Let $l$ is tangent line to $\omega$ in $D$. Do projective transformation such $l$ goes to an infinitely distant line. Our circle $\omega$ go to parabola, and $A, B, C$ is parabola's poins. $E$ goes to point where intersect line through $C$ parallel $AB$ and line through $B$ parallel $AC$. $F$ goes to point where intersect parabola and line through $A$ parallel $BC$. We want to prove that $EF$ is parallel to parabola's axe. Choose koordinates system such our parabola has equation $y=x^2, A(a;a^2), B(b;b^2), C(c;c^2)$. Then we have, that $E$ has abscissa $x(E)=c+(b-a)=b+c-a$, and line EF has equation $y=\frac{c^2-b^2}{c-b}x+V=(c+b)x+V$, and line $AF$ has equation $y=(b+c)x+W=(b+c)x-a(b+c-a)$ because $f(a)=a^2$ and $AF$ parallel $BC$. But also must be true equation $x^2=(b+c)x-a(b+c-a)$, whence $ x(F)=b+c-a$, $E$ and $F$ have equal abscissa and $EF$ is parallel to parabola's axe. We are done

Do exist other solutions, without this unusual method?

By Ceva sinEDB/sinEDC=sinEBD/sinEBC.sinECB/sinECD=DB'/B'A'.BA'/BD. C'A'/C'D.CD/CA'=A'B/A'C.A'C'/A'B'.CD/BD.DB'/DC'=A'B/A'C.AC/AB=FB/FC (Menelaus AB'C' gives A'C'/A'B'=BC'/BA.AC/CB' and BC'/DC'.B'D/B'C=C'D/C'A.B'A/B'D=sinBAD/sinDAC=DB/DC) so D,E,F collinear, done!

Is this a valid Solution? , Please check anyone. BMO Shortlist 2015 G2 wrote: Let $ABC$ be a triangle with circumcircle $\omega$ . Point $D$ lies on the arc $BC$ του $\omega$ and is different than $B,C$ and the midpoint of arc $BC$. Tangent of $\Gamma$ 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. (Saudi Arabia) As the problem is purely projective , we can take a Projective Transformation $\mathcal P$ which fixes $\omega$ and maps $AD\cap BC$ to the center of the circle $\omega$. So $ABDC$ must be a rectangle. So we just now have to deal with this problem after this transformation. Transformed Problem wrote: $ABDC$ is a rectangle. $AB\cap DD=C',\qquad $ $AC\cap DD=B'$ and $BC\cap DD=A'$. Let $AA'\cap \omega=F$ and $BB'\cap CC'=E$. Then $\overline{D-E-F}$. Claim 1:- $BCB'C'$ is a cyclic quadrilateral. $$\angle ABC=90^\circ-\angle ACB=90^\circ-\angle CBD=90^\circ-\angle CDB'=\angle CB'C'\implies CBC'B'\text{ is a cyclic quadrilateral.}$$ Claim 2:- $F$ is the Miquel Point of $BCB'C'$. Note that $\angle CFA'=\angle ABC=\angle AB'C'\implies B'CFA' \text{ is a cyclic quadrilateral}$. Hence $\odot(ABC)\cap\odot(CB'A')$ is the Miquel Point of $BCB'C'$. As we proved that $F$ is the Miquel Point of a cyclic quadrilateral $BCB'C'$, hence, $\angle EFA=90^\circ$. Also $\angle DFA=90^\circ$. Hence, $\overline{D-E-F}$. So Projecting back we also get that $\overline{D-E-F}$ . $\blacksquare$

@amar_04 Yes, your solution is correct. Here is my projective approach. Use projective transformation, that fixes $\omega$, sends $BC$ to diameter of $\omega$ and $D$ to the midpoint of arc $BC$. Then $2|BC|=|B'C'|$ from homothety. Denote $X=AD\cap BC$ and $Y=DE\cap BC$. It sufficies to check, that $|BX|=|CY|$, then problem holds from symmetry. Using homothety centered at $A$ we get $2|BX| = |B'D|$. Using homothety centered at $E$ we get $2|CY| = |B'D'|$, hence $|CY|=|BX|$. $\square$

I revisited this problem. And my new projective sollution is this one. I really liked, that this problem is easier to solve on parabola, then on circle. As standart procedure is to projectively transform parabola to circle, but here I did it the other way around. Proof: Use projective transformation, that sends line $A'B'C'$ to infinity such, that direction of $BC$ is perpendicular to direction of point $D$ at infinity ($BC$ is perpendicular to parabola axis of symmetry). After this transformation we get this problem: Transformed problem wrote: Let $ABC$ be a triangle on parabola $\omega$, such that $BC$ is perpedicular to the axis of symmetry of $\omega$. Denote $\ell$ and $k$ lines parallel to $AB$ resp. $AC$. Denote $F=\ell\cap k$. Parallel to $BC$ through $A$ intersects $\omega$ again at $E$. Prove that $EF$ is perpendicular to $BC$. As $ABCF$ is a parallelogram this problem falls trivialy by symmetry.

I'm not sure this is correct, because I've started studying projective transformations in the last few days. Take a projective map which fixes the circumcircle of ABC and sends D the point such that ABCD is an harmonic quadrilateral. Now AA' is tangent to the circle, thus A=F and the conclusion falls because (C', B'; D, A') =-1 and so E lies on AD.

This is sadly not correct. As projective transformation preserves cross ratio (it's one of few things it preserves). So you can't create harmonic quad from non-harmonic one. Not just that, but no two different points get mapped to single point (projective transformations of whole plane are bijective). And I don't see how you got that $A=F$ for harmonic quad. (I misread it as $A'=F$, yes for harmonic $A=F$).

A more straightforward (maybe), but more tedious solution: Let $DF$ intersect $AC'$ at $T$ (if $DF \parallel AC'$, the result can be proven in a similar manner, with much simpler calculations). From Menalaus' Theorem in $\triangle{AA'C'}$ for line $FDT$, we have: $$\frac{FA'}{FA} \cdot \frac{TA}{TC'} \cdot \frac{DC'}{DA'}=1 \Longleftrightarrow$$$$\frac{FA'}{FA} \cdot \frac{DC'}{DA'} \cdot TA=TC'$$From the reverse of Menalaus' Theorem in $\triangle{BB'C'}$ for points $EDT$, it suffices to prove that $$\frac{EB}{EB'} \cdot \frac{DB'}{DC'} \cdot \frac{TC'}{TB}=1 \Longleftrightarrow$$$$\frac{EB}{EB'} \cdot \frac{DB'}{DC'} \cdot \frac{TA}{TB} \cdot \frac{FA'}{FA} \cdot \frac{DC'}{DA'}=1 \Longleftrightarrow$$$$\frac{EB}{EB'} \cdot \frac{DB'}{DA'} \cdot \frac{TA}{TB} \cdot \frac{FA'}{FA}=1$$From $\triangle{TDA} \sim \triangle{TBF}$ and $\triangle{TBD} \sim \triangle{TFA}$, we obtain: $$\frac{TA}{TF}=\frac{DA}{BF}$$$$\frac{TB}{TF}=\frac{BD}{FA}$$Dividing these two relation, we have: $$\frac{TA}{TB}=\frac{DA}{BF} \cdot \frac{FA}{BD}$$Also, from Menalaus' Theorem in $\triangle{ABB'}$ for line $CEC'$, we get: $$\frac{EB}{EB'} \cdot \frac{CB'}{CA} \cdot \frac{C'A}{C'B}=1 \Longleftrightarrow$$$$\frac{EB}{EB'}=\frac{CA}{CB'} \cdot \frac{C'B}{C'A}$$So, the result is equivalent to: $$\frac{CA}{CB'} \cdot \frac{C'B}{C'A} \cdot \frac{DB'}{DA'} \cdot \frac{DA}{BF} \cdot \frac{FA}{BD} \cdot \frac{FA'}{FA}=1 \Longleftrightarrow$$$$\frac{CA}{CB'} \cdot \frac{C'B}{C'A} \cdot \frac{DB'}{DA'} \cdot \frac{DA}{BF} \cdot \frac{FA'}{BD}=1 \Longleftrightarrow$$$$\frac{A'F}{BF} \cdot \frac{C'B}{C'A} \cdot \frac{DB'}{CB'} \cdot \frac{DA}{BD} \cdot \frac{CA}{DA'}=1$$Now from tangent $C'D$, $\triangle{B'CD} \sim \triangle{B'DA}$ and $\triangle{A'FB} \sim \triangle{A'CA}$, we get: $$\frac{C'B}{C'A}=\frac{DB^2}{DA^2}$$$$\frac{B'C}{B'D}=\frac{CD}{DA}$$$$\frac{A'F}{BF}=\frac{A'C}{AC}$$Hence, it suffices to prove that: $$\frac{A'C}{AC} \cdot \frac{DB^2}{DA^2} \cdot \frac{DA}{CD} \cdot \frac{DA}{BD} \cdot \frac{CA}{A'D}=1 \Longleftrightarrow$$$$\frac{A'C}{A'D}=\frac{CD}{BD}$$which is true, because $\triangle{A'CD} \sim \triangle{A'DB}$. The proof is complete.

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