First see that $AFED$ is cyclic, now let $Y$ be the second intersection of the incircle and $AFED$, furthermore let $X=MI\cap (ABC)$ and $Z$ the second intersection of line $DI$ and the incircle. It's known that $X$ is the point of contact of the $A-$ mixtilinear incircle and $(ABC)$. Now see that $\angle YDB=\angle YZD=\angle YAE$ this implies that $AYIZ$ is cyclic, but $IY=IZ$ then $AI$ bisects $\angle YAZ$ and since $AE$ is bisector of $\angle BAC$ we get that $AY, AZ$ are isogonal with respect to $\angle BAC$. Now the homothety centered at $A$ that sends the incircle to the $A-$ excircle sends $Z$ to $Z'$ which is the contact point of the $A-$ excircle and $BC$ but we know that inversion centered at $A$ radius $\sqrt{bc}$ followed by a reflection across the $A-$ bisector swaps the $A-$ mixtilinear incircle and the $A-$ excircle which means that this inversion swaps $X$ and $Z'$ then $AX, AZ$ are isogonal with respect to $\angle BAC$, then $A, Y, X$ are collinear. It's known that since $X$ is the point of contact of the $A-$ mixtilinear incircle and $(ABC)$ we have $\angle BXA=\angle CXD=\angle C$, then $\angle XDY=\angle DXC+\angle DCX+\angle YDB=\angle C+\angle BAX+\angle YAE=\angle C+\angle \frac{A}{2}=\angle AED=\angle XYD$, then $XD=XY$ and $IX$ is the perpendicular bisector of $YD$. Finally let $N$ be the midpoint of $EF$, this is also the center of $(AFED)$, and by radical axis $NI \perp YD$, then $M, N, I$ are collinear

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Let $MN$ - diameter of $(ABC)$. By Shooting Lemma $NI^2=NB^2=NE\cdot NA$ $\implies$$\frac{NI}{NA}=$$\frac{NE}{NI}=$$\frac{NI-NE}{NA-NI}=$$\frac{IE}{IA}$ With $DF\parallel MN$ it would be $\frac{MF}{MA}=$$\frac{NI}{NA}=$$\frac{IE}{IA}$ By Menelaus we are done.

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See the solution to this problem on my Youtube channel. It's similar to zuss's solution but slightly different and with a little more explanation. https://www.youtube.com/watch?v=1ftR8sRi8qU

dgreenb801 wrote: See the solution to this problem on my Youtube channel. It's similar to zuss's solution but slightly different and with a little more explanation. https://www.youtube.com/watch?v=1ftR8sRi8qU nice solution!

Who can solve this problem by using harmonic quadrilateral?

$MA\cap BC=T$ Let $J$ be $A$-excenter, $N$ be the midpoint of the arc $BC$ not containing $A$.$TI\cap MJ=K$ Let $P$ be the midpoint of $BC$. Claim: $MJ\parallel EF$ Proof: We have $TI\perp EF$ since $I$ is the orthocenter of $TEF$. \[EP.ET=EA.EN=EB.EC=EI.EJ\]Thus $IPJT$ is cyclic. Also $NJ^2=NP.NM$ hence \[\angle KTP=\angle ITP=\angle IJP=\angle NJP=\angle JMP=\angle KMP\]Which gives that $MKPT$ is cyclic. So $TK\perp MJ. \square$ So we have \[-1=(J,I;E,A)\overset{M}{=}(MJ\infty, MI\cap EF;E,F)\]Which gives that $MI$ passes through the midpoint of $EF$ as desired.$\blacksquare$