First, angle chasing gives $\angle BDX=\angle CDY,$ so DDIT at point $D$ on complete quadrilateral $XSYTBC$ gives that the reflection $P$ of $S$ over $BC$ lies on line $DT.$ Next, second isogonality lemma gives $AS,AT$ are isogonal, and first isogonality lemma gives that the reflection $Q$ of $P$ over the midpoint of $BC$ lies on the isogonal to $AS,$ which is $AT.$ Next, angle chasing gives $\angle TBC=\angle TCB.$ Since $BCPQ$ is an isosceles trapezoid, we have $TPQ$ is isosceles, so the bisector of $\angle PTQ$ is parallel to $BC$ and thus reflection of $PT$ back over $BC$ finishes.

Unfortunately I trig bashed this. It’s not very hard to carry out, though

Invert three times and then it’s a trivial bary

Let $M$ be the midpoint of $BC$, and $S', T'$ the reflections of $S, T$ over $M$. Claim: $S'$ lies on $AT$ Proof: Since $\triangle ABX \overset{-}{\sim} \triangle AYC$, reflection along the angle bisector of $\angle ABC$ is an involution on the pencil of lines through $A$ with pairs $(AB, AC)$, $(AX, AY)$ and $(A\infty_{BS}, A\infty_{CS})$. By DDIT from $A$ onto complete quadrilateral $XYSTBC$, $AS$ is sent to $AT$. By DDIT from $A$ onto $BC\infty_{BX}\infty_{CY}SS'$, $(AS, AS')$ is also an involutive pair, so $AS'\equiv AT$. $\blacksquare$ Claim: $T'$ lies on $DS$ Proof: Using the circles $(ACDX)$, $(ABDY)$ we obtain $$\angle BDT=\angle DAY=\angle DAC-\angle YAC = \angle BAD-\angle BAX=\angle XAD=\angle TCB,$$so $T$ lies on the perpendicular bisector of $BC$. Furthermore $$\angle CDY=\angle BAY=\angle XAC=\angle XDB,$$so by DDIT from $D$ onto $XSYTBC$, $\angle CDT=\angle SDB$. The conclusion follows. $\blacksquare$ Finally, since $STS'T'$ is a parallelogram (and $T'\neq S$) it follows that $DST'\parallel ATS'$.

Ooh center spam? We first note that it's not too hard to show that there exists unique $X,Y$ satisfying all conditions. Now, the key claim is as follows: Claim. Circles $\odot(AXB)$ and $\odot(AYC)$ are both tangent to $AD$. Proof. Redefine $X$ as the intersection of $\odot(ADC)$ and the circle $\omega_B$ passing through $A$ and $B$ and tangent to $AD$. Redefine $Y$ (and $\omega_C$) similarly. We claim that $\triangle ABX\sim\triangle ACY$. To prove this, we use the circumcenter spam technique. Let $O_B$ and $O_C$ be the circumcenter of $\triangle ADB$ and $\triangle ADC$. Let $S_B$ and $S_C$ be the circumcenter of $\omega_B$ and $\omega_C$. Note that $O_BS_B$ and $O_CS_C$ meet at $O$, the circumcenter of $\triangle ABC$. Moreover, since $O_BO_C\parallel S_BS_C\perp AD$, it follows that $O_BO_CS_CS_B$ is an isosceles trapezoid. Thus, $\angle S_CO_BO_C = \angle S_BO_CO_B$, and each side is equal to $\angle DAX$ and $\angle DAY$, respectively, so $\angle BAX = \angle CAY$. $\blacksquare$ Back to the main problem. By the claim, let $\angle DAX = \angle DAY = \angle ABX = \angle ACY = \theta$. Then, \begin{align*} \angle XSY &= \angle A + 2\theta \\ \angle XDY &= \angle BDY + \angle CDX - 180^\circ = 180^\circ - (\angle BAY + \angle CAX) \\ &= 180^\circ - (\angle A + 2\theta), \end{align*}so $XSYD$ is cyclic. Moreover, $$\angle XTY = 180^\circ - \angle YBC - \angle XCB = 180^\circ - 2\theta = 180^\circ - \angle XAY,$$so $AXTY$ is also cyclic. Finally, \begin{align*} \measuredangle(AT,AX) &= \measuredangle TYX = \measuredangle BYD + \measuredangle DYX \\ &= \measuredangle BAD + \measuredangle DSX = \measuredangle BXA + \measuredangle DSX \\ &= \measuredangle(DS,AX). \end{align*}

First, note that $\angle{}BAX=\angle{}CAY$, so by the second isogonality lemma in $\triangle{}ABC$ we see that $\overline{AS}$ and $\overline{AT}$ are isogonal in $\angle{}CAB$. Then, note that $$\angle{}BDX=\angle{}CAX=\angle{}BAY=\angle{}CDY,$$so by the second isogonality lemma in $\triangle{}DBC$ we see that $\angle{}BDS=\angle{}CDT$. Also, we have that $$\angle{}TBC=\angle{}DAY=\angle{}DAX=\angle{}TCB,$$so $T$ lies on the perpendicular bisector of $\overline{BC}$. Let $M$ be the midpoint of $\overline{BC}$, and construct $D’$ and $S’$ as the reflections of $D$ and $S$ over $M$, respectively. Since $\angle{}ABS=\angle{}ACS$, we see that $S’$ lies on $\overline{AT}$ by the first isogonality lemma. Then, since $$\angle{}S’D’C=\angle{}SDB=\angle{}TDC=\angle{}TD’C,$$we see that $D’$ lies on $\overline{TS’}$. Since $SDS’D’$ is a parallelogram, this implies that $\overline{DS}\parallel\overline{AT}$.

Define $B'$ and $C'$ as the reflections of $B$ and $C$ about the perpendicular bisector of $AD$. Redefine $X$ as the second intersection of $ADC$ with $BC'$ and $Y$ as the second intersection of $ADB$ with $CB'$. We show that these points satisfy the problem. $$\angle AXB=180^{\circ}-\angle AXC'=180^{\circ}-\angle DAC=180^{\circ}-\angle DAB=180^{\circ}-\angle AYB'=\angle AYC$$$$\angle XBA=\angle C'BB'-\angle ABB'=\angle B'CC'-\angle ACC'=\angle YBA$$Thus $S$ is the intersection of $BC'$ and $CB'$ so $SA=SD$. By the Isogonal Lines Lemma $AS$ and $AT$ are isogonal, finishing the question.

Very nice problem from Michael Ren!. Found the finish after english class lol Claim 1: $TB=TC$ Proof: $\angle TBC=\angle DAY=\angle DAX=\angle TCB$ Claim 2: Given $BSCS'$ parallelogram, then $A,T,S'$ are colinear. Proof: By first isogonality lemma (or DDIT, can be proved with Little cotangent Lemma spam) we get $AS,AT$ isogonal in $\angle BAC$, but by second isogonallity lemma (or DDIT again, but can be proved by LoS only) we get $AS,AS'$ isogonal in $\angle BAC$ proving the claim as we get that $A,T,S'$ lie on a line isogonal to $AS$ The finish: Now let $BTCT'$ a rhombus, again by first isogonallity lemma (or DDIT) on $\angle BDC$ we get $S,D,T'$ colinear since $\angle YDC=\angle BAY=\angle CAX=\angle XDB$ so we get $DX,DY$ isogonal. And since $TST'S'$(T lol) is a parallelogram we get $AT \parallel SD$ as desired thus we are done .

It suffices to show that $\angle ADS=\angle DAT.$ Note that $\angle DAT=\angle SAD$ by the 2nd Isogonality Lemma, so it suffices to show that $\angle ADS=\angle SAD,$ or $SA=SD.$ Note that $$\angle XDB=180^\circ-\angle CDX=\angle XAC=\angle BAY=180^\circ-\angle YDB=\angle CDY=\theta.$$Also note that $$\frac{BX}{BD}=\frac{BX}{BA}\frac{BA}{BD}=\frac{CY}{CA}\frac{CA}{CD}=\frac{CY}{CD}.$$Thus $\triangle BXD$ and $\triangle CYD$ are a counterexample to SSA similarity. This implies that \begin{align*} &\angle DBX+\angle YCD=2(90^\circ-\theta)\\ &\implies (\angle B-\angle XBA)+(\angle C-\angle ACY)=2(90^\circ-\theta)\\ &\implies \angle XBA=\angle ACY = \frac{1}{2}(\angle B+\angle C-2(90^\circ-\theta))=\theta-\frac{1}{2}\angle A. \end{align*}We also have $$\angle XAD=\angle BAD-\angle BAX=\frac{1}{2}\angle A-(\angle A-\theta)=\theta-\frac{1}{2}\angle A.$$Thus $\overline{AD}$ is tangent to $(ABX),$ and similarly it is tangent to $(ACY).$ Now do a $\sqrt{bc}$ inversion at $A$ and then reflect across $\overline{AD}.$ Then $B$ and $C$ swap, and $\overline{BC}$ and $(ABC)$ swap $D^*$ is the arc midpoint of $BC$ $Y^*$ is the point on $\overline{CD^*}$ such that $\overline{BY^*}\parallel\overline{AD^*}$ $X^*$ is the point on $\overline{BD^*}$ such that $\overline{CX^*}\parallel\overline{AD^*}$ Perform a negative homothety at $D^*$ by a factor of $\tfrac{AC}{AB}.$ Let $A'$ be the image of $A.$ We have $A'X^*=\tfrac{AC}{AB}(AB)=AC.$ Hence $ACX^*A'$ is an isosceles trapezoid. This implies that the homothety mapped the center of $(ABY^*)$ to the center of $(ACX^*),$ so $D^*$ is collinear with these centers. It follows that $D'*$ is equidistant from the intersection points of these circles, which are $A$ and $S^*$. So $AD^*=D^*S^*,$ implying that $SA=SAD,$ as desired.

trig bashing works. just invert

First note that $AX$ and $AY$ are isogonal wrt $\angle A$ since $\angle BAX = \angle CAY$. Then DDIT on quadrilateral $BXYC$ from point $A$ gives that $(AB, AC)$, $(AX, AY)$, and $(AS, AT)$ are involutions, so $AS$ and $AT$ are isogonal. Then note that $\measuredangle{BDX} = \measuredangle{CDX} = \measuredangle{XAC}$ and similarly $\measuredangle{CDY} = \measuredangle{YAB}$, so $BX$ and $BY$ are isogonal wrt $\angle BDC$. So then DDIT on $BXYC$ from point $D$ gives that $DT$ and $DS$ are isogonal similarly. Net let $R$ be the point so that $BSCR$ is a parallelogram. We can show that $R - T - A$ or that $AR$ and $AS$ are isogonal. Then since $\angle ABS = \angle ACS$ by the first isogonality lemma we have that $AR$ and $AS$ are isogonal as desired. Then we have $\angle DAY = \angle YBD$ and $\angle DAY = \angle DAX = \angle DCX$ so $T$ lies on the perpendicular bisector of $BC$. Let $P$ be the point so that $BTCP$ is a rhombus. Notice that $\angle BDP = \angle BDT = \angle CDS$(by the isogonality) so $\angle BDS + \angle BDP = 180^\circ$ meaning $P - D - S$. Then since $TSRP$ is a parallelogram(reflection of $TS$ over the midpoint of $BC$) we have $TP \parallel SR \implies AT \parallel SD$ as desired.

We define a transformation $\mathcal{P}$ that is reflection across $AD$ followed by a homothety so that $\mathcal{P}(B)=C$. We can do this because $AD$ is the angle bisector. Let $\mathcal{P}(C)=B_1$ and $\mathcal{P}(D)=D_1$. Since $\triangle ABX \sim \triangle ACY$, $\mathcal{P}(X)=Y$, so $Y$ lies on $\mathcal{P}((ACD))=(AB_1D_1)$. Therefore, $Y$ is the Miquel Point of $BDD_1B_1$. Claim 1: $\textcolor{red}{S}$ and $\textcolor{red}{T}$ are isogonal. Without loss of generality, let $AB<AC$ so $AB_1>AC$, and we have \[\angle BYC=180^\circ-\angle BB_1C=180^\circ-\angle C\]Similarly, $\angle BXC=180^\circ-\angle B$. This implies that $\angle BXC+\angle BYC=180^\circ+\angle A$, together with the fact that $AX$ and $AY$ are isogonal implies that $X$ and $Y$ are isogonal. This clearly implies $S$ and $T$ isogonal. $\blacksquare$ Claim 2: $\textcolor{red}{SA=SD}$ Let $CS$ intersect $(ABD)$ at $B_2$ and $BS$ intersect $(ACD)$ at $C_2$. We have \[\angle ADB_2=\angle AYB_2=180^\circ-\angle AYC=-180^\circ+\angle AYB+\angle BYC=\angle ADB-\angle C=\frac12 \angle A=\angle BAD\]so $AB_2BD$ is an isosceles trapezoid. Similarly, $AC_2CD$ is an isosceles trapezoid. Therefore, $BB_2C_2C$ is an isosceles trapezoid. $S$, being the intersection of the diagonals of said trapezoid, must be on the common perpendicular bisector of the bases of the trapezoids, in particular $AD$. $\blacksquare$. Now we finish, noting that $AS$ and $AT$ are isogonal, so $\angle TAD=\angle SAD=\angle SDA$ and we are done.

Let rays $BX$ and $CY$ meet $(ADC)$ and $(ADB)$ at $C'$ and $B'$ respectively. We will show that $BB' \parallel AD \parallel CC'$. Suppose that this is not the case. Since \[ \angle BB'C = \angle BB'Y = \angle BAY = \angle XAC = \angle XC'C = \angle BC'C, \]we see that $BB'C'C$ is cyclic. Therefore, if the above three lines are not parallel, then they are concurrent at a point, say $P$. Now we have \[ \angle PBA = \angle B'BA = \angle B'YA = \angle AXC' = \angle ACC' = \angle ACP, \]and since $AD$ is the bisector of $\angle A$, we must have $\triangle PAB \cong \triangle PAC$. This gives $AB = AC$, which is impossible since we assumed $\triangle ABC$ to be scalene. Hence $BB' \parallel AD \parallel CC'$, so $B'$ and $C'$ are reflections of $B$ and $C$ over the perpendicular bisector of segment $AD$. In particular, $S$ lies on the perpendicular bisector of $AD$, so $AS = SD$. Now by DDIT on point $A$ and quadrilateral $SXDY$, we know that $AS$ and $AT$ are isogonal in $\angle A$. Therefore, $\angle TAD = \angle DAS = \angle SDA$ and so $AT \parallel DS$.

First we claim that given points $A$, $B$, and $C$, points $X$ and $Y$ are uniquely determined.

Then we claim we can construct $X$ as the intersection (other than $A$) of $(ACD)$ and the circle through $B$ tangent to $AD$ at $A$, and $Y$ as the intersection (other than $A$) of $(ABD)$ and the circle through $B$ tangent to $AD$ at $A$.

Now we claim $(AXTY)$ is cyclic since \[ \angle XTY = \angle BTC = 180^\circ - \angle YBD - \angle XCD = 180^\circ - \angle YAD - \angle XAD = 180^\circ - \angle XAY,\]and $(SXYD)$ is cyclic since \[\angle SXD + \angle SYD = \angle AXD + \angle AYD - \angle AXS - \angle AYS = 360^\circ - \angle ACD - \angle ABD - \angle AXS - \angle AYS = 360^\circ - \angle ACD - \angle ABD - \angle BAD - \angle CAD = 180^\circ.\] To finish, let $DS \cap XY = P$ and $AT \cap XY = Q$. It suffices to show $\angle SPX = \angle AQX$. We have \[ \angle SPX = 180^\circ - \angle XSD - \angle SXY = 180^\circ - \angle XYD - \angle SXY = 180^\circ - \angle XYT - \angle BYD - \angle SXY, \]and \[ \angle AQX = 180^\circ - \angle XAT - \angle AXY = 180^\circ - \angle XYT - \angle AXS - \angle SXY. \]It now suffices to show $\angle AXS = \angle BYD$, which is true since $\angle AXS = 180^\circ - \angle AXB = \angle BAD = \angle BYD$.

According to the problem condition, $\measuredangle BAX = \measuredangle YAC$ and $\measuredangle BDX = \measuredangle YDC$. This means that both $A$ and $D$ lie on the QL-Cu1 (the locus of all points which exist an isogonal conjugate) of quadriilateral $\mathcal{Q} = \{BX, XC, CY, YB\}$. In this case, one would be tempted to ask if $A$ and $D$ are, in fact, isogonal conjugates. Surprisingly, this condition (namely $\measuredangle BSA = \measuredangle DSC$) translates directly to $DS \parallel AT$ because $AS$ and $AT$ are isogonal in $\angle BAC$ by DDIT. Hence, we are left to prove that $X, Y$ are isogonal conjugates with respect to $\triangle ABC$. To prove this, we first prove that $X$ and $Y$ are uniquely determined given $\triangle ABC$. After this, we construct $X$ and $Y$ which satisfy the problem condition, then prove that they are in fact isogonal. For the first part, proofs have already been presented above, like by SHZhang (#15). Then, we define $X$ as the point such that $(AXDC)$ is concyclic and $(AXB)$ is tangent to $AD$, and $Y$ similarly. We use $\sqrt{bc}$ inversion. $X^* = C\infty_{AD} \cap BD^*$ and $Y^* = B\infty_{AD} \cap CD^*$. Using DDIT with $\mathcal{Q}^* = \{BX^*, X^*C, CY^*, Y^*B\}$ with respect to $A$, we get that $AX^*$ and $AY^*$ are isogonal in $\angle BAC$, proving that $\measuredangle BAX = \measuredangle YAC$. This proves that $X$ and $Y$ satisfy the similarity condition, and also that they are isogonal conjugates with respect to $\triangle ABC$. Hence, we are done with the proof.

This problem was proposed by me. I discovered the configuration by degenerating the quadrilateral in IMO 2018 P6 into $ABDC$, but I don't know if this was noticed by anyone taking the contest.

No DDIT because DDIT is skull.

After proving that $AD$ is tangent to $(ABX),(ACY)$ we can continue like this: $\angle ACY = \angle ABX = \angle XAD = \angle XCD = \angle BCX$. Analogously, $\angle ABX = \angle CBY$ and so $X,Y$ are isogonal conjugates in $ABC$. Let $K=CY \cap AX, M=DX \cap BY$. From the cyclic quadrilaterals we have that: $\angle AYT = \angle AYB = \angle ADB = 180 - \angle ADC = 180 - \angle AXC = 180 - \angle AXT => AXTY$ is cyclic. Using that $X,Y$ are isogonal conjugates in $ABC$ we get that: $\angle XDY = \angle XDA + \angle YDA = \angle XCA + \angle YBA = \angle YCB + \angle XBC = \angle SCB + \angle SBC =180 - \angle BSC = 180 - \angle XSY => SXDY$ is cyclic. From the above we conclude that: $\angle MYK = 180 - \angle BYC = \angle YBC + \angle YCB = \angle XBA + \angle YCB = \angle YCA + \angle YCB =\angle ACB = \angle ACD = 180 - \angle AXD = 180 - \angle KXM => KXMY$ is cyclic. From the cyclic quadrilaterals we get that: $\angle XKM = \angle XYM = \angle XYT = \angle XAT = \angle KAT => KM \parallel AT$ and $\angle XDS = \angle XYS = \angle XYK = \angle XMK => KM \parallel DS$. Hence, $DS \parallel AT$, as needed.

isogonality Note that $AX$ and $AY$ are isogonal in $\angle BAC$, so by the second isogonality lemma, $AS$ and $AT$ are isogonal in $\angle BAC$. Also as $\angle TBC = \angle YAD = \angle XAD = \angle TCB$ we have $TB = TC$. Furthermore, as $\angle AXT + \angle AYT = \angle ADC + \angle ADB = 180^\circ$, $AXTY$ is cyclic, so if $E = AT \cap BC$ then \[ \frac{BE}{EC} = \frac{\sin \angle BTE}{\sin \angle CTE} = \frac{\sin \angle ATY}{\sin \angle ATX} = \frac{AY}{AX} = \frac{AC}{AB} \]so $BD = EC$. Now since $\angle ABS = \angle ACS$, by the first isogonality lemma there is a point $S'$ on $ATE$ such that $BSCS'$ is a parallelogram, whence by symmetry we have $DS \parallel ES'$ which finishes.

By Isogonality Lemma $S$ and $T$ are isogonal conjugates. $DS \parallel AT \iff SA = SD$. Define $L = CY \cap (ABD)$, $K = BX \cap (ADC)$. $\angle LBK = \angle LBA + \angle ABX = \angle LYA + \angle ACY = \angle AXK + \angle ACY = \angle KCA + \angle ACY = \angle KCL \implies KLBC$ is cyclic. By Radical Axis Theorem, either $BL, AD, KC $ are concurrent or parallel. If they intersect at $P$, $\angle APB = \angle LBA - \angle BAD = \angle KCA - \angle DAC = \angle APC$. Combining this with $\angle BAD = \angle CAD \implies PBA \cong PCA \implies AB = AC $ contradiction. Thus, $BL, AD, KC$ are parallel. It can be deduced that $ADBL$ and $ADCK$ are isosceles trapezoids and collinearity of $B,D,C$ implies collinearity of $K,L,A$. This clearly shows $S$ is on the perpendicular bisector of $AD$, as desired.

I will present my hybrid solution (last part by Baricentric Coordinates)

Initially, observe that triangles $AXB$ and $AYC$ must be oppositively similar as points $X$ and $Y$ are inside $ \triangle ABC$. We start by doing a very bold Claim which might seem a little bit unintuitive; so bellow I will let some possible motivation to conjecture Claim 1.

Claim $1$: Circles $(ABX) $ and $(ABY)$ are tangent to $AD$. In order to prove Claim $1$, we first show the following Lemma: Points $X$ and $Y$ are uniquely defined.

We now are able to prove Claim 1. Proof of Claim 1: Let $X'$ be the interscetion of $(ACD) $ and the circle passing through $A$ and $B$ that is tangent to $AD$, and define $Y'$ similarly. Now by the Lemma, in order to prove Claim 1, we only need to show that $ \triangle ABX'$ and $ \triangle ACY'$ are oppositively similar. But note that by definition \[\measuredangle AX'B = \measuredangle (AD, AB) = -\measuredangle (AD,AC) = - \measuredangle AY'C \]and therefore it suffies to show that $\measuredangle BAX' = \measuredangle Y'AC$, which is equivalent to showing that $AX'$ and $AY'$ are isogonal with respect to $ \angle BAC$. Now let $U$ and $V$ be the images of $X'$ and $Y'$ with respect to the $-\sqrt{bc}$ inversion; we show $U $ and $V$ are isogonal conjugates. Indeed, let $M$ be arc midpoint of $BC$ not containing $A$, then $D$ is mapped to $M$ in the $-\sqrt{bc}$ inversion, and therefore we have that : $(ACD)$ is mapped to line $BM$; Circle through $A,B$ tangent to $AD$ is mapped to line $l_C$, the line through $C$ parallel to $AD$; and therefore $U = BM \cap l_C$; similarly, if we define $l_B$ in a similar fashion, then $V = CM \cap l_B$. Now a simple angle chasing shows that $BM,l_B$ are isogonal with respect to $\angle BAC$; and similarly for $CM$ and $l_C$; therefore $U$ and $V$ indeed isogonal conjugates and we are done $\square$. We now finish by Bary bashing. Note that as $AX$ and $AY$ are isogonal conjugates, it follows by the First Isogonality Lemma that $AS$ and $AT$ are isogonal too. So easy angle chasing shows that $AT$ is parllel to $DS$ if and only if $AS = DS$. So it suffies to show that $BX$, $CY$ and the perpendicular bisector of $AD$ are concurrent. So we perform a $-\sqrt{bc}$ inversion; note that it maps $BX$ to $(ACU)$; $CY$ to $(ABV)$; the perpendicular bisector of $AD$ to $\omega$, the circle with center at $M$ passing through $A.$ So it shuffies to show that $(ACU)$, $(ABV)$ and $\omega$ are coaxial. Now we are going to compute the equation of these circles in Barycentric coordinates, and show thta they are coaxial via their equations. We first compute $U$ and $V$. Recall from the proof of Claim 1 that $ U = BM \cap l_C$; we compute $M$, as $M$ is the arc midpoint of arc $BC$(not containing $A$) we must have that if $M = (x_M,y_M,z_M)$: $M$ is on the $A$ internal angle bisector, so $ y_M : z_M = b : c$ $M$ is on $(ABC)$ and so $ a^2y_Mz_M + b^2x_Mz_M + c^2x_My_M = 0$ These two relations easily yelds $M = ( -a^2 : b(b+c) :c(b+c))$. We now compute $U = (x_U,y_U,z_u)$; we have: $U \in BM$ and so $x_U : z_U = -a^2 : c(b+c)$; $U \in l_C$ so, by definition $CU \parallel AD$, then as the point of infinity of $AD$ has coordinates $(-b-c :b :c)$ it follows that $x_U:y_U = -b -c : b$ and therefore $ U = ( -a^2 (b+c) : a^2 b : c {(b+c)}^2)$. We now compute equation of circle $ACU$; it is well known the equation must have the form \[- a^2yz - b^2xz - c^2xy +(x+y+z)(u_1x +v_1y +w_1z) \]for some constants $u_1,v_1,w_1$; as $A$ and $C$ are in this circle, we must have $u_1 = w_1 = 0$. Now plugin $U$ into the above we get \[ a^2bc(b+c) ( -a^2(b+c) + b(b+c)^2 + a^2c) + a^2b v_1 ( -a ^2(b+c) + a^2b + c(b+c)^2) =0\iff \]\[ c(b+c) b ( {(b+c)}^2 - a^2) + v_1 c ( {(b+c)}^2 - a^2) \iff v_1 = -b(b+c)~(*) \]similarly we conclude that $(ABV)$ has equation \[- a^2yz - b^2xz - c^2xy -c(b+c)z(x+y+z) \]we now only need to compute the equation of $\omega $. We now that $\omega $ has equation \[- a^2yz - b^2xz - c^2xy +(x+y+z)(ux +vy +wz) \]for some constants $u,v,w$. As $A$ is on the circle $u = 0$. Now let $ N = (x, y,z)$ be the reflection of $A$ over $M$(which lies on $\omega$ by definition); then (dividing the above equation by ${(x+y+z)}^2$) \[ \frac{-a ^2yz -b^xz - c^2xz }{ {(x+y+z)}^2} + \frac{vy +wz}{x+y+z} =0~(**) \]now the trick is to notice that the first fraction equals to the power of $N$ with respect to $(ABC)$; but this should be equal to $AN \cdot NM = 2 AM^2$ by power of point. Now one can compute the displacement vector \[\vec{AM} = \left( \frac{- (b+c)^2}{(b+c)^2 - a^2} , \frac{b(b+c)}{(b+c)^2 - a^2} , \frac{c(b+c)}{(b+c)^2 - a^2} \right) \]And then we can easily compute \[ AN ^2 = 2AM^2 = 2\frac{bc(b+c)^2}{(b+c)^2 -a^2}\]therefore, by $(**)$ we have that $v,w$ must satisfy \[ vy + wx =- 2(x+y+z) \cdot \frac{bc(b+c)^2}{(b+c)^2 -a^2}\]but we can easily compute \[(x:y:z) = \vec N = 2\vec M - \vec A = 2( -a ^2: b(b+c) : c (b+c) ) - (1,0,0) = ( -a^2 -(b+c)^2 : 2b(b+c) : 2c(b+c) ) \]and putting in the above equation yelds \[ vb + wc = bc(b+c) ~(***)\]finally we know that the center of $\omega$ lies on $AI$ and therefore the external angle bisector of $ \angle BAC$, which have equation $cy + bz =0$, must be tangent to $\omega$. Now we know that the circle centered at $A$ with radius $0$ has equation \[ -a^2yz -b^2xz -c^2xy +(x+y+z)(c^2y + b^2z) = 0 \]and therefore \[ v -c^2 : w -b^2 = c : b \]because the left hand side represents the equation of the radical axis of $\omega$ and the radius $0$ circle centered at $A$. Now from the above equation and $(***) $ it follows that $v = w = -bc$. So now we computed all the relevant circles; namely $(ACU), (ABV) $ and $\omega$ have equations \[ -a^2yz -b^2xz -c^2xy + (x+y+z)(u_1x + v_1y + w_1z) = 0 \]\[ -a^2yz -b^2xz -c^2xy + (x+y+z)(u_2x + v_2y + w_2z) = 0 \]\[ -a^2yz -b^2xz -c^2xy + (x+y+z)(u_3x + v_3y + w_3z) = 0 \]where $p_1:=(u_1,v_1,w_1) = ( 0,-b(b+c), 0), p_2 = (u_2,v_2,w_2) = ( 0,0 ,-c(b+c))$ and $p_3 = (u_3,v_3,w_3) = (0,-bc, -bc)$. And since the equation of the radical axis of two circles is obtained by subtracting there equations, then we only need to show that $ p_1 -p_3 $ and $p_2 -p_3$ are linearly dependent, i.e., $(0,-b^2, bc)$ and $(0, bc, -c^2)$ which is clearly true and we are finally done.

Note that $AX$ and $AY$ are isogonal, so by the Second Isogonality Lemma, $AS$ and $AT$ are isogonal. Also, note that \[\angle TBC=\angle YBD=\angle YAD=\angle XAD=\angle XCD=\angle TCB,\]so $TB=TC$. Now let $T'$ be the isogonal conjugate of $S$. Then \[\angle T'BC=\angle SBA=\angle SCA=\angle T'CB,\]so $T'$ lies on the perpendicular bisector of $BC$ too. But since $T'$ also lies on the isogonal of $AS$ with respect to $\angle BAC$, it is exactly $T$. Now note that if we reflect $(ABD)$ about $AC$ then send it through the spiral similarity centered at $A$ sending $C$ to $B$, $Y$ will get sent to $X$. Since $A$ is still on this new circle, it has one other intersection with $(ACD)$, which will be $X$. Therefore, $X$ is unique, so it suffices to demonstrate one configuration that works for each $\triangle ABC$ (since that will be the one in the problem). Indeed, we claim that \begin{align*} T &= (a^2:(b+c)c:(b+c)b) \\ S &= ((b+c)bc:b^3:c^3) \\ Y &= (a^2(b+c)c:a^2b^2:(b+c)^2bc). \end{align*}We may check that $S$ and $T$ are isogonal conjugates, $BTY$ collinear, $CSY$ collinear, and $T$ lies on the perpendicular bisector of $BC$ (this has equation $0=a^2(z-y)+(c^2-b^2)x$, which this satisfies since $0=a^2(b+c)(b-c)+(c-b)(b+c)a^2$). Let $S'$ be the point such that $BSCS'$ is a parallelogram. Then $ATS'$ are collinear by the First Isogonality Lemma. But note that if $E$ is the point on $BC$ such that $BD=EC$, then $SD\parallel S'E$ (say by parallelogram lemma), so $SD\parallel AT$ is equivalent to showing that $ATE$ is collinear. But that's obvious since $E=(0:c:b)$. Since $AX$ and $AY$ will be isogonal by the Second Isogonality Lemma and \[\angle XBA=\angle SBA=\angle TBC=\angle TCB=\angle SCA=\angle YCA,\]we have the similarity in the original problem. Thus if we can show that $Y$ lies on $(ABD)$, we will similarly get that $X$ lies on $(ACD)$, and that will demonstrate that this is indeed the configuration in the problem, which will complete the problem since we already showed $AT\parallel SD$ in this configuration. So let's compute $(ABD)$. \[a^2yz+b^2xz+c^2xy+(ux+vy+wz)(x+y+z)=0\]$A$ gives $u=0$, and $B$ gives $v=0$. $D=(0:b:c)$ gives \[a^2bc+wc(b+c)=0\Longrightarrow w=-\frac{a^2b}{b+c}.\]Recall that $Y=(a^2(b+c)c:a^2b^2:(b+c)^2bc)$. Thus we want: \[a^4b^3c(b+c)^2+a^2b^3c^2(b+c)^3+a^4b^2c^3(b+c)= \frac{a^2b}{b+c}(b+c)^2bc(a^2bc+a^2c^2+a^2b^2+(b+c)^2bc)\]\[a^2b^2c(b+c)\Big(a^2b(b+c)+bc(b+c)^2+a^2c^2\Big)= a^2b^2c(b+c)\Big(a^2bc+a^2c^2+a^2b^2+(b+c)^2bc\Big)\]which is true. $\blacksquare$