# Show this quad is cyclic

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.

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.

• Please attach the blog link, so that others can also see it. Sep 11, 2020 at 6:52
• @TeresaLisbon there you go: artofproblemsolving.com/community/… Sep 11, 2020 at 14:11
• 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. Sep 11, 2020 at 14:40
• 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 Sep 11, 2020 at 14:44

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 $$\angle AIF' = \angle AEF’$$.

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

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

• 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 Sep 7, 2020 at 12:43
• @hellofriends I missed that. Give me some time to think of it.
– Mick
Sep 7, 2020 at 14:14
• @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.
– Mick
Sep 7, 2020 at 15:38
• 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 Sep 7, 2020 at 16:04
• @Mick why wouldn't you edit your answer to say it's wrong??..... Sep 8, 2020 at 19:32