Let the tangents of the circumcircle $\odot(ABC)$ at $B,C$ meet at $Y.$ $AS \equiv AY$ is the A-symmedian of $\triangle ABC.$ If $W \equiv AY \cap BC,$ then $NW$ cuts $MY$ at its midpoint $M',$ due to $AD \parallel MY.$ Since $M'(D,N,A,\infty)=-1=M'(S,W,A,Y)$ $\Longrightarrow$ $M',S,D$ are collinear. If $J \equiv MS \cap DY,$ then $JW \parallel YM$ as $S$ is on the D-median $DM'$ of $\triangle DMY.$ Thus from the complete $MWJY,$ it follows that $(M,W,D,L)=-1.$ Let $F$ be the reflection of $D$ on $M.$ Since $M',S,D$ are collinear, then by symmetry $M',R,F$ are collinear and since $UR \parallel MY,$ then $T \in YF.$ By symmetry, $Z \equiv MR \cap YF$ is the reflection of $J$ across $YM$ and therefore the projection $V$ of $Z$ on $BC$ is the reflection of $W$ on $M$ $\Longrightarrow$ $(M,V,F,U)=(M,W,D,L)=-1$ $\Longrightarrow$ $(M,V,F,U)=(Y,Z,F,T)=A(K,M,F,T)=-1.$ Since $AF \parallel MN,$ it follows that $AT$ is the A-median of $\triangle AKM,$ i.e. $AT$ bisects $\overline{MK}.$

Let $ V $ be the projection of $ S $ on $ BC $ and $ J $ $ \equiv $ $ MN $ $ \cap $ $ UR, $ $ P $ $ \equiv $ $ AT $ $ \cap $ $ MN $. From $ VS $ $ \stackrel{\parallel}{=} $ $ UR $ $ \stackrel{\parallel}{=} $ $ RT $ we know $ VRTS $ is a parallelogram. From $ AD $ $ \parallel $ $ UR $ $ \Longrightarrow $ $ J $ is the midpoint of $ UR $, so $ 2MJ $ $ \stackrel{\parallel}{=} $ $ VR $ $ \stackrel{\parallel}{=} $ $ ST $ $ \Longrightarrow $ $ KP $ $ \parallel $ $ ST $. Let $ X $ be the point on $ AM $ s.t. $ XP $ $ \perp $ $ BC $ and $ Y $ $ \equiv $ $ XP $ $ \cap $ $ BC $. Since $ \triangle KPX $ and $ \triangle STR $ are homothetic (with center $ A $), so $ XK $ $ \parallel $ $ RS $ $ \Longrightarrow $ $ XK $ $ \parallel $ $ BC $ $ \equiv $ $ MY $, hence combine $ XP $ $ = $ $ YP $ ($\because $ $ XY $ $ \parallel $ $ AD $) we get $ XKYM $ is a parallelogram $ \Longrightarrow $ $ KP $ $ = $ $ MP $.

I think we can prove a more general statement: Let $\triangle ABC$ be a triangle and let $M$ be a variable point on $BC.$ Let $AM$ meet $\odot(ABC)$ for a second time at $R.$ Let $U$ be the projection of $R$ onto $BC$ and let $T$ be the reflection of $U$ in $R.$ Let $D$ be the projection of $A$ onto $BC$ and let $N$ be the midpoint of $\overline{AD}.$ Let $\tau$ be the line passing through $R$ parallel to $BC.$ Point $S$ lies on $\tau$ and satisfies $MR = MS.$ Prove that if $K \equiv AS \cap MN$, then $AT$ bisects $\overline{MK}.$

My solution to the generalization: Let $R'$ be the reflection of $R$ in $M$ and let $H$ be the projection of $AD$ onto $\tau.$ Denote $V \equiv AT \cap MN, X \equiv AT \cap BC, Y \equiv AS \cap BC$, and let $D'$ be the reflection of $D'$ in $M.$ Note that $-1 = A(U, T; R, \infty) = (U, X; M, D).$ Then from $\triangle MUR \sim \triangle MDA$, we infer $XM : XD = UM : UD = RM : RA.$ Note that $M$ is the circumcenter of $\triangle RR'S$, which implies that $\angle R'SR = 90^{\circ}.$ Therefore, $\triangle RR'S \sim \triangle RAH$, implying that $\tfrac{RM}{RA} = \tfrac{RR'}{2RA} = \tfrac{RS}{2RH} = \tfrac{MY}{2MD}.$ Hence, $\tfrac{XM}{XD} = \tfrac{MY}{2MD}.$ It follows that \[\frac{XY}{XM} = \frac{YM}{XM} - 1 = 1 + 2\left(\frac{MD}{XD} - 1\right) = 1 + \frac{2XM}{XD} = 1 + \frac{MY}{MD} = \frac{D'Y}{D'M}.\] Because $AD' \parallel MN$, we obtain $-1 = A(Y, M; X, D') = (K, M; V, \infty).$ $\square$

Define $E\equiv\overline{AS}\cap\overline{BC}$, $P\equiv\overline{AT}\cap\overline{MN}$, $P'$ as the midpoint of $\overline{KM}$, and $F$ as the reflection of $M$ over $\overline{AD}$. To begin, observe that $[AMT]=[DMR]$ and $[ANT]=[NDU]$, so $$\frac{[AMT]}{[ANT]}=\frac{[DMR]}{[NDU]}=\frac{UR}{DN}\cdot\frac{DM}{DU}.$$Furthermore, because $\triangle ADM\sim\triangle RUM$ and $\overline{AF}\parallel\overline{MS}$, $$\frac{UR}{DN}\cdot\frac{DM}{DU}=\frac{2RM}{AR}=\frac{2SE}{AS}=\frac{2EM}{FM}=\frac{EM}{DM}.$$[asy][asy] size(9cm); defaultpen(fontsize(9pt)); pair A=(-56.8,69.1), B=(-80,-40), C=(80,-40), D=(-56.8,-40), E=(-33.8,-40), M=(0,-40), N=(-56.8,14.5), K=(-42.4,0.7), U=(24.1,-40), R=(24.1,-86.1), S=(-24.1,-86.1), T=(24.1,-132.3), F=(-113.8,-40); draw(A--B--C--cycle, linewidth(0.5)); draw(circumcircle(A,B,C), linewidth(0.5)); draw(A--R--U, linewidth(0.5)); draw(S--R--T, linewidth(0.5)); draw(A--D, linewidth(0.5)); draw(A--S, linewidth(0.5)); draw(M--N, linewidth(0.5)); draw(A--T, linewidth(0.5)); draw(D--R, linewidth(0.4)+grey); draw(M--T, linewidth(0.4)+grey); draw(N--T, linewidth(0.4)+grey); draw(N--U, linewidth(0.4)+grey); draw(A--F--B, linewidth(0.4)+grey); draw(S--M, linewidth(0.4)+grey); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, SW); dot("$E$", E, SW); dot("$M$", M, (-0.6,-1)); dot("$R$", R, SE); dot("$S$", S, SW); dot("$T$", T, SE); dot("$U$", U, SE); dot("$N$", N, SW); dot("$K$", K, SW); dot("$F$", F, SW); [/asy][/asy] Now, we apply Menelaus on $\triangle MND$ to obtain $$\frac{MK}{KN}\cdot\frac{NA}{AD}\cdot\frac{DE}{EM}=1\implies \frac{MK}{KN}=\frac{2EM}{DE};$$thus, $$\frac{MP'}{P'N}=\frac{EM}{DM}=\frac{[AMT]}{[ANT]}=\frac{MP}{PN},$$which immediately implies the desired conclusion.

Why $[AMT]=[DMR]$? Blast_S1 wrote: To begin, observe that $[AMT]=[DMR]$

@above: Sorry for the confusion, $[AMT]=[ART]-[MRT]=[DUR]-[MUR]=[DMR]$. Hope that helps

This is quite a terrible solution that I somehow hacked up. Let $H$ be the reflection of $D$ about $M$, let $V$ be the reflection of $U$ about $M$, let $I=AS\cap BC$, let $G=AT\cap BC$, and let $F=AT\cap MK$. We wish to show that $F$ is the midpoint of $MK$. We see that $\widehat{SB}=\widehat{RC}$, so $AS$ is the $A$-symmedian. Quiz 2 problem 2 actually implies then that $K$ is the symmedian point of $ABC$, but we won't be needing this fact. We need to first prove a lemma and a claim. Lemma: $(DV;IM)=-1$. Proof of Lemma: By $\sqrt{bc}$ inversion, we have that $AM\cdot AS=AI\cdot AR$, so \[\frac{SA}{AI}=\frac{RA}{MA}.\]This actually implies that \[\frac{SI}{AI}=\frac{MR}{MA}.\]But from similar triangles we have that $SI/AI=IV/ID$, and $MR/MA=MU/MD=MV/MD$, so we have \[\frac{IV}{ID}=\frac{MV}{MD},\]or $(DV;IM)=-1$. $\blacksquare$ Claim: $(GH;IM)=-1$. Proof of Claim: Note that $(UT;R\infty)=-1$, so porjection through $A$ implies that $(UG;MD)=-1$. The key to crack this claim is to invert about $M$ with radius $MB=MC$. For a point $P$, we will denote its image under this map by $P'$. We have $(UG;MD)=-1$, so $(U'G';\infty D')=-1$, so \[2D'=U'+G'.\]From the lemma, we have $(DV;IM)=-1$, so $(D'V';I'\infty)=-1$, so \[2I'=D'+V'.\]Adding these two equations, we learn that \[2D'+2I'=D'+G'+2M,\]or \[2I'=(2M-D')+G',\]or $2I'=H'+G'$. Thus, $(G'H';I'\infty)=-1$, so $(GH;IM)=-1$, as desired. $\blacksquare$ Note that $\triangle NDM\sim\triangle ADH$ by SAS, so $AH\parallel MK$. Thus, projecting $(GH;IM)$ through $A$, we get that \[(F\infty;KM)=-1.\]Thus, $F$ is the midpoint of $KM$, as desired. $\blacksquare$