Use Pascal’s theorem 5 times and you’re done BACQPY EFQPYE FFYPQE YFBQEC QFBYEC I refuse to write anything more than this.

Like you more complicated

YQ and BC meets on EF at S We can finish with polar of S = DT

Here's my full solution. Let $Z = QT \cap (O)$. We have $-1 = (P, Q; E, F) \overset{T}{=} (Y, Z; F, E)$ so $D, Y, Z$ are collinear. [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 O = (-1.3,-0.46); pair A = (0.72,3.8); pair B = (-5.44112,-2.71367); pair C = (2.79864,-2.79004); pair D = (3.86348,7.12335); pair E = (-3.27948,3.81897); pair F = (3.40671,-0.73362); pair T = (1.80627,0.35610); pair P = (-2.98671,-4.86261); pair Q = (1.63718,3.22794); pair Z = (2.04968,-3.77777); pair Y = (2.94353,1.59436); pair K = (6.47695,-2.82414); pair U = (2.43343,2.41914); pair V = (0.19937,-4.92988); import graph; size(10cm); draw(circle(O, 4.71465), linewidth(1)); draw(A--B, linewidth(1)); draw(B--C, linewidth(1)); draw(C--A, linewidth(1)); draw(D--A, linewidth(1)); draw(E--F, linewidth(1)); draw(D--P, linewidth(1)); draw(Q--Z, linewidth(1)); draw(P--Y, linewidth(1)); draw(D--Z, linewidth(1)); draw(C--K, linewidth(1)); draw(Q--K, linewidth(1)); draw(K--P, linewidth(1)); draw(V--D, linewidth(1)); dot("$A$", A, dir((-10, 20))); dot("$B$", B, dir((-20, -20))); dot("$C$", C, dir((20, -20))); dot("$D$", D, dir((7.652, 19.665))); dot("$E$", E, dir((-40.052, 18.103))); dot("$F$", F, dir((25.329, -38.638))); dot("$T$", T, dir((-50, -10))); dot("$P$", P, dir((-13.329, -51.739))); dot("$Q$", Q, dir((-10, 30))); dot("$Z$", Z, dir((13.032, -46.223))); dot("$Y$", Y, dir((20, 16.564))); dot("$K$", K, dir((8.305, 16.414))); dot("$U$", U, dir((26.657, 8.086))); dot("$V$", V, dir((12.063, -47.012))); [/asy][/asy] Let $U, V = DT \cap (O), K = QY \cap PZ$, then $KU, KV$ are tangent to $(O)$. Notice $-1 = (D, T; U, V) \overset{A}{=} (B, C; U, V)$ and so $B, C, K$ are collinear, $BQ \cap CY$ lies on polar of $K$ wrt $(O)$. Therefore $BQ, CY, DT$ are concurrent, or $X, C, Y$ are collinear. $\blacksquare$

$QR\cap (O)=R, QC\cap BY=S, BP\cap CR=K$ Pascal at $PQCABY$ gives that $D,S,T$ are collinear. $FEEYPQ$ gives that $T,D,QF\cap EY$ $QRYEFF$ gives that $T,RY\cap FF, YE\cap FQ$ are collinear hence $T,D, RY\cap FF$ are collinear so $R,Y,D$ are collinear. $BPQCRY$ gives that $K, D, S$ are collinear. $QBPYCR$ gives that $QB\cap YC, K, T$ are collinear so $QB\cap YC$ is on the line $DSXTK$ and this gives that $C,X,Y$ are collinear as desired.$\blacksquare$

Let R,S = DT ⋂ (O), L = DT ⋂ BC, K = BC ⋂ EF. Notice that -1 = (D,T;R,S) = A(B,C;R,S). Therefore, RCSB is a harmonic quadrilateral. hence, KS,KR are tangent to (O). Notice that EQFP is also harmonic quadrilateral and -1 = (K,T;F,E), then -1 = (K,T;F,E) = Y(Q,P;F,E). hence, Q,Y,K are collinear. so, QRYS is also harmonic quadrilateral. Let BY ⋂ CQ = G. Then, B(Q,R;Y,S) = B(X,R;G,S) = -1 = C(G,R;Y,S) = C(G,R;X,S). Hence, C,X,Y are collinear as required.

Even using Pascal $3$ times kills the problem $FEECAB \implies EC \cap FB \in DT$ and $EFFYPQ \implies QE \cap YF \in DT.$ Finally $ECYFBQ \implies DT , BQ , CY$ are concurrent.

Let $I=XA \cap (ABC)$ and $H=AP \cap BQ$ where $H$ lies on $EF$ because by Brokard $H$ lies on the polar of $D$ wrt $(ABC)$. Thus \[-1=(D, DQ\cap EF; Q, P)\overset{H}{=}(D,T; X, HP\cap AX)\overset{A}{=}(B, C; I, P).\] Pascal on $CYPIAC$ gives that it suffices to show $AT$ passes through the intersections of the tangents at $B$, $C$ which is trivial by pascal

Applying Pascal to $PQEFFY$, $EQBFYC$ and $EECABF$, we obtain that $BQ$ and $CY$ meet on $DT$ as desired.

Seungjun_Lee wrote: For a triangle $ABC$, $O$ is the circumcircle and $D$ is a point on ray $BA$. $E$ and $F$ are points on $O$ so that $DE$ and $DF$ are tangent to $O$ and $EF$ cuts $AC$ at $T(\neq C)$. $P(\neq B,C)$ is a point on the arc $BC$ not containing $A$, and $DP$ cuts $O$ at $Q (\neq P)$. Let $BQ$ and $DT$ meets on $X (\neq Q)$, and $PT$ cuts $O$ at $Y (\neq P)$. Prove that $C,X,Y$ are collinear. Let $Z$, $C'$ are second insections of lines $DC$, $DY$ with $(O)$, respectively $\hspace{0.5cm}$$X'$ is the insection of lines $QZ$, $YC$ We easy to see that all $AEBF$, $YEZF$, $QEPF$, $C'ECF$ are harmonic quadrilateral Which means $T(YZ, EF) = T(AB, EF) = T(CC', EF) = T(PQ, EF)$. Lead to lines $AC$, $QZ$, $BC'$, $YP$ concur at T Use Pascal Theorem for $\binom{C\ P\ B}{Q\ C'\ Y}$, we have: $T$, $X'$, $D$' are colinear Which mean $X' \equiv X$, done

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Let $Z=YF\cap QE$ and $W=CE\cap FB$. Pascal's on $FFEQPY$ gives that $D$, $T$, and $Z$ are collinear. Pascal's on $FBACEE$ gives that $W$, $D$, and $T$ are collinear. Let $X'=CY\cap BQ$. Pascal's on $CYFBQE$ gives that $X'$, $Z$, and $W$ are collinear. Since $X'\in DT$ we have that $X=X'$, so we are done.

We apply Pascal on $FFEQPY$, $EQBFYC$ and $FEECAB$ to obtain the desired result.

1 new point + pascal: Let $Z$ be the second intersection of $DY$ with $O$. Claim : $Q$, $T$ and $Z$ are collinear. Proof: Note that $-1 = (Y, Z; E, F) \overset{T}{=} (P, ZT \cap O; F, E)$. But $-1 = (P, Q; E, F)$, so $Q = ZT \cap O$. Now apply Pascal's theorem on hexagon $YCABQZ$ to obtain that $YC \cap BQ$, $CA \cap QZ$ and $AB \cap ZY$ are collinear. In particular, it shows that points $BQ \cap YC$, $D$ and $T$ are collinear. But $DT \cap BQ = X$, so $YC$ passes through $X$.