$\textbf{Problem:}$Let $ABC$ be a triangle with circumcircle $\omega$ . Point $D$ lies on the arc $BC$ not containing $A$ of $\omega$ and is different than $B,C$ and the midpoint of arc $BC$. Tangent of $\omega$ on $D$ intersects lines $BC$,$CA$,$AB$ at $A'$,$B'$,$C'$, respectively. Lines $BB'$ and $CC'$ intersect at $E$. Line $AA'$ intersects again the circle $\omega$ at $F$. Prove that points $D,E,F$ are collinear.
I tried to use menelaus theorem on bunch of triangles and in few ways I restated the problem to apply it.But all those attempts failed.I also tried to chase cross ratios but that didn't work out either.
Any help or solution will be appreciated.
Thanks @oldboy for the diagram.
Projective solution (or in modern contest languague, the method of moving points).
Let us fix circle and points $B,C,D$ on it as tangent at $D$ and let us move $A$ on the circle. Then also $E,F$ and $B',C'$ moves, but not $A'$. Then composition of projective maps $B'\mapsto A$ and $A\mapsto C'$ is also projective and this map induces projective map of pencil from $(B)$ to $(C)$: $BB'\mapsto CC'$.
This means that intersection of $BB'$ and $CC'$, that is point $E$, describes some conic (which goes through points $B$, $C$ and $D$). Now let line $DE$ meet circle at $F'$. Since conic and circle meet at $D$ we see that map $E\mapsto F'$ is well defined and it is projective from conic to circle. This also means that composition of projective maps $A\mapsto B'$, $B'\mapsto E$ and $E\mapsto F'$ i.e. $A\mapsto F'$ is projective map on circle it self.
We want to prove that this is actually an involution of circle $A\mapsto F$ with center at $A'$. By fundamental theorem of projective geometry we have to find 3 particular situation for $A$ when $F=F'$ which means that $F=F'$ is always true. But this is obviously true when $A\in\{B,C,D\}$ and we are done.
Any way here is an Euclidean geometry solution: https://artofproblemsolving.com/community/c6h2205298p16643760