Let $XY\cap AB=D$. Applying Pascal to $XXYLSK$ gives us $S, K, D$ are collinear. Similarly, applying Pascal to $YYXKRL$ gives us $R, L, D$ are collinear. Finally, applying Pascal to $XRLYSK$ gives us the desired result.

Solution with using Pascal twice

Projective time... Consider the composition of transformations $P\rightarrow K\rightarrow R\rightarrow Q\rightarrow S\rightarrow L\rightarrow P_1$ (where $Q$ is our desired intersection point) We claim that $P=P_1$ in all cases, which completes the proof Note that all transformations are projective (as inversion also preserves cross ratio). So, it's sufficient to prove this in 3 cases The first case $P=A$ is not hard to prove, just draw the points appropriately (note that we turned the "second intersection" points to "inverted" points, in order to apply projective stuff) $P=B$ is similar with the one above Let the second tangents from $A$ and $B$ to $\Gamma$ intersect at $R_1$ and $S_1$ respectively. Let $AR_1\cap BS_1 = D$. Applying Newton's theorem on the quadrilateral (which may not be convex) $CADB$ gives us the third case, which is $CD\cap AB=XR_1\cap YS_1$

$YS\cap XR=T$.Pascal to $XXKSLY$ $\implies XY,KS,AB$ concurrent.(1) Pascal to $YYSKRX$ $\implies AT,SK,XY$ concurrent.(2) (1) and (2) $\implies A,T,B$ collinear $\implies AB,XR,YS$ concurrent.$\blacksquare$.

Solved with v4913, DottedCaculator, amuthup, squareman, ETS1331, rjiangbz, CyclicISLscelesTrapezoid, nihao4112, bluelinfish, AI216, PROA200, PieAreSquared, cocohearts, hliu1. By Pascal's on cyclic hexagons $YYLRKX$ and $XXRLSY$, we're done.

CT17 wrote: Solved with v4913, DottedCaculator, amuthup, squareman, ETS1331, rjiangbz, CyclicISLscelesTrapezoid, nihao4112, bluelinfish, AI216, PROA200, PieAreSquared, cocohearts, hliu1. By Pascal's on cyclic hexagons $YYLRKX$ and $XXRLSY$, we're done. incredible.

CT17 wrote: Solved with v4913, DottedCaculator, amuthup, squareman, ETS1331, rjiangbz, CyclicISLscelesTrapezoid, nihao4112, bluelinfish, AI216, PROA200, PieAreSquared, cocohearts, hliu1. By Pascal's on cyclic hexagons $YYLRKX$ and $XXRLSY$, we're done. fifteen people. for... this?

sdash314 wrote: CT17 wrote: Solved with v4913, DottedCaculator, amuthup, squareman, ETS1331, rjiangbz, CyclicISLscelesTrapezoid, nihao4112, bluelinfish, AI216, PROA200, PieAreSquared, cocohearts, hliu1. By Pascal's on cyclic hexagons $YYLRKX$ and $XXRLSY$, we're done. fifteen people. for... this? Maybe they were brain storming several problems (?) The funniest part is that the solution is exactly the same as $3$ of the $4$ solutions already given above them.

Let the intersection point of $RL$ and $XY$ be a point say $K$. By the Pascal theorem on the $KRLYYX$ circle, you get $A, P, K$ are collinear. Again by the Pascal theorem on the $LSYXXR$ circle, you get $B, K$ and the intersection point of $XR$ and $YS$ are collinear. So, the intersection point of $YS$ and $XR$ on the segment $AB$. $\square$

By Pascal on $XYYKRL$ we have that L, R, and D are colinear and then applying Pascal to $XRLYSK$, we are done.