[asy][asy]import graph; size(6cm); real lsf=0.2; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-13.31,xmax=14.17,ymin=-6.54,ymax=9.19; pen zzttqq=rgb(0.6,0.2,0); pair A=(-5,6), B=(-6,-1), C=(5,-1), M=(-0.5,-1), D=(-1.97,-1), Q=(-5.56,2.11), P=(2.43,0.8), L=(-0.5,-4.38), H=(-2.47,-4.06); D(A--B--C--cycle,zzttqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR((-0.5,1.79),6.17)); D(CR((-1.23,3.48),4.54)); D(A--L); D(P--Q); D(M--(-1.56,1.45)); D(M--L); D(H--(-1.56,1.45)); D(H--M); D(H--L); D(A); MP("A",(-5.23,6.16),NE*lsf); D(B); MP("B",(-6.34,-1.22),N*lsf); D(C); MP("C",(5.17,-0.92),NE*lsf); D(M); MP("M",(-0.4,-0.86),NE*lsf); D(D); MP("D",(-1.87,-0.86),NE*lsf); D(Q); MP("Q",(-5.95,1.83),NE*lsf); D(P); MP("P",(2.48,0.36),N*lsf); D(L); MP("L",(-0.4,-4.24),NE*lsf); D((-1.56,1.45)); MP("N",(-1.46,1.58),NE*lsf); D(H); MP("H",(-2.75,-3.84),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] By some angle chasing, we see the problem is equivalent to showing that $AD \parallel MN$. Claim: $QB = PC$. Proof: Use Power of a Point to get $BM \cdot BD = AB \cdot QB$ and $CM \cdot CD = AC \cdot QC$. Now use the angle bisector theorem. Now notice that the vector $\vec{MN}$ is half of $\vec{BQ} + \vec{CP}$, which must be parallel to the angle bisector since they have the same magnitude.

Let $X$ be the midpoint of arc $BAC$. Then $\angle DAX=\angle XMD=90$, so $ADMX$ is cyclic with diameter $XD$. Then clearly $X,N,D,H$ are collinear, so $\angle LHD=\angle LHX=90$, and $H$ lies on $\Omega$. Also $X$ is the spiral symmetry center that brings $QP$ to $BC$, so it brings $NP$ to $MC$, and $\triangle XNM\sim\triangle XPC$. Finally $\angle LHD=\angle LMD=90$, so $DMLH$ is cyclic. Then \[ \angle NMX=\angle PCX=\angle ACX=\angle ALX=\angle DLM=\angle DHM=\angle NHM \] and we are done.

$CP*CA=CM*CD$ $BM*BD=BQ*BA$ $BA/CA=BD/CD$ $BM=MC$ so $CP=BQ$ if $K,L$ are midpoints of $BP$ and $CQ$ resp. than $KNLM$ is a rhombus because $CP=BQ$ and $MN$ is the bisector of $\angle KNL$ because $KN\parallel BA$ and $NL\parallel CA$ we have $MN\parallel AL$ and since $\angle LMD=\angle LHD=90$ $LMDH$ is cyclic so $\angle LPH=\angle LDH=\angle MNH$ so $ML$ is tangent to $\odot MHN$

See as angle LMN is right angle,$L,H,D,M$ cyclic.So the problem reduces to show that $AD||MN$. The circum-circle of $ADM$ is $-a^2yz-b^2zx-c^2xy+(\frac{a^2c}{2(b+c)}y+\frac{a^2b}{2(b+c)}z)(x+y+z)=0$. So, $Q=(a^2,2(b+c)c-a^2,0)$,$P=(a^2,0,2(b+c)b-a^2)$. The equation of a line parallel to $AD$ which passes through $M$ is nothing but $(b-c)x+(b+c)y-(b+c)z=0$......(i) The point $N=(a^2(b+c),b(2(b+c)c-a^2),c(2(b+c)b-a^2))$ obviously lies on (i). So done.

[asy][asy] import graph; size(10cm); 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 = -4.250413030748777, xmax = 9.900436046904861, ymin = -2.944494629373918, ymax = 5.464993173930597; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((1.207861222844678,4.566748395930078)--(-1.523481853650540,1.787798822706289)--(5.426607357771634,0.3037408472691735)--cycle, zzttqq); /* draw figures */ draw((1.207861222844678,4.566748395930078)--(-1.523481853650540,1.787798822706289), zzttqq); draw((-1.523481853650540,1.787798822706289)--(5.426607357771634,0.3037408472691735), zzttqq); draw((5.426607357771634,0.3037408472691735)--(1.207861222844678,4.566748395930078), zzttqq); draw(circle((1.961846242895856,1.093929125186070), 3.553725805082958)); draw((1.207861222844678,4.566748395930078)--(1.219746001150586,-2.381449174929492)); draw(circle((1.958780377377184,2.886325987887929), 1.840570304577901)); draw((0.2656345589186853,3.608099001743732)--(3.631274245051219,2.117909884017938)); draw((1.219746001150586,-2.381449174929492)--(-0.1124443826196421,-1.791601241767472)); draw((1.948454401984952,2.863004442880835)--(-0.1124443826196421,-1.791601241767472)); draw((-1.523481853650540,1.787798822706289)--(3.631274245051219,2.117909884017938)); draw((0.2656345589186853,3.608099001743732)--(5.426607357771634,0.3037408472691735)); draw((1.053896195700340,1.952854353362114)--(1.948454401984952,2.863004442880835)); draw((1.948454401984952,2.863004442880835)--(2.846120958345160,1.955919924506453)); draw((2.846120958345160,1.955919924506453)--(1.951562752060547,1.045769834987731)); draw((1.951562752060547,1.045769834987731)--(1.053896195700340,1.952854353362114)); draw((1.948454401984952,2.863004442880835)--(1.951562752060547,1.045769834987731)); draw((1.951562752060547,1.045769834987731)--(1.219746001150586,-2.381449174929492)); /* dots and labels */ dot((1.207861222844678,4.566748395930078),dotstyle); label("$A$", (1.267440262117152,4.654776741054082), NE * labelscalefactor); dot((-1.523481853650540,1.787798822706289),dotstyle); label("$B$", (-1.470532511051765,1.874896221357074), W * labelscalefactor); dot((5.426607357771634,0.3037408472691735),dotstyle); label("$C$", (5.486153412612117,0.3941558440310300), NE * labelscalefactor); dot((1.213614270113243,1.203344550474227),dotstyle); label("$D$", (1.267440262117152,1.288187769963736), NE * labelscalefactor); dot((1.207861222844678,4.566748395930078),dotstyle); dot((1.219746001150586,-2.381449174929492),dotstyle); label("$L$", (1.281409510959851,-2.301909182609787), NE * labelscalefactor); dot((1.951562752060547,1.045769834987731),dotstyle); label("$M$", (2.007810450780176,1.134526032694052), NE * labelscalefactor); dot((0.2656345589186853,3.608099001743732),dotstyle); label("$Q$", (0.3175313408136504,3.690898570907883), NE * labelscalefactor); dot((3.631274245051219,2.117909884017938),dotstyle); label("$P$", (3.684120311904003,2.196188944739140), NE * labelscalefactor); dot((1.948454401984952,2.863004442880835),dotstyle); label("$N$", (2.007810450780176,2.950528382244861), NE * labelscalefactor); dot((-0.1124443826196421,-1.791601241767472),dotstyle); label("$H$", (-0.05963837793921063,-1.701231482373750), NE * labelscalefactor); dot((1.053896195700340,1.952854353362114),dotstyle); label("$X$", (1.113778524847468,2.042527207469456), NW * labelscalefactor); dot((2.846120958345160,1.955919924506453),dotstyle); label("$Y$", (2.901842376712883,2.042527207469456), N * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By PoP, we have that \[BQ \times AB = BD \times BM \implies BQ = \dfrac{\dfrac{ac}{b+c} \times \dfrac{a}{2}}{c} = \dfrac{a^2}{2(b+c)}.\] Note that this expression is symmetric in $b,c.$ Similarly we get that $CP= \dfrac{a^2}{2(b+c)},$ so $BQ=CP.$ Let $X,Y$ be the midpoints of $BP,CQ$ respectively. Since \[XN= MY = \dfrac{1}{2} BQ = \dfrac{1}{2} CP = XM = NY,\] $\triangle NXM \cong \triangle NYM.$ Consequently, $MN$ bisects $\angle XNY.$ However, since $XN \parallel AB$ and $NY \parallel AC,$ $MN \parallel AD.$ Since $\angle LBC = \angle LAC = \angle LAB = \angle LCB,$ $\triangle LBM \cong \triangle LCM \implies LM \perp BC,$ which in turn implies quadrilateral $HLMD$ is cyclic. It follows that $\angle HML = \angle HDL = \angle HNM,$ so we're done. $\blacksquare$

First,let $R$ be the midpoint of the arc $BAC$.Now,from PoP we easy get that $BQ=CP$. Then,from an easy angle chase we obtain that it is enough to prove that $MN$ is parallel to $AD$ and this is true by an simple angle chase using that $RQB$ and $RMN$ are similar(cause $RQN$ is similar to $RMN$),so we are finished.

Let $\angle{CAM}=\theta,\angle{APQ}=x$.Then we have $\angle{QMB}=\angle{AMB}-\angle{AMQ}=C+\theta-x$.So $\angle{PMC}=A-C-\theta+x$.Also by PoP we have $BQ \cdot AB=BD \cdot BM$ and $CP \cdot AC=CM \cdot CD$ and equating these we get $BQ=CP$.Now sine rule in $\triangle{BQM}$ and $\triangle{CPM}$ gives $\frac{BM}{\sin{\angle{BQM}}}=\frac{BQ}{\sin(C+\theta-x)}$ and $\frac{CM}{\sin{\angle{CPM}}}=\frac{CP}{\sin(A-C-\theta+x)}$.Dividing these two relations and using the fact that $\sin{\angle{BQM}}=\sin{\angle{CPM}}$ we get $\sin(C+\theta-x)=\sin(A-C-\theta+x)$.Note that these angles cannot be supplementary so $C+\theta-x=A-C-\theta+x \implies x=C+\theta-\frac{A}{2}$.Now $\angle{HDL}=\angle{ADN}=\angle{QDN}-\angle{QDA}=\angle{QDN}-\angle{QPA}=90-\frac{A}{2}-x=90-C-\theta$. $\angle{LMD}=\angle{LHD}=90^{\circ} \implies LMDH$ is cyclic $\implies \angle{HML}=\angle{HDL}=90-C-\theta$.Let $AM$ meet $\Omega$ at $X$.Then $\triangle{XBC} \sim \triangle{MQP}$ so $\triangle{XMC} \sim \triangle{MNP}$.Thus $\angle{MNP}=\angle{XMC}=C+\theta$.Finally $\angle{HNM}=90-\angle{MNP}=90-C-\theta=\angle{HML}$ and we are through.

We must prove that $ MN || AD $, but it's true : Lemma: (from SHARYGIN'S book ) Let $ ABCD $ be a convex quadrilateral, and $ M \in AD , N \in BC $, such that $ frac{AB}{CD}=frac{AM}{MD}=frac{BN}{NC} $ and $ BA \cap CD=P$.Then angle bisector of $ \angle BPC $ and $ MN $ are parallel. Proof : (hint) Let $ ABNX $ and $ DCNY $ are parallelogramm ... But in the our problem we have $ frac{BQ}{CP}=frac{QN}{NP}=frac{BM}{CM}=1 $, so we are done.

Sorry to revive this topic . can you explain more how did you got this ? Now notice that the vector $\vec{MN}$ is half of $\vec{BQ} + \vec{CP}$, which must be parallel to the angle bisector since they have the same magnitude.[/quote]

Here is a solution by Ferid and Murad (me) A little angle-chasing gives that we need to prove $MN\parallel AD$. Let $MN\cap AC=K$. We will prove that $\frac{CM}{CD}=\frac{CK}{AC}$ Let's use Barycentric coordinates It's easy to get $K=(\frac{b+c}{2b}:0:\frac{b-c}{2b})$ The rest is clear by Distance formula.

This question was nice . Thank you for it HINT: (attachment figure) Prove $L,M,F$ are aligned. Prove $D,N,F$ are aligned. Prove $H$ lies on $\Omega$. Prove $\angle DHM=\angle DLM$ Prove $NM \parallel DL$ by Thales theorem. GOOD LUCK

v_Enhance wrote: Now notice that the vector $\vec{MN}$ is half of $\vec{BQ} + \vec{CP}$, which must be parallel to the angle bisector since they have the same magnitude. Why does this work?

It suffices to show that $\measuredangle{NHM}=\measuredangle{NML}$. However we have that $\measuredangle{NML}=\measuredangle{NMD}+\measuredangle{DML}=\measuredangle{NMH}+\measuredangle{HML}$. We also have that $\measuredangle{NHM}=\measuredangle{NMH}+\measuredangle{HNM}$. We must now show that $\measuredangle{HNM}=\measuredangle{HML}$, but since $LM$ bisects $BC$, we have that $\angle{BML}=90$, so $DMLH$ is cyclic, so we have that $\measuredangle{HML}=\measuredangle{HDL}$, implying it suffices to show that $AD\parallel MN$. We show this by using barycentrics on $\triangle{ABC}$. We have that $A=(1,0,0)$, $D=(0:b:c)$ and $M=(0,\frac{1}{2},\frac{1}{2})$. Plugging in these point into the general circle formula $$-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$$we get that the equation of the circle passing through these points is $$-a^2yz-b^2xz-c^2xy+\frac{a^2}{2(b+c)}(cy+bz)(x+y+z)=0$$We let $P=(n:0:m)$, where $n+m=b$. Plugging in and simplifying yields $$P=\bigg(\frac{a^2}{2(b+c)}:0:b-\frac{a^2}{2(b+c)}\bigg)$$Exploiting symmetry, we find that $$Q=\bigg(\frac{a^2}{2(b+c)}:c-\frac{a^2}{2(b+c)}:0 \bigg)$$Now we have that $N$ is the midpoint of $PQ$, but homogenizing both and taking the average seems like a pain, so instead, we find that $N=\frac{1}{2}(cP+bQ)$, which evaluates to $$N=\bigg(\frac{a^2}{4}:\frac{b}{2}(c-\frac{a^2}{2(b+c)}):\frac{c}{2}(b-\frac{a^2}{2(b+c)})\bigg)=\bigg(\frac{a^2}{4bc},\frac{1}{2}-\frac{a^2}{4c(b+c)},\frac{1}{2}-\frac{a^2}{4b(b+c)}\bigg)$$We now easily compute $$\vec{NM}=\bigg(-\frac{a^2}{4bc},\frac{a^2}{4c(b+c)},\frac{a^2}{4b(b+c)}\bigg)=\bigg(-1:\frac{b}{b+c}:\frac{c}{b+c}\bigg)=\vec{AD}$$implying that $AD\parallel MN$ as desired. $\square$

Delray wrote: v_Enhance wrote: Now notice that the vector $\vec{MN}$ is half of $\vec{BQ} + \vec{CP}$, which must be parallel to the angle bisector since they have the same magnitude. Why does this work? If $\vec v$ and $\vec w$ are two unit vectors then $\vec v + \vec w$ is parallel to the angle bisector of the two vectors (at the origin).

Finishing Evan's Solution, one can instead of using vectors construct the isotomics of $Q, P$ (reflect over the midpoints of $X, Y$ of $AB, AC$) and denote them $Q'P'$. Then, if $R$ is the midpoint of $Q'P'$ we see that the midpoint of $NR$ is the midpoint of $XY$ ($NXRY$ is a parallelogram), which is in fact the midpoint of $AM$, so $ARMN$ is a parallelogram, completing the proof. (If you look closely, this is really just Evan's solution).

[asy][asy]size(7cm); pair A=(1,7),B=(0,0),C=(10,0),D,M,L,P,Q,Nn,H,X; D=extension(bisectorpoint(B,A,C),A,B,C);M=(B+C)/2; L=extension(A,D,M,bisectorpoint(B,C)); P=abs(D)*abs(M)/abs(A)^2*A; Q=C+abs(D-C)*abs(M-C)/abs(A-C)^2*(A-C); X=2*circumcenter(A,B,C)-L; Nn=(P+Q)/2;H=foot(L,Nn,D); D(MP("A",A,N)--MP("P",P,W)--MP("B",B,SW)--MP("D",D,SW)--MP("M",M,SE)--MP("C",C,SE)--MP("Q",Q,E)--A); D(MP("X",X,N)--MP("N",Nn,NE)--MP("H",H,S)--MP("L",L,S)--X--P--Q--X,heavycyan); draw(B--H--C^^P--D--Q--M--P^^A--M^^H--M--Nn^^X--A--D--L^^B--X--C,heavymagenta); draw(circumcircle(A,B,C)^^circumcircle(A,P,Q),orange); dot(A^^B^^C^^D^^L^^M^^P^^Q^^X^^H^^Nn); [/asy][/asy] First of all, let $X$ be the midpoint of the arc $BAC$; we have $\angle XMD=\angle XAD=90^{\circ}$, so $X$ lies on $\odot (ADM)$. Next, note that $D$ is the midpoint of arc $PDQ$, so $PD=QD$. Also $X$ is the center of the spiral similarity mapping $PB\mapsto QC$; but $XB=XC$, so this is actually a rotation, implying $XP=XQ$. Thus $XND$ is the perpendicular bisector of $PQ$. Now $\angle LHX=90^\circ$ immediately gives $H\in\Omega$. Now we claim that $\triangle HBC\sim\triangle MQP$. The fact that $\angle BHC=\angle QMP$ is trivial (both are $180^\circ-\angle BAC$); we need to show $\tfrac{HB}{HC}=\tfrac{MQ}{MP}$. To prove this, we rudely ruin the whole synthetic aura built up so far, and shamelessly resort to sine rule chase. Note that:$$\frac{HB}{HC}=\frac{\sin\angle BXD}{\sin\angle CXD}=\frac{BD}{CD} $$and $$\frac{MQ}{MP}=\frac{\sin\angle MAQ}{\sin\angle MAP}=\frac{MC\cdot \frac{\sin\angle AMC}{AC}}{MB\cdot \frac{\sin\angle AMB}{AB}}=\frac{AB}{AC}=\frac{BD}{CD}.$$This proves our claim. To finish things off, note that $\angle MND$ and $\angle HML$ play similar roles in these similar triangles, so they are equal, which implies the required tangency. $\blacksquare$

@above, for not ruining all the synthetic aura, observe that if $J$ is the intersection of $XA, BC$, then from $$-1 { = } (B, C; D, J) \stackrel{X}{=} (B, C; H, A) \Longrightarrow AHBC \text{ is harmonic}$$we know $AH$ is the $A-$symmedian in $ABC$, from which $\measuredangle MQP = \measuredangle MAB = \measuredangle HAC = \measuredangle HCB$ and a similar relation completely remove the need of a sine rule chase.

Let $X$ be the midpoint of arc $BAC$. Note that $\angle DAX=\angle XMD=90^{\circ}$ so we get $X \in \odot(ADM)$. Thus, $X$ is the center of a spiral similarity mapping $QP \mapsto BC$. Note that $D \mapsto L$ and $N \mapsto M$ under this similarity. As $X, M, L$ are collinear, we get $X, N, D$ are also collinear. As $X$ is the spiral center for $NM \mapsto DL$, which is actually a homothety, $NM \parallel AD$. Also, $\angle LHD=\angle LMD=90^{\circ}$ so $H$ lies on the circle $\odot(DML)$. Evidently, $$\angle NMH=\angle ADN=\angle LDH=\angle LMH,$$so $LM$ is tangent to circle $\odot(HMN)$. $\blacksquare$

Solution: Let $\Omega\cap \odot (ADM) = R $. Claim: $H$ lies on $\Omega $. Proof of the claim: Suppose $NH\cap \Omega = H'$. Observe that $A$ is the center of the spiral similarity that takes $BD\rightarrow CL $. So, $\measuredangle ARM = \measuredangle ADM = \measuredangle ACL = \measuredangle ARL $. Thus, $L $, $M $, and $R $ are collinear. Again, $\measuredangle QMD = \measuredangle CMP $. But, $LM\perp BC$. So, $LM $ bisects $\angle PMQ $. Thus, $R $ is the Antipode of $D $ in $\odot (APQ) $. Hence, $R $, $N $, and $D $ are collinear. This gives $\measuredangle RH'L = \measuredangle RAL = 90^{\circ } $. So, $H = H'$. Back to the main problem: Now, it's simply spiral sims! Observe that $R $ is the center of the spiral similarity that takes $QN\rightarrow BM$. Trivial angle chasing shows that $R $ is also the center of the spiral similarity that takes $QB\rightarrow DL $. Similarity ratios yields $\frac {RN}{RM} = \frac {RD}{RL} $. So, $MN||DL $. Thus, $\angle MNH = \angle HDL = \angle HML$. This implies the result.

Woahh, I really loved the finishing part of the problem! [asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.2, xmax = 0.7, ymin = -1, ymax = 0.7; /* image dimensions */ draw((-0.12017963832166582,-0.7342605547929758)--(-0.11413561744406524,-0.6870466377339899)--(-0.16134953450305117,-0.6810026168563893)--(-0.16739355538065176,-0.7282165339153752)--cycle, linewidth(0.75) + blue); /* draw figures */ draw((-0.33243263824787167,0.4518565763842871)--(-0.6050023172689867,-0.34795642939451693), linewidth(0.5)); draw((-0.6050023172689867,-0.34795642939451693)--(0.5827518787672822,-0.3558560071212102), linewidth(0.5)); draw((0.5827518787672822,-0.3558560071212102)--(-0.33243263824787167,0.4518565763842871), linewidth(0.5)); draw(circle((-0.00948011773094358,-0.10455423302774991), 0.6433438577765354), linewidth(0.5)); draw(circle((-0.062165105619678994,0.093793746967782), 0.4486128943785326), linewidth(0.5)); 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[/asy][/asy] Let $K$ denote the midpoint of the major arc $\widehat{BC}$, $O$ denote the center of $\odot(ABC)$ and $O_1$ denote the center of $\odot(ADM)$. Firstly, note that $\overline{K-O-M-L}$ are collinear as they all lie on the perpendicular bisector of $\overline{BC}$. Now note that $\measuredangle DAK=\measuredangle LAK=90^\circ=\measuredangle KMD\implies ADMK$ is cyclic. Now as $A=BQ\cap PC$, we get that $A$ is the center of the spiral similarity mapping $QB\mapsto PC$. Now further note that as $KB=KC$, we also get that $KQ=KP$. Now moreover, we have that $H\in\odot(ABC)$, which gives that $\overline{K-O_1-D-H}$ are collinear. Thus we also have that $\measuredangle LDM=90^\circ=\measuredangle LHD\implies LHDM$ is cyclic. Thus our problem condition is equivalent to proving $\measuredangle LMH=\measuredangle MND\iff\measuredangle LDH=\measuredangle MND\iff\measuredangle ADN=\measuredangle MND$, which is thus finally equivalent to proving that $MN\parallel AD$. Now note that the spiral similarity centered at $K$ that mapped $QP\mapsto BC$ also maps their midpoints, that is, $N\mapsto M$. Not only this, but it also maps the centers of the circles, that is, $O_1\mapsto O$. Thus this gives $k$, which is the scale of the dilation of the spiral similarity $=\dfrac{AO_1}{AO}=\dfrac{AN}{AM}\implies\dfrac{AO_1}{AN}=\dfrac{AO}{AM}$, which thus implies $NM\parallel O_1O$ using Thales Theorem. Thus it finally suffices to prove that $AD\parallel OO_1$. This however simply follows from the fact that we have $\measuredangle LAK=90^\circ\implies LA\perp KA$ and that $OO_1\perp AK$ (the latter one due to Radical Axis) which on combining gives us that $AD\parallel OO_1$ and we are done.

Let $K$ be the antipode of $L$, hence the midpoint of arc $BAC$. Since $\angle KAD=\angle KMD=90^\circ$, $K$ lies on $(ADM)$. Therefore, it is the center of the spiral similarity sending $\overline{QP}$ to $\overline{BC}$, which also sends $N$ to $M$, and sends $D$ to $L$ as well since $D$ is the midpoint of arc $PQ$ (and $L$ is the midpoint of arc $BC$). Thus it follows that $\frac{KN}{KD}=\frac{KM}{KL}$, so $\triangle KMN \sim \triangle KLD \implies \measuredangle KMN=\measuredangle KLD=\measuredangle MLD$. Furthermore, since $DMLH$ is clearly cyclic, it follows that $\measuredangle MLD=\measuredangle MHD=\measuredangle MHN$, implying the desired tangency. $\blacksquare$

Let $K$ be the antipode of $L$ wrt $(ABC)$, so that $K$, $M$, and $L$ are collinear. Then since $\overline{KM} \perp \overline{KD}$ and $\overline{KA} \perp \overline{AD}$, $K$ also lies on $(ADM)$, and thus must be the center of the spiral sim which sends $\overline{BC}$ to $\overline{QP}$. Evidently this spiral similarity also takes $M$ to $N$ and $L$ to $D$ (the latter because $\overline{KL}$ and $\overline{KD}$ are diameter in their respective circles). So, $K$ is also the center of the spiral similarity taking $\overline{MN}$ to $\overline{LD}$, and thus $\triangle KMN \sim \triangle KLD$. To finish, note that $HDML$ is cyclic since $\overline{LH} \perp \overline{HD}$ and $\overline{LM} \perp \overline{MD}$, so $$\measuredangle KMN = \measuredangle KLD = \measuredangle MLD = \measuredangle MHD = \measuredangle MHN,$$and $\overline{KML}$ is indeed tangent to $(HMN)$.

First, note that since $\angle QAD=\angle PAD$, $D$ is the arc midpoint of $PQ$. Hence, $ND\perp PQ$, and as a result $PQ\parallel HL$. Next, consider the antipode of $L$ on $(ABC)$, which we call $T$. Since $$\angle TAD=\angle TMD=90,$$$T$ also lies on $(ADM)$. This also means that it is the antipode of $D$ on $(ADM)$ since $\angle TAD=90$, and hence $T$ also lies on $ND$. Note that this further means that $H$ is on $(ABC)$, since line $DN$ goes through the antipode of $L$. Thus, $DMLH$ is cyclic since $\angle DML=\angle DHL=90$, so $$\angle DHM=\angle DLM.$$As such, it suffices to show that $$NM\parallel AL$$since then we would have $$\angle TMN=\angle TLA=\angle THM$$by cyclic $DMLH$ which would imply the circumcircle tangency since $T,M,L$ are collinear. Unfortunately, $NM$ is a terrible line to angle chase with, so we turn to barycentric coordinates to prove that $AL\parallel NM$. We know that $D=(0:b:c)$ and $M=(0:1:1)$. Obviously, the equation of $(ADM)$ has $u=0$. Plugging in $D$ and $M$ into this equation reveals that $$2(v+w)=a^2$$and $$vb+wc=\frac{a^2bc}{b+c}.$$Solving this system of equations yields $$v=\frac{a^2c}{2(b+c)}, w=\frac{a^2b}{2(b+c)}.$$Thus, intersecting this circle with $AB$ and $AC$ yields $$P=(\frac{a^2}{2b(b+c)},0,1-\frac{a^2}{2b(b+c)})$$and symmetrically $$Q=(\frac{a^2}{2c(b+c)},1-\frac{a^2}{2b(b+c)},0).$$Let $m_b=2b(b+c)$ and $m_c=2c(b+c)$ ($m$ for "messy"). Then, we rewrite this as $$P=(\frac{a^2}{m_b},0,1-\frac{a^2}{m_b})$$and $$Q=(\frac{a^2}{m_c},1-\frac{a^2}{m_c},0).$$Thus, we have $$N=(\frac{a^2}{2}(\frac{1}{m_b}+\frac{1}{m_c}), \frac{1}{2}-\frac{a^2}{2m_c},\frac{1}{2}-\frac{a^2}{2m_b}).$$Thus, $$N-M=(\frac{a^2}{2}(\frac{1}{m_b}+\frac{1}{m_c}), -\frac{a^2}{2m_c},-\frac{a^2}{2m_b}).$$If we multiply this by $$m_bm_c\frac{2}{a^2},$$this becomes $$(m_b+m_c,-m_b,-m_c).$$However, expanding out the definition and dividing by $2(b+c)$, this is $$(b+c,-b,-c).$$Hence, $N-M$ is a real multiple of $(b+c,-b,-c)$. However, $$A-D=(\frac{b+c}{a+b+c},\frac{-b}{a+b+c},\frac{-c}{a+b+c})$$is also a real multiple of $(b+c,-b,-c)$, hence $NM\parallel AD$ and we are done.

Solved with r00tsOfUnity. We first note that $\angle NHL = 90^\circ$, so $DHLM$ is cyclic. Thus \[ \measuredangle LMH = \measuredangle LDH. \]We wish to prove that this is equal to $\measuredangle MNH,$ so it suffices to show that $AD \parallel MN.$ To do so, we will first prove that $BQ = CP.$ By Power of a Point on $B,$ we get \[ BQ \cdot AB = BD \cdot BM, \]so \[ BQ = \frac{BD \cdot BM}{AB}. \]Similarly, \[ CP = \frac{CD \cdot CM}{CA}. \]However, by the Angle Bisector Theorem, $\frac{BD}{AB} = \frac{CD}{CA},$ and since $BM = CM,$ this implies the claim. Thus if we set the circumcenter of $\triangle ABC$ (or any point on that matter) to be the origin, then the vector $b - q - (p - c) = b + c - p - q$ is parallel to the angle bisector of $\angle BAC.$ But this implies that the vector $\frac{b+c-p-q}{2} = m - n$ is parallel to $AD,$ so we are done.

The key step is to notice that $(ADM)$ passes through $K$, where $K$ is the top point of $(ABC)$ wrt $BC$, as $\angle KAD = \angle KMD = 90$. Claim 1: Point $H$ lies on $(ABC)$. The previous result tells us $KD$, the diameter of $(ADM)$, passes through $N$ and $KL$, the diameter of $(ABC)$, passes through $M$. We finish by noting $\angle KHL = 90$. ${\color{blue} \Box}$ Claim 2: $\angle KMN = \angle MHN$, which gives the desired result. Note $K$ is the center of spiral similarity sending $QB \rightarrow PC$, and by the Gliding Principle, it is also the center of spiral similarity mapping $NM \rightarrow PC$. Thus $\triangle KMN \sim \triangle KCP$. This congruency along with the cyclicity of $DMLH$ then gives the angle equality \[\angle KMN = \angle KPC = \angle KLA = \angle MHN. \quad \blacksquare\]

Let $R$ be the arc midpoint of $(BAC).$ We have $\measuredangle ADM=\measuredangle DCA+\measuredangle CAD=\measuredangle BLA+\measuredangle LAB=\measuredangle ABL=\measuredangle ARM$ so $ARMD$ is cyclic. Furthermore it is the miquel point of $BCPQ,$ and the spiral similarity taking $BC$ to $QP$ takes $M$ to $N.$ Also notice $LHDM$ is cyclic with diameter $LD.$ Thus \[\measuredangle HNM=\measuredangle RNM=\measuredangle RPC=\measuredangle RPA=\measuredangle RDA=\measuredangle HDL=\measuredangle HML.\]

First note that $LMDH$ is cyclic as $\angle LHD = \angle LMD = 90^{\circ}$. Let $L'$ be the antipode of $L$. Then notice that $L' \in (ADM)$ since $\angle L'AL = 90^{\circ}$ and $L' - O - M - L$ with $\angle L'MD = 90^{\circ}$. So it follows that $L'$ is the Miquel point of quadrilateral $PQBC$. So it follows that the spiral similarity at $L'$ sends $MN \to BQ$ by the gliding lemma(mean geometry theorem?), from which we get $\angle L'MN = \angle L'BQ = \angle L'LD = \angle MHD$ which gives our tangency as desired.

Observe that as $DHLM$ is cyclic, what we want to show is $\angle HML = \angle HNM = \angle MHD$, or rather $\triangle NMH$ is isosceles. This is however also equivalent to showing $NM \parallel DL$. Therefore it suffices to show $NM \parallel DL$. Also take note that $DN \perp QP$, as $AD$ bisects $\angle QAP$, so $D$ is the midpoint of minor arc $\hat{QP}$. From this, we make the following claim: Claim: $X$ is the midpoint of major arc $\hat{BC}$. Proof: If $X' = LM \cap (ABC)$, then \[\angle X'AD = 90^{\circ} = \angle X'MD \implies AX'MD \text{ is cyclic, so that } X = X'. \phantom{c} \square\] Now observe the spiral similarity at $X$ sending $QD \stackrel{X}{\mapsto} BL, DP \stackrel{X}{\mapsto} LC$, and $QP \stackrel{X}{\mapsto} BC$. From these spiral similarites, we find that $D \stackrel{X}{\mapsto} L$ and $N \stackrel{X}{\mapsto} M$ under $X$. Hence $X$ is also the midpoint of major arc $\hat{QP}$. Therefore, the spiral similarity taking $N \mapsto M$ takes $D \mapsto L$. Thus $\triangle XNM \sim \triangle XDL \implies NM \parallel DL$, as desired. $\blacksquare$

Let $L'$ be the midpoint of major arc $BC$. Since $\angle L'AD = \angle DML' = 90^{\circ}$, it follows that $L'$ lies on $(ADM)$. Since lines $AL'$ and $AD$ are the external and internal angle bisectors of $\angle QAP$ respectively, it follows that $L'$ lies on line $DN$. Since $\angle LHN = \angle LHL' = 90^{\circ}$, it follows that $H$ is the second intersection of $(DML)$ and $(ABC)$. Since $\angle QL'P = \angle BAC$, it follows that $\tfrac{L'N}{L'D} = \tfrac{L'M}{L'L}$. Since $DHLM$ is a cyclic quadrilateral, $L'D \cdot L'H = L'M \cdot L'L$. Multiplying these two equations together gives us \[ L'N \cdot L'H = L'M^2,\]so $(HMN)$ is tangent to line $ML$ as desired.

Let $T$ be the midpoint of arc $BAC$ Thus, $TD$ is diameter of circle $(APDMQ)$ Let $TD \cap PQ = N' \implies TN' \perp PQ$ Spiral Similarity centered at $T$ takes $P \rightarrow B$ $Q \rightarrow C $ $D \rightarrow L$ $ \implies N' \rightarrow M$ So, $N \equiv N' \implies MN \parallel DL $ Also, $H \in (ABC)$ as $\angle LHT = 90$ Now, $\angle LMH = \angle LDH = \angle MNH $ $\implies LM$ is tangent to $(MNH)$ $\blacksquare$

Note that $\angle DML=90^{\circ}$ so $DHLM$ is cyclic. Then we want to prove \[\angle HML=\angle HNM\]\[\angle HDL=\angle DNM\]\[\angle ADN=\angle DNM,\]so we want to prove $AD$ and $NM$ are parallel. Note that by power of a point, \[(BQ)(BA)=(BD)(BM)\Longrightarrow BQ=(BM)\cdot\frac{BD}{BA}=(CM)\cdot\frac{CD}{CA}=CP.\]Now we claim that the concyclicity is actually irrelevant: for any $Q$, $P$ in this orientation such that $BQ=CP$, we have $AD\parallel NM$. In fact, if we vary $Q$ linearly (not in the projective sense, just literally $Q=B+t\overrightarrow{BA}$), $P$ will also vary linearly, so $N$ will also vary linearly. Thus $N$ will always lie on a line, and we claim that this is exactly the line through $M$ parallel to $AD$. To do this, it suffices to check two points $Q$. If $Q=B$, then $P=C$ and $N=M$. If $Q=A$, then $P$ is the point on $AC$ such that $BA=PC$, and their midpoint $Z$ has barycentric coordinates $(b+c:0:b-c)$. Thus the displacement vector \[\overrightarrow{ZM}=\left(-\frac{b+c}{2b},\frac12,\frac{c}{2b}\right)\]will be parallel to \[\frac{1}{b+c}(-b-c,b,c)\]which is exactly the displacement vector $\overrightarrow{AD}$. $\blacksquare$

$ $

Again spiral very nice. Also stealing/using Dr. Chen's diagram from #2. Claim. $\odot ABC \cap \odot AQP = X$ is the antipode of $D$ wrt $\odot AQP$ and the antipode of $L$ wrt $\odot ABC$. Proof. Let $X'$ be the antipode of $L$ wrt $\odot ABC$. Then since $\angle LMD = \angle LAX'$, $MDAX'$ is cyclic and $X'$ lies on $\odot AQP$. Thus $X=X'$. Since $\angle DMX =90^{\circ}$, $DX$ is a diameter of $\odot AQP$. But $\overleftrightarrow{DN}$ is also a diameter of $\odot AQP$ as well, so $X$ is the antipode of $D$ wrt $\odot AQP$ as well. $\Box$ We now show that it is sufficient to prove that $NM \parallel DL$ through a simple angle chase (noting that $HDML$ is cyclic). $\angle HML = \angle HDL$, but we seek $\angle HNM = \angle HML$, so it is sufficient to prove that $\angle HDL = \angle HNM$, which is $NM \parallel DL$. Now consider the spiral similarity centered at $X$ sending $QP$ to $BC$. Since $N$ and $M$ are midpoints of $QP$ and $BC$, this spiral similarity also sends $N$ to $M$. Similarly, since $D$ and $L$ are both arc midpoints of arcs $QP$ and $BC$, $D$ is sent to $L$. Now the spiral itself implies that \[ \frac{DX}{DN} = \frac{LX}{LM} \implies \frac{XN}{XD} = \frac{XM}{XL}. \]This then implies that $\triangle NXM \sim \triangle DXL$ (by SAS) since $XND$ and $XML$ are collinear, so $NM \parallel DL$, as desired. $\blacksquare$