$\textbf{Problem:}$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.

enter image description here

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.

  • $\begingroup$ Take a photo of your sketch and add it to the question. $\endgroup$ Oct 25, 2020 at 15:26
  • 2
    $\begingroup$ Added picture, the statement is correct $\endgroup$
    – Oldboy
    Oct 25, 2020 at 20:45
  • 1
    $\begingroup$ @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 $\endgroup$
    – Dr. Mathva
    Oct 26, 2020 at 19:43
  • 1
    $\begingroup$ @Yesit'sme you have an aops acc ? $\endgroup$ Oct 28, 2020 at 8:12
  • 1
    $\begingroup$ @SunainaPati yes $\endgroup$ Oct 29, 2020 at 13:50

1 Answer 1


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.