Let $ O $ be the circumcenter of $ \triangle ABC $. From Easy - Circles and Midpoints, we get $ \angle NXT = 90^{\circ} $, and $ OX=OM $. Similarly, $ OY=ON $. Thus, $ MNYX $ becomes an isosceles trapezoid, and hence we have $ KM=KX $. Also, from the construction in the link, we get $ XT = AM $ and $ \angle KXT = \angle OXT - \angle OXY = 90^{\circ} - \angle OMN = \angle AMN $. Therefore, $ \triangle KMA \cong \triangle KXT \implies KA=KT $

First we will prove that $MX \parallel AT$. Let $X'$ be the second point where the parallel line to $AT$ passing through $M$ meets the circumcircle of $AMT$. We will see that $X'$ lies on the perpendicular bisector of $AC$. From this it follows that $X' = X$ and hence $MX \parallel AT$. We claim that $\triangle MBT$ and $\triangle X'TC$ are congruent. Since $AMX'T$ is a trapezoid inscribed in a circle, it is an isosceles trapezoid, and hence $MB = MA = X'T$. Also, we have $BT = TC$ because $T$ is the midpoint of the arc $BC$. Finally, $\angle X'TC = \angle X'TA + \angle ATC = \angle BAT + \angle ABC$ (using again that $AMX'T$ is an isosceles trapezoid for the first part, and inscribed angles in $\omega$ for the second part). But we also have $\angle BAT = \angle TBC$, because the chords $BT$ and $TC$ of $\omega$ have equal lengths. Therefore, $\angle X'TC = \angle TBC + \angle ABC = \angle MBT$. We have proven that $\triangle MBT \equiv \triangle X'TC$. In particular $MT = X'C$. But $MT = X'A$, because they are the diagonals of an isosceles trapezoid. Then $X'A = X'C$, that is, $X'$ lies on the perpendicular bisector of $AC$, as wanted. So far we know $MX \parallel AT$, and moreover that $AMXT$ is an isosceles trapezoid. We can prove in an analogous way that $ANYT$ is an isosceles trapezoid. It follows that the segments $MX$, $AT$ and $NY$ all have the same perpendicular bisector. But then $K$, which is the intersection point of $MN$ and $XY$, belongs to this perpendicular bisector. In particular $KA = KT$, and we are done. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -4.300000000000004, xmax = 11.32000000000001, ymin = -3.020000000000004, ymax = 6.300000000000007; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((2.540000000000003,5.000000000000000)--(0.4000000000000005,0.8400000000000009)--(5.720000000000006,0.6800000000000008)--cycle, zzttqq); /* draw figures */ draw((2.540000000000003,5.000000000000000)--(0.4000000000000005,0.8400000000000009), zzttqq); draw((0.4000000000000005,0.8400000000000009)--(5.720000000000006,0.6800000000000008), zzttqq); draw((5.720000000000006,0.6800000000000008)--(2.540000000000003,5.000000000000000), zzttqq); draw(circle((3.099747970952588,2.081620034173432), 2.971575241504656)); draw((xmin, -12.51783780793794*xmin + 36.79530803216237)--(xmax, -12.51783780793794*xmax + 36.79530803216237)); /* line */ draw((xmin, -2.472454225932990*xmin + 6.554507712121498)--(xmax, -2.472454225932990*xmax + 6.554507712121498)); /* line */ draw((xmin, 3.330360015372642*xmin-10.91438686348901)--(xmax, 3.330360015372642*xmax-10.91438686348901)); /* line */ draw(circle((5.312765398507484,2.258409145960098), 3.899301009939554)); draw(circle((0.5621914115686578,1.878904789071374), 3.694991492297976)); draw((xmin, 0.7361111111111118*xmin-0.2001388888888885)--(xmax, 0.7361111111111118*xmax-0.2001388888888885)); /* line */ draw((xmin, -0.5144230769230775*xmin + 3.676201923076927)--(xmax, -0.5144230769230775*xmax + 3.676201923076927)); /* line */ draw((2.540000000000003,5.000000000000000)--(3.099747970952588,2.081620034173432)); draw((3.099747970952588,2.081620034173432)--(0.4000000000000005,0.8400000000000009)); draw((xmin, 0.1918009229459768*xmin + 0.6836954941066192)--(xmax, 0.1918009229459768*xmax + 0.6836954941066192)); /* line */ draw((xmin, -0.03007518796992487*xmin + 2.964210526315793)--(xmax, -0.03007518796992487*xmax + 2.964210526315793)); /* line */ draw((1.470000000000002,2.920000000000003)--(1.623769685397414,0.9951360184175412)); draw((4.130000000000004,2.840000000000003)--(4.237333239206217,1.496419920216036)); draw((0.5621914115686578,1.878904789071374)--(10.27832614694408,2.655087935430257)); /* dots and labels */ dot((2.540000000000003,5.000000000000000),dotstyle); label("$A$", (2.620000000000003,5.120000000000005), NE * labelscalefactor); dot((0.4000000000000005,0.8400000000000009),dotstyle); label("$B$", (0.4800000000000005,0.9600000000000011), NE * labelscalefactor); dot((5.720000000000006,0.6800000000000008),dotstyle); label("$C$", (5.800000000000006,0.8000000000000009), NE * labelscalefactor); dot((1.470000000000002,2.920000000000003),dotstyle); label("$M$", (1.560000000000002,3.040000000000004), NE * labelscalefactor); dot((4.130000000000004,2.840000000000003),dotstyle); label("$N$", (4.220000000000004,2.960000000000003), NE * labelscalefactor); dot((3.010417678247103,-0.8886121982839235),dotstyle); label("$T$", (3.100000000000004,-0.7600000000000009), NE * labelscalefactor); dot((3.099747970952588,2.081620034173432),dotstyle); label("$O$", (3.180000000000004,2.200000000000003), NE * labelscalefactor); dot((1.623769685397414,0.9951360184175412),dotstyle); label("$X$", (1.700000000000002,1.120000000000001), NE * labelscalefactor); dot((4.237333239206217,1.496419920216036),dotstyle); label("$Y$", (4.320000000000005,1.620000000000002), NE * labelscalefactor); dot((10.27832614694408,2.655087935430257),dotstyle); label("$K$", (10.36000000000001,2.780000000000003), NE * labelscalefactor); dot((5.312765398507484,2.258409145960098),dotstyle); label("$O_1$", (5.400000000000007,2.380000000000003), NE * labelscalefactor); dot((0.5621914115686578,1.878904789071374),dotstyle); label("$O_2$", (0.6400000000000007,2.000000000000000), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] This lemma took me so long to prove I'm not even sure why D:< Lemma: $AMXT$ is an isoceles trapezoid. By symmetry, $ANYT$ would also be an isoceles trapezoid. Proof: Let $X'$ be on the circumcircle of $AMT$ such that $X'M||AT$. Then, $AM = X'T$ and we already know $OT = OA$. But we also have $\angle TON = 180^{\circ} - \angle C$, so $X'OT \cong MOA$ since $\angle MOA = \angle C$. So we have $MO = X'O$. Therefore, $AOX' \cong TOM$. Notice that $\angle TOC = \angle A$. We also have $\angle TAB = \angle A/2$ so $\angle XMA = 180^{\circ} - A/2$ implying that $\angle X'MO = 90^{\circ} - A/2$ so $\angle XOM = A$. Therefore, $\angle X'OC = \angle MOT$, so $X'OC \cong MOT \cong X'OA$, so $X'C = X'A$, so $X'$ lies on the perpendicular bisector of $AC$. Hence, $X' = X$. $\Box$ Therefore, $XMNY$ is an isoceles trapezoid; the perpendicular bisector of $XM$ and $NY$ and $AT$ are the same line. Furthermore, $MN \cap XY$ lies on the perpendicular bisector, so $KA = KT$, as desired.

Claim: $AT \parallel MX$ Proof: Let the line passing through $M$ parallel to $AT$ meet $(AMT)$ at $X'.$ Since $(AMTX')$ is cyclic and every cyclic trapezoid is isoceles, we must have that $BM = AM = TX'.$ Now, some trivial angle chasing (too lazy to post it now; will complete it later) yields $\triangle BTM \sim \triangle CTX',$ implying $X=X'.$ $\blacksquare$ Now, since $ATMX$ is cyclic, it must be isoceles. Similarly, so must be $ATNY.$ It follows that $AT, MX, NY$ share the same perpendicular bisector, so $K$ must also lie on this perpendicular bisector. $\blacksquare$

Claim:$AT\parallel NY$ Proof:Let point $N'$ be on circumcircle of $\triangle {ANT}$ such that $AT\parallel NN'$ and let $TN'$ intersects with $AC$ inside $\omega$ at $R$ (if outside its same), intersects with $\omega$ at $P$. Since $ANN'T$ is isosceles trapezoid, we have $\triangle{ART}$ , $\triangle{NN'R}$ and $\triangle{PRC}$ are isosceles which means $AN=NC=TN'=N'P$ , thus we have $N'$ is midpoint of $TP$. Since $AB\parallel TP$ and $ABTP$ is isosceles trapezoid, we conclude $MN'\perp TP$ $\implies$ $N'=Y$.

AnonymousBunny wrote: $\triangle BTM \sim \triangle CTX',$ implying $X=X'.$ $\blacksquare$ How does this imply $X=X'$ ?

Claim: $AMXT$ is an isoceles trapezoid. I will prove this indirectly. Let $X'$ be the point on $\odot AMT$ such that $MX' ||AT$ $\implies TAMX'$ is an isosceles trapezoid. Thus we have $OA=OT, OM=OX'$ and $AM=TX'$ $\implies \triangle OAM \cong OTX'$ $\implies \angle TOX' = \angle AOM= \angle C$ But we know $\angle TOX = 180- \angle TON = \angle C$ $\implies X=X'$ Hence the claim. The remaining proof is trivial.

Can anyone barycentric/complex bash this?

Let $\tau, \ell$ be the perpendicular bisectors of $\overline{AT}, \overline{AC}$, respectively. Let $Q$ lie on $\omega$ such that $TQ \parallel AC$ and let $X'$ be the midpoint of $\overline{TQ}.$ Since $ACTQ$ is an isoceles trapezoid, $O, N, X'$ are collinear on $\ell.$ Meanwhile, from $\widehat{AQ} = \widehat{TC} = \widehat{TB}$, it follows that $ATBQ$ is an isoceles trapezoid as well. Then from symmetry in $\tau$, we see that $ATX'M$ is an isoceles trapezoid too. Hence, $X'$ lies on the circumcircle of $\triangle AMT \implies X' \equiv X.$ Therefore, $\tau$ is the perpendicular bisector of $\overline{MX}$, and analagous arguments show $\tau$ is the perpendicular bisector of $\overline{NY}$ as well. It follows that $XYNM$ is an isoceles trapezoid, so by symmetry, $K \in \tau.$ Thus, $KA = KT$, as desired. $\square$

It suffices to show that $AMXT$ is an isosceles trapezoid, which implies that segments $AT, MX$ and $NY$ have the same perpendicular bisector by symmetry. Let $X'$ be the point where the perpendicular bisector of $BC$ hits $(AMT)$ inside $\triangle ABC$. Suppose the external angle bisector of $A$ hits $(AMT)$ at $F$; from $\angle TAF=90^{\circ}$ and $\angle TX'N=90^{\circ}$, it follows that $X, N, F$ are collinear. Thus $\angle XTA=\angle XFA=\angle NFA=90^{\circ}-\angle NAF=\angle TAN=\frac{\angle A}{2}$. But $\angle MAT=\angle BAT=\frac{\angle A}{2}$, so we're done$.\:\blacksquare\:$

Complex bash because why not. Let $Y'$ be the reflection of $N$ over the perpendicular bisector of $AT$. We will show that $Y=Y'$. We know that $Y'$ clearly lies on the circumcircle of $\triangle ANT$. Now we can complex bash to obtain that $Y'M\perp AB$. Let the circumcircle of $\triangle ABC$ be the unit circle. WLOG, let $t=\bar{a}$. This implies $y'=\bar{n}=\frac{1}{n}$. Additionally, because $T$ is the midpoint of arc $BC$, $\frac{1}{a}=\sqrt{bc} \Rightarrow a^2bc=1$. Also notice that $n=\frac{a+c}{2}$ and $m=\frac{a+b}{2}$. We have that $\frac{y'-m}{a-b}=\frac{\overline{\frac{a+c}{2}}-\frac{a+b}{2}}{a-b}=\frac{1}{2}\cdot \frac{\frac{1}{a}+\frac{1}{c}-a-b}{a-b}=\frac{1}{2}\cdot \frac{a^2b+abc-b-a}{a-b}=\frac{1}{2}\cdot \frac{a+c-\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}-\frac{1}{b}}=-\overline{\left(\frac{y'-m}{a-b}\right )}$ So indeed, $Y'M\perp AB \implies Y'=Y$. Similarly we can define $X'$, and show that $X'=X$. The result quickly follows. $\Box$

Darn, this problem trolled me. D: Let the perpendicular bisector of $AT$ be $\overline{p}$. Observe that $\odot(AMT)$ is symmetric wrt $\overline{p}$. Now, I claim $\overline{OM}$ and $\overline{ON}$ are symmetric with respect to $\overline{p}$. Let the reflection of $N$ wrt $\overline{p}$ be $N'$, and similarly for $M$. We will show $M, O, N'$ are collinear, which proves the claim, as we can use the same argument to show that $N, O, M'$ are collinear. To that end, we use complex numbers. Denote the complex number associated with each point by the corresponding lowercase letters. Set $\overline{p}$ as the real axis. As a result, we have $t = \frac{1}{a} = \sqrt{bc} \implies a^2bc=1$ by definition of $T$. So we simply want \[ \frac{0-\frac{a+b}{2}}{0-\frac{2}{a+c}} = \overline{\left(\frac{0-\frac{a+b}{2}}{0-\frac{2}{a+c}}\right)} \]This easily expands to $\frac{(a+b)(a+c)}{4} = \frac{(a+b)(a+c)}{4a^2bc}$, which is true. Therefore, $M$ and $X$ are related by reflection about $\overline{p}$, and similarly for $N$ and $Y$. So $K$ lies on $\overline{p}$, and $KA=KT$ as desired.

I think this problem is about the same as the configuration given in IMO 2007 P4! First let us investigate some properties of $X, Y$. We claim that $TY \parallel AM$ as well as $TX \parallel AN$. Let $Y'$ be a point such that $TY'\parallel AM$ and $\angle MY'T =90^{\circ}$. If suffices to prove that $TY'=AN$, then $ANY'T$ cyclic $\Rightarrow Y=Y'$, because we already have $\angle NAT=\angle MAT=\angle Y'TA$. Let $O$ be the circumcenter of $\triangle ABC$, then let $MO\cap AT=U$, $NO\cap AT=V$. Then we have $\angle AUM=90- \angle ABC=\angle AVN$, and $\angle MAU=\angle NAV$, thus $\triangle AMU \sim \triangle ANV$. Then $\frac{AU}{AM}=\frac{AV}{AN}$. But $\angle OUV=\angle OVU$, so $AU=TV$, so $\frac{AU}{AM}=\frac{AV}{AN}=\frac{UT}{AN} \Rightarrow \frac{AU}{UT}=\frac{AM}{AN}$. Back to the original problem, $TY'\parallel AM$ implies $\frac{AM}{AU}=\frac{TY'}{TU}$, so $TY'=AM$, so $Y=Y'$, and thus $TY\parallel AM$. Likewise $TX\parallel AN$. In particular, $ANYT, AMXT$ are isosceles trapezoids, thus so as $MNXT$. This concludes that $MN\cap XT=K$ is on the perpendicular bisector of $MX, NY, AT$, thus $KA=KT$.

Let $L$ be the midpoint of $\overline{TA}$; it's clear that $L$ is the midpoint of arc $MN$ containing $O$ of $\odot(MON)$. So, $\overline{ON}, \overline{OM}$ are symmetric in $\overline{OL}$. Claim: $AMXT$ is an isosceles trapezoid. (Proof) Redefine $X$ to be the reflection of $M$ in line $\overline{MO}$. Note that $A, M, X, T$ are the vertices of an isosceles trapezoid and $X$ lies on $\overline{ON}$ establishing the claim. Same reasoning applies to $ANYT$. $\square$ Evidently, $\overline{MN}, \overline{XY}$ are symmetric in line $\overline{LO}$ so $KA=KT$ as desired. $\square$

Let $Z $ be the foot of the $\perp$ from $O $ to $AT $. Claim: $MX||AT||NY $. Proof of the claim: Since, $OT\perp BC $, so, $\angle TOY = \angle ABC = \angle AON $. But, $OA = OT $. Thus, $\Delta TOY$ and $\Delta AON $ are congruent. So, $AN = NT\implies ANYT $ is an isosceles trapezoid. Similarly, $AMXT $ is also an isosceles trapezoid. So, $MX||AT||NY $. Main problem: It is easy to see that $A$, $M$, $Z$, $O$, and $N $ are concyclic. As $MX||AT||NY $, so, $\angle OYT = \angle OXT = 90^{\circ} $. So, $T $, $X $, $Z $, $O $, and $Y $ are also concyclic. Again, $TY = AN $ and $TX = AM $. Also, $\angle BAC = \angle XTY $. So, $\Delta AMN\cong \Delta TXY $. So, $MNYX $ is cyclic (isosceles trapezoid). Hence, by the Radical Axis Theorem, $MN $, $XY$, and $OZ $ are concurrent. This implies the result.

Let $(AMT)$ meet $NX$ again at $X_1$.

, $\angle$ $TAX_1$ = $90^{\circ}$ $\Rightarrow$ $\angle$ $TXN$ = 90$^{\circ}$ Also, $\angle$ $XNA$ = $90^{\circ}$ $\Rightarrow$ $TX$ $\parallel$ $AC$ Now, Let $TX$ $\cap$ $AB$ = $P$ $\Rightarrow$ $\angle TPA$ = $180^{\circ} - \angle BAC$ $\Rightarrow$ $\angle TXM - \angle BMX$ = $180^{\circ} - \angle BAC$ And, $\angle TXM$ = $180^{\circ} - \angle MAT$ = $180^{\circ} - \frac{\angle BAC}{2}$ $\Rightarrow$ $\angle BMX$ = $\frac{\angle BAC}{2}$ $\Rightarrow$ $MX$ $\parallel$ $AT$ Similarly, We get $TY$ $\parallel$ $AB$ and $NY$ $\parallel$ $AT$ $\Rightarrow$ $AMXT$ and $ANYT$ are isosceles cyclic trapeziums. $\Rightarrow$ $MX$, $NY$, $AT$ share the same perpendicular bisector, and $MN$ and $XY$ meet on this line only $\Rightarrow$ $K$ lies on the perpendicular bisector of $AT$ $\Rightarrow$ $KA$ = $KT$. Q.E.D.

Claim: $ANYT$ is an isosceles trapezoid Proof: Let $Y'$ be a point such that $TY' \parallel AB$. Let $O$ be the circumcenter of $\triangle ABC$. Then $\angle OTY' = \angle OTA + \angle ATY' = \angle OAT + \angle TAB = \angle OAT + \angle TAC = \angle OAC$ And $\angle OY'T = \angle ONA = \pi /2$. Since $OT=OA \Longrightarrow \triangle OAN \cong \triangle OTY'$. So $AN = TY' $ but $\angle NAT = \angle Y'TA$ $ANY'T$ is an isosceles trapezoid. So $ANY'T$ cyclic. But $Y' = (ANT) \cap OM$. Hence $Y'=Y$. Similarly $AMXT$ is an isosceles trapezoid. $AT \parallel MX \parallel NY$. Hence $K,O$ lie on the perpendicular bisector of $AT$. Thus $KA=KT$

Solution found with Makorn and dantaxyz. Let $O$ be the circumcenter of $\triangle ABC$. Let $R$ be the midpoint of $BC$. Then $M,O,R,B$ are concyclic since $\angle OMB=\angle ORB=90$, and similarly $N,O,R,C$ are cyclic. Hence, $\angle TOM=180-B$. Since $\angle ANO=\angle AMO=90$, $A,M,O,N$ are concyclic, so $\angle YON=180-\angle MON=A$. Then, \begin{align*} \angle TOX &= 180-\angle TOY-\angle YON \\ &= \angle TOM - \angle YON \\ &= (180-B)-(180-\angle MON) \\ &= (180-B)-A =C. \end{align*}But we also have $\angle AOM=\frac12 \angle AOB=C$. Hence, $\angle AOM=\angle TOX$. Let $D$ be the foot of $O$ to $AT$. Then \[\angle DOX=\angle DOT-\angle TOX, \ \angle DOM = \angle DOA-\angle AOM.\]We know that $OD$ is the perpendicular bisector of $AT$ since $T$ is the midpoint of arc $BC$. So $\angle DOT=\angle DOA$, and hence $\angle DOX=\angle DOM$. Consider the mapping which reflects a point over $OD$. Clearly, $T\to A$. Let $X\to X'$. Then $OD$ is the perpendicular bisector of $AT$ and $XX'$, so $ATXX'$ is an iscoceles trapezoid (with axis of symmetry $OD$). But $\angle DOX=\angle DOM$, and this condition forces $M=X'$. So $OD$ is the perpendicular bisector of $MX,AT$. Similarly, $OD$ is the perpendicular bisector of $AT,NY$. Hence, $MXYN$ is an iscoceles trapezoid and $OD$ is its axis of symmetry. Then $K\in DO$ so $KA=KT$.

Nice and cute problem ig We redefine $X$ as the intersection of the parallel from $N$ to $AT$ from $(ANT)$, similarly redefine $Y$, Let $T_A = (AO) \cap AT $ It is easy to see that $\triangle OXT \equiv \triangle ONA$ CLAIM 1: $(OT_AXYT)$ is cyclic

CLAIM 2: $O-X-M, O-Y-N$

CLAIM 3: $MNXY$ is cyclic

now apply radax $(AMN),(TXY),(MNXY)$ to get the desired conclusion.

Lemma: In triangle $ABC$ let the bisector of $\angle A$ meet $BC$ at $D,$ let $M$ be the midpoint of $AD.$ If $Z$ is on line $CM$ and in $\triangle ABC$ such that $\angle AZB=90^\circ,$ then $AD$ is tangent to $(ACZ).$ Proof: Let $P$ be a point on the circle with diameter $AB$ such that $PB\parallel AD.$ If $PA$ intersects $BC$ at $E$ then we see $(B,C;D,E)=-1,$ and projecting through $P$ gives $(\infty,PC\cap AD,D;A)=-1,$ and as such we have $P,M,C$ collinear. Now we have \[\angle DAC=\angle BAD=\angle ABP=\angle AZP=\angle AZM=180-\angle AZC,\]implying the tangency. Now in our original problem, inversion at $A$ with radius $\sqrt{\frac{AB\cdot AC}2}$ followed by reflection across the bisector of $\angle A$ lets us apply our lemma twice, and inverting back gives $MX\parallel AT\parallel NY.$ Since $MXTA,NYTA$ are cyclic, it follows that $MX,AT,NY$ share a perpendicular bisector. Now $K$ is clearly on this bisector, done.

Note that both $(AMT)$ and $(ANT)$ are symmetric about the perpendicular bisector of $\overline{AT}$, which we will denote by $\ell$, which passes through $O$. Now we claim that $AMXT$ is an isosceles trapezoid. Indeed note that the $\ell$ is also the perpendicular bisector of $\overline{MX}$. To see this first note that if we let $R$ and $S$ be the arc midpoints of $\widehat{AC}$ and $\widehat{AB}$ we have that $\overline{AT} \perp \overline{RS}$ which can be verified easily with complex numbers. Then the reflection of $S$ about $O$ say $S' = \overrightarrow{MO} \cap (ABC)$ satisfies $\overline{S'R} \parallel \overline{AT}$ and in fact $S'RAT$ is a cyclic isosceles trapezoid. However we can also easily note that $\overline{S'R} \parallel \overline{MX}$ due to symmetry. Thus we have $\overline{MX} \parallel \overline{AT}$ and it follows easily that $MXAT$ is an isoscleles trapezoid as claimed. By symmetry so is $NYTA$ and hence so is $NYMX$. However then it is clear that $K$ lies on the perpendicular bisector of $\overline{AT}$ and hence we're done.

I will show that both $AMXT$ and $ANYT$ are isosceles trapezoids, which implies the result by symmetry. Let $C'$ be the point on $(ABC)$ such that $AC'CT$ is an isosceles trapezoid. Then $MT=MC'$; redefining $X$ to be the reflection of $M$ over the perpendicular bisector of $\overline{AT}$, it follows that $XA=XC$, hence $\overline{XN} \perp \overline{AC}$, as needed.

The main claim is that $TY \perp YM.$ To see why, we phantom-point $Y'$ as the intersection of the perpendicular bisector of $AB$ and the line through $T$ parallel to $AB.$ We prove that $ANY'T$ is an isosceles trapezoid, which proves our claim. First, $$\measuredangle NAT = \measuredangle CAT = \measuredangle TAB = \measuredangle ATY.$$Next reflect $T$ across point $Y'$ to get $Z,$ which lies on $(ABC)$ since $Z$ and $T$ are reflections across the perpendicular bisector of $AB.$ Thus $ABTZ$ is an isosceles trapezoid -> $AZ = BT = CT$ -> $AZCT$ is an isosceles trapezoid -> $ZT = AC$ -> $\frac{ZT}{2} = TY = \frac{AC}{2} = AN,$ proving our claim. We similarly see that $\angle TXN = 90^\circ$ and $AM = TX.$ This claim implies $3$ things: 1. $AM = TX$ and $AN = TY.$ 2. $TXOY$ is cyclic, implying that $\measuredangle XTA = \measuredangle XOY = \measuredangle NOM = \measuredangle NAM.$ This implies that $\triangle AMN \cong \triangle TXY,$ so $MN = XY.$ 3. $MX \parallel AT \parallel NY,$ so $MXYN$ is an isosceles trapezoid. Therefore, the perpendicular bisector of $AT$ is just that of $MX.$ Since $MNYX$ is isosceles, $MN \cap XY = K$ lies on this perpendicular bisector, and we are done.

Claim: $\angle TYM = 90^{\circ}$ Let the line parallel to AB trough T be l. Let $S_{AB} \cap l = Y'$. If we prove that $ANY'T$ is an isosceles trapezoid our claim will be shown. Now $\angle NAT = \angle CAT = \angle TAB = \angle ATY$. After this reflect T across point Y' to get P which lies on (ABC) since P and T are reflections across $S_{AB}$ $\Rightarrow$ ABTP is an isosceles trapezoid $\Rightarrow$ AP = BT = CT $\Rightarrow$ APCT is an isosceles trapezoid $\Rightarrow$ PT = AC, $\frac{PT}{2} = TY = \frac{AC}{2} = AN$, which is the end of the proof for our claim. Similarly we see that $\angle TXN = 90^{\circ}$ and AM = TX. From $\angle TYM = \angle TXN = 90^{\circ}$, we have that AM = TX and AN = TY. Also TXOY is cyclic, from which we have that $\angle XTA = \angle XOY = \angle NOM = \angle NAM$. This means $\triangle AMN \cong \triangle TXY$, so MN = XY. From $TY \perp YM$, and $TX \perp XN$, $MX \parallel AT \parallel NY$ $\Rightarrow$ MXYN is an isosceles trapezoid. From this being true, it follows that $S_{AT} \equiv S_{MX}$. Since MNYX is isosceles, $MN \cap XY = K$ and $K \in S_{AT}$, so KA = KT. We are ready.

let $O$ be the circumcenter of $ABC$, and denote the perpendicular bisector of $AT$ as $l$ let $NO$ intersect the circumcircle of $ANT$ at $P$, and let $MO$ intersect the circumcircle of $AMT$ at $Q$ since $T$ is the midpoint of arc $BC$ of $\omega$, $BAT=CAT$, and $NPT=NAT=MAT=MQT$, we can get that $l$ is the angle bisector of $NOY$ then, $NO=YO$, $MO=XO$, $MN=XY$, $MNY=XYN$, and $NK=YK$, so $K$ is on $l$ $T$ is the reflection of $A$ over $l$, so $KA=KT$

Notice the perpendicular bisector of $AT$, say $\ell$, is both the line connecting the centers of $(AMT)$, $(ANT)$ and the axis of symmetry between line $MOY$ and $NOX$. Consequently, $AMXT$ and $ANYT$ are isosceles trapezoids, so segment $MN$ is simply the reflection of segment $XY$ over $\ell$, and thus their intersection $T$ must lie on $\ell$. $\blacksquare$

Yay!! do $\sqrt{\frac{bc}{2}}$ inversion. If $D$ is feet of perpendicular from $A$ to $BC$ and $J=AT \cap MN$. Then $X' = CJ \cap (ADB)$ and $Y' = BJ \cap (ACD)$. now using well known inversion we have $$AJ^2=JX'\cdot JC=JY' \cdot JB$$So $X',Y',B,C$ cyclic. Now back to original problem we have $M,N,X,Y$ cyclic, with $MX \parallel NY$. As $O = MY \cap NX$ we have $J -O - K$. $K$ will lie on perpendicular bisector of $AT$.

Proving $MX \parallel AT \parallel NY$ suffices as this implies that the perpendicular bisector of $AT$ is also the perpendicular bisector of $NY$ and of $MX$ which implies the result. To show that $MX \parallel AT$, let $X'$ denote the second intersection of the parallel from $M$ to $AT$ with $(AMT)$, it suffices to prove that $X'N$ is the perpendicular bisector of $AC$, we have that $MB=AM=TX'$, we also have that $CT=BT$, finally we have $CTX'=\frac{BAC}{2}+ABC$ and we have that $MBT=ABC+\frac{BAC}{2}$ which gives us that $\triangle CX'T \equiv \triangle MBT$ and thus $CX'=MT=AX'$ proving the result. A similar argument proves $NY \parallel AT$.

Define $X'$ as a point on $(AMT)$ such that $MX' \parallel AT$. Claim: $X = X'$ Proof: \[\angle TBM = \angle TBC + \angle B = \frac{\angle A}{2}+\angle B = \angle CTX'\]since $\angle CTX' = \angle ATC + \angle ATX' = \angle B + \frac{\angle A}{2}$. By construction, $MB = MA = X'T$ and $TB = TC$, so $\Delta TBM \cong \Delta CTX'$. Therefore, $CX' = MT = X'A$, so $\Delta AX'C$ is isosceles. Thus, $\overline{X'N} \perp \overline{AC}$, so $X = X'$. $\square$ Due to this, the perpendicular from $O$ to $\overline{AT}$ must also be perpendicular to $\overline{MX}$ and $\overline{NY}$, which finishes as then $\Delta KAT$ is isosceles.

101st post, greetings to my Taiwanese fellas

Let $O$ denote the circumcenter of $\Delta{ABC}$. It is obvious that lines $OM$ and $ON$ are the perpendicular bisectors of $AB$ and $AC$. The key claim is the following: Claim: $\Delta{AMN}$ and $\Delta{TXY}$ are symmetric about line $OK$. Proof. Let $X'$ and $Y'$ be the feet of the perpendiculars from $T$ to lines $ON$ and $OM$ respectively. In triangles $\Delta{AMO}$ and $\Delta{TX'O}$, we have: $OA = OT = R$ $\measuredangle{TX'O} = 90^{\circ} = - \measuredangle{AMO}$ $\measuredangle{X'OT} = \measuredangle{NOL} = \measuredangle{NCL} = \measuredangle{ACB} = - \measuredangle{MOA}$ Hence by the SAA test, $\Delta{AMO} \cong \Delta{TX'O}$. Similarly, $\Delta{ANO} \cong \Delta{TY'O}$. This implies quadrilaterals $\square{AMON}$ and $\square{TX'OY'}$ are congruent. Further, $\square{MX'Y'N}$ is an isosceles trapezoid and line $OK$ is the perpendicular bisector of each of $MX'$, $NY'$, and $AT$. It is not hard to see that $X = X'$ and $Y = Y'$: Since $\square{AMX'T}$ and $\square{ANY'T}$ are both isosceles trapezoids, they must be cyclic. Now, since $T$ is the reflection of $A$ across line $OK$, we obviously must have $KA = KT$.

Phantom the points: It suffice to show that $AMXT, ANYT$ are isosceles trapezoids (we would have $MX, AT, NY$ to have common perpendicular bisector which would pass though $K$). Redefine $X$ on $(AMT)$ such that $MX \parallel AT$. We will prove that: $XA = XC$ (which implies $XN \perp AC$). Reflecting with respect to the perpendicular bisector of $AT$, it suffice to show that $MT=MC'$ where $C' = YT \cap (ABC)$ (with $C' \neq T$) which is true as: $AB \parallel C'T$. Hence we are done,

Claim: We have that $AMXT$ and $ANYT$ are isosceles trapezoids. Proof: Let $X'$ be the point on $(AMT)$ such that $AMX'T$ is an isosceles trapezoid. It suffices to show that $X'$ lies on the perpendicular bisector of $\overline{AC}$. From $\angle X'TA = \angle MAT = \angle TAC$, we have $\overline{X'T} \parallel \overline{AC}$. So, if $N'$ is the foot from $X'$ to $\overline{AC}$ and $R$ is the foot from $T$ to $\overline{AC}$, we have \[AN' = AR - N'R = AR - AM = AN,\]since $AR$ is the average of $AB$ and $AC$ by, say, Simson line. The claim is hence proven. The claim thus follows from symmetry, as lines $MN$ and $XY$ are symmetric about the perpendicular bisector of $\overline{AT}$.