Define $W$ as the point on $\Omega$ such that $\overline{BW} \parallel \overline{DD}$. Let $X = \overline{WD} \cap \overline{AC}$. By Pascal on $ABWDDC$, lines $EX$, $BW$, and $CD$ concur at a point $Y$. The point $Z = \overline{BW} \cap \overline{XF} \cap \overline{AD}$ is defined similarly. Triangles $DYZ$ and $TEF$ are homothetic, and the homothety center is $\overline{EY} \cap \overline{FZ} = X$. Thus $X$ lies on $\overline{DTW}$. As $DB=DW$ and $TK=TD$, that homothety should map $B$ to $K$, as needed. [asy][asy] import graph; size(11cm); pen yqqqqq = rgb(0.50196,0.,0.); pen zzttqq = rgb(0.6,0.2,0.); pen ttffqq = rgb(0.2,1.,0.); pen qqzzff = rgb(0.,0.6,1.); pen ffdxqq = rgb(1.,0.84313,0.); pen xfqqff = rgb(0.49803,0.,1.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw((-0.88361,-0.46821)--(-0.48947,0.87201)--(0.99165,-0.12888)--(0.,-1.)--cycle, linewidth(0.6) + zzttqq); draw(circle((0.,0.), 1.), linewidth(0.6) + yqqqqq); draw((-0.88361,-0.46821)--(-0.48947,0.87201), linewidth(0.6) + zzttqq); draw((-0.48947,0.87201)--(0.99165,-0.12888), linewidth(0.6) + zzttqq); draw((0.99165,-0.12888)--(0.,-1.), linewidth(0.6) + zzttqq); draw((0.,-1.)--(-0.88361,-0.46821), linewidth(0.6) + zzttqq); draw((-0.88361,-0.46821)--(0.99165,-0.12888), linewidth(0.6) + yqqqqq); draw((0.,-1.)--(0.31009,0.18597), linewidth(0.6) + ttffqq); draw((0.31009,0.18597)--(-1.03999,-1.), linewidth(0.6) + qqzzff); draw((0.31009,0.18597)--(2.28072,-1.), linewidth(0.6) + qqzzff); draw((-1.03999,-1.)--(-0.88361,-0.46821), linewidth(0.6) + yqqqqq); draw((0.99165,-0.12888)--(2.28072,-1.), linewidth(0.6) + yqqqqq); draw((0.31009,0.18597)--(0.48947,0.87201), linewidth(0.6) + ttffqq); draw((-1.03999,-1.)--(2.13107,0.87201), linewidth(0.6) + ffdxqq); draw((2.28072,-1.)--(-3.11056,0.87201), linewidth(0.6) + ffdxqq); draw((2.13107,0.87201)--(-3.11056,0.87201), linewidth(0.6) + xfqqff); draw((-3.11056,0.87201)--(-0.88361,-0.46821), linewidth(0.6) + yqqqqq); draw((2.13107,0.87201)--(0.99165,-0.12888), linewidth(0.6) + yqqqqq); draw((-1.03999,-1.)--(2.28072,-1.), linewidth(0.6) + qqzzff); dot("$A$", (-0.88361,-0.46821), dir((-7.580, -4.254))); dot("$B$", (-0.48947,0.87201), dir((1.427, 3.388))); dot("$C$", (0.99165,-0.12888), dir((7.084, -1.217))); dot("$D$", (0.,-1.), dir((-1.389, -6.694))); dot("$E$", (-1.03999,-1.), dir((-1.757, -7.348))); dot("$F$", (2.28072,-1.), dir((-2.079, -7.348))); dot("$T$", (0.31009,0.18597), dir((-5.898, 2.632))); dot("$K$", (0.62019,-1.), dir((-1.574, -6.039))); dot("$W$", (0.48947,0.87201), dir((1.356, 3.388))); dot("$X$", (0.18983,-0.27397), dir((1.220, 2.496))); dot("$Y$", (2.13107,0.87201), dir((1.435, 2.733))); dot("$Z$", (-3.11056,0.87201), dir((1.146, 2.733))); [/asy][/asy] Remark: The motivation from this solution came from extending $\overline{DT}$ to $\Omega$ and realizing from the diagram that $\overline{BW} \parallel \overline{DD}$. After this, trying Pascal is natural, leading to $Y$ and $Z$, and the rest of the solution flows from there.

Let $T'$ be the reflection of $T$ across $EF$. Notice that $B,E,F,T'$ are concyclic, and $$ \measuredangle T'BF = \measuredangle T'EF = \measuredangle FET = \measuredangle(DD,CD) = \measuredangle DBC = \measuredangle DBF,$$so $T'\in BD$. Therefore, if we let $B'$ the the point on $\odot(ABC)$ such that $BB'\parallel EF$, we get that $D,B',T$ are colinear. Now, let $X = BK\cap \Omega$. It suffices to show that there is an involution swapping $(A,C)$, $(D,B')$, $(B,X)$. Projecting this involution to $EF$, we need an involution swapping $(E,F)$, $(D,\infty)$, and $(BB\cap DD, K)$. Letting $Y=BB\cap DD$, we notice that $\angle YBD = \angle YDB = \angle T'DK = \angle T'KD$ or $B,K,T',Y$ are concyclic. Hence $$DE\cdot DF = DB\cdot DT' = DY\cdot DK,$$implying the conclusion.

[asy][asy] unitsize(1.5cm); pair B, E, F, D, A, C, T, K, P, Q; B=(2,3sqrt(5)); E=(0,0); F=(8,0); D=(4,0); A=abs(E-D)*abs(E-D)/abs(E-B)/abs(E-B)*(B-E)+E; C=abs(F-D)*abs(F-D)/abs(F-B)/abs(F-B)*(B-F)+F; T=extension(E,E+D-C,F,F+A-D); P=extension(E,T,A,C); Q=extension(F,T,A,C); K=2*foot(T,E,F)-D; draw(circle(extension(D,D+(0,1),(B+D)/2,(B+D)/2+(B-D)*(0,1)),abs(extension(D,D+(0,1),(B+D)/2,(B+D)/2+(B-D)*(0,1))-D))); draw(circumcircle(T,D,K)); draw(D--B--E--F--B--A--D--C--A--D--T--E--T--F--T--K--Q--K--B); label("$B$", B, NW); label("$E$", E, SW); label("$F$", F, SE); label("$D$", D, S); label("$A$", A, W); label("$C$", C, dir(0)); label("$T$", T, N); label("$K$", K, SE); label("$P$", P, NW); label("$Q$", Q, NE); [/asy][/asy] Let $P$ be the intersection of $ET$ and $AC$, and let $Q$ be the intersection of $FT$ and $AC$. We have \begin{align*} \angle PAD&=\angle CDF\\ &=\angle PED, \end{align*}so $APDE$ is cyclic. Similarly, $CQDF$ is cyclic. If these two circles intersect at $D$ and $D'$, then \begin{align*} \angle DD'E+\angle DD'F&=\angle BAD+\angle BCD\\ &=180^{\circ}, \end{align*}so $D=D'$. Now, we have \begin{align*} \angle QFD&=\angle QCD\\ &=\angle TPQ\\ &=180^{\circ}-\angle EPQ, \end{align*}which means $EFQP$ is cyclic. Now, we get \begin{align*} \angle PQT&=\angle TEF\\ &=\angle PDF \end{align*}since $DT$ is tangent to the circumcircle of $APDE$. Since \begin{align*} \angle DKT&=\angle KDT\\ &=180^{\circ}-\angle DCF\\ &=\angle DQT \end{align*}we have that $DKQT$ is cyclic. Therefore, $DKQTP$ is a cyclic pentagon. Now, since we have \begin{align*} \angle BDE&=\angle BCD\\ &=180^{\circ}-\angle DCF\\ &=\angle TQD\\ &=\angle TKD. \end{align*}Therefore, $BD\parallel TK$. We also have \begin{align*} \angle QKD&=180^{\circ}-\angle DPQ\\ &=\angle APD\\ &=180^{\circ}-\angle BEF. \end{align*}This means that $QK\parallel BE$, so $BAD$ and $KQT$ are homothetic. The center of homothety is the point of concurrency of $BK$, $AQ$, and $TD$, which means that $BK$, $AC$, and $DT$ concur.

Let $FT$ and $ET$ meet $AC$ at $X$ and $Y$, respectively. We begin with some Claims. Claim 1: $DXCF$ and $DEAY$ are cyclic. Proof: Note that $\angle DFT=\angle EDA=\angle DCX$, which implies that $DXCF$ is cyclic. Similarly $DEAY$ is cyclic, too $\blacksquare$ Let $ET$ meet $CB$ at $P$ and $FT$ meet $AB$ at $Q$. Claim 2: $FDQB$ and $DEBP$ are cyclic. Proof: Note that $\angle DBQ=\angle DCA=\angle XFD=\angle QFD,$ and so $FDQB$ is cyclic. Similarly $DEBP$ is cyclic, too. $\blacksquare$ Claim 3: $EDTQ$ and $FDTP$ are cyclic. Proof: Note that $\angle DET=\angle DEY=\angle DAC=\angle DBF=\angle DQT,$ and so $EDTQ$ is cyclic. Similarly $FDTP$ is cyclic, too $\blacksquare$ Now, we have all we need for Claim 4: Points $D,K,X,T,Y$ are concyclic. Proof: Note that $\angle DXT=180^\circ-\angle DXB=180^\circ-\angle DCF=\angle DCB=\angle DAE=\angle DYE,$ and so $DXTY$ is cyclic. Moreover, $\angle DKT=\angle FDT=\angle TPB=\angle DCB=180^\circ-\angle DCF=180^\circ-\angle DXF=\angle DXT,$ and so $K \in (D,X,T,Y)$, as desired $\blacksquare$ Now, note that $\angle KYT=\angle KDT=\angle DKT=\angle DXT=\angle DCB=\angle YPB,$ hence $KY \parallel CB$. In addition, $\angle DKT=\angle DYE=\angle DAE=\angle DCB=\angle EDB,$ hence $KT \parallel DB$. Trivially $DC \parallel YT$, and so triangles $KYT$ and $DCB$ have pairwise parallel sides, which means that $KB,YC \equiv AC,TD$ concur (an easy way to see this is via Desargues' theorem, or alternatively some easy Thales applications). Thus, we are done.

Let $X=\overline{TE}\cap\overline{AC}$ and $Y=\overline{TF}\cap\overline{DC}.$ $\textbf{Claim: }$ $AXDE$ and $CYDF$ are cyclic. $\emph{Proof: }$ Since $\overline{TF}\parallel\overline{AD},$\[\angle YFD=\angle ADE=\frac{1}{2}\widehat{AD}.\]Hence \begin{align*} \angle CFY &=\angle CFD-\angle YFD\\ &=\frac{1}{2}(\widehat{AB}-\widehat{CD})-\frac{1}{2}\widehat{AD}\\ &=\frac{1}{2}(\widehat{AB}-\widehat{CD}). \end{align*}Now from $\triangle CYF,$ \begin{align*} \angle CYF &= 180^\circ-\angle YCF-\angle CFY\\ &= \angle YCB-\angle CFY\\ &= \frac{1}{2}\widehat{AB}-\frac{1}{2}(\widehat{AB}-\widehat{CD})\\ &=\frac{1}{2}\widehat{CD}\\ &=\angle CDF, \end{align*}implying that $CYDF$ is cyclic. $CYDF$ is cyclic by the same reasoning, so we are done. $\blacksquare$ $\textbf{Claim: }$ $TXDKY$ is cyclic $\emph{Proof: }$ Write \begin{align*} \angle TXD &= 180^\circ-\angle DXE\\ &= 180^\circ-\angle DAE\\ &= \angle DAB\\ &= 180^\circ-\angle DCB\\ &= \angle DCF\\ &= \angle DYF\\ &= 180^\circ-\angle DYT, \end{align*}so $TXDY$ is cyclic. Next, since $\angle TYX=\angle TEF$ and $\angle TXY=\angle TFE,$ we know $\overline{XY},\overline{EF}$ are reflections about the angle bisector of $\angle ETF.$ Thrfore, the circumcenter of $\triangle TXY,$ say $O,$ lies on the $T$-altitude of $\angle TEF.$ Thus $OD=OK,$ so $K$ lies on $TXDY,$ as desired. $\blacksquare$ Finally, observe the following: \[\angle TYX=\angle CYF=\angle CDF=\angle DAC,\]\[\angle TXY=\angle AXE=\angle ADE=\angle DCA,\]\[\angle KYX=180^\circ-\angle KDX=\angle XDE=180^\circ-\angle XAE=\angle BAC,\]\[\angle KXY=\angle KDY=180^\circ-\angle FCY=\angle BCA.\]These imply that $YKXT\sim ABCD,$ where corresponding sides are parallel. Hence, these quadrilaterals are perspective from a point, so $\overline{AY},\overline{BK},\overline{CX},\overline{DT}$ concur.

Let $M\equiv TE\cap AC$ and $N\equiv TF\cap AC$, by angle chasing it's easy prove that $TMDKN$ is cyclic. By ratio's lemma in $\triangle ABC$ and $\triangle ADC$ we get: $$\frac{AL}{LC}=\frac{\sin{\angle ABL}\cdot \sin{\angle ACB}}{\sin{\angle CBL}\cdot \sin{\angle BAC}}=\frac{\sin{\angle ADL}\cdot \sin{\angle ACD}}{\sin{\angle CDL}\cdot \sin{\angle CAD}}$$, then $$\frac{\sin{\angle ABL}}{\sin{\angle CBL}}=\frac{\sin{\angle ADL}\cdot \sin{\angle ACD}\cdot \sin{\angle BAC}}{\sin{\angle CDL}\cdot \sin{\angle CAD}\cdot \sin{\angle ACB}} ...(\star)$$ $\triangle AEF$ and $\triangle TEF$ we get: $$\frac{EK}{KF}=\frac{\sin{\angle EBK}\cdot \sin{\angle BFE}}{\sin{\angle FBK}\cdot \sin{\angle BEF}}=\frac{\sin{\angle ETK}\cdot \sin{\angle TFE}}{\sin{\angle FTK}\cdot \sin{\angle TEF}}$$, then $$\frac{\sin{\angle EBK}}{\sin{\angle FBK}}=\frac{\sin{\angle ETK}\cdot \sin{\angle TFE}\cdot \sin{\angle BEF}}{\sin{\angle FTK}\cdot \sin{\angle TEF}\cdot \sin{\angle BFE}}...(\star\star)$$ , but by angle chasing we get $\angle ACB=\angle FTK$, $\angle BAC=\angle ETK$, $\angle ACD=\angle TFE$, $\angle CAD=\angle REF$, $\angle ADL=\angle BEF$ and $\angle CDL=\angle BFE$. Finally, by $(\star)$ and $(\star\star)$ we get $$\frac{\sin{\angle ABL}}{\sin{\angle CBL}}=\frac{\sin{\angle EBK}}{\sin{\angle FBK}}\quad \text{and}\quad \angle ABL+\angle CBL=\angle EBK+\angle FBK$$, then $\angle ABL=\angle EBK$, hence $B$, $L$ and $K$ are collinear.

Let $BD$ meet $\odot (BEF)$ again at $L$ and $DT$ meet $\Omega$ again at $G.$ Let $AD\cap BG=X,$ $CD\cap BG=Y,$ $AC\cap DG=Z.$ Pascal on both $CBGDDA,$ $ABGDDA$ gives $Z=XF\cap YE.$ Observe that $\measuredangle EFL=\measuredangle EBD=\measuredangle ADE=\measuredangle TFE$ and similarly $\measuredangle FEL=\measuredangle TEF$ so $L,T$ are symmetric wrt $EF.$ Thus $DB,DG$ are also symmetric and therefore $BG\parallel EF.$ Finally $\measuredangle XYD=\measuredangle FDC=\measuredangle LEF$ and $\measuredangle YXD=\measuredangle LFE$ implying $$LEFDK\stackrel{-}{\sim} DYXGB\implies (EDKF)=(YGBX)\implies Z\in BK\text{ } \blacksquare$$

Let $P = AC\cap BD, G = ET \cap AC, J = AC\cap FT$ and $Q = BD\cap FT$ and $H = BD\cap ET$. Firstly, note that $EF$ is tangent to $(ABC)$ so $AGDE$ and $DJCF$ are cyclic quadrilaterals (by angle chasing) From the existence of the miquel point of the triangle $BEF$ we have that $(ADE)$ and $(DCF)$ are tangent to each other in $D$. Then: $$Pot_{(DCF)}(T) = TD^2 = TK^2 = TJ \cdot TF \Rightarrow \angle TDJ = \angle TKJ = \angle EFT$$ So $DKTJ$ is cyclic, so $\angle KTJ = \angle JDK$, but notice that $ABCD$ is cyclic and $AD\parallel FT \Rightarrow \angle ADB = \angle ACB = \angle DQJ$, so $QJCB$ is cyclic $\Rightarrow Pot_{(ABC)}(F) = FD^2 = FB\cdot FC = Pot_{(QJCB)}(F) =FQ \cdot FJ$, so $FD$ is tangent to $(QDJ) \Rightarrow \angle FDJ = \angle DQF = \angle FTK$ so that $KT$ is parallel to BD. Note that $\angle TGJ = \angle ADE = \angle ABD = \angle ACD = \angle JFD = \angle TDJ$, so $GDKJT$ is cyclic. Take $I = BK \cap AC, Z = (GDKJT) \cap BK$, then $\angle ZJQ = \angle TKB = \angle KBD \Rightarrow BQJCZ$ is cyclic. Now notice that using the smaller arc $ZT$ again we get $\angle ZKT = \angle ZGT = \angle ZGH$, so $GHZB$ is cyclic, but $AGHB$ is also cyclic since $\angle CGH = \angle ACD = \angle ABH$, so we get that $AGHZB$ is also a cyclic pentagon. So we have: 1. $I$ is in $AG$ which is the radical axis of $(AGHZB)$ and $(AED)$. 2. $I$ is in $BZ$ which is the radical axis of $(AGHZB)$ and $(BQZJC)$. 3. $I$ is in $CJ$ which is the radical axis of $(BQZJC)$ and $(DCF)$. So $I$ is on the radical axis of $(AED)$ and $(DCF)$, but note that $\angle TGJ=\angle GCD=\angle JCD=\angle JFK\Rightarrow EGJF$ is cyclic, so using power of point we will have $TJ\cdot TF=TG\cdot TE \Rightarrow T$ is also on the radical axis of $(AED)$ and $(DCF)$ , so this axis is $DT \Rightarrow$ lines $AC,DT,$ and $BK$ are concurrent. https://cdn.discordapp.com/attachments/983535514353291295/989231948616564837/unknown.png

A bit sketchy solution, similar to that in #7, though sine bash is mostly omitted due to laziness. I now regret I didn't see the homothety so as to avoid the bash at the end. Introduce the intersection of $TE$ and $AC$ to be $E'$ and similarly define $F'$. Easy angle chasing gives that $AE'DE, CF'DF, EE'F'F, TE'DF'$ are cyclic and that $TD$ is tangent to both $(AED), (DCF)$. Note that $\angle TKD=\angle TDK= \angle EAD= \angle DCB= \angle DF'T$, hence $K$ also lies on $(TE'F')$. Let $AC$ meet $DT$ at $P$. To prove $B-P-K$, we will prove $\frac{sin \angle ABP}{sin \angle CBP}=\frac{sin \angle EBK}{sin \angle FBK}$, which is equivalent to $\frac{AP}{PC}.\frac{BC}{BA}=\frac{KE}{KF}.\frac{BF}{BE}$, due to ratio lemma in $ABC$ and $EBF$. Now the idea is to write all the ratios in terms of $sin \angle EBD, sin \angle FBD$; $sin \angle BAC, sin \angle ACB$ and $sin \angle BEF, sin \angle BFE$. I will give only a sketch how to do so. Calculate $\frac{AP}{PC}$ with ratio lemma in triangle $ADC$, $\frac{KE}{KF}$ with ratio lemma and $\frac{TE}{TF}$ with LoS in triangle $TEF$, $\frac{BF}{BE}$ with LoS in $BEF$ and $\frac{BC}{BA}$ with LoS in $ABC$.

Let $ET$ meet $AC$ at $P$ and $FT$ meet $AC$ at $Q$. Claim $: EAPD$ and $FCQD$ are cyclic. Proof $:$ Note that $\angle PED = \angle CDF = \angle CAD = \angle PAD$. we prove the other one with same approach. Claim $: EPQF$ is cyclic. Proof $:$ Note that $\angle TPQ = \angle EPA = \angle EDA = \angle EFT = \angle DFQ$. Claim $: DPTQ$ is cyclic. Proof $:$ Note that $\angle PDQ = \angle 180 - \angle PDE - \angle QDF = \angle 180 - \angle A - \angle C = \angle B = \angle 180 - \angle D = \angle PTQ$. Claim $: DPTK$ is cyclic. Proof $:$ Note that $\angle DKT = \angle KDT = \angle 180 - \angle PDT - \angle ADP - \angle EDA = \angle 180 - \angle AEP - \angle PED - \angle EDA = \angle DAE = \angle DCB = \angle DQT = \angle 180 - \angle DPT$. Note that we need to prove $KT || DB$ for proving our collinearity. Claim $: KPTQ$ and $BCDA$ are homothetic. Proof $:$ Note that $\angle TPQ = \angle EPA = \angle EDA = \angle DCA$ and $TQP = \angle FQC = \angle FDC = \angle DAC$ so $DAC$ and $TQP$ are similar. $\angle KQP = \angle EDP = \angle BAC$ and $KPQ = \angle KDQ = \angle ACB$ so $KQP$ and $BAC$ are similar. Now that $KPTQ$ and $BCDA$ are homothetic we have $KT || BD$ as wanted. we're Done.

Let $TE \cap AC=P$ and $TF \cap AC=Q$. Claim 1 : $AEDP, CFDQ, PQEF$ are cyclic. Proof : $\angle PED= \angle TEF= \angle CDF = \angle CAD=\angle PAD$. Hence $AEDP$ is cyclic. Similarly, $CFDQ$ is cyclic too. $\angle FEP =\angle DEP= \angle DAP= \angle DAQ=\angle AQT=\angle PQT=180-\angle PQF$. Hence $PQEF$ is cyclic. $ \blacksquare$ Claim 2 : $(AEDP)$ and $ (CFDQ)$ are tangent at $D$, with $DT$ being the common tangent. Proof : $\angle ADC= 180-\angle ABC= 180 -\angle EBF =\angle BEF +\angle BFE=\angle AED +\angle CFD$. Thus gives $AEDP$ and $ CFDQ$ are tangent at $D$. Radical axis theorem on $(AEDP), (CFDQ), (PQEF)$ gives, the radical axis of $(AEDP)$ and $ (CFDQ)$ passes through $T$. But as the two circles are tangent at $D$, the radical axis will be the common tangent at $D$. Hence, $DT$ is the radical axis of $(AEDP)$ and $ (CFDQ)$. $\blacksquare$ Claim 3 : $TPDQ$ is cyclic. Proof : $\angle TDQ=\angle DCQ= \angle DCP =\angle TPC =\angle TPQ$. Hence, $TPDQ$ is cyclic. $\blacksquare$ Let $DF \cap (TPDQ)= K'$. Claim 4 : $K'=K$. Proof : $\angle TDK'= \angle TDF =180- \angle DCF$. (Due to the tangency) $\angle TK'D =\angle TQD =180-\angle DQF =\angle DCF$. Hence, $\angle TDK'=\angle TK'D \implies TD=TK'$ and as $K'$ lies on $EF$, $K'\equiv K$.$ \blacksquare$ Hence note that $K$ lies on $(TPDQ)$. Let $ BK\cap AC=X$. Claim 5 : $X$ lies on $DT$. Proof : First we prove that $X$ is the centre of a homothety mapping $\triangle BAC$ to $\triangle KQP$. Reim's theorem on $(AEBD)$ and $(QKBD)$ gives $AE \parallel QK \implies AB \parallel QK$ . Similarly, we get $BC \parallel KP$. Also lines $CA \equiv PQ$. Hence, as $BK, AQ, CP$ concur at $X$, a homothety at $X$ maps $\triangle BAC$ to $\triangle KQP$. Hence, $\frac{XA}{XC}=\frac{XQ}{XP} \implies XP \cdot XA = XQ \cdot XC$. Hence, $X$ has equal powers wrt $(AEDP)$ and $ (CFDQ)$. Hence, $X$ lies on their radical axis, which is just $DT$! $\blacksquare$ Hence, $BK, AC, DT $ concur at $X$ and hence we are done !

solved with awang11 [asy][asy] size(5cm); defaultpen(fontsize(8pt)); pair A,B,C,D,E,F,T,K,P,Q,g; B=dir(120); A=dir(210); C=dir(-30); D=dir(-80); E=intersectionpoint(A--(2A-B),D--D+2*dir(190)); F=intersectionpoint(C--(2C-B),D--D+2*dir(10)); T=intersectionpoint(E--E+3*(C-D),F--F+3*(A-D)); P=intersectionpoint(A--C,E--T); Q=intersectionpoint(A--C,F--T); g=intersectionpoint(D--T,A--C); K=intersectionpoint(g--(2g-B),E--F); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--E--F--C--cycle); draw(E--T--F); draw(circumcircle(A,P,D),blue); draw(circumcircle(D,Q,C),blue); draw(circumcircle(T,P,Q),gray); draw(T--D); dot("$A$",A,W); dot("$B$",B,dir(B)); dot("$C$",C); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F); dot("$T$",T,N); dot("$P$",P,N); dot("$Q$",Q,N); dot("$K$",K,dir(K)); dot(g,red); [/asy][/asy] As shown, let $TE, TF$ intersect $AC$ again at $P,Q$. Due to $$\measuredangle APE = \measuredangle ACD = \measuredangle ABD = \measuredangle ADE$$we have $ADEP$ cyclic, and similarly $CDFQ$ cyclic. In addition, we have $TDPQ$ cyclic because$$\measuredangle PTQ = \measuredangle CDA = \measuredangle CBA = \measuredangle PAE + \measuredangle FCQ = \measuredangle PDE + \measuredangle FDQ = \measuredangle PDQ.$$The point $K$ clearly also lies on this circle (e.g. the tangent at $T$ is parallel to $EF$). Finally, by Reim we have $KQ \parallel AE$ and $KP \parallel CF$, implying the cyclic quadrilaterals $ABCD$ and $QKPT$ are homothetic through the desired concurrency.

Let $\measuredangle$ denote directed angles $\!\!\!\mod 180^\circ$. Let $\overline{AC}$ intersect $\overline{ET}$ at $P$ and $\overline{FT}$ at $Q$. Claim: $AEDP$ and $CFDQ$ are cyclic. Proof: We have $\measuredangle APE=\measuredangle ACD=\measuredangle ADE$, so $AEDP$ is cyclic. Similarly, $CFDQ$ is cyclic. Claim: $DPTQ$ is cyclic. Proof: Since $\overline{CD} \parallel \overline{ET}$ and $\overline{AD} \parallel \overline{FT}$, we know that $\measuredangle CDA=\measuredangle ETF$. Thus, we have \begin{align*}\measuredangle PDQ=\measuredangle PDE=\measuredangle FDQ=\measuredangle PAE+\measuredangle FCQ=\\ \measuredangle CAB+\measuredangle BCA=\measuredangle CBA=\measuredangle CDA=\measuredangle ETF,\end{align*}so $DPTQ$ is cyclic. Claim: $\overline{DT}$ is tangent to the circumcircles of $AEDP$ and $CFDQ$. Proof: We have $\measuredangle TDP=\measuredangle TQP=\measuredangle DAP=\measuredangle DET$, so $\overline{DT}$ is tangent to the circumcircle of $AEDP$. Similarly, $\overline{DT}$ is tangent to the circumcircle of $CFDQ$. Claim: $DPTQK$ is cyclic. Proof: We have $\measuredangle TKD=\measuredangle FDT=\measuredangle DQF=\measuredangle DQT$, so $K$ is on the circumcircle of $DQT$, which means $DPTQK$ is cyclic. Let $\overline{DT}$ intersect $\Omega$ at $B' \neq \Omega$. We have \[\measuredangle TKD=\measuredangle TPD=\measuredangle EPD=\measuredangle EAD=\measuredangle BAD=\measuredangle BDE,\]so $\overline{TK} \parallel \overline{BD}$. Since $\overline{CD} \parallel \overline{ET}$, $\overline{AD} \parallel \overline{FT}$, and $\overline{PQ} \parallel \overline{AC}$, $CDA$ and $PTQ$ are homothetic. Since $D$ and $K$ are on the circumcircle of $PTQ$, $B$ and $B'$ are on the circumcircle of $CDA$, $\overline{TD} \parallel \overline{DB'}$, and $\overline{TK} \parallel \overline{BD}$, $B'CDAB$ and $DPTQK$ are homothetic. Therefore, lines $AC$, $BK$, and $DT$ are concurrent, so we are done.

Let $X=\overline{ET}\cap\overline{AC}$ and $Y=\overline{FT}\cap\overline{AC}.$ Then, $\measuredangle EDA=\measuredangle DCA=\measuredangle EXA$ so $EDXA$ is cyclic. Similarly, $FCYD$ is also cyclic. Hence, $$\measuredangle DYT=\measuredangle DYF=\measuredangle DCF=\measuredangle DCB=\measuredangle DAE=\measuredangle DXE=\measuredangle DXT,$$and $DXTY$ is cyclic. Since \begin{align*}\measuredangle DKT&=\measuredangle TDK=\measuredangle TDX+\measuredangle XDE=\measuredangle TYX+\measuredangle XAE=\measuredangle FYC+\measuredangle CAB\\&=\measuredangle FDC+\measuredangle CDB=\measuredangle EDB=\measuredangle DCB=\measuredangle DCF=\measuredangle DYF=\measuredangle DYT,\end{align*}$DTYK$ is cyclic and $DXTYK$ is a cyclic pentagon. Notice in the above angle chase we also showed $\measuredangle EDB=\measuredangle EKT,$ so $\overline{BD}\parallel\overline{KT}.$ Also, $$\measuredangle DKY=\measuredangle DXY=\measuredangle DXA=\measuredangle DEA,$$so $\overline{AB}\parallel\overline{KY}.$ Therefore, $\triangle ABD$ and $\triangle YKT$ are homothetic since their corresponding sides are parallel, so $\overline{AY},$ $\overline{BK},$ and $\overline{DT}$ concur. $\square$

I will just make a sketch since my solution is the same as everyone else's sol. ISL 2020 G4 wrote: Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $BA$ and $BC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{AD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $AC,DT,$ and $BK$ are concurrent. Sketch. Let $P=AC\cap TE$, and $Q=AC\cap TF$. By angle chasing, we can show that $APDE$, and $CQDF$ are cyclic and tangent at $D$, with common tangent $TD$. We can also show that $DPTQK$ is cyclic. Then by Reim's theorem on $(AEDP)$, and $(PDKQ)\implies AE\parallel QK$, also Reim on $(CQDF)$, and $(QKDP)\implies KP\parallel CF$. Thus we have that quadrilaterals $ABCD$, and $QKPT$ are homothetic $\implies AQ, BK, CP, DT\implies AC,DT,$ and $BK$ are concurrent.$\blacksquare$

Let line $AC$ cut the lines $TE$ and $TF$ in points $X$ and $Y,$ respectively

Lemma 1: Points $T,X,Y,K$ and $D$ are concyclic.

Lemma 2: Point $T$ lies on the reflection of line $DB$ with respect to $EF.$

For finish we use the same homothety we used in last part of the proof of the Lemma. Consider the homothety that sends $D$ to $T,$ $C$ to $T$ and $Y$ to $A.$ Since the center of this homothety is the intersection of line $TL$ with line $AC,$ it is sufficient to show that $B$ is mapped to $K.$ In order to show this, let $J$ be the second intersecion of $DT$ wtih $\Omega;$ then by the Lemma 1, $D$ is the image of $J$ and taking the inverse of our homothety show that $K$ is the image of $B,$ if and only if $DB = DJ$ which is an consequence of Lemma 2 .

Solution. We begin by claiming that if $B'$ lies on $\Omega$ with $BB' \parallel FE$, then $D, T,$ and $B'$ are collinear. Let $ET$ and $FT$ meet $AB$ and $BC$, respectively, at $Q$ and $P$. Then quadrilateral $BPTQ$ is cyclic, as $\angle BPT = \angle BAD = 180^{\circ} - \angle BCD = 180^{\circ} - \angle BQT$. Moreover, quadrilateral $BPDF$, as $\angle BPF = \angle BAD = \angle BDF$ by Angle by Tangent; similarly quadrilateral $BQDE$ is cyclic. Next, $\angle QTF = 180^{\circ} - \angle STR = 180^{\circ} - \angle CDA = \angle CBA$. But as $QBED$ is cyclic, we have $\angle CBA = \angle QDF$, so quadrilaterals $QTDF$ and $PTDE$ are cyclic, the latter following from similar methods. Finally, we obtain $\angle TDF = 180^{\circ} - \angle FQT = \angle BQE = \angle BDE$, implying that $DT$ and $DB$ are isogonal with respect to straight angle $\angle FDE$, which implies that $D, T,$ and $B'$ are collinear. Now, let $CD$ and $BB'$ meet at $L$, and let $ET$ and $BB'$ meet at $M$. Let $DB'$ and $AC$ meet at $U$. By Pascal's Theorem applied to hexagon $BB'DDCA$, we obtain that $L, U,$ and $E$ are collinear. We want to show that $B, U,$ and $K$ are collinear. Note that triangles $TKD$ and $B'DB$ are similar, so $\frac{KD}{BB'} = \frac{TD}{B'D}$. Thus, as $D, U, B'$ and $E, U, L$ are collinear triplets, it suffices to demonstrate that $\frac{KD}{ED} = \frac{BB'}{B'L}$, or equivalently that $\frac{TD}{B'D} = \frac{BB'}{B'L}$. We have $\frac{TD}{B'D} = \frac{ML}{B'L}$ by parallel lines, and thus we just want $ML = BB'$. But this follows from $MLDE$ being a parallelogram. We are done.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.01) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -17.517866068993314, xmax = 16.862187920264457, ymin = -14.520337387842513, ymax = 7.479255853297635; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-3.4917584884379496,6.523139237915429)--(-8.249521229063657,-3.3016408214766497)--(4.881903935063295,-3.182696752961007)--cycle, linewidth(1) + zzttqq); /* draw figures */ draw((-3.4917584884379496,6.523139237915429)--(-8.249521229063657,-3.3016408214766497), linewidth(1) + zzttqq); draw((-8.249521229063657,-3.3016408214766497)--(4.881903935063295,-3.182696752961007), linewidth(1) + zzttqq); draw((4.881903935063295,-3.182696752961007)--(-3.4917584884379496,6.523139237915429), linewidth(1) + zzttqq); draw(circle((-1.709513763174707,-0.4043239615511177), 7.153051281844129), linewidth(1)); draw((-8.249521229063657,-3.3016408214766497)--(-2.06123620481089,-7.548722744145873), linewidth(1)); draw((-2.06123620481089,-7.548722744145873)--(4.881903935063295,-3.182696752961007), linewidth(1)); draw((-3.4917584884379496,6.523139237915429)--(-2.06123620481089,-7.548722744145873), linewidth(1)); draw((-0.794956088461727,-1.2920676847127956)--(-2.06123620481089,-7.548722744145873), linewidth(1)); draw((-0.794956088461727,-1.2920676847127956)--(-0.14934643118693514,-7.642846065055185), linewidth(1)); draw((-3.4917584884379496,6.523139237915429)--(-0.14934643118693514,-7.642846065055185), linewidth(1) + linetype("2 2")); draw((-0.794956088461727,-1.2920676847127956)--(-10.114232491381776,-7.1522695781635655), linewidth(1)); draw((-0.794956088461727,-1.2920676847127956)--(9.12373506718392,-8.099364656100821), linewidth(1)); draw((-3.4917584884379496,6.523139237915429)--(-10.114232491381776,-7.1522695781635655), linewidth(1)); draw((-10.114232491381776,-7.1522695781635655)--(9.12373506718392,-8.099364656100821), linewidth(1)); draw((9.12373506718392,-8.099364656100821)--(-3.4917584884379496,6.523139237915429), linewidth(1)); draw(circle((-0.9536921022935714,-4.516409522835409), 3.228246801462852), linewidth(1)); draw(circle((3.4894976089213263,-8.672131944862295), 5.663275916213557), linewidth(1)); draw(circle((-6.057673719286227,-6.7398865029370505), 4.077465986639081), linewidth(1)); draw((-3.9284677977303013,-3.262500844743557)--(-0.14934643118693514,-7.642846065055185), linewidth(1)); draw((-0.14934643118693514,-7.642846065055185)--(1.9978816179854324,-3.2088201435142487), linewidth(1)); draw((-3.9284677977303013,-3.262500844743557)--(-2.06123620481089,-7.548722744145873), linewidth(1)); draw((-2.06123620481089,-7.548722744145873)--(1.9978816179854324,-3.2088201435142487), linewidth(1)); /* dots and labels */ dot((-3.4917584884379496,6.523139237915429),dotstyle); label("$B$", (-3.3774378615845566,6.750993456349548), NE * labelscalefactor); dot((-8.249521229063657,-3.3016408214766497),dotstyle); label("$A$", (-8.141487708286649,-2.9895161028311246), NE * labelscalefactor); dot((4.881903935063295,-3.182696752961007),dotstyle); label("$C$", (4.9975797033184834,-2.86813903667311), NE * labelscalefactor); dot((-2.06123620481089,-7.548722744145873),dotstyle); label("$D$", (-1.9512573342278796,-7.237713418361636), NE * labelscalefactor); dot((-10.114232491381776,-7.1522695781635655),linewidth(4pt) + dotstyle); label("$E$", (-9.992487967196379,-6.903926486427096), NE * labelscalefactor); dot((9.12373506718392,-8.099364656100821),linewidth(4pt) + dotstyle); label("$F$", (9.245777018849012,-7.84459874915171), NE * labelscalefactor); dot((-0.794956088461727,-1.2920676847127956),linewidth(4pt) + dotstyle); label("$T$", (-0.6767981395687213,-1.047483044302891), NE * labelscalefactor); dot((-0.14934643118693514,-7.642846065055185),linewidth(4pt) + dotstyle); label("$K$", (-0.0395685422391421,-7.389434751059154), NE * labelscalefactor); dot((-3.9284677977303013,-3.262500844743557),linewidth(4pt) + dotstyle); label("$G$", (-3.8022575931376097,-3.0198603693706283), NE * labelscalefactor); dot((1.9978816179854324,-3.2088201435142487),linewidth(4pt) + dotstyle); label("$H$", (2.1148743820656257,-2.959171836291621), NE * labelscalefactor); dot((-1.1887281102798304,-3.2376843620673754),linewidth(4pt) + dotstyle); label("$L$", (-1.0712736045822702,-2.9895161028311246), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $G=ET\cap AC$ and $FT\cap AC$. Claim: $AGDE$ and $FCHD$ are cyclic. Proof: By the given parallel and tangent condition: \[\angle GED=\angle CDF=\angle CAD=\angle GAD,\]\[\angle HFD=\angle ADE=\angle ACD=\angle HCD,\]implying the desired claim $\square$ Claim: $\triangle DAC\sim \triangle TEF$ Proof: We can say: \[\angle GAD=\angle GED,\]\[\angle HCD=\angle HFD,\]so by AA, we are done $\square$ Claim: $THDG$ is cyclic Proof: We can say: \begin{align*} \angle GDH&=180-\angle DGH-\angle DHG\\ &=180-\angle AED-\angle CFD\\ &=\angle ABC\\ &=180-\angle ADC\\ &=180-\angle GTH\text{ }\square\\ \end{align*} Claim: $K$ lies on $(THDG)$ Proof: We can say: \begin{align*} \angle TKD&=\angle TDK\\ &=180-\angle EDT\\ &=\angle DET+\angle DTE\\ &=\angle DAH+\angle DHG\\ &=\angle THD\text{ }\square\\ \end{align*} Claim: $\triangle KTG$ is homothetic to $\triangle BDC$ Proof: Clearly, $TG\parallel CD$, by the given condition. Also, we can say: \[\angle DKG=\angle DHG=180-\angle DHC=\angle DFB,\]implying that $KG\parallel BC$. Also: \begin{align*} \angle BDE&=\angle BDA+\angle ADE\\ &=\angle BCA+\angle AGE\\ &=\angle BCA+\angle GCD\\ &=180-\angle DCF\\ &=\angle THD\\ &=\angle TKD,\\ \end{align*}meaning that $TK\parallel BD$, implying the homothetic relationship $\square$ By this homothety, $BK$, $DT$, $AC$ are concurrent at a point.

Define $P = \overline{AC} \cap \overline{TE}$ and $Q = \overline{AC} \cap \overline{TF}$. Claim: $APDE$ and $CQDF$ are cyclic. Proof: \[\angle CDF = \angle CAD \implies \angle TEF = \angle CAD \implies \text{APDE cyclic}\]$CQDF$ follows similarly. Claim: $TPDQ$ cyclic. Proof: By the previous claim, we must have $\Delta TEF \sim \Delta DAC$. Thus \[\angle ETF = \angle ADC = 180^{\circ} - \angle ABC = 180^{\circ} - (360^{\circ} - \angle AEP - \angle CFQ - (360^{\circ} - \angle ETF))\]\[\implies 2 \angle ETF = 180^{\circ} + \angle AEP + \angle CFQ = 180^{\circ} + \angle ADP+\angle CDQ\]\[\implies 2 \angle ETF = 180^{\circ} + \angle ADC - \angle PDQ\]so we're done since $\angle ETF = \angle ADC$ by similar triangles. Claim: $\overline{DT}$ is tangent to $AEDP$ and $CFDQ$. Proof: \[\angle TDP = \angle TQP = \angle CQF = \angle PED = \angle PAD\]which implies $\overline{DT}$ is tangent to $APDE$. $CFDQ$ follows similarly. Claim: $K \in (TPDQ)$. Proof: \[\angle TKD = \angle TDK = \angle TDQ+\angle QDK = \angle TDQ+180^{\circ} - \angle DQF - \angle QFD\]Due to the previous claim, $\angle TDQ = \angle QFD$, so \[\angle TKD = 180^{\circ} - \angle DQF = \angle TQD\]Claim: $\overline{QK} \parallel \overline{EB}$. Proof: \[\angle QKD = 180^{\circ} - \angle QPD = \angle APD = 180^{\circ} - \angle BED\]Claim: $\overline{TK} \parallel \overline{BD}$. Proof: \[\angle TKD = 180^{\circ} - \angle TPD = \angle EPD = \angle EAD = 180^{\circ} - \angle BAD = 180^{\circ} - \angle BDF = \angle BDE\]where the second to last equality comes from $D$ being tangent. The last two claims and the fact that $\overline{TQ} \parallel \overline{AD}$, implies that $\Delta BAD$ and $\Delta KQT$ are homothetic, which implies the desired result.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.276270569275933cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.532399712494254, xmax = 23.74387085678168, ymin = -7.14104595325827, ymax = 6.97566791428178; /* image dimensions */ pen ttqqqq = rgb(0.2,0.,0.); pen vwryqq = rgb(0.33725490196078434,0.09411764705882353,0.); pen ffzzqq = rgb(1.,0.6,0.); pen ttffcc = rgb(0.2,1.,0.8); pen wwttqq = rgb(0.4,0.2,0.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); pen qqwwzz = rgb(0.,0.4,0.6); pen ffzztt = rgb(1.,0.6,0.2); pen svsvsv = rgb(0.1450980392156863,0.1450980392156863,0.1450980392156863); /* draw figures */ draw(circle((7.903367043426526,-1.2951814396192736), 5.037347092213034), linewidth(0.5) + ttqqqq); draw((3.2300941064266873,-3.175443597865302)--(12.595385088485273,-3.128165998274964), linewidth(1.) + vwryqq); draw((12.595385088485273,-3.128165998274964)--(10.922881571912448,2.736864906802049), linewidth(1.) + ffzzqq); draw((6.474703725140418,3.535324419026468)--(10.922881571912448,2.736864906802049), linewidth(1.) + ttffcc); draw((3.2300941064266873,-3.175443597865302)--(6.474703725140418,3.535324419026468), linewidth(1.) + wwttqq); draw(circle((9.314235715780503,-1.4101089193310572), 4.448048269194426), linewidth(1.) + cqcqcq); draw((9.314235715780503,-1.4101089193310572)--(10.922881571912448,2.736864906802049), linewidth(1.) + qqwwzz); draw((3.2300941064266873,-3.175443597865302)--(12.84102175793724,1.3004101486579263), linewidth(1.) + qqwwzz); draw((6.474703725140418,3.535324419026468)--(12.595385088485273,-3.128165998274964), linewidth(1.) + qqwwzz); draw((9.314235715780503,-1.4101089193310572)--(18.713388138866275,-3.0972812666566414), linewidth(1.) + ttffcc); draw((7.373724796069257,5.394753613375204)--(9.314235715780503,-1.4101089193310572), linewidth(1.) + ffzztt); draw((7.373724796069257,5.394753613375204)--(18.713388138866275,-3.0972812666566414), linewidth(1.) + vwryqq); draw((12.595385088485273,-3.128165998274964)--(18.713388138866275,-3.0972812666566414), linewidth(1.) + vwryqq); draw((6.474703725140418,3.535324419026468)--(7.373724796069257,5.394753613375204), linewidth(1.) + vwryqq); draw((9.314235715780503,-1.4101089193310572)--(12.595385088485273,-3.128165998274964), linewidth(1.) + vwryqq); /* dots and labels */ dot((7.903367043426526,-1.2951814396192736),dotstyle); label("$O$", (7.999129289798491,-1.0787849145422848), NE * labelscalefactor); dot((6.474703725140418,3.535324419026468),dotstyle); label("$A$", (6.563894344236758,3.741033932493392), NE * labelscalefactor); dot((10.922881571912448,2.736864906802049),dotstyle); label("$D$", (10.998127683509574,2.9484414998697472), NE * labelscalefactor); dot((3.2300941064266873,-3.175443597865302),dotstyle); label("$B$", (3.3078389453504373,-2.9638696191606826), NE * labelscalefactor); dot((12.595385088485273,-3.128165998274964),dotstyle); label("$C$", (12.690419634246544,-2.92102678496481), NE * labelscalefactor); dot((7.373724796069257,5.394753613375204),linewidth(4.pt) + dotstyle); label("$E$", (7.463593862350082,5.561854385817981), NE * labelscalefactor); dot((18.713388138866275,-3.0972812666566414),linewidth(4.pt) + dotstyle); label("$F$", (18.795523507158393,-2.92102678496481), NE * labelscalefactor); dot((9.314235715780503,-1.4101089193310572),linewidth(4.pt) + dotstyle); label("$T$", (9.39152140116435,-1.2287348342278392), NE * labelscalefactor); dot((12.84102175793724,1.3004101486579263),linewidth(4.pt) + dotstyle); label("$K$", (12.926055222323843,1.4703637201121398), NE * labelscalefactor); dot((9.819901082500111,-0.10653981331882342),linewidth(4.pt) + dotstyle); label("$X$", (9.905635411514822,0.056550191648341304), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Let $E_1$ be the intersection of lines $ET$ and $AC$, and define $F_1$ similarly. We have $\angle DAC = \angle CDF = \angle TED$, so $AEDE_1$ is cyclic; similarly, $CFDF_1$ is cyclic. Furthermore, since $\angle DAE$ and $\angle DCE$ are complementary, it follows that a homothety centered at $D$ sends $(AEDE_1)$ to $(CFDF_1)$. Furthermore, we have $\angle TE_1 F_1 = \angle AE_1 E = \angle ADE = \angle TFE$, so $E_1F_1 FE$ is cyclic. So, by the radical axis theorem, $\overline{DT}$ is the radical axis (common internal tangent) of $(AEDE_1)$ and $(CFDF_1)$. Some angle chasing (left as an excercise to the reader) shows $TE_1DKF_1$ cyclic; letting $X$ be the second intersection between line $BK$ and $(TE_1DKF_1)$, some more angle chasing shows $BAE_1X$ and $BCF_1X$ are cyclic; line $BK$ is the radical axis of these two circles. So, both line $BK$ and $DT$ pass through the point $P$ on line $AC$ for which $PE_1 \cdot PB = PF_1 \cdot PC$.

Make the problem $A-$centric, because why not. For clarity, we solve the following problem: amuthup wrote: Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $AB$ and $AC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{BD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $BC,DT,$ and $AK$ are concurrent. All angles are directed. Define $X=TE\cap BC$, $Y=TF\cap BC$ and $A\ne Z=(AXB)\cap (AYC)$. Claim 1: $BXDE$ and $CYDF$ are cyclic. Proof. $\angle FDC=\angle DBC=\angle DBX$ because of tangency condition. Since $ET\parallel DC$, $\angle FDC=\angle DET=\angle DEX$. Hence, $\angle DBX=\angle FDC=\angle DEX$. So, $BXDE$ is cyclic. Similarly, $CYDF$ is cyclic. Claim 2: $TZXDY$ is cyclic. Proof. We have \[\angle YDX=180^\circ-\angle FDY-\angle XDE=\angle EBX+\angle YCF-180^\circ=\angle BAC\]Since $TE\parallel CD$ and $TF\parallel BD$. So, $\angle XTY=\angle CDB=180^\circ-\angle BAC$. So, $\angle XTY+\angle YDX=180^\circ$ which means $TXDY$ is cyclic. Note that $\angle AZX=180^\circ-\angle CBA$ and $\angle YZA=180^\circ-\angle ACB$. So, \[\angle XZY=180^\circ-\angle AZX+180^\circ-\angle YZA=\angle CBA+\angle ACB=180^\circ-\angle BAC=\angle XTY,\]and it follows that $Z$ lies on $XYT$. Claim 3: $TD$ is the common tangent of circumcircles $(BDE)$ and $(CDF)$. Proof. Note, \[\angle TDX=\angle TYX=\angle FYC=\angle FDC=\angle FEX=\angle DEX\]which means $TD$ is tangent to $(BDE)$. Similarly, we get $TD$ is tangent to $(CDF)$. We now involve $K$ into the diagram (aka mess). Claim 4: $K$ lies on circumcircle $(TZXDY)$ Proof. $TD$ is tangent to $(BDE)$ which means $\angle TDE=180^\circ-\angle EXD$. So, \[\angle TKD=\angle KDT=\angle EXD=\angle EBD=\angle ACD=\angle TYD\]which means $TYKD$ is cyclic. Claim 5: $A, Z, K$ are collinear. Proof. We have, \[\angle AZX=\angle EBX=\angle KDX=180^\circ-\angle KZX\]and $\angle AZX+\angle KZX=180^\circ$ gives the result. Claim 6: $AK, BC, TD$ are concurrent. Proof. Let $J$ be the radical center of circles $(ABX), (BDE), (CDF)$. Then, $J$ lies on lines $BX, TD$. So, $J=TD\cap BC$. Note that $BC$ is also the radical axis of $(ACY), (CYF)$. Hence, $J=BC\cap TD$ is the radical center of $(BDE), (CDF), (ACY)$ as well. So, $J$ has equal power from all four circles. Hence, $J$ also lies on the radical axis of $(ABX), (ACY)$ which is $AK$. So, $J$ lies on $TD$, $BC$ and $AK$, as desired. The proof is complete. $\blacksquare$

let point $G$ on $\Omega$ exist so that $BG$ and $EF$ are parallel, and let $BG$ and $AD$ intersect at $H$ angle chasing gives that $DBG=DGB=BDE=GDF=EAD=BCD=HAB$, so $DTG$ are collinear with $DB=DG$ and $DT=KT$, so triangles $BDW$ and $DTK$ are similar more angle chasing yields that $TFK=HDE=DHB$, so since $HBD=FKT$, triangles $HBD$ and $FKT$ are similar then, we can just map a homothety for our sets of similar triangles and we are done

I don't see why we can't move some points, fellas (correct me if i'm wrong tho)

Let $TE$ and $TF$ intersect $AC$ at $G$ and $H$, respectively. Claim: $EAGD$ is concyclic $$\angle GAD=\angle CAD=\angle TEF=\angle GED$$Claim: $FCHD$ is concyclic $$\angle HCD=\angle ACD=\angle TFE=\angle HFD$$Claim: $TGHD$ is concyclic $$180^{\circ}-\angle GDH=\angle GDE+\angle HDF=\angle BAC+\angle BCA=\angle ADC=\angle ETF=\angle GTH$$Claim: $K$ lies on $(TGHD)$ $$\angle TKD=\angle TDK=\angle TDA+\angle ADE=\angle TGH+\angle DTH=\angle TGD$$Claim: $KGHT\sim ABCD$ $$\angle GKE=\angle GHD=\angle CHD \Rightarrow GK|| BC$$$$\angle HKF=\angle HGD=\angle AGD \Rightarrow HK|| BA$$As $GH$ lies on $AC$ the claim follows. Since $KGHT$ and $ABCD$ are homothetic about a $180^{\circ}$ angle lines $BK$, $TD$, and $AC$ concur.

Ratio lemma never fails! I claim that this is one of the most motivated solutions to this problem. First, note that if $P = BD\cap (BEF)$, then $\measuredangle PEF = \measuredangle DBC= \measuredangle FDC = \measuredangle FET$ so $T$ is the reflection of $P$ over $EF$ by symmetry. In particular, $PDTK$ is a rhombus. Now we can directly finish from here. Let $w,x,y,z = \frac{1}{2}\widehat{BA}, \frac{1}{2}\widehat{BC}, \frac{1}{2}\widehat{DA}, \frac{1}{2}\widehat{DC}$. By angle chase $\angle FPK = w$ and $\angle EPK = x$, so $$\frac{KE}{KF} = \frac{PE}{PF}\cdot \frac{\sin{\angle KPE}}{\sin{\angle KPF}} =\frac{\sin{y}}{\sin{z}}\cdot \frac{\sin{x}}{\sin{w}}$$ Then if $X = BK\cap AC$, we have $$\frac{XA}{XC} =\frac{\sin{y}}{\sin{z}}\cdot \frac{\sin{x}}{\sin{w}}\cdot \frac{BF}{BE}\cdot \frac{BA}{BC} = \frac{\sin{y}}{\sin{z}}\cdot \frac{\sin{(w+2y)}}{\sin{(x+2z)}}.$$ On the other hand, if $X' = DT\cap AC$, we have $$\frac{X'A}{X'C} = \frac{DA}{DC}\cdot \frac{\sin{\angle TDA}}{\sin{\angle TDC}} = \frac{\sin{y}}{\sin{z}}\cdot \frac{\sin{(w+2y)}}{\sin{(x+2z)}}$$ and we're done.

Solved with mathhater420 Let $ET$ and $FT$ meet $AC$ at $P$ and $Q$ respectively. Note that $\angle PED=\angle CDF=\angle CBD=\angle CAD$, so $PAED$ is cyclic. Similarly, $QCFD$ is cyclic. Claim 1: $TPDQ$ is cyclic. We have $$\angle PDQ=180-\angle PDE-\angle QDF=180-\angle BAC-\angle BCA=\angle ABC$$$$=\angle ABD+\angle CBD=\angle ACD+\angle CAD=\angle TPQ+\angle TQP.$$ Claim 2: $TD$ is tangent to both $(PAED)$ and $(QCFD)$. We have $$\angle TDP=\angle TQP=\angle PAD,$$and similarly for $\angle TDQ=\angle QCD$. This means that the radical axis of the two circles is line $DT$. Let $AC$ and $BK$ meet at $N$. Thus, it suffices to show that $N$ lies on the radical axis, or in other words, $NP\cdot NA=NQ\cdot NC$. Claim 3: $K$ also lies on $(TPDQ)$. We have $$\angle TKD=\angle TDK=\angle EPD.$$ Claim 4: $PK$ is parallel to $BC$, and similarly $QK$ is parallel to $BA$. We have $$\angle PKD=\angle PQD=\angle CFD.$$ Due to Claim 4, triangles $BAC$ and $KQP$ are homothetic. Furthermore, $N$, which is the intersection of $BK$ and $APQC$, is the center of (negative) homothety. Thus, $NP\cdot NA=NQ\cdot NC$, as desired.

Let $TE \cap AC = P$, $TF \cap AC = Q$, let $BA \cap TF = X$, $CA \cap TE = Y$. Our goal will be to show the existence of a homothety from $ABCD$ to $QKPT$, which finishes since it implies $BK$, $DT$, and $ACPQ$ are concurrent. Step 1: Doxxing $TD$. Note that $\angle XFD = \angle XFE = \angle ADE = \angle ABD = \angle XBD$, so $(XBFD)$ is cyclic. Now $\angle TED = \angle YED = \angle YEF = \angle CDF = \angle CBD = \angle FBD = \angle FXD = \angle TXD$, so $(TXED)$ is cyclic. Step 2: Figuring out $TPQ$. Note that $\angle APE = \angle YPC = \angle PCD = \angle ACD = \angle ADE$, so $(APED)$ is cyclic. Symmetrically, $(CQFD)$ is cyclic. Thus $\angle PDQ = 180 - \angle DPQ - \angle DQP = \angle APD + \angle CQD - 180 = 180 - \angle AED - \angle CFD = \angle ABC$. Now since $\angle PTQ = 180 - \angle ABC$, we see that $(TDPQ)$ is cyclic. Step 3: Finish. Observe $(TXED)$ cyclic implies $\angle TDE = 180 - \angle TXE = \angle TXB = \angle DAB$, so by the condition that $K$ is on segment $DF$ we see that $\angle TDK = \angle TDF = 180 - \angle DAB = \angle DCB$. By isosceles properties, we see that $\angle TKD = \angle DCB$ as well. Now notice $\angle TQD = \angle TQP + \angle PQD = \angle PQD + \angle CQF = 180 - \angle DQF = 180 - \angle DCF = \angle DCB$, so $(TPQKD)$ is cyclic. Now we show the final homothety exists. It is sufficient to show that the quadrilaterals are similar and that each side is parallel to the corresponding similar side. First observe we can see that $TPQ$ is similar to $DCA$ by angles we have already found, then we can see that $\angle KTP = 180 - 2\angle DCB + \angle DTP $, we find $\angle DTP = \angle DQP = 180 - \angle DQC = \angle DFC = 180 - \angle DCF - \angle CDF = \angle DCB - \angle DBC$. Substituting this back in, we get $\angle KTP = 180 - \angle DCB - \angle DBC = \angle BDC$. We see that the unique point $K'$ such that $ABCD$ is similar to $QK'PT$ satisfies $\angle PTK' = \angle CDB$, and lies on the circumcircle of $PQT$, which is exactly $K$. Thus the similarity is done, for the parallel sides just see that we already have two given by the problem statement and the rest follow trivially from the similarity, so the homothety exists and we are done.

Let $P$ and $Q$ be the intersections of $\overline{ET}$ and $\overline{FT}$ with side $AC$ respectively. We start off with making some minor observations. First, since \[\measuredangle QFD = \measuredangle ADE = \measuredangle ACD = \measuredangle QCD\]we have that quadrilateral $DQCF$ is cyclic. Similarly, we can show that quadrilateral $APDE$ is also cyclic. Now, we can further note the following. Claim : Points $P$ , $Q$ , $D$ , $T$ and $K$ are concyclic. Proof : First, note that, \[\measuredangle TQD = \measuredangle FQD = \measuredangle FCD = \measuredangle BCD = \measuredangle EAD = \measuredangle EPD\]so quadrilateral $DPTQ$ must be cyclic. Further, \begin{align*} \measuredangle TKD &= \measuredangle KDT + \measuredangle FDT \\ &= \measuredangle FDQ + \measuredangle QDT \\ &= \measuredangle BCQ + \measuredangle QPT \\ &= \measuredangle BCQ + \measuredangle APE \\ &= \measuredangle BCQ + \measuredangle ADE \\ &= \measuredangle BCA + \measuredangle ACD \\ &= \measuredangle BCD\\ &= \measuredangle FCD \\ &= \measuredangle FQD \\ &= \measuredangle TQD \end{align*}which implies that indeed point $K$ also lies on this circle, which proves the claim. With these observations in hand, we are now in a position to prove our key claim. Claim : Quadrilaterals $ABCD$ and $QKPT$ are homothetic. Proof : We are already given that $TP \parallel CD$ and $TQ \parallel AD$, so we focus on the other two pairs of sides. Note that, \[\measuredangle FKQ = \measuredangle DPQ = \measuredangle DEA\]so we must have $KQ$ is parallel to $AB$. Similarly, we have that $KP$ is parallel to $BC$. Since pairs of polygons with all pairs of sides, being pairwise parallel are considered homothetic, the claim is proved. Now, we know that lines joining corresponding vertices of homothetic polygons concur, at the center of homothety. Thus, lines $\overline{AQ} , \overline{CP},\overline{BK}$ and $\overline{DT}$ must concur. It is clear that lines $\overline{AQ}$ and $\overline{CP}$ are both just the line $\overline{AC}$, so we have that lines $AC$ , $DT$ and $BK$ are concurrent, as desired.

Beautiful problem! [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-13.56,1.39); pair B = (-9.64,9.51); pair C = (-0.66,0.09); pair D = (-6.51117,-3.75196); pair E = (-16.28414,-4.25287); pair F = (2.55932,-3.28706); pair T = (-5.67034,2.71630); pair X = (-8.47125,0.87718); pair Y = (-2.29611,0.25488); pair K = (-4.17259,-3.63210); pair Z = (-5.94254,0.62234); import graph; size(10cm); pen xfqqff = rgb(0.49803,0,1); pen cczzqq = rgb(0.8,0.6,0); pen ffxfqq = rgb(1,0.49803,0); pen ffqqff = rgb(1,0,1); pen yqqqqq = rgb(0.50196,0,0); pen wwzzff = rgb(0.4,0.6,1); draw(circle((-6.86567,3.16446), 6.92551), linewidth(0.4) + linetype("2 2") + xfqqff); draw(B--A, linewidth(0.4) + red); draw(A--C, linewidth(0.4) + red); draw(C--B, linewidth(0.4) + red); draw(A--D, linewidth(0.4) + cczzqq); draw(D--C, linewidth(0.4) + cczzqq); draw(B--E, linewidth(0.4) + red); draw(B--F, linewidth(0.4) + red); draw(E--F, linewidth(0.4) + ffxfqq); draw(T--E, linewidth(0.4) + cczzqq); draw(T--F, linewidth(0.4) + cczzqq); draw(T--D, linewidth(0.4) + blue); draw(T--K, linewidth(0.4) + ffqqff); draw(B--K, linewidth(0.4) + blue); draw(B--D, linewidth(0.4) + ffqqff); draw(X--D, linewidth(0.4) + yqqqqq); draw(Y--D, linewidth(0.4) + yqqqqq); draw(circle((-11.44336,-3.11081), 4.97368), linewidth(0.4) + linetype("4 4") + green); draw(circle((-5.50064,-0.59454), 3.31519), linewidth(0.4) + wwzzff); draw(circle((-1.93351,-4.34703), 4.61617), linewidth(0.4) + linetype("4 4") + green); draw(Y--K, linewidth(0.4) + red); dot("$A$", A, dir((8.000, 20.000))); dot("$B$", B, dir((8.000, 20.000))); dot("$C$", C, dir((8.000, 20.000))); dot("$D$", D, dir((7.117, 20.196))); dot("$E$", E, dir((8.414, 16.287))); dot("$F$", F, dir((8.068, 15.706))); dot("$T$", T, dir((7.034, 15.370))); dot("$X$", X, dir((7.125, 15.282))); dot("$Y$", Y, dir((7.611, 15.512))); dot("$K$", K, dir((7.259, 16.210))); dot("$Z$", Z, dir((8.254, 16.766))); [/asy][/asy] Let $ET$ and $FT$ intersect $AC$ at points $X$ and $Y,$ respectively. Claim 1: $AXDE$ and $CYDF$ are cyclic. Proof: We have that $$\measuredangle DAX = \measuredangle DAC = \measuredangle EDC = \measuredangle DET = \measuredangle DEX,$$so $AXDE$ is cyclic. We can similarly obtain that $CYDF$ is cyclic, proving our claim. Claim 2: $TXDY$ is cyclic. Proof: This is once again an angle chase: \begin{align*} \measuredangle TXD &= \measuredangle EXD = \measuredangle EAD = \measuredangle BAD \\ &= \measuredangle BCD = \measuredangle FCD = \measuredangle FYD = \measuredangle TYD. \end{align*} Claim 3: $TD$ is a common tangent to the circumcircles of $AXDE$ and $CYDF.$ Proof: We have that $$\measuredangle TDX = \measuredangle TYX = \measuredangle TYA = \measuredangle DAY = \measuredangle DAX,$$so $TD$ is tangent to $(AXDE)$ and is similarly tangent to $(CYDF),$ as claimed. Claim 4: $(TXDY)$ actually contains $K.$ Proof: We yet again angle chase: $$\measuredangle TKD = \measuredangle KDT = \measuredangle EDT = -\measuredangle TDE = -\measuredangle DXE = \measuredangle TXD.$$ Claim 5: $TK \parallel BD.$ Proof: Note that $$\measuredangle TKD = \measuredangle TXD = \measuredangle EXD = \measuredangle EAD = \measuredangle BAD = \measuredangle BDE,$$implying the claim. Now, let $Z = BK \cap TD$ be the center of a negative homothety sending $BD$ to $TK.$ Since $TY \parallel AD$, $TK \parallel BD,$ and $YK \parallel AB$ by Reim, we see that $\triangle BAD$ and $\triangle KYT$ are homothetic, and the center of homothety is just $Z.$ Therefore, the homothety sends $A$ to $Y,$ and in particular $A,Z,Y$ are collinear, as desired.

Let $\{S,U\}=ET,FT\cap AC$, we have \begin{align*} &\angle SED= \angle CDF = \angle SAD\implies \odot(ASDE)\\ & \angle UFD=\angle SAD=\angle UCD\implies \odot(CUDF) \end{align*}and \begin{align*} \angle STU&=180^{\circ}-(\angle TEF+\angle TFE)\\&=180^{\circ}-(\angle CAD+\angle ACD)\\&=\angle ABC\\&=180^{\circ}-(\angle ACB+\angle CAB)\\&=180^{\circ}-(\angle UDF+\angle SDE)\\&=180^{\circ}-\angle SDU \end{align*}So $\odot(STUD)$ is cyclic, now we show $DT$ is tangent to $\odot(CUDF)$ and $\odot(ASDE)$ \begin{align*} &\angle SDT= \angle SUT= \angle CSF = \angle CUF = \angle CDF=\angle TED\\ &\angle UFD = \angle UCD = \angle CST = \angle UDT \end{align*}We angle chase more to get $K\in \odot(STUD)$ \begin{align*} \angle DKT &=180^{\circ}-\angle TDF\\&=180^{\circ}-(180^{\circ}-\angle DCF)\\&=180^{\circ}-\angle DAE\\&=180^{\circ}-\angle DSE\\&=\angle DST \end{align*}If $\triangle KST$ and $\triangle BCD$ are homothetic we'll be done, so we prove \begin{align*} \angle TKD&=\angle TDU+\angle UDF=\angle TSU+\angle ACB=\angle ADE+\angle ADB=\angle BDE\implies KT\parallel BD\\ \angle SKE&=\angle SUD=180^{\circ}-\angle CUD =\angle CFD\implies KS\parallel BC \end{align*}and we are done! [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(9); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.301059888544868, xmax = 5.427247847678922, ymin = -3.9302067049047666, ymax = 7.930620330584966; /* image dimensions */ /* draw figures */ draw(circle((-1.7641834099733615,-0.43395414750665956), 2.7297207616073), linewidth(0.5)); draw((-1.8986143502387,6.4055425990452)--(-4,-2), linewidth(0.5)); draw((0.36842333045447384,-2.137880180828441)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((-1.8986143502387,6.4055425990452)--(2.0450554562137,-2.190799580619854), linewidth(0.5)); draw((-4,-2)--(2.0450554562137,-2.190799580619854), linewidth(0.5)); draw((-1.8986143502387,6.4055425990452)--(-2.7477736566214856,-0.5200043091637574), linewidth(0.5)); draw((-2.7477736566214856,-0.5200043091637574)--(2.0450554562137,-2.190799580619854), linewidth(0.5)); draw((0.7169072967555019,0.7042744505058862)--(-3,2), linewidth(0.5)); draw((-2.7477736566214856,-0.5200043091637574)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((0.36842333045447384,-2.137880180828441)--(-3,2), linewidth(0.5)); draw((-4,-2)--(0.4401007829004468,1.307652427730301), linewidth(0.5)); draw((-2.7477736566214856,-0.5200043091637574)--(0.4401007829004468,1.307652427730301), linewidth(0.5)); draw(circle((-1.0709534122302529,0.24925606117104787), 1.8448543707750797), linewidth(0.5)); draw(circle((-0.33230478570055555,3.6735207021679015), 3.1491696202301815), linewidth(0.5)); draw(circle((1.2496916833029261,-0.8034932059595474), 1.599131798330576), linewidth(0.5)); draw((-0.23573917726046134,-1.3957074129133287)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((-2.512254266719884,1.400837363648374)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((-4,-2)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((-2.512254266719884,1.400837363648374)--(0.4401007829004468,1.307652427730301), linewidth(0.5)); draw((0.4401007829004468,1.307652427730301)--(-0.23573917726046134,-1.3957074129133287), linewidth(0.5)); /* dots and labels */ dot((-3,2),linewidth(3pt) + dotstyle); label("$A$", (-3.3737381706765612,1.9371173073321764), NE * labelscalefactor); dot((-4,-2),linewidth(3pt) + dotstyle); label("$B$", (-4.335853129672412,-2.3056519380757203), NE * labelscalefactor); dot((0.36842333045447384,-2.137880180828441),linewidth(3pt) + dotstyle); label("$C$", (0.30122552516004736,-2.5737823364843977), NE * labelscalefactor); dot((0.7169072967555019,0.7042744505058862),linewidth(3pt) + dotstyle); label("$D$", (0.790169192846463,0.8803680900744474), NE * labelscalefactor); dot((-1.8986143502387,6.4055425990452),linewidth(3pt) + dotstyle); label("$E$", (-2.2381270715339183,6.432244574771769), NE * labelscalefactor); dot((2.0450554562137,-2.190799580619854),linewidth(3pt) + dotstyle); label("$F$", (2.130821184889861,-2.4949204545994923), NE * labelscalefactor); dot((-2.7477736566214856,-0.5200043091637574),linewidth(3pt) + dotstyle); label("$T$", (-3.042518266759957,-0.8545933113934656), NE * labelscalefactor); dot((-1.3505221927790012,-0.026271895669134612),linewidth(3pt) + dotstyle); label("$R$", (-1.197150230653162,-0.20792587993724349), NE * labelscalefactor); dot((0.4401007829004468,1.307652427730301),linewidth(3pt) + dotstyle); label("$K$", (0.5693559235687269,1.4008565105148214), NE * labelscalefactor); dot((-2.512254266719884,1.400837363648374),linewidth(3pt) + dotstyle); label("$S$", (-2.916339255744108,1.195815617614068), NE * labelscalefactor); dot((-0.23573917726046134,-1.3957074129133287),linewidth(3pt) + dotstyle); label("$U$", (-0.45584854093504795,-1.7693911412583652), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]