Doesn't seem hard but it got me stuck:

  • $I$ is the incenter of $\triangle ABC$
  • $D$ the contact point of the incircle with $BC$
  • $M,M'$ are the intersection of the circumcircle of $\triangle ABC$ with the perpendicular bisector of $BC$, $M'$ on the arc $BAC$
  • $E = AD \cap (ABC)$
  • $F,F' = M'E \cap (BCI)$

Show that $AIEF'$ lies on a circle. enter image description here

I saw this problem in a blog. We know that quadrilateral $BFCF'$ is harmonic because $BM'$ and $CM'$ are both tangent lines.

Therefore line $F'F$ is symedian of $\triangle BF'C$. The blog said that this, along with the fact that $\angle DIM = \angle DEF$ implies the quad $AIEF'$ is cyclic and I don't see this implication (even tho I do see that $FF'$ is symedian and the angle equality).

If you guys can come up with any other ideas that will be cool too.

A couple facts:

  • $E$ is midpoint of $FF'$

  • the bisector of $\angle BF'C$ meets both: $(BCI)$ and $MM'$ at the same point.

  • 1
    $\begingroup$ Please attach the blog link, so that others can also see it. $\endgroup$ Sep 11, 2020 at 6:52
  • $\begingroup$ @TeresaLisbon there you go: artofproblemsolving.com/community/… $\endgroup$ Sep 11, 2020 at 14:11
  • $\begingroup$ Thanks, +1. It is baffling how one can understand so many skipped steps! The context is way out of my reach, I've struggled for some time just to get what you found! Now I will try to proceed. Naturally collinearity of $I,D,F'$ finishes it, but I can't see that from where we are now. $\endgroup$ Sep 11, 2020 at 14:40
  • $\begingroup$ this is part IV of a series of theorems so it may seen kinda random but it also may use previous results and just like you I can't see it $\endgroup$ Sep 11, 2020 at 14:44

2 Answers 2


OP mentions this blog post as the inspiration for the question. That post might be saying that $AIEF'$ is cyclic because of "radical axis", but if so it makes the error of assuming that $I,D,F'$ are collinear.

The blog post refers to a thread on the problem, of which this is one entry. The entry defines $F'$ as collinear with $I,D$, and in a later step (after showing $AIEF'$ cyclic) demonstrates that $F'$ is collinear with $E,M'$.

Assuming $I,D,F'$ collinear the "radical axis" method amounts to computing powers of the point $D$ with respect to the circles $(ABC)$ and $(BIC)$ and the three lines $AE,BC,IF'$, giving us $$AD\cdot DE=BD\cdot DC=ID\cdot DF'.$$ By the intersecting chords theorem $AD\cdot DE=ID\cdot DF'$ implies $AIEF'$ is cyclic.

After this, we can angle chase (as in the thread entry) to show that $F'$ is collinear with $E,M'$ giving us the harmonic quad setup.


Note: AS pointed out, the following answer is invalid, as it assumes $I,D,F'$ are collinear.

The object is to show $\angle AIF' = \angle AEF’$.

enter image description here

$\angle AIF' = (black) + [grey] = (0.5\angle A+ \angle C + 0.5\angle B) +[90^0 – 0.5\angle B] = 0.5\angle A + \angle C + 90^0$

$\angle AEF’ = 90^0 + (purple) = 90^0 + (blue + [brown]) = 90^0 + (0.5\angle A + [\angle C])$

  • $\begingroup$ thanks, but when you say that $[grey] = [90^0 - 0.5 \angle B]$ you are assuming $I,D,F'$ are collinear which is what I want to prove $\endgroup$ Sep 7, 2020 at 12:43
  • $\begingroup$ @hellofriends I missed that. Give me some time to think of it. $\endgroup$
    – Mick
    Sep 7, 2020 at 14:14
  • $\begingroup$ @hellofriends All I can think of is the following:- $\angle DIM = \angle DEF$ because they both are equal to $\angle AMM’$. Let the circle (AEF’) cut IB produced at K. Then, $\angle AKF’ = \angle DEF$. Next, suppose that F’D produced cuts the (AEF’K) at some point J. Then, the exterior angle at J = $\angle AKF’ = \angle DIM$. From which, we can try to get a contradiction unless I is J. Since the post is therefore not that convinving, I will have it deleted shortly. $\endgroup$
    – Mick
    Sep 7, 2020 at 15:38
  • $\begingroup$ I think it is a good approach! Please don't delete your answer because your mistake is very common and that way people will avoid it $\endgroup$ Sep 7, 2020 at 16:04
  • $\begingroup$ @Mick why wouldn't you edit your answer to say it's wrong??..... $\endgroup$ Sep 8, 2020 at 19:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .