i got that the concurrency is equivalent to prove that $AB:AE=CB:CE$ ...

Who has any solution this problem? I have some idea but i haven't finished to prove yet!

we have that $AB,DE,FC$ are collinear.

Any solution to this?

Consider an inversion centered at $D$ with radius $1$ : Points $ E' , F' , A' ,B' , C' $ lie on a line in that order . By Inversion distance formula the conditions are equivalent to $A'B' \cdot C'F' =2B'C' \cdot F'A' $ , $ \; E'B'=2B'C' \;$ and $ E'F'=2F'A' $ For brevity let $F'A'=x$ and $B'C'=y$ , now $E'B'=2y , E'F'=2x , E'A'=3x , E'C'=3y , A'B'=2y-3x , C'F'=3y-2x \;\;\; $ and $ \; A'C'=3(y-x) $ $\Longleftrightarrow y>x$ The condition $A'B' \cdot C'F' =2B'C' \cdot F'A' $ is now $2y^2-5xy+2x^2=0$ $ \Longleftrightarrow y=2x$ Line $AD $ stays the same , line $BE$ goes to the $(BED) $ , line $ CF$ goes to $(CFD)$ $A'B' \cdot A'E' = A'F' \cdot A'C' $ is equivalent to $3x(2y-3x)=3x(y-x) $ wich is true , so $A'$ has equal powers wrt $(BED) \;$ and $(CFD)$ $ \Longleftrightarrow A'D$is the radical axis of $(BED) \;$ and $(CFD)$ $\Longleftrightarrow AD, BE,CF$ are concurrent .q.e.d

Since $(AC; BF) = (DB; CE) = -2$ it follows $\overline{BC}$, $\overline{AD}$, $\overline{EF}$ are concurrent; similarly for the other two triples. Consider a projective transform sending the Pascal line of $ABCDEF$ to infinity; then $\triangle ACE$ and $\triangle BDF$ are equilateral and isosceles trapezoids induce equal lengths $s = FA = BC = DE$, $t = AB = CD = EF$, $d = AD = BE = CF$. By Ptolemy \[s^2 + td = FB \cdot AC = AC \cdot BD = t^2 + sd \implies d = s + t.\]Now $(AC; BF) = -2$ gives $td = t(s + t) = 2s^2$, whence $s = t$. Then $ABCDEF$ is a regular hexagon and the concurrency follows.

CantonMathGuy wrote: Consider a projective transform sending the Pascal line of $ABCDEF$ to infinity There's a small, but important, configuration issue in here; since for such a projective transformation to exist, we need a null intersection of $(ABCDEF)$ and that line.

WizardMath wrote: CantonMathGuy wrote: Consider a projective transform sending the Pascal line of $ABCDEF$ to infinity There's a small, but important, configuration issue in here; since for such a projective transformation to exist, we need a null intersection of $(ABCDEF)$ and that line. Please correct me if I'm wrong, but I don't think it's possible for the Pascal line to intersect the circumcircle if $ABCDEF$ is convex and non-self-intersecting.

CantonMathGuy wrote: Please correct me if I'm wrong, but I don't think it's possible for the Pascal line to intersect the circumcircle if $ABCDEF$ is convex and non-self-intersecting. That's true, but the problem didn't mention that it was convex and non-self-intersecting. Sorry for being a bit pedantic about these issues though. (I don't know whether it is the general assumption or not, to assume convexity)