# A geometry problem about colinearity

$$\textbf{Problem:}$$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.

• Take a photo of your sketch and add it to the question. Oct 25, 2020 at 15:26
• Added picture, the statement is correct Oct 25, 2020 at 20:45
• @Yes it’s me, thanks for fixing that :) I’ll have a look at the problem in a while; let’s see if can come up with some approach. It looks difficult Oct 26, 2020 at 19:43
• @Yesit'sme you have an aops acc ? Oct 28, 2020 at 8:12
• @SunainaPati yes Oct 29, 2020 at 13:50

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

• Awesome! [+1] What is the motivation behind using the Moving Points technique? Nov 2, 2020 at 12:16
• Well it is pretty much projective configuration, so... @Dr.Mathva Nov 2, 2020 at 18:54
• Nice solution! Other way to go is by noting that the problem is purely projective ;) Nov 6, 2020 at 8:18