Let $(J)$ be the incircle of $\triangle A_0BC$, $(J)$ is tangent to $BC,CA_0,BA_0$ at $K',U,V$, respectively. $I'$ is the center of $(\omega')$ ,$(\omega')$ is tangent to $CA_0,BA_0$ at $X,Y$, respectively. $I$ is the center of $(\omega)$ and $(I)$ is tangent to $CA,BA$ at $E,F$, respectively. $E',F'$ are symmetry to $E,F$ through $A$, respectively. We have: $$BK'=\dfrac{BC+BA_0-CA_0}{2}=\dfrac{BY+AY+BC-CX-XA_0}{2}=\dfrac{BF'+BC-CX}{2}=\dfrac{BF'+BC-CE'}{2}$$$$=\dfrac{BA+AF'+BC-AE'-AC}{2}=\dfrac{AB+BC-AC}{2}=BK$$$\Rightarrow K \equiv K'$ $\Rightarrow (I)$ is tangent to $(J)$ at $K$. Because $A_0$ is the homothetic center of $(J)$ and $(I')$, then: $$\dfrac{AI'}{AJ}=\dfrac{I'Y}{JV}=\dfrac{KI}{KJ} \Rightarrow II' \parallel A_0K$$The above equality is true because $K$ is the homothetic center of $(J)$ and $(I)$. But we have $AI=AI'$ so $AJ$ bisects $A_0K$. Therefore, we just need to prove $\overline{A,J,M}$. Let $G$ be the symmetry point of $K$ through $J$, $Z$ is the midpoint of $A_0K$. $\Rightarrow ZJ \parallel A_0G$. Let $A_0G \cap BC=L$, then it is easy to prove that $\overline{A,G,L}$ and $JM \parallel AG$. $\Rightarrow \overline{A,J,M}$. Q.E.D

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[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ 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 = -6.544653865195542, xmax = 17.45224398846036, ymin = -4.751583952603447, ymax = 11.720154295376325; /* image dimensions */ /* draw figures */ draw(circle((4.86,0.44), 2.0286941612771505), linewidth(1)); draw(circle((7.967455678067624,5.3452482571528925), 2.0286941612771505), linewidth(1)); draw(circle((5.117531968523624,1.954893932491891), 3.5653223991773912), linewidth(1)); draw((11.729994623426101,9.821276910966443)--(5.117531968523624,1.954893932491891), linewidth(1)); draw((7.967455678067624,5.3452482571528925)--(4.86,0.44), linewidth(1)); draw((5.117531968523624,1.954893932491891)--(4.52,-1.56), linewidth(1)); draw((11.729994623426101,9.821276910966443)--(-5.324507327869696,0.1135662457378519), linewidth(1)); draw((-5.324507327869696,0.1135662457378519)--(12.920131678633394,-2.9880223853676813), linewidth(1)); draw((3.3538050985009225,5.053408308452726)--(12.920131678633394,-2.9880223853676813), linewidth(1)); draw((11.729994623426101,9.821276910966443)--(4.52,-1.56), linewidth(1)); draw((11.729994623426101,9.821276910966443)--(7.464025193339071,-2.0604842828676433), linewidth(1)); draw((4.392621473674303,2.414121909388988)--(12.920131678633394,-2.9880223853676813), linewidth(1)); draw((5.675326746934564,6.374857655400438)--(7.464025193339071,-2.0604842828676433), linewidth(1)); draw((-5.324507327869696,0.1135662457378519)--(9.505143038470829,3.624524764146081), linewidth(1)); draw((1.0697589327346875,-0.9734590185648957)--(6.413727839033812,2.8926241285764465), linewidth(1) + linetype("4 4")); /* dots and labels */ dot((4.86,0.44),dotstyle); label("$I$", (4.554956553176656,0.3697052234816341), NE * labelscalefactor); dot((4.52,-1.56),dotstyle); label("$K$", (4.450440447542267,-1.9714555427286888), NE * labelscalefactor); dot((-5.324507327869696,0.1135662457378519),dotstyle); label("$C$", (-5.666718577866668,-0.3619075159590918), NE * labelscalefactor); dot((7.464025193339071,-2.0604842828676433),dotstyle); label("$B$", (7.439601068685815,-2.5985521765350255), NE * labelscalefactor); dot((6.413727839033812,2.8926241285764465),linewidth(4pt) + dotstyle); label("$A$", (6.498956117976307,3.066220748848881), NE * labelscalefactor); dot((11.729994623426101,9.821276910966443),linewidth(4pt) + dotstyle); label("$A_0$", (11.99650327434521,9.964283720718582), NE * labelscalefactor); dot((1.0697589327346875,-0.9734590185648957),linewidth(4pt) + dotstyle); label("$M$", (0.8341831925923788,-1.4697782356836195), NE * labelscalefactor); dot((5.117531968523624,1.954893932491891),linewidth(4pt) + dotstyle); label("$D$", (4.680375879937924,1.6866081544749407), NE * labelscalefactor); dot((7.967455678067624,5.3452482571528925),dotstyle); label("$I'$", (7.79495582784274,5.595510505201105), NE * labelscalefactor); dot((3.3538050985009225,5.053408308452726),linewidth(4pt) + dotstyle); label("$F$", (2.987214968660809,5.240155746044181), NE * labelscalefactor); dot((8.473129610845193,0.7501166614073307),linewidth(4pt) + dotstyle); label("$E$", (8.714697557425371,0.6832535403848023), NE * labelscalefactor); dot((4.392621473674303,2.414121909388988),linewidth(4pt) + dotstyle); label("$H$", (4.241408236273487,2.5645434418038118), NE * labelscalefactor); dot((6.844567295497547,0.8608237750431357),linewidth(4pt) + dotstyle); label("$G$", (6.9797302038944995,0.9549954150342148), NE * labelscalefactor); dot((12.920131678633394,-2.9880223853676813),linewidth(4pt) + dotstyle); label("$J$", (12.999857888435352,-2.828487608930682), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $I$ and $I'$ be the centre of $\omega$ and $\omega'$ respectively. Obviously, $A$ is the midpoint of $II'$. Let $\Omega$ be the incenter of $\bigtriangleup A_0BC$. Claim 01. $\Omega$ touches $BC$ at $K$. Proof. This is the consequence of Pithot on concave quadrilateral $A_0BAC$ and degenerate quadrilateral $ABKC$: \[ A_0B - A_0C = AB - AC = KB - KC. \text{ } \blacksquare \] Now, let $D$ be the center of $\Omega$. We know $D, I', A_0$ are collinear and by the previous claim, $D, I, K$ are also collinear. Claim 02. $I'I \parallel A_0K$. Proof. The claim follows from the following: \[ \frac{DA_0}{I'A_0} = \frac{r_{\Omega}}{r_{\omega'}} = \frac{r_{\Omega}}{r_{\omega}} = \frac{DK}{IK}. \text{ } \blacksquare \] From this claim, we obtain $DA$ bisects $A_0K$. We are left to prove that $M \in AD$. Claim 03. Let $\omega$ touches $AB, AC$ at $G, H$. Then $D \in GH$. Proof. Let $\Omega$ touches $A_0B, A_0C$ at $E, F$. By Menelaus, $EF, GH, BC$ concur at a point $J$. We know that $\overline{J-G-H} \perp AI$. By La Hire, $A_0K$ is the polar of $J$ with respect to $\Omega$, so $JD \perp A_0K$. Since $A_0K \parallel AI$, therefore $JD \parallel \overline{J-G-H} \implies D \in GH$. To finish the proof of the problem, we use the well-known lemma that says $AM, GH, IK$ concur. This implies $D \in AM$, as desired. $\square$

It seems that these problems are somehow connected.(picture in last post) https://artofproblemsolving.com/community/c6h1174705p5660092

A very creative problem. But maybe it's a bit easy for a P8, because one can solve it quickly with just homothety. Here's my proof:

Attachments:

Let $\omega_0$ be the incircle of $A_0BC$. The first observation is that $\omega_0$ touches $BC$ at $K$. Indeed, if $E'$, $F'$ are the reflections of $E$, $F$ over $A$, then \[ BK = \frac{BC+AB-AC}{2} = \frac{BC+BF'-CE'}{2} = \frac{BC+A_0B-A_0C}{2}. \]Hence Monge's theorem on $\omega$, $\omega'$ and $\omega_0$ implies that $AI\parallel A_0K$. Let $J$ be the incenter of $A_0BC$ and $N$ be the midpoint of $A_0K$. Then $J=A_0I'\cap KI$ and $J$, $A$, $N$ are collinear. On the other hand, it is well-known that $J$ lies on $MN$ (a homothety at $K$ transforms this into the famous intouch-extouch lemma), so $M$, $A$, $N$ are collinear, as desired.

Denote by $I,I',J$ centers of $\omega, \omega '$ and incircle $\gamma$ of $A_0BC$ respectively. $$|BK|=\frac{|AB|+|BC|-|CA|}{2}=\frac{|A_0B|+|BC|-|CA_0|}{2} \implies K=\gamma \cap BC\implies J=KI\cap A_0I',$$and by Monge for $\omega, \omega ', \gamma$ we obtain $A_0K\parallel II'$ so line $AJ$ bisects $A_0K.$ Thus the Gauss line $AJ$ of $A_0BKC$ passes through $M$ as required.

Define $K', B',C'$ (reflections across $A$). $L$ is the midpoint of $A_0K$. By Brianchon's Theorem on $C'CA_0BB'K'$, $A_0K' \parallel B'C \parallel BC' \parallel AM$. Also by Thales Theorem $A_0K' \parallel AL$ implying $A-M-L$.

Let $I'$ be the reflection of $I$ over $A$. Note that if $BB', CC'$ are tangents to $\omega'$ then $BB' = BA + AE, CC' = CA + AF \implies A_0B - A_0C = BK - CK \implies$ the incircle $\Omega$ of $\triangle{A_0BC}$ is tangent to $BC$ at $K$. By Monge on $\omega, \omega', \Omega$, $AK \parallel AI$, so if $I_2$ is the center of $\Omega$ then $A_0, I', I_2$ and $K, I, I_2$ are collinear $\implies \frac{A_0K}{I'I} = \frac{I_2K}{I_2I}$. Through a bit of simplifying we get that $\frac{M_2K}{AI} = \frac{I_2K}{I_2I}$ if $M_2$ is the midpoint of $A_0K$. Note that $M_2, A, M$ are collinear if and only if $KI \cap AM = I_2$, AKA $I_2 \in AM$. By the Iran Lemma, if $E, F$ are the tangency points of $\omega$ then $AM, EF, IK$ concur; call this point $Y$. It suffices to show that $\angle{YBI} = \angle{ABI'}$; note that by inversion about $\omega$, $\angle{YBI} = \angle{IZM_B}$ if $Z, M$ are the projection from $A$ onto $IK$ and the midpoint of $FK$, respectively, and by angle chasing, $\triangle{ZFK} \sim \triangle{BIA}$, so $\angle{IZM_B} = \angle{BAM_B} = \angle{ABI'}$ and we are done. $\square$

Let $\omega_0$ denote the incircle of $\triangle A_0BC$ and let $J$ denote its incenter. Let $\omega' $ be tangent to $BA_0,CA_0$ at $R,S$ respectively. Let $N$ denote midpoint of $A_0K$. Denote by $\bullet '$ the reflection of $\bullet$ over $A$. Then we have: $$ A_0B+BC-A_0C = BC+BR-CS = BC+ BF'-CE'= BC+AB-AC$$So $\omega_0 \cap BC = K$ and hence $K= AI' \cap JK$. Now by Monge on $\{\omega, \omega', \omega_0\}$, we get that $II' \parallel A_0K$. This means $JA$ bisects $A_0K$. However $J$ is known to lie on $MN$ by a famous lemma, so we're done. $\square$