. Let $K,L$ be the midpoints of $AC,AB$ respectively and $PQ$ intersects $BC,KL$ at $M,V$. $S$ is $A-$dumpty point and $(DSV)$ meets $PQ$ at $W$. Prove that $(DAS)$ and $(DWM)$ are tangent to each other. Since the radical axises of $(AO),(BOC)$ and nine point circle are concurrent, we conclude that $SO,PQ,KL$ meet at $V$ which is equavilent to the collinearity of $S,O,V$. Note that $\measuredangle ASV=\measuredangle ASO=90$. Let $l$ be the tangent to $(DAS)$ at $D$. If $E$ is the midpoint of $AD$, then $A,E,S,V$ are concyclic. \[\measuredangle (WD,l)=\measuredangle (SD,l)+\measuredangle WDS=\measuredangle SAD+\measuredangle WVS=\measuredangle SVE+\measuredangle WVS=\measuredangle WVK=\measuredangle WMD\]As desired.$\blacksquare$ \[\]