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.
 A: 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.
A: Note: AS pointed out, the following answer is invalid, as it assumes $I,D,F'$ are collinear.
The object is to show $\angle AIF' = \angle AEF’$.

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