Let $AB\cap{\omega_1}=\{X'\}$ and $AC\cap{\omega_2}=\{Y'\}$. Then $\angle{X'DY}=\angle{ABC}=\angle{YY'X'}$ thus $X'DY'Y$ is cyclic. Analogously, $XX'DY'$ is cyclic too. Consequently $X,Y,Y',D,X'$ all lie on a circle. From here we obatin that $XY$ and $X'Y\"$ are antiparallel. Therefore $XY||BC$.

We can also use Menalaus and some quick computations: Let $ R$ be the second point of intersection of the two circles, which lies on $ AD$. Applying Menelaus to collinear points $ X, D, F$ in triangle $ ABS$ gives us: $ \frac {AX}{XB} = \frac {AD}{SD} * \frac {SF}{BF} ... (1)$. Applying Menelaus to collinear points $ E, D, Y$ in triangle $ ACS$ gives us: $ \frac {AY}{CY} = \frac {AD}{SD} * \frac {SE}{CE} ... (2)$. To prove $ XY$ is parallel to $ BC$, we must prove that $ \frac {AX}{XB} = \frac {AY}{CY}$ - using $ (1), (2)$ that is equivalent to proving: $ \frac {SF}{BF} = \frac {SE}{CE}$, which is true since: $ SF*CE - SE*BF = SF(SC+SE) - SE(BS+SF) = SF*SC - SE*SB = SR*RD - SR*RD = 0$. Here we have used Power of a point to see that $ SF*SC = SR*RD = SE*SB$.

We use barycentrics on $\triangle DEF$. Let $D = (1,0,0)$, $E = (0,1,0)$ and $F = (0,0,1)$ and define $a=EF$, $b=FD$ and $c=DE$. Set $D_1 = (u:m:n)$ and $A = (v:m:n)$, where $D_1$ is the second intersection of $\omega_1$ and $\omega_2$. Let $L = \tfrac{-a^2mn-b^2nu-c^2um}{u+m+n}$. (Guess what? $u$ never appears again, because the degree of freedom described by $u$ is encoded in $L$.) Then, we find that \[ (DED_1) : -a^2yz-b^2zx-c^2xy + \left( x+y+z \right)\left( \frac{L}{n} z \right) = 0. \] Now we obtain $B$ by setting $x=0$ and dividing through by $z \neq 0$ (so that $B \neq E$) to give $(y+z) \frac{L}{n} = a^2y$; thus, we find that \[ B = (0 : L : na^2 - L). \] Then computing $X = DF \cap AB$, which is nice since $DF : y=0$, we find that \[ X = \left( -\det \left[ \begin{array}{cc} v & m \\ 0 &n \end{array} \right] : 0 : \det \left[ \begin{array}{cc} m & n \\ L & na^2 - L \end{array} \right] \right) \] That is, \[ X = \left( -vL: 0 : mna^2-mL-nL \right). \] Similarly, $Y = \left( -vL : mna^2-mL-nL : 0 \right)$. So $XY \parallel EF$ and we're done.

Suppose $B=(-1,0),F=(a,0),E=(b,0),C=(1,0)$ and eq of circles are $x^2+y^2+2g_1x+2h_1y=1-2g_1$ and $x^2+y^2+2g_2x+2h_2y=1+2g_2$. Now certainly we've $g_1=-\frac {1+a}{2},g_2=\frac {1-b}{2}$. Now equation of line $AD$ be $2x(g_1-g_2)+2y(h_1-h_2)+c_1-c_2=0$ so intersection point of $AD,BC$ is $S=(-\frac {c_1-c_2}{h_1-h_2},0)$. Now so finally we get $\frac {BF}{SF}=\frac {SE}{ES}$, and now just by menalaus we're done.

$\omega_1$ and $\omega_2$ intersects each other at $D,E$,and $AB,AC$ at $K$ and $L$,now $AE.AD=AL.AC$ and $AE.AD=AK.AB$,hence we get $KBCL$ is cyclic.so $\angle FDL=\angle BKL$ and $\angle EDK=\angle CLK$ ,so $KLYDX$ is cyclic. so $\angle DLY=\angle DXY$,also $\angle DLY=\angle EFD$,hence done.

Let $w_1 \cap AB =G$, $w_2 \cap AC=H$. Let $Y'$ be the point on $AC$ such that $XY' || BC$. Notice $\angle Y'HD = \angle DFB = \angle DXY'$ and so $XDY'H$ is cyclic. Notice also $\frac{AX}{AY'} = \frac{AB}{AC} = \frac{AH}{AG}$ since $GHBC$ is cyclic. Therefore $GHDXY'$ is cyclic. So $\triangle GHD$'s circumcircle passes through $X$, and through $Y$ analogously. Then $Y'=Y$ so we're done.

Dear Mathlinkers, another approach withe the Reim's theorem... Sincerely Jean-Louis

jayme, Let $AB\cap{\omega_1}=\{B,X'\}, AC\cap{\omega_2}=\{C,Y'\}$. Then $\angle{X'DY}=\angle{ABC}=\angle{YY'X'}$, so $X'DY'Y$ is cyclic. Analogously, $XX'DY'$ is cyclic too.By Reim's theorem on $\{(XYD), \omega_1 \}$, $XY \parallel BE$, as required. Alternatively, we may apply Reim's theorem on $\{(XYD), \omega_ 2\}$, so $XY \parallel FC$, as required. This solution is, of course, only a breath away from Sailor's solution.

Here is another solution: We work in the projective plane. Denote by $D' \in AD$ the second intersection of $\omega_1$ and $\omega_2.$ Let $R = DX \cap D'B$ and $S = DY \cap D'C.$ Then using directed angles, we have that \begin{align*} \measuredangle RD'S &= \measuredangle BD'C = \measuredangle BD'D + \measuredangle DD'C \\ &= \measuredangle BED + \measuredangle DFC = \measuredangle FED + \measuredangle DFC \\ &= -\measuredangle EDF = -\measuredangle SDR = \measuredangle RDS. \end{align*} Therefore, we have that $\measuredangle RD'S = \measuredangle RDS$, implying that $R, S, D, D'$ are concyclic. It follows that \[\measuredangle D'RS = \measuredangle D'DS = \measuredangle D'DE = \measuredangle D'BE = \measuredangle D'BC.\] Then, since $\measuredangle D'RS = \measuredangle D'BC$, we conclude that $RS \parallel BC.$ Now, consider the lines $DD', XB, YC.$ Since these three lines all concur at one point (that is, $A$), we deduce that $\triangle DXY$ and $\triangle D'BC$ are in perspective. By Desargue's Theorem, it follows that $R = DX \cap D'B, \; S = DY \cap D'C, \; XY \cap BC$ are collinear. Therefore, $XY \cap BC$ lies on $RS$, which is a line parallel to $BC.$ It follows that $XY \parallel BC$, as desired. $\square$

If I'm not mistaken, this problem follows directly from angle chasing. Suppose $\omega_1 \cap AB=M$ and $\omega_2 \cap AC=P$; we claim $BPMC$ is cyclic. This follows quickly from Power of a Point; let the second intersection of the two circles be $D'$, then $AD\cdot AD'=AM\cdot AB= AP \cdot AC$. Now, we will prove that $XMDPY$ is cyclic. It suffices to show that $XMDY$ is cyclic, since the total conclusion follows from symmetry. Remark that $\angle XDM=180^{\circ}-\angle MDE=\angle B$ and $\angle XPM=180^{\circ}-\angle MPC=\angle B$ by the previously established cyclicity. To prove the parallelism, we check that $\angle YXE=\angle FED$. Angle chasing yields $$\angle YXD=\angle YMD=180^{\circ}-\angle BMD=\angle BED=\angle FED$$ as desired$.\;\blacksquare$

Let the second intersections of $\omega_1$ and $\omega_2$ with sides $AB$ and $AC$ be $X'$ and $Y'$. We claim that $XX'DYY'$ is cyclic. This is true since $X'Y'D=180-\angle B-\angle DY'C=180-\angle B-\angle DFE=\angle X'XD$. However, we know that $X'Y'CB$ is cyclic. So $XY$ is antiparallel to $X'Y'$, which is antiparallel to $BC$. It follows that $XY\parallel BC$. Other configurations can be handled similarly.$\Box$

We shall use the configuration depicted below. Other cases are "similarly" handled. [asy][asy]import graph; size(10cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-3.715082517158455,xmax=8.57101545322826,ymin=-0.8000376867959583,ymax=6.082084462204611; pen ubqqys=rgb(0.29411764705882354,0.,0.5098039215686274), uuuuuu=rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666), yqqqqq=rgb(0.5019607843137255,0.,0.), qqffff=rgb(0.,1.,1.); pair A=(0.4149689204417927,5.0446348930406835), B=(-0.8700481876639894,0.45559198934964207), C=(4.982241647714512,0.4482712566770508), D=(1.0490909090909089,2.889090909090912), X_1=(1.8477622637606381,0.17420124349384086), F=(1.537023192408008,0.4525809412653818), X=(0.5330076578212042,5.4661738956848795), Y=(-0.004570323506018968,5.466846361472242), T=(1.6558161096067847,3.7958842185943857); D(A--B--C--cycle,blue); D(A--B,blue); D(B--C,blue); D(C--A,blue); D(CR((0.5886467082123572,1.2787144125300975),1.6749093476359163),gray); D(CR((3.2616521993693435,2.0650642018842325),2.3610267845639625),gray); D(Y--X,blue); D(CR((0.26247397537495465,4.071780606014076),1.420394705571974),yellow); D(T--D,yqqqqq); D((-0.2304653826081835,2.739665029583861)--T); D(A--X_1,linetype("2 2")+qqffff); D(Y--A); D(X--A); D(F--X,yqqqqq); D((2.0452777254599024,0.4519451566971582)--Y,yqqqqq); D((-0.2304653826081835,2.739665029583861)--D,yqqqqq); D(A,linewidth(3.pt)+blue); MP("A",(0.6,5.017183344460874),NE*lsf,fp+blue); D(B,linewidth(3.pt)+blue); MP("B",(-0.8064720015300348,0.5509562685505738),NE*lsf,fp+blue); D(C,linewidth(3.pt)+blue); MP("C",(5.0425371227937825,0.5509562685505738),NE*lsf,fp+blue); D(D,linewidth(3.pt)+red); MP("D",(1.2120719082230216,2.8238049228394453),NE*lsf,fp+red); D((2.0452777254599024,0.4519451566971582),linewidth(3.pt)+ubqqys); MP("E",(2.149820653698851,0.4873800824166193),NE*lsf,fp+ubqqys); MP("\omega_1",(-1.3468695836686484,1.7589038050957084),NE*lsf,fp+gray); D(X_1,linewidth(3.pt)+uuuuuu); MP("\omega_2",(3.3418741437104984,3.6820834356478307),NE*lsf,fp+gray); D(F,linewidth(3.pt)+ubqqys); MP("F",(1.545846885426283,0.4873800824166193),NE*lsf,fp+ubqqys); D(X,linewidth(3.pt)+ubqqys); MP("X",(0.5922040934169649,5.557580926599487),NE*lsf,fp+ubqqys); D(Y,linewidth(3.pt)+ubqqys); MP("Y",(0.05180651127835134,5.557580926599487),NE*lsf,fp+ubqqys); D((-0.2304653826081835,2.739665029583861),linewidth(3.pt)+uuuuuu); MP("S",(-0.45680297779328494,2.8238049228394453),NE*lsf,fp+uuuuuu); D(T,linewidth(3.pt)+uuuuuu); MP("T",(1.720681397294658,3.8887060405831826),NE*lsf,fp+uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $S$ and $T$ be the second intersections of $\omega _1$ with $AB$ and $\omega _2$ with $AC$ respectively. Note that $A$ lies on the radical axis of the circles $\omega _1$ and $\omega _2$, so $A$ has same power wrt these circles. Thus we have $AS\cdot AB=AT\cdot AC$, so that $STCB$ is cyclic. Now we have the following equality of angles:\begin{align*} \angle SXD=&\angle BXF\\ =&\angle XFC-\angle XBF\\ =&\angle DFC-\angle SBC\\ =&\angle DTA-\angle STA\qquad\text{[since quadrilaterals DFCT and SBCT are cyclic]}\\ =&\angle STD\end{align*}Thus points $S,X,D,T$ are concyclic. Similarly it may be shown that $S,Y,D,T$ are concyclic; thus it follows that $S,T,X,Y$ are concyclic (all lie on the circumcircle of $\triangle SDT$). Then by power of point we have $$AY\cdot AT=AX\cdot AS\implies \frac{AX}{AY}=\frac{AT}{AS}\quad (1).$$From cyclic quadrilateral $SBCT$, we also have $$AS\cdot AB=AT\cdot AC\implies \frac{AB}{AC}=\frac{AT}{AS}\quad (2).$$From $(1)$ and $(2)$, we obtain that $\frac{AX}{AY}=\frac{AB}{AC}$. By the converse of Thales' Theorem, this implies $XY||BC$, as desired. $\blacksquare$

JuanOrtiz wrote: Let $w_1 \cap AB =G$, $w_2 \cap AC=H$. Let $Y'$ be the point on $AC$ such that $XY' || BC$. Notice $\angle Y'HD = \angle DFB = \angle DXY'$ and so $XDY'H$ is cyclic. Notice also $\frac{AX}{AY'} = \frac{AB}{AC} = \frac{AH}{AG}$ since $GHBC$ is cyclic. Therefore $GHDXY'$ is cyclic. So $\triangle GHD$'s circumcircle passes through $X$, and through $Y$ analogously. Then $Y'=Y$ so we're done. That's exactly my solution too Nice and simple

We can also use barycentrics on $\triangle ABC$. Let $Z=\omega _1\cap \omega _2$. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1), D=(p:q:r),Z=(s:q:r)$ for some $p,q,r,s$. Now if $\omega _1$ has the standard equation $a^2yz+b^2zx+c^2xy=(x+y+z)(ux+vy+wz)$, we get that $v=0$ and $u,w$ satisfy these two equations: \[a^2qr+b^2rp+c^2qp=(p+q+r)(pu+rw)\text{ and }a^2qr+b^2rs+c^2qs=(s+q+r)(su+rw).\]To find $E$ we need to solve for $w$. One way to do so is to subtract the two equations, factor out $(p-s)$, then plug this equation into one of the originals; computation is not bad and is omitted. We get $w=\tfrac{a^2qr(p+q+r+s)+b^2prs+c^2pqs}{(p+q+r)(s+q+r)r}.$ Now if $E=(0,e,1-e)$ we get $a^2e(1-e)=w(1-e)\implies E=(0:w:a^2-w)$. Let's not plug $w$ into this; as we shall see, we're almost done. Now we need to find $Y$. Let $Y=(y,0,1-y)$, then \[\begin{vmatrix} y & 0 & 1-y \\ p & q & r \\ 0 & w & a^2-w \end{vmatrix}=0\iff w(p-y(p+q+r))+a^2qy=0.\]After multiplying by $r$, this is symmetric in $q$ and $r$, so we are done.

Nice problem. Here's my solution: Note that the problem states that $A$ lies on the radical axis of $\omega_1$ and $\omega_2$. This motivates us to consider the intersection of $\omega_1$ with $\overline{AB}$ and $\overline{AC}$. Call these intersections as $K$ and $L$, and let $\odot (DXY)=\gamma$. By the converse of Reim's Theorem on $\omega_1$ and $\gamma$, it suffices to show that $K$ lies on $\gamma$. Now, $$\angle KXD=\angle CFD-\angle FBX=\angle ALD-\angle ALS=\angle KLD$$This means that $X$ lies on $\odot (DKL)$. Similarly we can show that $Y$ lies on $\odot (DKL)$. Together these two facts give that $K$ lies on $\gamma$. Hence, done. $\blacksquare$

Here is another solution using barycentric coordinates. We take $\triangle ABC$ as the reference triangle (Unlike the solution by @v_Enhance , which takes $\triangle DEF$). Let $A=(1,0,0), B=(0,1,0), C=(0,0,1)$. Let $\omega_1: -a^2yz-b^2zx-c^2xy + \left( x+y+z \right)\left( s_1 x +t_1 z \right) = 0$ and $\omega_2: -a^2yz-b^2zx-c^2xy + \left( x+y+z \right)\left( s_2 x +t_2 y \right) = 0$. As their radical axis passes through $A$, $s_1=s_2=s$ and thus the equation of radical axis is $t_1z=t_2y$. Thus, let $D=(x_0:t_1:t_2)$. By putting $x=0$ in $\omega_1$ , we obtain $E=(0:t_1:a^2-t_1)$. Let $Y=(x_1,0,1-x_1)$. $$\begin{vmatrix} x_1 & 0 & 1-x_1\\ 0 & t_{1} & a^2-t_1 \\ x_0 & t_{1} & t_{2}\\ \end{vmatrix}=0\implies x_1=\frac{x_0}{x_0+t_{1}+t_{2}-a^2}$$Similarly, if $X=(x_2,1-x_2,0),\ x_2=\frac{x_0}{x_0+t_{1}+t_{2}-a^2}$. Therefore, $x_1=x_2$, whence, $XY\parallel BC$. $\blacksquare$

Just angle chasing?! I feel like this might be too simple... Please let me know if I messed up. Let $w_{1}$ intersect $BC$ at $Z (\not\equiv B)$, and $w_{2}$ intersect $BC$ at $W$. $BZWC$ cyclic by Pop. So, $\angle ACB = \angle AZW = \angle XDW$. So, $XZDW$ cyclic. Similarly, $YXDW$ cyclic, which means that $\frac{AY}{AX} = \frac{AC}{AB}$, completing our proof.

Moving points work too by showing the stronger result: Stronger USA TST 2004/4 wrote: Let $ABC$ be a triangle and $Z \in \overline{BC}$ be a point. Let $E \in \overline{BZ},F \in \overline{ZC}$ be two points such that $ZE \cdot ZB=ZF \cdot ZC.$ Let $D$ be an arbitrary point in $\overline{AZ}.$ Let $X=AB \cap FD$ and $Y=AC \cap ED.$ Then $XY \parallel BC.$ Move $D$ on $AZ.$ We need to show $X \mapsto D \mapsto Y$ is the same map as $X \mapsto X\infty_{BC} \cap AC,$ so need to check it for three positions. The case $D=A$ is obvious. The cases when $D=\infty_{AZ}$ and $D=AZ \cap E\infty_{AC}$ are easy by using the hypothesis $ZE \cdot ZB=ZF \cdot ZC.$

Seems like a new (possibly cleaner? :O) bary sol. Let $D=(1,0,0), B=(0,1,0), C=(0,0,1), BC=a, DC=b, DB=c$. In the circle equation, let $\omega_1: (0,0,w)$, $\omega_2: (0,v,0)$, so that the radical axis of these two circles is $(t:w:v)$ and so we can set $A=(x:w:v)$. Plugging $x=0$ into $\omega_1$, we get $$a^2yz=(y+z)wz \implies (a^2-w)y=zw \implies E=(0:w:a^2-w)$$Intersecting the cevians $DE: (t:w:a^2-w)$ and $CA: (x:w:t)$ we get $Y=(x:w:a^2-w)$. Similarly, $X=(x:a^2-v:v)$ and we're clearly done. They're all cevians

[asy][asy] size(8cm); defaultpen(fontsize(9pt)); import geometry; pair A = dir(110), B = dir(180+30), C = dir(-30), D = (-0.2, 0.05), D1 = 1.3*D - 0.3*A; path w1 = circumcircle(B, D, D1), w2 = circumcircle(C, D, D1); pair E = intersectionpoints(line(B, C), w1)[1], F = intersectionpoints(line(B, C), w2)[0], G = intersectionpoints(line(A, B), w1)[0], H = intersectionpoints(line(A, C), w2)[0], X = extension(D, F, A, B), Y = extension(D, E, A, C); draw(A--B--C--cycle, linewidth(0.9)); draw(w1^^w2, heavygray); draw(circumcircle(B, G, H), dotted); draw(circumcircle(G, H, D), dashed); draw(E--Y^^F--X, gray); string[] names = {"$A$", "$B$", "$C$", "$D$", "$D_1$", "$E$", "$F$", "$G$", "$H$", "$X$", "$Y$"}; pair[] points = {A, B, C, D, D1, E, F, G, H, X, Y}; pair[] ll = {A, B, C, D, D1, E, F, G, H, X, Y}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] Let $\omega_1, \omega_2$ intersect the sides $AB,AC$ again at $G,H$ as shown. Since \[ AD \cdot AD_1 = AG \cdot AB = AH \cdot AC, \]$BGHC$ is cyclic. The key observation is that $XYHDG$ is cyclic, which would clearly imply the problem (since then $\measuredangle AXY = \measuredangle GHY = \measuredangle ABC$). But this follows from \[ \measuredangle XDH = \measuredangle FDH = \measuredangle BCH = \measuredangle BGH = \measuredangle XGH. \]

Use barycentrics on triangle $DEF,$ and let $P=(p:m:n)$ and $A=(a_1:m:n).$ Notice that $$\omega_1:-a^2yz-b^2zx-c^2xy+(x+y+z)(\lambda z)$$where $\lambda=\frac{a^2mn+b^2pn+c^2pm}{(p+m+n)n}.$ Letting $B=(0,b_1,1-b_1),$ we find that $-a^2b_1(1-b_1)+\lambda(1-b_1)=0$ or $b_1=\frac{\lambda}{a^2}.$ Hence, $B=(0:\lambda:a^2-\lambda).$ Let $X=(x_1:0:x_2)$ and notice that \begin{align*}\begin{vmatrix}x_1&0&x_2\\a_1&m&n\\0&\lambda&a^2-\lambda\end{vmatrix}&=x_1\begin{vmatrix}m&n\\ \lambda&a^2-\lambda\end{vmatrix}+x_2\begin{vmatrix}a_1&m\\0&\lambda\end{vmatrix}\\&=-x_1(\lambda m+\lambda n-ma^2)+x_2a_1\lambda\\&=0\end{align*}so $X=(a_1:0:m+n-ma^2/\lambda).$ Similarly, $Y=(a_1:m+n-na^2/(\lambda n/m):0)$ so we are done. $\square$

Let $AB \cap \omega_1=X_1$ and $AC \cap \omega_2=Y_1$ Also let our second our intersection be $K$ Its trivial to see that $BCY_1X_1$ is cyclic by Pop $$AK.AD=AX_1 \cdot AB=AY_1 \cdot AC$$Now angle chase finishes the problem $$\angle X_1DY_1=\angle ABC=\angle YY_1X_1 \implies XX_1Y_1Y \text{is cyclic} \implies XY||BC$$

This works right?

People like to overcomplicate this. Quote: Let $D$ be inside $\triangle ABC$ and $E$ on $AD$ different to $D$. Let $\omega_1$ and $\omega_2$ be the circumscribed circles of $\triangle BDE$ and $\triangle CDE$ respectively. $\omega_1$ and $\omega_2$ intersect $BC$ in the interior points $F$ and $G$ respectively. Let $X$ be the intersection between $DG$ and $AB$ and $Y$ the intersection between $DF$ and $AC$. Show that $XY$ is $\|$ to $BC$. Let $P=\overline{DE}\cap\overline{BC}$. By PoP, $\frac{PB}{PG}=\frac{PC}{PF}$. Let $Y'$ be the intersection of line through $X$ parallel to $\overline{BC}$ and $\overline{DF}$. Then considering following composition of homotheties, \begin{align*} \overline{BC}\overset{P}{\rightarrow}\overline{GF}\overset{D}{\rightarrow}\overline{XY'}\overset{\overline{BX}\cap\overline{CY'}}{\rightarrow}\overline{BC}, \end{align*}thus $P,D,\overline{BX}\cap\overline{CY'}$ are collinear, hence $Y\equiv Y'$. We are done.

Let $\omega_1$ meet $AB$ at $P$ and $\omega_2$ meet $AC$ at $Q$ and Let our circles meet at $S$. Note that $AP.AB = AS.AD = AQ.AC$ so $PQCB$ is cyclic so by simple angle chasing we just need to prove $PQYX$ is cyclic. we have $\angle PDY = \angle ABC = \angle AQP \implies PQYD$ is cyclic. $\angle QDX = \angle ACB = \angle APQ \implies QPXD$ is cyclic so now we have $PQYDX$ is cyclic. we're Done.

Let $\omega_1$ and $\omega_2$ meet again at $T$, $K = BT \cap \overline{DFX}$, and $L = CT \cap \overline{DEY}$. Observe $$\measuredangle KDL = \measuredangle FDE = \measuredangle DFE + \measuredangle FED = \measuredangle DFC + \measuredangle BED$$$$= \measuredangle DTC + \measuredangle BTD = \measuredangle BTC = \measuredangle KTL$$so $DKTL$ is cyclic. This yields $$\measuredangle DFE = \measuredangle DFC = \measuredangle DTC = \measuredangle DTL = \measuredangle DKL$$which means $\overline{BEFC} \parallel KL$. It's clear that $ABC$ and $DKL$ are centrally perspective at $T$. Now, Desargues Theorem implies $AB \cap DK = X$, $AC \cap DL = Y$, and $BC \cap KL = \infty_{BC}$ are collinear, which finishes. $\blacksquare$ Remark: This is the first time I've successfully applied Desargues .

Let $K=\overline{AD}\cap\overline{BC}$. Then apply DDIT to $A,EFXY$, and project onto $\overline{BC}$, to obtain the involutive pairs $i:(E,X),(F,Y),(K,\overline{XY}\cap\overline{BC})$. Because $KE\cdot KX=KF\cdot KY$, $i$ must be an inversion at $K$ with that product as its power. Hence $K\overset{i}\leftrightarrow\infty_{BC}$, implying $\overline{XY}\cap\overline{BC}=\infty_{BC}$ as desired.

Let $AD$ intersect $BC$ at $P$. Note that the radical axis of $\omega_1$ and $\omega_2$ is line $AP$, so we have $$PB\cdot PE=PC\cdot PF.$$Save this equation for now. We will employ barycentric coordinates. Suppose that $D=(p,q,r)$, $E=(0,e,1-e)$, and $F=(0,1-f,f).$ The equation for line $ED$ is $$[q(1-e)-er]x+[p(e-1)]y+epz=0,$$which we see $(p,q,r)$ and $(0,e,1-e)$ satsifies. Intersecting this with line $AC,$ we see that $Y$ satisfies $$[q(1-e)-er]x+epz=0.$$Note that $$\frac{AY}{CY}=\frac{\text{z coordinate of point Y}}{\text{x coordinate of point Y}}=\frac{er-q(1-e)}{ep}.$$By symmetry, we then have $$\frac{AX}{BX}=\frac{fq-r(1-f)}{fp}$$Therefore, $XY\parallel AB$ is equivalent to $$\frac{er-q(1-e)}{ep}=\frac{fq-r(1-f)}{fp}$$$$fer-fq+feq=feq-re+fer$$$$fq=re.$$ Now, we return to the equation $$PB\cdot PE=PC\cdot PF.$$Note that $P=(0:q:r)$, so $$PB=\frac{r}{q+r}a$$and $$PE=(\frac{q}{q+r}-e)a,$$so $$PB\cdot PE=a^2(\frac{r}{q+r})(\frac{q}{q+r}-e).$$Similarly, $$PC\cdot PF=a^2(\frac{q}{q+r})(\frac{r}{q+r}-f).$$Thus, we have $$(\frac{r}{q+r})(\frac{q}{q+r}-e)=(\frac{q}{q+r})(\frac{r}{q+r}-f)$$$$r(q-eq-er)=q(r-fq-fr)$$$$qr-eqr-er^2=qr-fqr-fq^2$$$$eqr+er^2=fqr+fq^2$$$$er(q+r)=fq(q+r)$$$$er=fq,$$as desired.

We use barycentric coordinates with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Let $D=:(p,q,r)$ and let the other intersection of $\omega_1$ and $\omega_2$ be $W=(t:q:r)$. Then $\omega_1$ is given by \[ -a^2yz-b^2zx-c^2xy+(x+y+z)(u_1x+wz)=0. \]It follows that \begin{align*} pu_1+rw&=a^2qr+b^2rp+c^2pq\\ tu_1+rw&=\frac{a^2pr+b^2rt+c^2tq}{t+q+r}. \end{align*}In particular, note that if $\omega_2$ is given by \[ -a^2yz-b^2zx-c^2xy+(x+y+z)(u_2x+vy)=0, \]we have $u_1=u_2$ by symmetry (the only thing that is different is $rw$ being replaced with $qv$). Thus $rw=qv$. Letting $x=0$ in $\omega_1$ gives $y=\frac{w}{a^2}$ so $E=\left(0,\frac{w}{a^2},1-\frac{w}{a^2}\right)$. Let $X=:(x_1,0,x_2)$ so \[ \begin{vmatrix} x_1 & 0 & x_2\\ p & q & r\\ 0 & \frac{w}{a^2} & 1-\frac{w}{a^2} \end{vmatrix} =0. \]Solving for $x_2$, we get $x_2=1-\frac{pw}{qa^2}$. Letting $Y=:(y_1,y_2,0)$, we similarly get $y_2=1-\frac{pv}{ra^2}$. Since $rw=qv$, we have $x_2=y_2$ and $x_1=y_1$. Thus $\overrightarrow{XY}=(0,y_2,-x_2)$ is parallel to $\overrightarrow{BC}=(0,-1,1)$. $\square$

Suppose that $\omega_1$ intersects $AB$ at $U,$ $\omega_2$ intersects $CA$ at $V,$ $\omega_1$ intersects $\omega_2$ at the second point $S$. We have $\overline{AU} \cdot \overline{AB} = \overline{AD} \cdot \overline{AS} = \overline{AV} \cdot \overline{AC}$. Then $B, C, U, V$ lie on a circle. We also have $\angle{UXD} = 180^{\circ} - \angle{ABC} - \angle{XFB} = \angle{UVC} - \angle{DVC} = \angle{UVD}$. Then $X, U, V, D$ lie on a circle. Similarly, $Y, U, V, D$ lie on a circle. So $X, Y, U, V$ lie on a circle. Hence $\angle{AYX} = \angle{AUV} = \angle{ACB}$ or $XY \parallel BC$

Rather standard. We let $M$ and $N$ be the second intersections of circles $\omega_1$ and $\omega_2$ with $\overline{AB}$ and $\overline{AC}$ respectively. It is not hard to see that $BMCN$ is cyclic since, \[AB \cdot AM = AD \cdot AP = AN \cdot AC\]Further, we can prove the following key claim. Claim : Points $M$ , $N$ , $D$ , $X$ and $Y$ lie on a circle. Proof : Simply note that, \[\measuredangle MXD = \measuredangle BXF = \measuredangle BFX + \measuredangle XBF = \measuredangle EFD + \measuredangle MBF = \measuredangle CFD + \measuredangle MBC = \measuredangle CND + \measuredangle MNC = \measuredangle MND \]from which it is clear that $MXND$ is cyclic. A similar argument show that $MYND$ is also cyclic, proving the claim. Now, the result follows by Reim's Theorem.