[asy][asy] size(10cm); pair T = (0,0); pair A = 3*dir(15); pair B = 4*dir(-15); pair E = 3*dir(-15); pair F = 4*dir(15); pair G = incenter(T,A,E); pair H = incenter(T,B,F); pair P = intersectionpoints(circumcircle(A,E,G),circumcircle(B,F,H))[0]; pair Q = intersectionpoints(circumcircle(A,E,G),circumcircle(B,F,H))[1]; pair X = 0.6*P + 0.4*Q; pair C = 2*foot(circumcenter(A,E,G),A,X)-A; pair D = 2*foot(circumcenter(B,F,H),B,X)-B; pair Y = extension(A,D,B,C); dot('$T$', T, dir(180)); dot('$A$', A, dir(120)); dot('$B$', B, dir(240)); dot('$X$', X, dir(225)); dot('$C$', C, dir(C)); dot('$D$', D, dir(D)); dot('$Y$', Y, dir(Y)); draw(circumcircle(A,E,G),deepgreen); draw(circumcircle(B,F,H),deepgreen); draw(T--B,deepgreen); draw(T--F,deepgreen); draw(A--Y,mediumred); draw(B--Y,mediumred); draw(A--C,mediumred); draw(B--D,mediumred); draw(circumcircle(A,B,C),lightred); draw(1.2*P-0.2*Q--2.3*Q-1.3*P,heavygreen); [/asy][/asy] Here is a solution using mOvInG pOiNtS. Let $\ell$ be the radical axis of $\Gamma_1$ and $\Gamma_2$, so $X\in \ell$ by the Radical Axis Theorem. We restate the problem as follows: Problem'. Let $\Gamma_1$ and $\Gamma_2$ be circles common with common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Let $\ell_1$ touch $\Gamma_1$ at $A$, $\ell_2$ touch $\Gamma_2$ at $B$, and the radical axis of $\Gamma_1$ and $\Gamma_2$ be $\ell$. Choose $X\in l$, and let $\overline{AX}$ and $\overline{BX}$ meet $\Gamma_1$ and $\Gamma_2$ again at $C$ and $D$, respectively. Define $Y=\overline{AD}\cap\overline{BC}$. Show that $T$, $X$, $Y$ are collinear. Fix $\Gamma_1$ and $\Gamma_2$ (and hence $\ell, T, A, B$) and animate $X$ linearly on $\ell$. Then $C$ moves projectively on $\Gamma_1$ (it is the image of the perspectivity through $A$ from $\ell$ to $\Gamma_1$) and thus has degree $2$, and similarly for $D$. $\overline{AD}$ has degree at most $0+2=2$, and similarly for $\overline{BC}$. $Y=\overline{AD}\cap\overline{BC}$ has degree at most $2+2=4.$ The collinearity of $T,X,Y$ has degree at most $0+1+4=5.$ Thus it suffices to verify the problem for six different choices of $X$. We choose: $\ell\cap \ell_1$: here $Y$ approaches $A$ as $X$ approaches $\ell\cap \ell_1$. $\ell\cap\ell_2$: here $Y$ approaches $B$ as $X$ approaches $\ell\cap \ell_2$. $\ell\cap \overline{AB}$: here $Y$ approaches $\ell\cap \overline{AB}$ as $X$ approaches $\ell\cap \overline{AB}$. the point at infinity along $\ell$: here $Y=T$. the two intersections of $\Gamma_1$ and $\Gamma_2$: here $Y=X$. (The final two cases may be chosen because we know that there exists a choice of $A,B,C,D$ for which $ABCD$ is convex; this forces $\Gamma_1$ and $\Gamma_2$ to intersect.)

Consider the following generalization of the problem: Generalization wrote: Let $ABCD$ be a cyclic quadrilateral, $X=AC\cap BD$, and $Y=AD\cap BC$. Let $T$ be a point on line $XY$, $\Gamma_1$ be the circle through $A$ and $C$ tangent to $TA$, and $\Gamma_2$ be the circle through $B$ and $D$ tangent to $TB$. Then $\Gamma_1$ and $\Gamma_2$ are viewed at equal angles from $T$. To solve the original problem from here we can specialize to the case where $\Gamma_1$ is tangent to $TB$ as well, and use a phantom point argument. This formulation is asking for length computations so that is what we will do. We have \[\frac{[TAC]}{[TBD]}=\frac{[YAC]}{[YBD]}=\frac{YA\cdot YC}{YB\cdot YD}=\frac{AC^2}{BD^2},\]so \[\frac{AC^2}{TA\cdot AC\sin\angle TAC}=\frac{BD^2}{TB\cdot BD\sin\angle TBD}.\]The radii of $\Gamma_1$ and $\Gamma_2$ are $r_1=\frac{AC}{2\sin\angle TAC}$ and $r_2=\frac{BD}{2\sin\angle TBD}$. Hence \[\frac{r_1}{TA}=\frac{r_2}{TB},\]which is exactly what we want to prove. [asy][asy] size(8cm); pair A = dir(170); pair B = dir(5); pair C = dir(100); pair D = dir(45); pair X = extension(A, C, B, D); pair Y = extension(A, D, B, C); pair T = X+2.2*(Y-X); fill(circumcircle(A, C, C+2*A-2*foot(C, A, T)), rgb(66, 160, 255)+white); fill(T--A--2*foot(A, circumcenter(A, C, C+2*A-2*foot(C, A, T)), T)-A--cycle, rgb(66, 160, 255)+white); fill(circumcircle(B, D, D+2*B-2*foot(D, B, T)), rgb(66, 160, 255)+white); fill(T--B--2*foot(B, circumcenter(B, D, D+2*B-2*foot(D, B, T)), T)-B--cycle, rgb(66, 160, 255)+white); draw(A--T--B); draw(unitcircle); draw(circumcircle(A, C, C+2*A-2*foot(C, A, T)), blue); draw(circumcircle(B, D, D+2*B-2*foot(D, B, T)), blue); draw(A--X--B); draw(A--D); draw(B--C); draw(X--T, dotted); draw(2*foot(B, circumcenter(B, D, D+2*B-2*foot(D, B, T)), T)-B--T--2*foot(A, circumcenter(A, C, C+2*A-2*foot(C, A, T)), T)-A, blue); string[] names = {"$A$", "$B$", "$C$", "$D$", "$T$", "$X$", "$Y$"}; pair[] pts = {A, B, C, D, T, X, Y}; pair[] labels = {dir(210), dir(-45), dir(80), D, dir(-90), X, Y}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy]

See https://artofproblemsolving.com/community/c6h1952595p13480666 for moving points exposition!

First solution, by inversion (Brandon Wang). Let $\ell_1$ and $\ell_2$ intersect $\Omega$ again at $E$ and $F$ respectively. [asy][asy] size(9cm); defaultpen(fontsize(9pt)); pair T,A,B,O1,O2,P,Q,X,C,D,EE,F,Y,Z; T=(0,0); A=dir(40); B=0.4/A; draw(A--T--B); O1=2*circumcenter(T,A,1/A); O2=2*circumcenter(T,B,reflect(T,O1)*B); P=intersectionpoint(circle(O1,length(O1-A)),circle(O2,length(O2-B))); Q=reflect(T,O1)*P; real x=0.55; X=x*P+(1-x)*Q; C=2*foot(O1,X,A)-A; D=2*foot(O2,X,B)-B; EE=2*foot(circumcenter(A,B,C),T,A)-A; F=2*foot(circumcenter(A,B,C),T,B)-B; Y=extension(B,C,A,D); Z=extension(A,B,C,D); draw(circumcircle(A,B,C),gray); draw(C--A--Y--B--D,gray+dashed); draw(A--Z--D,gray); draw(circle(O1,length(O1-A))); draw(circle(O2,length(O2-B))); draw(A--T--(1/A)); draw(EE--T--F); draw(T--Y); draw(Z--F); real y=1.1; clip( (100,y)--(-100,y)--(-100,-100)--(100,-100)--cycle); dot("$T$",T,N); dot("$A$",A,dir(150)); dot("$B$",B,SW); dot("$P$",P,dir(-75)); dot("$Q$",Q,dir(70)); dot("$X$",X,N); dot("$C$",C,dir(-20)); dot("$D$",D,dir(20)); dot("$E$",EE,SW); dot("$F$",F,W); dot("$Y$",Y,E); dot("$Z$",Z,S); [/asy][/asy] The key claim is this: Claim. $\overline{AB}$, $\overline{CD}$, $\overline{EF}$ concur. Proof. Invert at $A$, using $\bullet'$ to denote the inverse, to obtain the following picture. [asy][asy] size(5cm); defaultpen(fontsize(9pt)); pair O,B,U,T,A,C,P,D,Q; O=(0,0); B=dir(10); U=1.8*B; A=intersectionpoint( (0,1)--(20,1),circle(U,length(B-U))); T=2*foot(U,(0,1),(20,1))-A; C=U+length(B-U)*dir(270)-(0.4,0); P=extension(A,T,B,C); D=2*foot(O,B,C)-B; Q=2*foot(U,B,C)-B; real x=1.8, y=1.3; draw(circle(O,1)); draw(circle(U,length(B-U))); draw(P--C); draw(foot(P+(-x,0),C,C+(1,0))--(C+(y,0))); draw(foot(C+(y,0),A,T)--(P+(-x,0))); label("$\ell_1'$",P+(-x,0),N); label("$\Gamma_1'$",foot(P+(-x,0),C,C+(1,0)),N); label("$\Gamma_2'$",dir(160),dir(160)); label("$\ell_2'$",U+(length(B-U),0),E); dot("$B'$",B,E); dot("$T'$",T,N); dot("$A$",A,N); dot("$C'$",C,S); dot("$E'$",P,N); dot("$D'$",D,SW); dot("$F'$",Q,NE); [/asy][/asy] The homothety at $B'$ sending $\Gamma_2'$ to $\ell_2'$ sends $\ell_1'$ to $\Gamma_1'$, so \[\frac{B'C'}{B'F'}=\frac{B'E'}{B'D'}\implies B'C'\cdot B'D'=B'E'\cdot B'F',\]whence $B'$ lies on the radical axis of $(AC'D')$ and $(AE'F')$. Inverting back gives the desired conclusion. $\blacksquare$ Let $Z=\overline{AB}\cap\overline{CD}$, and let $\ell$ be the polar of $Z$ with respect to $\Omega$. By Brokard's theorem on $ABCD$, $\ell=\overline{XY}$, but by Brokard's theorem on $ABEF$, $\ell=\overline{TX}$. Thus $T$, $X$, $Y$ are collinear, as desired. Second solution, by moving points. Since $ABCD$ is convex, $\Gamma_1$ and $\Gamma_2$ intersect at two points $P$ and $Q$; else, the radical axis intersects all four segments $AB$, $BC$, $CD$, $DA$, which is absurd. By radical axis theorem, $X$ lies on $\ell:=\overline{PQ}$. Animate $X$ on $\ell$. We will show that $\overline{AD}$, $\overline{BC}$, $\overline{TX}$ concur at a point $Y$. Then $C$ and $D$ move projectively on their respective circles, so $\overline{AC}$, $\overline{BD}$ each have degree $2$ and $\overline{TX}$ has degree $1$. The concurrence has degree $5$, so we need to verify the hypothesis for $6$ values of $X$. Take $X$ at infinity along $\ell$. Then $C$ and $D$ are the reflections of $A$ and $B$ in the line through the centers of $\Gamma_1$ and $\Gamma_2$, so $Y=T$. Take $X=\ell\cap\overline{AB}$. Then $A$, $B$, $C$, $D$ collinear, so the result is clear. Take $X=\ell\cap\overline{AT}$. Then $C=A$, so $Y=A$, which lies on $\overline{TX}$. The case $X=\ell\cap\overline{BT}$ follows analogously. Take $X=P$. Then, $C=D=P$, so $Y=P$, from which the conclusion is clear. The case $X=Q$ follows analogously. This completes the proof. Remark. I have not checked the details, but Edward Wan claims that we can instead move the center $O$ of $\Omega$ and show that $-1=T(AB;XZ)$, where $Z=\overline{AB}\cap\overline{CD}$. It can be shown that $O\to X$ and $O\to Z$ are projective, so this reduces the problem to three cases.

Excellent problem! a1267ab wrote: Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex. Suppose lines $AC$ and $BD$ meet at point $X$, while lines $AD$ and $BC$ meet at point $Y$. Show that $T$, $X$, $Y$ are collinear. Merlijn Staps Note that $X$ lies on the radical axis of $\Gamma_1, \Gamma_2$, which is the line through the midpoint of $\overline{AB}$, parallel to the external bisector of angle $ATB$. We prove the following equivalent result. (Phantom Version). In triangle $TAB$, a point $S$ lies on $\overline{AB}$ and points $E, F$ lie on $\overline{TA}, \overline{TB}$ respectively, such that $A,E,F,B$ lie on a circle $\gamma$, and $S, E, F$ are collinear. Point $X$ lies on the line $\ell$ passing through the midpoint $M$ of $\overline{AB}$, and parallel to the external bisector of angle $ATB$ such that $\overline{XT}, \overline{AF}, \overline{BE}$ concur. Lines $\overline{XA}$ and $\overline{XB}$ meet $\gamma$ again at $C$ and $D$ respectively. Then $S$ lies on $\overline{CD}$; $\overline{AD}, \overline{BC}, \overline{XT}$ concur; a circle $\Gamma_1$ passes through $A$ and $C$, touching $\overline{TA}, \overline{TB}$; a circle $\Gamma_2$ passes through $B$ and $D$, touching $\overline{TA}, \overline{TB}$. Proof. The first two claims rely only on the condition that $X$ lies on the polar of $S$ in $\gamma$. Consider homography that sends $\overline{ST}$ to the line at infinity, preserving $\gamma$. Clearly, $AEFB$ is now a rectangle, and $X$ lies on the perpendicular bisector of $\overline{AB}$ and the claims are trivial. Proofs for the other two cases are analogous, so we only do the third case. Define $\Gamma_1$ as the circle through $A$ touching $\overline{TA}$ and $\overline{TB}$, inside angle $ATB$. Let $C_1=\overline{XA} \cap \Gamma_1$ and $C_2=\gamma \cap \Gamma_1$, with $C_i \ne A$ (unless double roots). We prove $C_1=C_2$, and the conclusion shall follow. Move $S$ on line $\overline{AB}$; notice that $X$ moves on $\ell$ and $S \mapsto X$ is projective; since $(TA, TB; TS, TX)=-1$ by definition. Note that $X \overset{A}{\mapsto} C_1$ is projective; so $S \mapsto C_1$ is a projective map $\overline{AB} \mapsto \Gamma_1$. Note that $S \mapsto E$ is projective; since $\overline{SE}$ is always perpendicular to the $T$-diameter in $\triangle TAB$. Finally, $AEBC_2$ cyclic forces $E \mapsto C_2$ projective; so $S \mapsto C_2$ is a projective map $\overline{AB} \mapsto \Gamma_1$. Proving these maps coincide, requires checking $C_1=C_2$ for three points $S$ on $\overline{AB}$. $S=\infty$: then $X$ coincides with $M$; $\gamma$ degenerates into line $\overline{AB}$ and $C_1, C_2$ both coincide with the other intersection $\Gamma_1 \cap \overline{AB}$, than $A$. $S$ coincides with foot of internal bisector $\ell^{+}$ of $\angle ATB$: then $X$ goes to the infinity point on $\ell$, so $C_1, C_2$ both coincide with the reflection of $A$ in $\ell^{+}$. $S=A$; then $X$ maps to reflection of $B$ in $\ell^{+}$; so $C_1, C_2$ both coincide with $A$. The proof is complete. $\blacksquare$ And of course, see zacchro's reference for more discussion on moving points!

Currently, there are more moving points solutions than other solutions in this thread, and I also solved this problem with moving points... I'm scared for the future of geometry...

@above, The other USA TST geo also seems to be very very MMP friendly. So yeah, same thoughts here. And then geo is the one thing that I know a little bit so ;_;

Let $AB$ intersect $\Gamma_1$ at $M$ and $\Gamma_2$ at $N$. Lemma: $AN=BM$ (this is really nice if you haven't seen it before).

Let $\Omega$ intersect $l_1$ and $l_2$ again at $P$ and $Q$, respectively, as in Brandon Wang's solution. I claim that $AB, CD, PQ$ intersect. We prove this by Trig Ceva's. Note that $C$ is the center of spiral similarity sending $AP$ to $MB$, so $CB/MB=CP/AP$. Similarly, $DA/NA=DQ/BQ$, so \[\frac{AP\cdot CB}{CP}=\frac{BQ\cdot DA}{DQ}\]\[AP\cdot CB\cdot DQ = BQ\cdot DA\cdot CP\]And so $AB, CD, PQ$ concur at a point $R$. Now by Brokard's, $T,X,Y$ all lie on the polar of $R$ with respect to $\Omega$.

The finish to this solution is slightly cleaner than the one I wrote in-contest, which used Monge's theorem on another circle to prove OY'T' collinear.

USA Winter TST for IMO 2020 P2 wrote: Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex. Suppose lines $AC$ and $BD$ meet at point $X$, while lines $AD$ and $BC$ meet at point $Y$. Show that $T$, $X$, $Y$ are collinear. Merlijn Staps Solution (with Geometrix):- Let $l_1\cap\odot(ABCD)=E$ and $l_2\cap\odot(ABCD)=F$. Claim :- $AB,CD,EF$ are concurrent. We Invert around $B$ and let the map be $\Psi$ and let the inverse of any point or inverted image of any cline $X$ be $X'$. Now after this Inversion $(\Psi)$ this happens $$\begin{cases}\Psi: \ell_2\leftrightarrow \ell_2 \text{which contains point} B=\ell_2' \\ \Psi:\Gamma_2\leftrightarrow \text{A line parallel to} \ell_2'=\Gamma_2' \\ \Psi:\Gamma_1\leftrightarrow \text{A Circle tangent to} \ell_2' =\Gamma_1'\\ \Psi:\ell_1\leftrightarrow \text{Circle Externally tangent to} \Gamma_1' \text{and} \Gamma_2'=\ell_1' \\ \Psi:A\leftrightarrow \text{Point of tangency made by} \Gamma_1' \text{and} \ell_1'=A' \\ \Psi:C\leftrightarrow \text{An arbitary line} (\ell) \text {passing through} A'\cap \Gamma_1'=C '\\ \Psi:D\leftrightarrow \ell\cap \Gamma_2'=D' \\ \Psi:E\leftrightarrow \ell\cap \ell_1'=E' \\ \Psi:F\leftrightarrow \ell\cap \ell_2'=F' \end{cases}$$ So after this Inversion $(\Psi)$, the Inverted Statement is read as follows. Inverted Statement wrote: A line $\ell_2'$ contains a point $B$ and we draw a line $(\Gamma_2')$ which is $\|$ to $\ell_2'$. Let $\Gamma_1'$ be a circle tangent to $\ell_2'$ and $\ell_1'$ be another circle externally tangent to $\Gamma_1'$ and $\Gamma_2'$ at $A',D'$ respectively. Let an arbitary line $(\ell)$ passing through $A'$ intersect $\Gamma_1',\Gamma_2',\ell_2',\ell_1'$ at $C',D,'F',E'$ respectively. So now it suffices to show that $A'\in\text{ Radical Axis of} \odot(BC'D')$ and $\odot(BE'F')$. Now observe the Homothety $\mathcal H$ at $A'$ mapping $\Gamma_1'$ to $\ell _1'$, so by this Homothety $\mathcal H:D'\leftrightarrow F'$ and $\mathcal H:E'\leftrightarrow C'$. Hence, $$\frac{A'E'}{E'C'}=\frac{A'D'}{A'F'}\implies A'C'.A'D'=A'E'.A'F'$$So power of $A'$ WRT to the circles $\odot(BC'D')$ and $\odot(BE'F')$ are same $\implies A'\in\text { Radical Axis}$ of $\odot(BC'D')$ and $\odot(BE'F')$. So Inverting back we get that $AB,CD,EF$ are concurrent at a point $K$. To finish it off notice that by Brocard's Theorem on $\odot(ABCD)$ we get that $XY$ is the Polar of $K$ also it's clear that $T\in$ Polar of $K$. Hence, $\overline{T-X-Y}$. $\blacksquare$

amar_04 wrote:

Why not invert around $A $?

....................

The_Turtle wrote:

The finish to this solution is slightly cleaner than the one I wrote in-contest, which used Monge's theorem on another circle to prove OY'T' collinear. Seems a bit problematic if I'm not mistaken since you send real points to complex points-or at least it needs justification why the projective properties you ask for are maintained Edit:Yeah,the projectivity is "imaginary" so every real point becomes imaginary. If instead the circles did not intersect,the projectivity would be real and everything would be alright. To quote Hatton: ()" It is only if and when projective properties of real and of imaginary points have been shown to be identical,that the process of projecting real points to the circular points at infinity is justifiable as a means of obtaining a theorem for real points". To clarify I'm not saying that the solution is wrong. I'm saying that I don't know

Ignore $ABCD$ convex. The main idea is to show that $T$ lie on radical axis of $\odot(XAD)$ and $\odot(XBC)$. To do that, let $P=TA\cap\odot(XAD)$ and $Q=TB\cap\odot(XBC)$. We aim to show that $A,B,P,Q$ are concyclic. Let $DP$ cut $\Gamma_2$ again at $C'$ and $CQ$ cut $\Gamma_1$ again at $D'$. Observe that $$\measuredangle BDC' = \measuredangle XDP = \measuredangle XAP = \measuredangle CAP$$thus arc $BC'$ of $\Gamma_2$ have the same angle as arc $AC$ of $\Gamma_1$. Similarly arc $AD'$ of $\Gamma_1$ have the same angle as arc $BD$ of $\Gamma_2$. Consider the transformation $f$ defined by homothety at $T$ which sends $\Gamma_1\mapsto\Gamma_2$ followed by reflection across the angle bisector of $\angle ATB$. Clearly $f(A)=B$, $f(C)=C'$ and $f(D')=D$. This transformation is affine thus $f(\overline{CD'}) = \overline{DC'}$ which means $f(Q)=P$. Hence $\triangle TAB\stackrel{-}{\sim}\triangle TQP$ which implies the concyclicity.

any bary soln as this looks baryable!

The_Turtle wrote:

The finish to this solution is slightly cleaner than the one I wrote in-contest, which used Monge's theorem on another circle to prove OY'T' collinear. Cool solution. I think all these nonsense theorems are unnecessary. I'll omit the primes from your diagram. Let $P=\overline{TA}\cap\overline{BD}$ and $Q=\overline{TB}\cap\overline{AC}$, and let $d(X,\ell)$ denote the distance from $X$ to $\ell$. After the homography, we want to show that \[\frac{AC}{BD}=\frac{d(Y,\overline{AC})}{d(Y,\overline{BD})}\stackrel?=\frac{d(T,\overline{AC})}{d(T,\overline{BD})}=\frac{AQ}{BP},\]i.e.\ $DP/PB=CQ/QA$. Take the homothety at $T$ sending $\Gamma_2$ to $\Gamma_1$, so that $A$, $B'$, $C$, $D'$ are concyclic and lie on $\Gamma_1$, and $DP/PB=D'P'/P'B'$. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); real d = 5/2; real s = 3/2; real c = 3/4; path g1p = circle((0, -d), 1); path g1 = circle((c, -d), 1); path g2 = scale(s) * g1p; //draw(g1p, dotted); pair O = origin; pair T = extension((0, 0), (1, 0), s * (0, -d), (c, -d)); dot("$T$", T, dir(90)); real a = abs(T - (c, -d)); pair A = intersectionpoints(g1, circle(T, sqrt(a^2 - 1)))[0]; pair B = intersectionpoints(g2, circle(T, s*sqrt(a^2 - 1)))[1]; draw(A + (s-1 + 1.5) * (A - T) -- T -- B + 1.0 * (B - T)); pair C = (2 * c - A.x, A.y); pair D = (- B.x, B.y); draw(B -- D); pair Y = extension(A, D, B, C); pair M = s * intersectionpoints(O -- D, g1p)[0]; pair N = s * intersectionpoints(O -- B, g1p)[0]; draw(M--B,gray); draw(N--D,gray); dot("$A$", M, dir(120)); dot("$C$", N, dir(60)); dot("$B$", B, dir(-30)); dot("$D$", D, SW); //draw(B -- O -- D, dotted); draw(M -- N); draw(D--M, dashed); draw(N--B,dashed); pair die=0.9(scale(s)*(c,-d)-(0,s)); clip( (die+(-100,0))--(die+(100,0))--(die+(100,100))--(die+(-100,100))--cycle); draw(g2); pair P=extension(T,A,B,D); pair Q=extension(T,B,M,N); draw(P--D); draw(N--Q); dot("$P$",P,NW); dot("$Q$",Q,E); [/asy][/asy] Again omit the primes. Just notice that \[\frac{DP}{PB}=\left(\frac{AD}{AB}\right)^2=\left(\frac{BC}{BA}\right)^2=\frac{CQ}{QA}.\](It is worth mentioning we may use Pascal theorem on $AACDBB$ as well.) $\square$ If that doesn't seem much simpler than Pascal + Desargues, here's an alternative using strictly elementary methods: note that $\measuredangle PAD=\measuredangle ACD$ and $\measuredangle ADP=\measuredangle ADB=\measuredangle DAC$, so $\triangle PAD\sim\triangle DCA$, and similarly $\triangle QBC\sim\triangle CDB$. Thus \[PD\cdot AC=AD^2=BC^2=QC\cdot BD\implies\frac{PD}{BD}=\frac{QC}{AC},\]as desired.

@below Yeah thanks for informing. I will fix my solution as soon as I get the time to.

That doesn't quite work, as $TX$ will intersect $\gamma$ (a cubic) in two points other than $T$. For this particular method (moving $X$ along $\ell$) I don't believe it's possible to do without checking at least 6 cases of $X$ (although $X \in AB$ can be hidden by using it to show $Y$ has degree $\le 3$.) EDIT: @TheUltimate123... you called Pascal a nonsense theorem?

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.cm); real labelscalefactor = 1.0; /* changes label-to-point distance */; defaultpen(fontsize(10pt)); pen dotstyle = black; /* point style */ real xmin = 4., xmax = 11., ymin = -1., ymax = 3.; /* image dimensions */ /* draw figures */ draw(circle((7.78,0.74), 1.019803902718557), linewidth(1.)); draw(circle((9.299520820490239,0.8869011879778688), 1.5428849251131946), linewidth(0.5)); draw((4.817527272727283,0.4536)--(8.633834637842673,2.2787904789682334), linewidth(0.5)); draw((4.817527272727283,0.4536)--(8.912795304303,-0.6067310516551885), linewidth(0.5)); draw((7.34,1.66)--(8.731841990792773,0.37394423298667867), linewidth(0.5)); draw((7.850427725877604,1.4166398927064516)--(8.912795304303,-0.6067310516551885), linewidth(0.5)); draw((7.34,1.66)--(8.605907646213748,1.0564445797363042), linewidth(0.5)); draw((8.605907646213748,1.0564445797363042)--(8.912795304303,-0.6067310516551885), linewidth(0.5)); draw((8.009713508732649,1.7335953421316597)--(8.195794072343277,-0.1911902541393934), linewidth(0.5)); draw((4.817527272727283,0.4536)--(8.605907646213748,1.0564445797363042), linewidth(0.5) + linetype("2 2")); draw(circle((7.893684292592922,2.1643370218071647), 0.7489473462329945), linewidth(0.5)); draw(circle((8.272753306502535,-0.21779870730589768), 0.7489473462329939), linewidth(0.5)); draw((8.633834637842673,2.2787904789682334)--(7.838118503380405,0.392131383232714), linewidth(0.5)); draw((4.817527272727283,0.4536)--(10.739584596250433,0.3330869466614837), linewidth(0.5)); /* dots and labels */ dot((7.34,1.66),dotstyle); label("$A$", (7.377293775285266,1.7550732363857555), NE * labelscalefactor); label("$\Gamma_1$", (6.9586594848748025,1.1461506321523538), NE * labelscalefactor); dot((4.817527272727283,0.4536),dotstyle); label("$T$", (4.855973617131337,0.5467424436100992), NE * labelscalefactor); dot((8.633834637842673,2.2787904789682334),linewidth(4.pt) + dotstyle); label("$E$", (8.671254309281245,2.3544814249280104), NE * labelscalefactor); label("$\Gamma_2$", (9.156489509529736,2.0593372680165413), NE * labelscalefactor); dot((8.009713508732649,1.7335953421316597),linewidth(4.pt) + dotstyle); dot((8.195794072343277,-0.1911902541393934),linewidth(4.pt) + dotstyle); dot((7.52438522000618,-0.24724925132851253),linewidth(4.pt) + dotstyle); label("$F$", (7.5580676734170575,-0.16683873322591836), NE * labelscalefactor); dot((8.083218799547728,0.9732691572506359),linewidth(4.pt)+dotstyle); label("$X$", (8.186019109032753,1.041492059549738), NE * labelscalefactor); dot((8.912795304303,-0.6067310516551885),linewidth(4.pt) + dotstyle); label("$B$", (9.013773274162533,-0.5664441922540882), NE * labelscalefactor); dot((7.850427725877604,1.4166398927064516),linewidth(4.pt) + dotstyle); label("$D$", (7.6768992343134355,1.4886695970336423), NE * labelscalefactor); dot((8.731841990792773,0.37394423298667867),linewidth(4.pt) + dotstyle); label("$C$", (8.823484960339595,0.38499737686060187), NE * labelscalefactor); dot((8.605907646213748,1.0564445797363042),linewidth(4.pt) + dotstyle); label("$Y$", (8.737855219119274,1.1651794635346477), NE * labelscalefactor); dot((7.838118503380405,0.392131383232714),linewidth(4.pt) + dotstyle); label("$H$", (7.47224066171082,0.46111270238977703), NE * labelscalefactor); dot((6.814053828530863,0.4129709587450996),linewidth(4.pt) + dotstyle); label("$J$", (6.854000912272186,0.4896559494632177), NE * labelscalefactor); dot((10.739584596250433,0.3330869466614837),linewidth(4.pt) + dotstyle); label("$I$", (10.62170952596636,0.44208387100748325), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The shortest solution is here guys!! First we claim $\angle{FCB}+\angle{ADE}=180$. This is just an angle chase. This implies $\odot ADE,\odot FCB$ are congruent. Now Since $YD.YA=YC.YB$, $Y$ lies on their radical axis. Similarly $T$ lies on their radical axis. Hence it remains to show $X$ lies on the radical axis. To show this, we demonstrate that $X$ is the centre of inversion with negative power that maps $\odot ADE$ to $\odot FCB$, which will imply the latter since they are congruent. Let $EX$ meet $\Gamma_2$ at $H$. We have $XA.XC=XD.XB=XH.XE$. Hence an inversion maps $A\mapsto C, D\mapsto B, E\mapsto H$. We need $FHCB$ cyclic, which is equivalent to $C,H,T$ collinear (Bcoz $AHCE$ is cyclic and radical axis). Let $HC$ meet $\Gamma_2$ at $I$ and $\Gamma_1$ at $J$. We can have $AC\parallel EI$ which implies $CF\parallel IB$. Hence we easily get $AJIE,JFBI$ cyclic. Hence by radical axis, $T,I,J$ are collinear. We are done.

Ah, it seems when I wrote the problem on paper, the notations of $X,Y$ and $C,D$ where mixed up. I'll change it here to avoid more confusion

Another short way. Suppose that $l_2,l_1$ touch $\Gamma_1,\Gamma_2$ at $A',B'$ respectively. Clearly $AA'BB'$ is cyclic. Let $XA'\cap\Gamma_1=C'(\neq A')$ and $A'C\cap C'A=Z$. Notice that $X$ is radical center of $\Gamma_1,\Gamma_2,\Omega$ we get $A'BDC'$ is cyclic. Hence from simple angle chasing, $AB'DC'$ is cyclic. Also $T$ is radical center of $(AA'BB'),(A'BDC'),(AB'DC')$ thus $T,C',D$ are collinear. Apply Desargues's on $(CA,BD,A'C')$ we have $T,Y,Z$ are collinear. Moreover, applying Pascal's on $(A'A'CAAC')$ means that $X,Z,T$ are collinear. Combining this results gives us that $T,X,Y(,Z)$ are collinear.

I looooove "synthetic" geometry. I loooove when the only good solutions are length bash and moving points. Yep. Let $M, N = AT \cap \Omega, BT \cap \Omega$, then it suffices to show $AB, CD, MN$ concur via the projective transformation mapping $\Omega$ to the polar of $AB \cap CD$ through $A$ then to $\Omega$ through $B$ is equivalent to antiversion through $AB \cap CD$ since a projective transformation is determined by three points. It suffices to show $\frac{AM}{MB} \cdot \frac{AN}{NB} = \frac{AD}{DB} \cdot \frac{AC}{CB}$. Let $P = AD \cap \Gamma_2, Q = BC \cap \Gamma_1$, then $PB \parallel AN, QA \parallel MB$ by Reim's theorem. Now notice $\frac{AN}{MB} = \frac{AT}{BT}$ and $\frac{AM}{NB} = \frac{\sin \angle ABM}{\sin \angle BAN}$. So letting $A_1 = AB \cap \Gamma_1, B_1 = AB \cap \Gamma_2$, it follows that the left hand side is equal to $\frac{r_{\Gamma_1} \sin \angle QAA_1}{r_{\Gamma_2} \sin \angle PBB_1} = \frac{QA_1}{PB_1}$. Now notice that by since $\triangle ADB \sim \triangle AB_1P, \triangle BCA \sim \triangle BA_1Q$, it follows that the right hand side is equal to $\frac{\frac{AB_1}{PB_1}}{\frac{BA_1}{QA_1}}$, so it suffices to show $AB_1 = BA_1$. Letting $A' = \ell_1 \cap \Gamma_2, B' = \ell_2 \cap \Gamma_1$, we have $AB_1 \cdot AB = AA'^2 = BB'^2 = AB \cdot BA_1$, so we are done. EDIT: Actually the author's solution is way too beautiful and finishes in one word. Wow. Skill issue.

@above I like synthetic geometry too <3 Let $A' = \ell_2\cap \Gamma_1$ and $B' = \ell_1 \cap \Gamma_2$. Define $T_1 = BY\cap \Gamma_1, T_2 = AY\cap \Gamma_2$, $G_1 = BX\cap AT_1, G_2 = AX\cap BT_2$, and $C' = A'X\cap \Gamma_1$ and $D' = B'X\cap \Gamma_2$. First, observe that \[\angle T_2BT = 180 - \angle BDT_2 = 180 - \angle ACT_1 = \angle TAT_1.\]Furthermore, since $T_1\in \Gamma_1, T_2\in \Gamma_2$, this means $\triangle TAT_1 \sim \triangle TBT_2$, so $T_1$ and $T_2$ are isogonal wrt $\ell_1, \ell_2$. Now, by second isogonality lemma, $Y = AT_2 \cap BT_1$ and $AT_1 \cap BT_2$ are isogonal wrt $\ell_1$ and $\ell_2$. It suffices to show $\frac{AG_1}{G_1T_1} = \frac{BG_2}{G_2T_2}$, since then $\triangle TAG_1 \sim \triangle TBG_2$, so $G_1$ and $G_2$ are isogonal wrt $\ell_1, \ell_2$. This means $X = BG_1 \cap AG_2$ is isogonal with $BG_2 \cap AG_1 = BT_2 \cap AT_1$, so $TX = TY$. Let $R_1, R_2 = \ell_2\cap AT_1, \ell_1\cap BT_2$ respectively. Now, \[\frac{AR_1}{BR_2} = \frac{AT}{BT} = \frac{AT_1}{BT_2}\]so showing $(T_2B;G_2R_2) = (T_1A;G_1R_1)$ suffices. We have \[(T_1A;G_1R_1) \overset{B}{=} (CA;X\ell_2\cap AG_2) \overset{A'}{=} (CA;C'A') = (C'A';CA)\]Similarly, $(T_2B;G_2R_2) = (DB;D'B')$. Observe that $D'CAB'$ is cyclic by radical axis, and so is $DC'A'B$, so $T,D',C$ and $T,D,C'$ are collinear. By homothety through $T$, we have \[(C'A';CA) = (TD\cap \Gamma_2 B; TD'\cap \Gamma_2 B') \overset{T}{=} (DB;D'B')\]Therefore, $(T_2B;G_2R_2) = (T_1A;G_1R_1)$ so $X,Y,$ and $T$ are collinear.

Let $A',B'$ be the other tangency points as shown in diag below. $XA'$ meet $\Gamma_1$ again at $D'$.Since $A'D'DB$ is cyclic $D,D'$ are antihomologous points so $T-D'-D$ are collinear.$E=A'C \cap AD'$. $M,N$ are midpoints of $AB', A'B$. $X$ lies on $MN$ as $MN$ is radax of $\Gamma_1,\Gamma_2$.Note that the inversion wrt $X$ with power $XC \cdot XA$ maps $E$ to the second intersection of $(AMXD'),(A'NXC)$ say $E'$.By radax theorem on $(AMXD'),(A'NXC), (AA'MN)$ we get $T-E'-E-X$ are collinear. By $DDIT$ on quad $ACD'A'$ with point $T$ we get that there exists an involution swapping $(TA,TB) ;(TC,TD);(TX,TX)$ hence by $DDIT$ on quad $ACDB$ with point $T$ we get $T-Y-X$ collinear.

Thanks @above. Let $\Omega$ intersect $\ell_1$ and $\ell_2$ again at $J$ and $L$ respectively. Inverting at $A$ ,the diagram becomes like the one attached below.(not elaborating how) Denote the images of points with their given notations only. Since originally $AGBH$ was an isosceles trapezium and hence cyclic,in the inverted picture points $G-B-H$ should be collinear. Now clearly in the inverted diagram $GC \parallel AT \implies \triangle BJH \sim \triangle BCG$ and by homothety reasons $\triangle BLG \sim \triangle BDH$. From the similarities we've $$\frac{BJ}{BC}=\frac{BH}{BG}=\frac{BD}{BL} \implies BJ.BL=BC.BD$$This implies $B$ lies on the radical axis of $(AJL)$ and $(ACD)$ [In inverted diagram ofc]. Or $AB,(AJL),(ACD)$ have a common point. Inverting back,we've $AB,JL,CD$ concurrent.Let they all intersect at $E$.Now by Brocards theorem $T,X,Y$ lie on the polar of $E$ w.r.t $\Omega$.Hence they are collinear $\blacksquare$

To get rid of the annoying tangency case, note that $\Gamma_1, \Gamma_2$ must intersect at two distinct points $P$ and $Q$, otherwise the radical axis of the two circle with intersect all four sides of $ABCD$, which is impossible considering the convexity of $ABCD$. By the radical axis theorem, $X \in \ell := \overline{PQ}$. It suffices to show that $\overline{AD}, \overline{BC}, \overline{TX}$ concur at point $Y$. Animate $X$ along $\ell$. Since $C$ and $D$ now move projectively on $\Gamma_1$ and $\Gamma_2$, respectively, we have $\deg(\overline{AD})=2$ and $\deg(\overline{BC})=2$. Noting that $\deg(\overline{TX})=1$, the degree of the concurrency is $\deg(\overline{AD})+\deg(\overline{BC})+\deg(\overline{TX})=2+2+1=5$, so it suffices to prove the problem for $6$ distinct positions of $X$. $X=P$: This yields $C=D=P$, from which $Y=P$, implying the conclusion. $X=Q$: Analogous to the previous case. $X=\ell \cap \overline{AB}$: We have $A, B, C, D$ are all collinear, implying the conclusion. $X$ is the point at infinity on $\ell$: In this case, $C$ is the antipode of $A$ on $\Gamma_1$ and $D$ is the antipode of $B$ on $\Gamma_2$ $X=\ell \cap \overline{AT}$: We have $C=A$, so $Y=A \in \overline{TX}$. $X= \ell \cap \overline{BT}$: Analogous to the previous case. The above cases complete the proof.

Call radius of $\Gamma_1,\Gamma_2$ , $R_1,R_2$ if we show that $\frac{\sin{\widehat{AXT}}}{\sin{\widehat{BXT}}}= \frac{\sin{\widehat{CXY}}}{\sin{\widehat{DXY}}}$ Then by Lemma 2 $T,X,Y$ are collinear. now by Lemma 1 in $\triangle YCD$ & point $X$ & $\widehat{XCY} = \widehat{XDY}$ We have : $$\frac{\sin{\widehat{CXY}}}{\sin{\widehat{DXY}}}= \frac{CY}{DY}= \frac{AC}{BD}$$now by Lemma 1 in $\triangle TAB$ & point $X$ & using law of sines in $\Gamma_1,\Gamma_2$ We have : $$\frac{\sin{\widehat{AXT}}}{\sin{\widehat{BXT}}}= \frac{\sin{\widehat{TAX}}}{\sin{\widehat{TBX}}} \cdot \frac{TB}{TA}= \frac{\sin{\widehat{TAX}}}{\sin{\widehat{TBX}}} \cdot \frac{R_1}{R_2}= \frac{AC}{BD}$$We are Done:)

kapilpavase wrote: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.cm); real labelscalefactor = 1.0; /* changes label-to-point distance */; defaultpen(fontsize(10pt)); pen dotstyle = black; /* point style */ real xmin = 4., xmax = 11., ymin = -1., ymax = 3.; /* image dimensions */ /* draw figures */ draw(circle((7.78,0.74), 1.019803902718557), linewidth(1.)); draw(circle((9.299520820490239,0.8869011879778688), 1.5428849251131946), linewidth(0.5)); draw((4.817527272727283,0.4536)--(8.633834637842673,2.2787904789682334), linewidth(0.5)); draw((4.817527272727283,0.4536)--(8.912795304303,-0.6067310516551885), linewidth(0.5)); draw((7.34,1.66)--(8.731841990792773,0.37394423298667867), linewidth(0.5)); draw((7.850427725877604,1.4166398927064516)--(8.912795304303,-0.6067310516551885), linewidth(0.5)); draw((7.34,1.66)--(8.605907646213748,1.0564445797363042), linewidth(0.5)); draw((8.605907646213748,1.0564445797363042)--(8.912795304303,-0.6067310516551885), linewidth(0.5)); draw((8.009713508732649,1.7335953421316597)--(8.195794072343277,-0.1911902541393934), linewidth(0.5)); draw((4.817527272727283,0.4536)--(8.605907646213748,1.0564445797363042), linewidth(0.5) + linetype("2 2")); draw(circle((7.893684292592922,2.1643370218071647), 0.7489473462329945), linewidth(0.5)); draw(circle((8.272753306502535,-0.21779870730589768), 0.7489473462329939), linewidth(0.5)); draw((8.633834637842673,2.2787904789682334)--(7.838118503380405,0.392131383232714), linewidth(0.5)); draw((4.817527272727283,0.4536)--(10.739584596250433,0.3330869466614837), linewidth(0.5)); /* dots and labels */ dot((7.34,1.66),dotstyle); label("$A$", (7.377293775285266,1.7550732363857555), NE * labelscalefactor); label("$\Gamma_1$", (6.9586594848748025,1.1461506321523538), NE * labelscalefactor); dot((4.817527272727283,0.4536),dotstyle); label("$T$", (4.855973617131337,0.5467424436100992), NE * labelscalefactor); dot((8.633834637842673,2.2787904789682334),linewidth(4.pt) + dotstyle); label("$E$", (8.671254309281245,2.3544814249280104), NE * labelscalefactor); label("$\Gamma_2$", (9.156489509529736,2.0593372680165413), NE * labelscalefactor); dot((8.009713508732649,1.7335953421316597),linewidth(4.pt) + dotstyle); dot((8.195794072343277,-0.1911902541393934),linewidth(4.pt) + dotstyle); dot((7.52438522000618,-0.24724925132851253),linewidth(4.pt) + dotstyle); label("$F$", (7.5580676734170575,-0.16683873322591836), NE * labelscalefactor); dot((8.083218799547728,0.9732691572506359),linewidth(4.pt)+dotstyle); label("$X$", (8.186019109032753,1.041492059549738), NE * labelscalefactor); dot((8.912795304303,-0.6067310516551885),linewidth(4.pt) + dotstyle); label("$B$", (9.013773274162533,-0.5664441922540882), NE * labelscalefactor); dot((7.850427725877604,1.4166398927064516),linewidth(4.pt) + dotstyle); label("$D$", (7.6768992343134355,1.4886695970336423), NE * labelscalefactor); dot((8.731841990792773,0.37394423298667867),linewidth(4.pt) + dotstyle); label("$C$", (8.823484960339595,0.38499737686060187), NE * labelscalefactor); dot((8.605907646213748,1.0564445797363042),linewidth(4.pt) + dotstyle); label("$Y$", (8.737855219119274,1.1651794635346477), NE * labelscalefactor); dot((7.838118503380405,0.392131383232714),linewidth(4.pt) + dotstyle); label("$H$", (7.47224066171082,0.46111270238977703), NE * labelscalefactor); dot((6.814053828530863,0.4129709587450996),linewidth(4.pt) + dotstyle); label("$J$", (6.854000912272186,0.4896559494632177), NE * labelscalefactor); dot((10.739584596250433,0.3330869466614837),linewidth(4.pt) + dotstyle); label("$I$", (10.62170952596636,0.44208387100748325), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The shortest solution is here guys!! First we claim $\angle{FCB}+\angle{ADE}=180$. This is just an angle chase. This implies $\odot ADE,\odot FCB$ are congruent. Now Since $YD.YA=YC.YB$, $Y$ lies on their radical axis. Similarly $T$ lies on their radical axis. Hence it remains to show $X$ lies on the radical axis. To show this, we demonstrate that $X$ is the centre of inversion with negative power that maps $\odot ADE$ to $\odot FCB$, which will imply the latter since they are congruent. Let $EX$ meet $\Gamma_2$ at $H$. We have $XA.XC=XD.XB=XH.XE$. Hence an inversion maps $A\mapsto C, D\mapsto B, E\mapsto H$. We need $FHCB$ cyclic, which is equivalent to $C,H,T$ collinear (Bcoz $AHCE$ is cyclic and radical axis). Let $HC$ meet $\Gamma_2$ at $I$ and $\Gamma_1$ at $J$. We can have $AC\parallel EI$ which implies $CF\parallel IB$. Hence we easily get $AJIE,JFBI$ cyclic. Hence by radical axis, $T,I,J$ are collinear. We are done. I can't understand first angle chase $\angle{FCB}+\angle{ADE}=180$. Could you explain more?

amazing! Let $\ell$ be the radical axis of $\Gamma_1$ and $\Gamma_2$. By radical center, $X$ lies on $\ell$. Also, let $A_1$ and $B_1$ be the other tangency points. Define $E=TC\cap \Gamma_1\neq C$, $G=TD\cap \Gamma_2\neq D$, $F=AE\cap \ell$, $K=FB_1\cap BG$. Claim: $FB_1\parallel AX$ and $BG\parallel A_1X$. Proof. Note that $-1=(AA_1;EC)\overset{A}{=}(TA\cap \ell, \infty; F,X)$ which is valid since $\ell \parallel AA_1$. Then, as $TA\cap \ell$ is the midpoint $M$ of $AB_1$, we obtain $AXB_1F$ is a parallelogram. Thus $FB_1\parallel AX$ and $B_1X\parallel AF$. This proves the claim by symmetry. $\blacksquare$ Now, triangles $AA_1X$ and $B_1BK$ are homothetic; hence $K$ lies on line $TX$. Therefore, it suffices to show $X,Y,K$ are collinear. Let $I=B_1K\cap (B_1BGD)\neq B_1$ and $J=B_1X\cap (B_1BGD)\neq B_1$. Claim: $I$ and $J$ lie on line $TC$. Proof. First, by radical axis, $AB_1CJ$ are cyclic. Then \[\measuredangle(CJ, JXB_1)=\measuredangle CJE=\measuredangle CAB_1=\measuredangle CEA=\measuredangle (CE,AE)=\measuredangle(CE,JXB_1)\]so $CE$ or equivalently $TC$. To finish, $-1=(XF;M\infty)\overset{B_1}{=}(JI;BB_1)$ so $IJ$ passes through $T$. Another way to do this is $EAB_1I$ is cyclic by radical axis and then \[\measuredangle(JI,FB_1I)=\measuredangle JIB_1=\measuredangle FAB_1=\measuredangle ECA=\measuredangle(EC,AC)=\measuredangle(EC,FB_1I)\]$\blacksquare$ Now, let $L=BI\cap B_1G$. To finish, Pascal on $BJIB_1DG$ gives $L=BJ\cap B_1D$ lies on the line between $T=JI\cap DG$ and $K=IB_1\cap GB$. Desuarge on triangles $ADB_1$ and $CBJ$ gives that they are perspective from $X$, so $Y=AD\cap CB$ must lie on the line between $L=DB_1\cap BJ$ and $T=AB_1\cap CJ$. Thus $T,L,Y$ are collinear and since line $TL$ is line $TXKL$ by the Pascal, we get $T,X,Y$ are collinear. Huzzah!

We use moving points. Note that by doing algebra, the reflection of a fixed point over a line of degree $1$ has degree $2$. For now suppose that $\Gamma_1$ and $\Gamma_2$ intersect, and let their centers be $O_1$ and $O_2$ respectively. Vary the center $O$ of $\Gamma$ with degree $1$ along the perpendicular bisector of $\overline{AB}$, so $\overline{OO_1}$ and $\overline{OO_2}$ are degree $1$. Thus $C$ and $D$, which can be obtained by reflecting $A$ and $B$ over the respective lines, are degree $2$. Then, $\overline{AD}$ and $\overline{BC}$ are degree $2$. Furthermore, since $C$ moves on the same circle as $A$ lies on and a similar result holds for $D$, $\overline{AC}$ and $\overline{BD}$ are degree $1$. When $O$ goes to infinity (along its line), $A,B,C,D$ becomes a line ($\Omega$ is the union of $\overline{AB}$ and the line at infinity, which is actually a circle), so $\overline{AD}$ and $\overline{BC}$ coincide (at least) once, as do $\overline{AC}$ and $\overline{BD}$. Thus $X=\overline{AC} \cap \overline{BD}$ has degree $1$ and $Y=\overline{AD} \cap \overline{BC}$ has degree $3$. Thus the collinearity condition has degree $4$, so it suffices to check $5$ cases. When $C$ and $D$ are the tangency points between $\Gamma_1$ and $\ell_2$ and $\Gamma_2$ and $\ell_1$ respectively, the result is clear by symmetry. When $\Omega$ is tangent to $\ell_1$ and thus $\Gamma_1$ at $A$, $A=C$ and $\overline{AC}=\ell_1$. Then $X$ lies on $\ell_1$ and $Y=A=C$, so the collinearity holds. The same is true when $\Omega$ is tangent to $\ell_2$ at $B$. When $\Omega$ passes through either choice of $\Gamma_1 \cap \Gamma_2$, we have $C=D$, so the collinearity holds since $X=Y=C=D$ (clearly it can't pass through both). This gives us a total of $1+2+2=5$ cases, so the problem always holds when $\Gamma_1$ and $\Gamma_2$ intersect. Instead of noticing that $\Gamma_1$ and $\Gamma_2$ always intersect, we use the following argument. Change the way we vary the problem, fixing $\ell_1,\ell_2,\Gamma_1$ and vary $B$ along $\ell_2$. Fix any line $\ell$ parallel to the bisector of the angle formed by $\ell_1$ and $\ell_2$ (that intersects $\Gamma_1$), and suppose we constrain the center of $\Omega$ to lie on $\ell$. Then the collinearity condition has finite degree—say, at most $9000$—but there are infinitely many choices of $B$ such that $\Gamma_1$ and $\Gamma_2$ intersect, hence the collinearity always holds. Since $\ell$ is chosen arbitrarily this finishes the problem. $\blacksquare$

Fix $\Gamma_1$ and $\Gamma_2.$ Since $XA \cdot XC = XD \cdot XB,$ it follows $X$ lies on the radical axis $\ell$ of $\Gamma_1$ and $\Gamma_2.$ Now, animate $X$ projectively on $\ell.$ Note $X \xmapsto{B} D$ and $X \xmapsto{A} C$ are projective maps, and in particular, we have $\deg(C)=\deg(D)=2.$ Then, we see $\deg(\overline{BC})=2$ and $\deg(\overline{AD})=2,$ so by Zack's Lemma, $\deg(Y)=2+2-1=3,$ where we subtract the one case when $\overline{AD}$ and $\overline{BC}$ coincide (observe that this is the same case as when $\overline{AC}$ and $\overline{BD}$ concur, that is, $\overline{ADXCB}$ is a line). Thus, the hypothetical collinearity $\overline{TXY}$ has degree $\deg(T)+\deg(X)+\deg(Y)=4.$ It remains to check five cases of $X.$ Take $X$ the first intersection of $\Gamma_1$ and $\Gamma_2.$ Then $D,X,C,Y$ all coincide, and $\overline{TXY}$ is collinear. Take $X$ the second intersection of $\Gamma_1$ and $\Gamma_2.$ The same argument from above follows. When $X=\ell \cap \ell_1,$ we have $Y=A,$ so $\overline{TXY}$ collinear in this case. When $X=\ell \cap \ell_2,$ we have $Y=B,$ so $\overline{TXY}$ collinear in this case. When $X=\infty_{\ell},$ we have $Y=T,$ and evidently we have $\overline{TXY}$ collinear. Note now that $\Gamma_1$ and $\Gamma_2$ must always intersect, otherwise by radical axis we have that $ABCD$ is never convex.

By Radical Axis Theorem note that $X$ must be on the radical axis of $\Gamma_1, \Gamma_2.$ Therefore we can move $X$ linearly on the radical axis. Let $\ell$ be the radical axis. Redefine $C,D,T$ with respect to $X$ by all the collinearities. Then $C,D$ moves projectively with respect to $X$ by perspectivity. Note that the maximum degrees of $T,X,Y$ are $0,1,4,$ respectively, as $Y$ is the intersection of two moving lines of maximum degree $2.$ Thus, we simply need to check that the problem is true for $5+1=6$ cases of $X.$ These are $\overline{AB}\cap \ell$ $\ell_1 \cap \ell$ $\ell_2 \cap \ell$ Point of infinity on $\ell.$ Both of $\Gamma_1 \cap \Gamma_2.\blacksquare$

Let $B'=\Gamma_2 \cap \ell_1$ and $D'=\Gamma_2 \cap \overline{XB'}$. Power of a point at $X$ gives $ACD'B'$ cyclic. An inversion at $T$ with radius $\sqrt{TA \cdot TB'}$ fixes $(ACD'B')$ and swaps $\Gamma_1$ and $\Gamma_2$, so it swaps $C$ and $D'$. Thus $\overline{CD'}$ passes through $T$. By Desargues on perspective triangles $ADB'$ and $CBD'$, we know $T$, $Y$, and $\overline{BD'} \cap \overline{B'D}$ are collinear. By Pascal on $BBDB'B'D'$, we know $T$, $X$, and $\overline{BD'} \cap \overline{B'D}$ are collinear. Thus $T$, $X$, and $Y$ are collinear, as desired. $\square$

Let $A'$ be the tangency point between $\Gamma_1$ and $\ell_2$ and let $B'$ be the tangency point between $\Gamma_2$ and $\ell_1$. Let $E$ and $F$ be the reflections of $A$ and $B$ over $X$, respectively. We will show that $X$ lies on the radical axis of $(A'BC)$ and $(AB'D)$ (which obviously passes through $T$ and $Y$). Firstly, note that $X$ lies on the midline of trapezoid $AB'BA'$ since it is the radical axis of $\Gamma_1$ and $\Gamma_2$. Therefore, $E$ lies on line $BB'$. Then, we have $\angle BEC = \angle A'AC = \angle BA'C$, so $A'BCE$ is cyclic. Similarly, $AB'DF$ is cyclic. We now have