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 $BAC$
- $E = AD \cap (ABC)$
- $F,F' = M'E \cap (BCI)$
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.