By alternate segment theorem we have that: $$ 180^{\circ}=\angle YCD+\angle CDY+\angle DYC=\angle YCB+\angle DXY+\angle BXD=\angle YCB+\angle BXY $$so $BXYC$ is cyclic. Thus, applying power of a point at $K$: $$ KT^2=KB \cdot KC=KX \cdot KY=KD^2 \Longrightarrow KT=KD $$Now let $AI$ meet $\Gamma$ at $M$, the midpoint of $\widehat{BC}$ not containing $A$. Let the tangent at $M$ meet $KT$ at $Q$. Observe that $QM \parallel KD$ so $\angle TKD=\angle TQM$ and also $KT=KD$, $QT=QM$ hence $\triangle TKD \sim \triangle TQM$. As $T,K,Q$ are collinear, this means that $T,D,M$ are collinear so $TD$ and $AI$ meet at $M$ which lies on $\Gamma$.

Sketch of a solution. Let $S$ be the reflection of $D$ to $K$. $BX\cap (I)=P,CY\cap(I)=Q$. We have $DP=DQ$. \[\angle DCY=180-\angle CYD-\angle YDC=\angle IDY-\angle CYD+90=\angle PDY\]\[\angle DYP=\angle DXB=\angle CYD\implies \angle YXD=\angle YPD=180-\angle DXB-\angle DCY\]Thus $BXCY$ is cyclic. \[KS^2=KD^2=KX.KY=KB.KC=KT^2\]$\implies (S,D;B,C)=-1\implies S,E,F$ are collinear. $(S,D;B,C)=-1$ and $\angle DTS=90\implies TD$ is the angle bisector of $\angle CTB$. $\implies AI$ and $TD$ intersect on $\Omega$ as desired.$\blacksquare$

same proof as #2 up to $KT = KD$, then use $KD^2 = KB \cdot KC$ and ratio lemma STP $$\frac{BK}{KC} = \left(\frac{BD}{DC}\right)^2$$to finish

neat

bruh this is the same as post 3 oops

Note $KD^2 = KX.KY$ hence we have $\measuredangle KYD = \measuredangle XDK \implies \measuredangle XYC = \measuredangle CBX$. which give us $B,C,X,Y$ cyclic. $$KD^2=KX.KY=KB.KC=KT^2$$ As $KT$ is tangent we get, $$\measuredangle KTD = \measuredangle BTD + \measuredangle KTB = \measuredangle BTD + \measuredangle TCD = \measuredangle TDB$$which implies $\measuredangle BTD = \measuredangle DTC$. Hence $TD$ pass pass through midpoint of arc $BC$.

We can use the method of false position. Let $T'$ be the second intersection of $MD$ with $\Omega$, and $K'$ be the intersection of the tangent at $T'$ with $BC$. Then, let a line through $K'$ intersect the two arcs at $X'$ and $Y'$. We wish to prove $\angle BX'D=\angle DY'C$. This is equivalent to proving $\angle BX'Y'=\angle DY'C+\angle CDY'$, since $\angle CDY'=\angle DX'Y'$. Hence, it is enough to prove $\angle K'X'B=\angle K'CY'$, or that $BX'Y'C$ is a cyclic quadrilateral. So, we wish to prove $K'X'\cdot K''Y'=KB\cdot K'C$. However, this is equivalent to $K'T'=K'D$. Note that $\angle K'T'D=\angle T'CM=\angle T'CD+\angle BCM=\angle T'CD+\angle CT'M=\angle T'DK'$, so we are done.

Note that \[ \angle BXY = \angle BXD + \angle DXY = \angle DYC + \angle CDY = 180^\circ - \angle BCY, \]so $BCXY$ is cyclic. Therefore, we have $KD^2 = KX \cdot KY = KB \cdot KC = KT^2$, and so $KD = KT$. Therefore, \[ \angle BTD = \angle KTD - \angle KTB = \angle KDT - \angle TCD = \angle DTC, \]so $TD$ is the angle bisector of $\angle BTC$.

Let lines $(DX)$ and $(DY)$ cut the incircle again at points $X'$ and $Y'$. The angle condition then gives $DX'=DY'$, or in other words $(X'Y')\parallel (BC)$. Therefore, we have that points $B,C,Y$ and $X$ are concyclic by Reim's theorem. Looking at powers from $K$ then gives $KD^2=KB\times KC=KT^2$. We thus consider the circle $\omega$ tangent to $(KT)$ and $(KD)$ at points $T$ and $D$. This circle is tangent to $\Omega$ at $T$ and tangent to $(BC)$ at $D$, so shooting lemma finishes. $\square$

Claim: $BXYC$ is cyclic Proof: We have $\angle BXY+\angle BCY=\angle BXD+\angle DXY+\angle DCY=\angle DYC+\angle YDC+\angle DCY=180^{\circ}$ $\blacksquare$ Thus $KD^2=KX\cdot KY=KB\cdot KC$ so $K$ is the center appolonius of points $S$ such that $\frac{SB}{SC}=\frac{DB}{DC}$. Since $KT^2=KB\cdot KC=KD^2$ we have $KT=KD$ so $T$ is on this circle. But then $\frac{TB}{TC}=\frac{BD}{CD}=\frac{BF}{CE}$ so $TBF\sim TCE$ meaning $T\in(AFIE)\cap (ABC)$ so $T$ is the sharky devil point. We are done.

Let $T'$ be the second intersection of $\Delta AEF$ and $\Delta ABC$. We claim that $T'=T$. Indeed, as $\measuredangle BXY = \measuredangle BXD + \measuredangle DXY = \measuredangle DYC + \measuredangle CDY = \measuredangle DCY = \measuredangle BCY$ we get that $BXYC$ is cyclic. Then $K$ is the radical center of $(AEF), \Omega$ and $(BXYC)$, so $K$ lies on $AT'$, hence $T'=T$. But now, $T$ is the $A$-sharkydevil point, so the desired conclusion is well-known (a quick proof is that since $T$ is the Miquel point of $BFEC$, we have $TB/TC=FB/EC=DB/DC$, so $TD$ bisects $\angle BTC$ and the claim follows).

Yay I actually have an original solution for once. Why is EGMO geo so easy. Invert about a circle centred at $D$ with radius $\sqrt{DB\cdot DC}$, followed by a flip about $D$. The incircle is sent to a parallel line to $BC$, and $\Omega$ is fixed by Power of a Point. Now the angle condition is reformulated as $\angle DCX'=\angle DBY'$, but since $X',Y'$ lie on a parallel line to $BC$, $Y'BCX'$ is an isosceles trapezium. Now as the center of $(Y'DX')$ lies on the perpendicular bisector of $X'Y'$ as it is a chord, by symmetry $K'=(Y'DX')\cap BC$ is the reflection of $D$ about the midpoint of $BC$. In fact, $(DK'T')$ being tangent to $\Omega$, implies $T'$ is the arc midpoint of $BC$ as the tangency point is fixed about a reflection across the perpendicular bisector of $BC$ (Else we would have two intersection points of $(DK'T')$ with $\Omega$), and $D,K'$ swap under this reflection. In fact as $\Omega$ is fixed $T'=TD\cap \Omega$, but as it is the arc midpoint by the Incenter-Excenter Lemma $A,I,T'$ are collinear and we are done,

Let $\vartheta=\angle BXD=\angle DYC$, $\omega=\angle KYD=\angle XDB$, and $k=\angle YKC=\angle XKD$. We first note that quadrilateral $BXYC$ is cyclic, since $\angle KBX=\angle BXD+\angle BDX=\angle XYC$. So, also let $\varphi=\angle KXB=\angle YCD$. Since $KD$ is tangent to the incircle of $\triangle ABC$ and $KT$ is tangent to $\Omega$ at $T$, from the power of point theorem we get \[ KD^2=KX\cdot KY=KB\cdot KC=KT^2, \]and so $KD=KT$. From triangles $\triangle KXD$ and $\triangle KYC$ we have \[ \frac{KX}{XD}=\frac{\sin\omega}{\sin k}. \quad \text{and}\quad \frac{KY}{YC}=\frac{\sin \varphi}{\sin k}. \]By the ratio lemma, we have \[ \frac{KB}{BD}=\frac{KX}{XD}\cdot \frac{\sin \varphi}{\sin \theta}=\frac{\sin\omega}{\sin k}\cdot \frac{\sin \varphi}{\sin \theta} \]and \[ \frac{KD}{DC}=\frac{KY}{YC}\cdot \frac{\sin \omega}{\sin \theta}=\frac{\sin \varphi}{\sin k}\cdot \frac{\sin \omega}{\sin \theta} \]Hence $\frac{KD}{DC}=\frac{KB}{BD}$, and so, \[ \frac{KD}{KB}=\frac{DC}{BD}. \] Also, triangles $\triangle KTB$ and $\triangle KCT$ are similar, since $\angle KTB= \angle BCK$, so \[ \frac{TC}{TB}=\frac{KT}{KB}=\frac{KD}{KB}=\frac{DC}{BD}. \]Therefore, $TD$ bisects angle $\angle BTC$, and so $TD$ and $AI$, as the angle bisector of $\angle BAC$, meet at the midpoint $M$ of arc $BC$.

Let $K'$ be the exsimilicenter of $(BXD)$ and $(DYC)$ which, by the angle condition, is also the exsimilicenter of $BD$ and $DC$. Consider the inversion centered at $K'$ sending $(BXD)$ to $(DYC)$, which has radius $K'D$. Then $X$ is sent to a point $Y'$. Notice $K'X\cdot K'Y'=K'D^2$, thus $(DXY')$ is actually the incircle and thus $Y\equiv Y'$. Hence $K$ is the exsimilicenter of $BD$ and $DC$. Now let $M=AI\cap \Omega$, which is clearly the arc midpoint of $\widehat{BC}$. Redefine $T=MD\cap \Omega$ and redefine $K=TT\cap BC$. We show that $K$ is the exsimilicenter of $BD$ and $DC$, or more precisely, that \[KD^2=KB\cdot KC=KT^2\implies KD=KT.\] Luckily this is easy: let $Z=MM\cap TT$. Then $\triangle ZMT$ is isosceles with vertex $Z$ and thus $\triangle KDT$ is isosceles with vertex $K$ by homothety. $\blacksquare$ hopefully not a fakesolve since i used phantom points twice

By the Ping-Pong Lemma on line $B\infty_{BC}CK$, it suffices to verify this for one choice of $X$ and $Y$. Let this be the point where $BX$ and $CY$ concur at $D'=2I-D$. Then, $(K,D;B,C)\stackrel{D'}=(K,DD'\cap XY;X,Y)\stackrel D=(D,D';X,Y)\stackrel{D'}=(D,\infty_{BC};B,C)$. If $M$ is the midpoint of arc $BC$ and $MD$ intersects $\Omega$ at $S$, then $$(D,\infty_{BC};B,C)\stackrel M=(S,M;B,C)\stackrel S=(SS\cap BC,D;B,C),$$so $KS$ is tangent to $\Omega$.

Here is my solution Let the incircle be $\omega$ , $AI \cap \Omega = \{Z\}$ So we need to show that points $T,D,Z$-are collinear $\textbf{Claim:}$ Points $B,X,Y,C$-are concyclic $\textbf{Proof:}$ From the triangle $\triangle CDY$ we have: $\angle CYD+\angle CDY+\angle YCD=180$ Combining with $\angle CYD=\angle BXD$ and $\angle CDY=\angle DXY$ because $CD$-is tangent to the incircle We get that: $\angle BXD+\angle DXY+YCB=180 \implies \angle BXY+\angle YCB=180$ which means that Points $B,X,Y,C$-are concyclic. Let $\odot(BXYC)=\Gamma$ $\textbf{Claim:}$ $KT=KD$ $\textbf{Proof:}$ By Power of the Point theorem we get: $KT^2=Pow(K,\Omega)=KB*KC=Pow(K,\Gamma)=KX*KY=Pow(K,\omega)=KD^2 \implies KT^2=KD^2 \implies KT=KD$ $\textbf{Claim:}$ Points $T,D,Z$-are collinear $\textbf{Proof:}$ $180-\angle CDT=\angle KDT \stackrel{KT=KD }{=} \angle KTD= \angle KTB + \angle BTD\stackrel{\Omega}{=} \angle TCB+\angle BTD \implies$ $180-\angle CDT=\angle TCB+\angle BTD \implies 180-\angle CDT-\angle TCB=\angle BTD \implies \angle CTD=\angle BTD$ $\angle CTD+\angle BTD=\angle BTC \stackrel{\Omega}{=}\angle BAC \equiv \angle A \implies \angle CTD=\angle BTD=\frac{\angle A}{2}$ Let $TD \cap \Omega=\{Z'\}$ So $\angle BTZ'=\frac{\angle A}{2}=\angle BAZ , \angle CTZ'=\frac{\angle A}{2}=\angle CAZ$ So $Z=Z' \implies$ Points $T,D,Z$-are collinear $\blacksquare$ This is my 50th post

First, I claim $BXYC$ is cyclic. Indeed, if $BX$ and $CY$ intersect the incircle again at $X',Y'$ then \[\measuredangle BXY=\measuredangle X'XY=\measuredangle X'Y'Y=\measuredangle BCY'=\measuredangle BCY\]as the angle condition gives $DX'=DY'$ or $X'Y'\parallel BC$. Thus $K$ is the point on $BC$ with $KD^2=KB\cdot KC$. Now, let $S$ be the $A$-sharkydevil point, the miquel point of $BFEC$. Lemma: $S,D,M$ are collinear. Proof. Note that $(BIC)$ and $(AI)$ are tangent, and $(a$ inverts to $BC\cap AI$. Thus $S$ inverts to the foot from $I$ to $BC$ as $(ABC)$ and $BC$ swap so $S$ and $D$ are swapped. $\blacksquare$ Now, let $T=EF\cap BC$ be with $(BC;TK)=-1$. AS $KD^2=KB\cdot KC$ we get $KD=KT$. But the above lemma showed that $SD$ bisects $\angle BSD$ hence $\angle TSD=90^{\circ}$ so $KS=KD$. Thus $S$ is this desired tangency point which finishes. Remark: Disappointing how $X,Y$ have nothing to do with the problem and $K$ is fixed. Also a very unoriginal idea.

Here's another attempt... Not a full solution but an outline for another Introduce midpoints of $\overline{DE}$ and $\overline{DF}$ as $P$ and $Q$. Imitate the angle chase as in above solution to show that $BCXY$ is cyclic. By an inversion around the incircle, we trivially get $PQXY$ is also cyclic. Now, the pairwise radical axis of $\odot(DEF)$, $\odot(PQXY)$ and $\odot(BXCY)$ will concur precisely at $K$. From here, it should be not be difficult to show that $T$ is precisely the Sharky-Devil Point.

OK PoP and angel chase! Some angle equalities first: By condition of tangency, $\angle YDC=\angle DXY$. Thus, $$\angle BXY = \angle BXD+\angle DXY = \angle DYC + \angle YDC = 180^o-\angle YCD = 180^o-\angle YCB \implies \text{BXYC is cyclic}$$By PoP, $KT^2=KB.KC$, $KD^2=KX.KY$ and $KB.KC=KX.KY$. Thus $KD=KT$. Now suppose $TD \cap \Omega \equiv M$. Then $$180^o- \angle BDM = \angle TDK = \angle KTM = \angle TCM = 180^o-\angle TBM \implies \angle TBM = \angle BDM$$Thus $BDM \sim TBM \implies \frac{DM}{BM} = \frac{BM}{TM} \implies MD.MT=MB^2$ and by Shooting Lemma, $M$ is the midpoint of arc $BC$ not containing $A$. $AI$ also passes through $M$ and we are done.

DottedCaculator wrote: By the Ping-Pong Lemma on line $B\infty_{BC}CK$, it suffices to verify this for one choice of $X$ and $Y$. Let this be the point where $BX$ and $CY$ concur at $D'=2I-D$. Then, $(K,D;B,C)\stackrel{D'}=(K,DD'\cap XY;X,Y)\stackrel D=(D,D';X,Y)\stackrel{D'}=(D,\infty_{BC};B,C)$. If $M$ is the midpoint of arc $BC$ and $MD$ intersects $\Omega$ at $S$, then $$(D,\infty_{BC};B,C)\stackrel M=(S,M;B,C)\stackrel S=(SS\cap BC,D;B,C),$$so $KS$ is tangent to $\Omega$. What is Ping-Pong Lemma?

Note $\angle{XDB}=\angle{XYD}; \angle{XBD}+\angle{XYC}=\angle{XBD}+\angle{XYD}+\angle{DYC}=\angle{XBD}+\angle{BXD}+\angle{BDX}=180^{\circ}$ Thus, we have $B,X,Y,C$ are concyclic. Moreover, we have $KT^2=KB\cdot KC=KX\cdot KY=KD^2, KT=KD, \angle{KTD}=\angle{KDT}$ Let $TD$ meet $(ABC)$ at $M$, $\angle{KTD}=\angle{KTB}+\angle{BTD}=\angle{TMB}+\angle{BTD}; \angle{TDB}=\angle{MBD}+\angle{TMB}, \angle{MBD}=\angle{BTM}\implies \widehat{BM}=\widehat{CM}$ Thus, $M$ is the midpoint of the minor arc $BC$. By Fact5, $AI$ meets $(ABC)$ at the midpoint of the minor arc $BC=M$. We are done.

The angle condition implies $BCXY$ cyclic as, \begin{align*} \angle KBX = \angle BXD + \angle XDK = \angle BYD + \angle KYD = \angle KYC \end{align*}Now note that $KT^2 = KB \cdot KC = KX \cdot KY = KD^2$ and hence $KD = KT$. However then, \begin{align*} \angle BTD = \angle KTD - \angle KTB = \angle KDT - \angle TCB = 180 - \angle TDC - \angle TCB = \angle DTC \end{align*}and so $\overline{TD}$ is an angle bisector and we're done.

After finding $BCXY$ cyclic with angle chasing, $KD² = KX * KY = KB * KC = KT²$ so $KD = KT$ Now, consider a circle that is tangent to $KT$ at $T$ and $KD$ at D (since $KD = KT$, we can say that such circle exists). Let $O_2$ be its center. Note that $OO_2$ is perpendicular to $KT$, and $IO_2$ is perpendicular to $KD$ because $T$ and $D$ are their common touch point, respectively. It's well known that $AI \cap (ABC)$ at point M, and $ID$ is parallel to $OM$ (here $M$ is the midpoint of minor arc $BC$). Let $TD$ intersect (ABC) again at $M'$. Since $\angle O_2DT = \angle O_2TD = \angle OTM'$, we get that $\triangle O_2TD$ $\sim$ $ \triangle OTM'$, So $ MO // ID // IO_2 // M'O$, and since both $M$ and $M'$ are on $(ABC)$, $M = M'$

$$\angle BXY=\angle BXD+\angle DXY=\angle DYC+\angle CDY=\angle DCY=\angle BCY$$so $BCXY$ cyclic (directed angles). Now $KD^2=KX\cdot KY=KB\cdot KC=KT^2$ and since $KT$ is the tangent, $TD$ is the (internal) angle bisector, so we are done.

Let $M = EF \cap BC$ and $K_1$ be midpoint of segment $MD$ Obviously, $(M,D;B,C) = -1$ , thus $K_1^2 = K_1B \cdot K_1C$ Let the circle centered at $K_1$ that passed through $D$ be $\omega$. Consider an inversion wrt $\omega$, we get that the inversion of $B$ is $C$, so $\omega$ and $\Omega$ are orthogonal circles. Since, $ID \perp K_1D$, $\omega$ and the incircle of $ABC$ are orthogonal as well. Consider an inversion of $X$ wrt $\omega$, call the point we got $X_1$. Since $\omega$ is orthogonal to the incircle, $X_1$ is on the incircle and we get that $$\angle{BXD} = \angle{DX_1C} = \angle{DYC}$$So, $X_1 = Y$, which implies $K = K_1$ And since $K$ is centre of $\omega$, $T$ is the intersection of $\omega$ and $\Omega$. Note that $\angle{MTD} = 90^\circ$($MD$ is diameter of $\omega$) and $(M,D;B,C) = -1$, so $TD$ bisect $\angle{BTC}$, which follows $TD$ pass through midarc of minor arc $\widehat{BC}$. Since, $AI$ bisect $\angle{BAC}$, we get that $TD, AI$ intersect on the midarc of minor arc$\widehat{BC}$, as desired.

Solution from Twitch Solves ISL: We redefine the points in the reverse order: let $M$ be the arc midpoint of minor arc $BC$, let $T$ be the second intersection of ray $TD$ with $\Omega$, and let $K$ be the point on line $BC$ such that $\overline{KT}$ is tangent to $\Omega$. Then we will show that if a line through $K$ meets the incircle again at $X$ and $Y$, then $\angle BXD = \angle DYC$. A phantom point argument will then show the originas result. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-3.15888,1.43930); pair B = (-4.5,-2.5); pair C = (2.5,-2.5); pair I = (-2.36683,-0.97282); pair D = (-2.36683,-2.5); pair E = (-1.49432,0.28055); pair F = (-3.81252,-0.48065); pair M = (-1.,-5.13543); pair T = (-3.97684,0.60428); pair K = (-6.16456,-2.5); pair X = (-3.60416,-1.86796); pair Y = (-0.85509,-1.18934); import graph; size(12cm); pen zzttqq = rgb(0.6,0.2,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen aqaqaq = rgb(0.62745,0.62745,0.62745); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw(circle((-1.,-1.49361), 3.64181), linewidth(0.6) + yqqqqq); draw(circle(I, 1.52717), linewidth(0.6) + yqqqqq); draw(A--M, linewidth(0.6) + yqqqqq); draw(K--B, linewidth(0.6)); draw(A--D, linewidth(0.6)); draw(T--M, linewidth(0.6) + yqqqqq); draw(T--K, linewidth(0.6) + qqwuqq); draw(B--X, linewidth(0.6) + gray); draw(X--D, linewidth(0.6) + gray); draw(D--Y, linewidth(0.6) + gray); draw(C--Y, linewidth(0.6) + gray); draw(K--Y, linewidth(0.6) + aqaqaq); dot("$A$", A, dir(130)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$I$", I, dir((3.539, 8.205))); dot("$D$", D, dir(225)); dot("$E$", E, dir((4.017, 7.482))); dot("$F$", F, dir((-23.362, 0.859))); dot("$M$", M, dir((-6.535, -33.120))); dot("$T$", T, dir((-20.887, 9.005))); dot("$K$", K, dir((-34.397, -11.544))); dot("$X$", X, dir((-42.204, 3.012))); dot("$Y$", Y, dir((8.881, -1.047))); [/asy][/asy] Claim: $KT = KD$. Proof. Note $\angle KTD = \frac{1}{2} \widehat{TM} = \frac{1}{2}(\widehat{TB} + \widehat{MC}) = \angle TDK$. $\blacksquare$ Hence, it follows that \[ KX \cdot KY = KD^2 = KT^2 = KB \cdot KC \]so quadrilateral $BXYC$ is cyclic. From this we deduce the inverse similarities \begin{align*} \triangle KXD &\overset{-}{\sim} \triangle KDY \\ \triangle KXB &\overset{-}{\sim} \triangle KCY. \end{align*}Hence, \[ \angle BXD = \angle XKD - \angle KXB = \angle YDK - \angle YCK = \angle DYC \]as needed.

Let $w$ be the incircle and let $S \equiv BX \cap w$. We have: $\angle DYS=\angle DYC$, $\angle DSY=\angle CDY$ which implies $\angle YCD=\angle YDS=\angle YXS$ $\implies$ $BXYC$ is cyclic. Then: $\frac{KB}{KC}=(\frac{BX}{CX}).(\frac{BY}{CY})=(\frac{BD}{DC}.\frac{sin\angle BXD}{sin \angle DXC}).(\frac{BD}{DC}.\frac{sin\angle BYD}{sin\angle DYC})=(\frac{BD}{DC})^2$. Then $(\frac{TB}{TC})^2=\frac{KB}{KC}$ thus we are done by ratio lemma.

Note that $$\measuredangle BXY = \measuredangle BXD + \measuredangle DXY = \measuredangle DYC + \measuredangle CDY = \measuredangle YCB,$$so $BXYC$ is cyclic. Then $K$ lies on the radical axis of the circumcircle and the incircle, so $KT = KD$, and $$\measuredangle BTD = \measuredangle KTD + \measuredangle BTK = \measuredangle TDC + \measuredangle DCT = \measuredangle DTC.$$Therefore, $TD$ bisects $\angle BTC$, so $TD$ and $AI$ meet at the midpoint of arc $\widehat{BC}$ on $\Omega$.

im like 20% confident that this finish works Claim: $BXYC$ is cyclic. Proof. Just angle chasing - we have \[ \measuredangle XBC =\measuredangle XBD = \measuredangle XDB + \measuredangle BXD = \measuredangle XYD + \measuredangle DYC = \measuredangle XYC \]which is enough. $\square$ Hence $KD^2 = KX \cdot KY = KB \cdot KC = KT^2$ so $KD = KT$. To finish, define $T'$ to be the point such that there exists a circle $\Gamma$ internally tangent to $\Omega$ at $T'$ and tangent to $\overline{BC}$ at $D$ (such a circle clearly exists and is unique by Shooting Lemma), and define $K' = \overline{T'T'} \cap \overline{BC}$. Then PoP gives $K'D^2 = K'T'^2 = K'B \cdot K'C$, so $K' = K$ and $T' = T$, and Shooting Lemma finishes.

See that it suffices to prove that $TD$ bisects $\angle BTC$. Note by angle chase that $BXYC$ is cyclic so $KX.KY=KB.KC=KT^2=KD^2$ So $\angle KTD=\angle KDT$ And by alternate segment theorem, $\angle KTB=\angle TCD$ So exterior angle property is now enough to imply that $\angle BTD=\angle DTC$.

Solved with @Jakkjdm Let $P$ be the $A$-Sharky-Devil point and $Q$ in $BC$ such that $QP$ is tangent to $\Omega$. Since It´s well-know that $PD$ and $AI$ concurr in the midpoint $R$ of the arc $BC$, it´s sufficient to prove that, if $M$ and $N$ are points in the incircle collinear with $Q$ in that order, will need to have $\angle BMD=\angle DNC$. A simple angle chasing gives us $\angle BMD=\angle DNC \iff BCNM$ is cyclic $\iff QM\cdot QN=QB\cdot QC=QD^2=QP^2 \iff QD=QP \iff \angle QPD=\angle QDP$ But, we know that $\angle QPD=\widehat{PB}+ \widehat{BR}=\widehat{PB}+ \widehat{CR}=\angle DPC+ \angle DCP=\angle PDQ$, so we´re done. $\blacksquare$

Easy angle chasing and KD²=KX×KY implies that BXYC is cyclic.And we get KD²=KT²=KX×KY=KB×KC.This implies that KD=KT.We note that <KTB=<KCT=α.And <CTD=β.And we get <TDB=α+β,<KTD=<KDT=α+β.This implies that <BTD=<DTC=α.We get TD is angle bisector of <BTC.Then TD and AI intersect on (ABC) because this point is midpoint of arc BC.We are done.

Underwhelming. Note that $BXYC$ is cyclic because $\angle BXY = \angle BXD + \angle DXY = \angle BXD + \angle CDY = 180 - \angle BCY$. Now Power of Point on $K$ wrt to the circumcircle and then the incircle gives us that $KT = KD$. Now it suffices to show that $\angle BTD = \angle DTC$ but this is trivial because $\angle BTD = \angle KTD - \angle BTK = \angle KDT - \angle TCD = \angle DTC$