Let $I_a$ be an $A-excenter$. If the lengths of the sides are $a,b,c,wlog-b \ge c$, we can easily see that $MD=\frac{b-c}{2}$, where $D$ denotes the contact point of $A-excircle$ and $BC$, also if $A'$ is foot of perpendicular from $A$ to $BC$, we have $A'K=\frac{(b-c)(b+c-a)}{2a}$, so $\Delta AA'K$~ $\Delta I_aDM$, because of well-known radius of the excircle and the length of $AA'$, so $I_aM//AK$. Now we get $<M I_a D=<K A A'=x$. Easy to see that in $\Delta I_aCB, I_aN$ is symmedian, let it crosses its circumcircle at $Q$. Also let $I_aM$ cuts the same circle at $L$. We should show that $P=L$, or that $NL,AK$ intersect on $\omega$, that is $<LNA+<KAN=\pi-\frac{\beta-\gamma}{2}$. It is well-known that $L,Q$ are symmetric wrt $MN$, so $<LNM=<QNM=<NI_aD=<NI_aC-\frac{\gamma}{2}=<BI_aM-\frac{\gamma}{2}=\frac{\beta-\gamma}{2}-x$, so $<LNA=\frac{\pi}{2}-(\beta-\gamma)+x$, $<KAN=\frac{\pi}{2}-\gamma-x+\frac{\pi}{2}-\frac{\alpha}{2}=\pi-\frac{\beta-\gamma}{2}-<LNA$, $Q.E.D$.

Lemma: Given triangle ABCD, AC cuts BD at E then EB/ED=AB/AD.CB/CD Proof: Let IK cuts (BIC) at J, JT cuts (O) at N'. Let Ia be the excenter which is diametrically opposite to I then let K' is the projection of Ia on BC We have KT.KA=KB.KC=KI.KJ so AITJ is cyclic hence TN'F=TAF=TJK so NF//IK so N=N' It is easy to see that NB, NC are two tangent of (BIC) then (HBJC) is harmonic quadliteral because N,H,J is collinear Therefore HB/HC=JB/JC=(KB/KC):(IB/IC) (lemma) Because K' symmetric with K wrt M then IaK'B~IKB and IaK'C~IKC so IaB/IaC=(IB/IC):(KB/KC)=HC/HB Apply the reverse of the lemma: HIa pass through the midpoint of BC which is M so H,M,Ia is collinear. Thus MN.MF=MB.MC=MH.MIa so NHIaF is concylic Finally TAF=TNF=HNF=HIaF then AK//HM (q.e.d)

nguyenhaan2209 wrote: Lemma: Given triangle ABCD, AC cuts BD at E then EB/ED=AB/AD.CB/CD Proof: Let IK cuts (BIC) at J, JT cuts (O) at N'. Let Ia be the excenter which is diametrically opposite to I then let K' is the projection of Ia on BC We have KT.KA=KB.KC=KI.KJ so AITJ is cyclic hence TN'F=TAF=TJK so NF//IK so N=N' It is easy to see that NB, NC are two tangent of (BIC) then (HBJC) is harmonic quadliteral because N,H,J is collinear Therefore HB/HC=JB/JC=(KB/KC):(IB/IC) (lemma) Because K' symmetric with K wrt M then IaK'B~IKB and IaK'C~IKC so IaB/IaC=(IB/IC):(KB/KC)=HC/HB Apply the reverse of the lemma: HIa pass through the midpoint of BC which is M so H,M,Ia is collinear. Thus MN.MF=MB.MC=MH.MIa so NHIaF is concylic Finally TAF=TNF=HNF=HIaF then AK//HM (q.e.d) Please rewrite your answer in Latex mode. Thank you

Let $K'$ be the point on $BC$ so that $(B, C; K, K') = -1.$ Then we know that $IK' \perp AK$, and so it suffices to show that $IK' \perp PM.$ Since $(B, C; K, K') = -1$, we have that $K'K \cdot KM = KB \cdot KC.$ Let $P'$ be the point where $K'I$ hits $(\triangle BIC)$ again. Then, we have that $K'I \cdot K'P = K'B \cdot K'C = K'K \cdot K'M,$ so that $IPMK$ is cyclic. We will show that $P = P'$, which would imply the problem because $\angle IP'M = 90.$ Notice that $\frac{BI}{CP'} = \frac{KB}{KP'}$ and $\frac{BP'}{CI} = \frac{KP'}{KC}.$ Multiplying these two allows us to find that $$\frac{BP'}{CP'} = \frac{\sqrt{b(s-b)}}{\sqrt{c(s-c)}}.$$ We also find from Sine Law that $\frac{BT}{TC} = \frac{\frac{BK}{KC}}{\frac{AB}{AC}} = \frac{b(s-b)}{c(s-c)} = \frac{BP'^2}{CP'^2}.$ Now, we claim that $\frac{BT}{TC} = \frac{BP^2}{CP^2},$ which would imply $P= P'$ and the problem. Indeed, we know that $2 \cdot \angle BTP = \angle BTC = 2(180 - \angle BPC)$, where the second part follows because the circumcenter of $\triangle BPC$ is on $\omega.$ This means that $\angle BPC = 180 - \angle BTP = \angle TBP + \angle PBT \Rightarrow \angle TBP = \angle CPT.$ Similarly, $\angle TCP = \angle TPB$ and so $\triangle TBP \sim \triangle TPC.$ This means that $\frac{TB}{TC} = \frac{TB}{TP} \cdot \frac{TP}{TC} = \frac{BP^2}{CP^2},$ and so we're done. $\square$

Here's a proof using only angel chasing. rerarura13 wrote: Let $ABC$ be an acute triangle with $AB<BC$. Let $I$ be the incenter of $ABC$, and let $\omega$ be the circumcircle of $ABC$. The incircle of $ABC$ is tangent to the side $BC$ at $K$. The line $AK$ meets $\omega$ again at $T$. Let $M$ be the midpoint of the side $BC$, and let $N$ be the midpoint of the arc $BAC$ of $\omega$. The segment $NT$ intersects the circumcircle of $BIC$ at $P$. Prove that $PM\parallel AK$. Let $KI$ meet $(BIC)$ again in $X,$ and $AI$ meet $(ABC)$ again in $S,$ so that $S$ is the arc midpoint. Then by radical axis, $AITX$ is cyclic, and so $$\measuredangle ATN=\measuredangle ASN=\measuredangle SIX=\pi-\measuredangle XIA \overset{(AITX)}{=} \pi-\measuredangle XTA,$$and so $N$ lies on $TX.$ Redefine $P$ as $I_AM \cap (BIC),$ where $I_A=AI \cap (ABC)$ is the $A$ excenter. By radical axis again, $NPSI_A$ is cyclic. Then $$\measuredangle TNS=\measuredangle TAS=\measuredangle TXI=\measuredangle PXI=\measuredangle PI_AS=\measuredangle PNS,$$and so $N,P,T$ are collinear. Hence, $P=NT \cap (BIC).$ Now, $PM \parallel AK$ follows obviously since then $$\measuredangle PI_AS=\measuredangle PNS=\measuredangle TAI_A.$$So done. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -12.191959748405909, xmax = 14.418181346666099, ymin = -12.233416928471323, ymax = 4.891170069022775; /* image dimensions */ pen qqzzff = rgb(0,0.6,1); pen ccwwff = rgb(0.8,0.4,1); /* draw figures */ draw(circle((-3,-1), 5), linewidth(0.7) + qqzzff); draw(circle((-3.004874281908244,-5.999997624137023), 5.392601209344099), linewidth(0.7) + ccwwff); draw((-7.543358383079728,-3.0875570897341134)--(1.5392796127962531,-3.0964113615539057), linewidth(0.7)); draw((-7.485558026209027,1.209019962225733)--(-3.004874281908244,-5.999997624137023), linewidth(0.7)); draw((-2.995125718091756,3.999997624137023)--(-4.888325032324366,-5.629711499899014), linewidth(0.7) + linetype("4 4") + red); draw((-7.485558026209027,1.209019962225733)--(-4.888325032324366,-5.629711499899014), linewidth(0.7)); draw((-3.004874281908244,-5.999997624137023)--(-2.995125718091756,3.999997624137023), linewidth(0.7)); draw((-5.851538673766596,-1.4199703705996423)--(-5.85316593836157,-3.089204785408001), linewidth(0.7)); draw((-3.916187068159175,-0.6849570025956403)--(-3.002039385141738,-3.09198422564401), linewidth(0.7)); draw((-5.85316593836157,-3.089204785408001)--(-5.860462996489554,-10.574465997202392), linewidth(0.7)); draw((-3.004874281908244,-5.999997624137023)--(-0.15820989004988917,-10.580024877674402), linewidth(0.7)); draw((-4.888325032324366,-5.629711499899014)--(-5.860462996489554,-10.574465997202392), linewidth(0.7)); draw((-3.002039385141738,-3.09198422564401)--(-0.15820989004988917,-10.580024877674402), linewidth(0.7)); draw(shift((-16.131716453382786,-5.9872008321945875))*xscale(11.24907326271081)*yscale(11.24907326271081)*arc((0,0),1,-24.06611258385175,39.770702869388344), linewidth(0.7) + linetype("2 2") + green); draw(shift((10.126842171474522,-1.0127967919424432))*xscale(14.046855489848465)*yscale(14.046855489848465)*arc((0,0),1,159.09237081614714,222.92918626938732), linewidth(0.7) + linetype("2 2") + green); /* dots and labels */ dot((-7.543358383079728,-3.0875570897341134),dotstyle); label("$B$", (-8.220674432886607,-3.1019901838695962), NE * labelscalefactor); dot((1.5392796127962531,-3.0964113615539057),linewidth(4pt) + dotstyle); label("$C$", (1.846660316009457,-3.17787461665022), NE * labelscalefactor); dot((-7.485558026209027,1.209019962225733),dotstyle); label("$A$", (-8.043610756398484,1.4004861611140813), NE * labelscalefactor); dot((-5.851538673766596,-1.4199703705996423),linewidth(4pt) + dotstyle); label("$I$", (-5.792372583906652,-1.0025208769390048), NE * labelscalefactor); dot((-5.85316593836157,-3.089204785408001),linewidth(4pt) + dotstyle); label("$K$", (-6.45003766800539,-3.532001969626464), NE * labelscalefactor); dot((-2.995125718091756,3.999997624137023),linewidth(4pt) + dotstyle); label("$N$", (-3.0605330038042022,4.486453094192782), NE * labelscalefactor); dot((-4.888325032324366,-5.629711499899014),linewidth(4pt) + dotstyle); label("$T$", (-4.780580146831671,-6.466200037143917), NE * labelscalefactor); dot((-3.004874281908244,-5.999997624137023),linewidth(4pt) + dotstyle); label("$S$", (-3.313481113072948,-6.744442957339538), NE * labelscalefactor); dot((-3.916187068159175,-0.6849570025956403),linewidth(4pt) + dotstyle); label("$P$", (-4.401157982928552,-0.4207402256208893), NE * labelscalefactor); dot((-3.002039385141738,-3.09198422564401),linewidth(4pt) + dotstyle); label("$M$", (-3.515839600487944,-3.683770835187712), NE * labelscalefactor); dot((-5.860462996489554,-10.574465997202392),linewidth(4pt) + dotstyle); label("$X$", (-6.197089558736645,-11.297508924176963), NE * labelscalefactor); dot((-0.15820989004988917,-10.580024877674402),linewidth(4pt) + dotstyle); label("$I_A$", (0.00013911834761619294,-11.120445247688842), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Please see @above diagram lazy to make one thanks @above We begin by noting that the radical center of circumcircles of $\triangle ABC$, $\triangle BIC$, $\triangle AIX$ is $BC \cap IX \cap AT_1$, where $T_1 = \odot (AIX) \cap \odot (ABC)$. $BC \cap IX = K$, so $A - K - $ are collinear which means that quadrilateral $AIXT$ is a cyclic quadrilateral. Also it can be seen that it suffices to prove that $P, M, I_A$ are collinear. $\angle TNS = \angle KXP = \angle II_A P$ and so $\triangle MSI_A \sim MPN$. Due to corresponding sides of similar angles property, we have that $MS \cdot MN = MP \cdot MI_A$, but $M, S, N$ are collinear and $M \in BC =$ $\mathrm{Radical}$ $\mathrm{Axis}$ of $\odot (ABC)$ and $\odot (BIC)$ which means that $P, M, I_A$ are collinear which proves our Theorem, because $\angle IAT = \angle IXT$ $($$AIXT$ cyclic$)$ $ = \angle II_AP = \angle AI_AM$ and we're done by converse of alternate angles property of parallel lines.

let $R=EF \cap BC$ $P'=RI \cap (BIC)$ $S=IK \cap (BIC)$ $I_a$ be the $A-$excenter we have $I_aS \parallel BC$ $-1=(B,C;K,R)=(B,C;S,P')$ $-1=(B,C;M,P_\infty)=(B,C;I_aM \cap (BIC),S)$ so $I_a-M-P'$ are collinear $MB.MC=MP.MI_a=ML.MN$ so $NPLI_a$ is cyclic it's well-known that $MI_a \parallel AK$ $\angle TNL=\angle TAL =\angle LI_aP'=\angle P'NL$ so $P'-N-L$ are collinear so $P'=P$

Redefine $P$ as intersection of $(BIC)$ and $\overline{I_AM}$. Let $L$ be the intersection of $\overline{AI}$ and $(ABC)$. Let $I_A$ be the $A$-excenter of $\triangle ABC$. I contend that $T,D,N$ are collinear. [asy][asy]import olympiad; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,I,K,T,M,P,IA,L,N; A=dir(130);B=dir(205);C=dir(335);I=incenter(A,B,C);K=foot(I,B,C);T=intersectionpoints(circumcircle(A,B,C),A--100K-99A)[1]; M=midpoint(B--C);L=intersectionpoints(circumcircle(A,B,C),A--100I-99A)[1];N=-L;P=intersectionpoints(N--T,circumcircle(B,I,C))[0]; IA=2L-I; draw(A--B--C--cycle,heavygreen);draw(circumcircle(A,B,C),royalblue);draw(incircle(A,B,C),royalblue); draw(A--IA,lightgreen);draw(A--T,lightgreen);draw(P--IA,orange);draw(N--L,orange);draw(N--T,lightgreen); draw(circumcircle(I,B,C),royalblue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$L$",L,dir(L)); dot("$I$",I,dir(I)); dot("$K$",K,dir(K)); dot("$T$",T,dir(T)); dot("$N$",N,dir(N)); dot("$M$",M,dir(M)); dot("$P$",P,dir(P)); dot("$I_A$",IA,dir(IA)); [/asy][/asy] Claim: $\overline{AK}\parallel \overline{I_AM}$. Proof. Deploy barycentric coordinates. Note that $K=(0:s-c:s-b)$, $I_A=(-a:b:c)$, $M=(0,\tfrac{1}{2},\tfrac{1}{2})$. All we need is $$ \begin{vmatrix} -a & s-c & s-b\\ 0 & \frac{1}{2} & \frac{1}{2}\\ -a& b& c \end{vmatrix}=0,$$which is true after expanding. $\square$ Note that $\triangle IPI_A\sim\triangle LTN$. By PoP, $MP\cdot MI_A=MB\cdot MC=ML\cdot MN$, which implies that $NPLI_A$ is cyclic quadrilateral. Hence, we have $\measuredangle PNL=\measuredangle PII_A=\measuredangle TNL$, we conclude that $T,P,N$ are collinear. $\blacksquare$

$\sqrt{bc}$ inversion and $(BIC)$ inversion aparently go by hand... Let $G$ the $A$-sharky devil point and let $N'$ the midpoint of the minor arc $BC$ on $\omega$, let $I_A$ the $A$-excenter and let $T'$ be the reflection of $T$ w.r.t. $MN$ and let $AI \cap BC=D$, now let $MI_A \cap (BIC)=P'$ and now we are going to prove that $P=P'$. By PoP $$MN \cdot MN'=BM^2=MP' \cdot MI_A \implies NP'N'I_A \; \text{cyclic}$$Now clearly $K,T'$ are inverses w.r.t. $\sqrt{bc}$ inversion so we can easly get that $N,T',I_A$ are colinear, its known that $G,K,N'$ are colinear and u can easly show by inversion w.r.t. $(BIC)$ that $T_1$ the inverse of $T$ w.r.t. $(BIC)$ lies on $GA'$ where $A'$ is the point on $\omega$ such that $AA' \parallel BC$ and also note that since $A,K,T$ are colinear u have $T_1,G,D,N$ cyclic which gives that $G,D,T'$ are colinear, now by angle chasing $$\angle MT_1N'=\angle DGN'=\angle N'NT'=\angle N'P'M \implies N'T_1P'M \; \text{cyclic}$$And inverting this w.r.t. $(BIC)$ u get $N,P',T$ colinear which means $P=P'$ thus we are done

Solved with erkosfobiladol. Let $Q$ be midpoint of arc $BC$ not containing $A$. $AI\cap BC=D, AI\cap (BIC)=\{I,J\}$. Let $JM\cap (BIC)=\{J,R\}$ We will prove that $T,R,N$ are collinear where $N$ is the midpoint of arc $BAC$. $\textbf{Claim:}\ \overline{JMR}\parallel \overline{AKT}.$ Since $DQ.DA=DB.DC=DI.DJ$, \[\frac{DA}{DJ}=\frac{DI}{DQ}=\frac{DK}{DM}\]$\textbf{Claim:} \ JQRN$ is cyclic. \[MR.MJ=MB.MC=MQ.MN\]$\textbf{Claim:}\ T,R,N$ are collinear. \[\angle QNR=\angle QJR=\angle AJR=\angle QAT=\angle QNT\]which gives the result as desired.$\blacksquare$