$E$ is the midpoint of line segments $BC$. $F$ is the reflection of $E$ in $(A, B, C)$. $EF \cap (A, B, C) = M$. $I$ is the incentre of $\triangle{ABC}$. $AF \cap (A, B, C) = G$, $GI \cap (A, B, C) = D$ ($D \not\equiv G \not\equiv A$). Prove that $MD$ passes through the midpoint of $IE$.
I have tried many things but all of them didn't work. Right now I don't have any great approaches to this problem.
Claim 1. $AE$ and $AF$ are symmetric with respect to $AI$.
Proof. Let $O$ be the circumcenter of $ABC$. By definition of $F$, $OE \cdot OF = OA^2$. Hence $\triangle OEA \sim \triangle OAF$. It follows that $\angle EAO = \angle OFA$. Hence $$\angle FAM = \angle OMA - \angle MFA = \angle MAO - \angle EAO = \angle MAE.$$
Claim 2. $\dfrac{BG}{CG}=\dfrac{BA}{CA}$.
Proof. By Claim 1., $\angle BAG = \angle EAC$. Also, $\angle AGB = \angle ACE$. Hence $\triangle ABG \sim \triangle AEC$. Similarly $\triangle GAC \sim \triangle BAE$. It follows that
$$\frac{AB}{BG} = \frac{AE}{EC} = \frac{AE}{EB} = \frac{AC}{CG},$$ so $\dfrac{BG}{CG}=\dfrac{BA}{CA}$.
Claim 3. Let $BI$, $CI$ intersect the circumcircle of $ABC$ at $X$, $Y$, respectively. Then $$\frac{XD}{YD}=\frac{XM}{YM}.$$
Proof. Since $\triangle BIG \sim \triangle DIX$, we have $$\frac{BG}{BI} = \frac{XD}{DI}.$$ Analogously, $$\frac{CG}{CI} = \frac{YD}{DI}.$$ Hence $$\frac{XD}{YD}=\frac{BG}{CG} \cdot \frac{CI}{BI}.$$ Analogously $$\frac{XM}{YM}=\frac{BA}{CA} \cdot \frac{CI}{BI}.$$ But by Claim 2. we have $\dfrac{BG}{CG}=\dfrac{BA}{CA}$, so $$\frac{XD}{YD}=\frac{BG}{CG} \cdot \frac{CI}{BI}=\frac{BA}{CA} \cdot \frac{CI}{BI} = \frac{XM}{YM}.$$
Claim 4. $MX$ and $MY$ bisect the segments $CI$, $BI$, respectively.
Proof. By Trefoil Lemma, $XI=CI$ and $MI=CI$. Hence $MX$ is perpendicular bisector of $CI$, hence it bisects $CI$. Analogously, $MY$ bisects $BI$.
Claim 5. Let $K$, $L$ be the midpoints of $CI$, $BI$. Let $DM$ instersect $KL$ at $Z$. Then $Z$ is the midpoint of $KL$.
Proof. Let $\ell$ be the tangent to the circumcircle of $ABC$ at $M$. Then $\angle MYD = \angle(\ell, MD) = \angle(KL, MD) = \angle LZM$. Hence $\triangle MYD \sim \triangle MZL$. Analogously, $\triangle MXD \sim \triangle MZK$. It follows that $$\frac{MZ}{ZL} = \frac{MY}{YD} = \frac{MX}{XD} = \frac{MZ}{ZK},$$ so $ZL=ZK$.
Claim 6. $Z$ is the midpoint of $IE$.
Proof. $KILE$ is a parallelogram since $E,K,L$ are the midpoints of $BC$, $CI$, $IB$. Hence midpoints of $KL$, $IE$ coincide. Hence the claim.
The problem is solved by Claim 6.
The following is a solution using projective geometry.
Let $X$ and $Y$ be the midpoints of arcs $AC$, $AB$ of the circumcircle of $ABC$. Then $X$ lies of $BI$ and $Y$ lies on $CI$. Moreover, due to well-known trefoil lemma, $MB=MI=MC$, $XC=XI$ and $YB=YI$. Hence $XM$ and $YM$ intersect $CI$ and $BI$ at their midpoints $K$ and $L$, respectively.
Using well-known properties of symmedians we see that $(AGBC)=-1$.
It is well-known that given a circle $\omega$ and a point $P$, the map $\phi \colon \omega \to \omega$ assigning to a point $T$ the unique point $T'\in \omega$ such that $T,P,T'$ are collinear is projective. Taking as $\omega$ the circumcircle of $ABC$ and $I$ as a $P$ we get that $$-1=(AGBC)=(\phi(A)\phi(G)\phi(B)\phi(C))=(MDXY).$$ Projecting $\omega$ through $M$ to the line $KL$ we obtain $$-1=(MDXY)=(\infty ZKL).$$ where $\infty$ is the point at infinity of the line $KL$ and $Z$ is the intersection of $KL$ with $DM$. This means that $Z$ is the midpoint of $KL$.
But $E,K,L$ are the midpoints of $IBC$, so $IKEL$ is a parallelogram, hence $Z$ is a midpoint of $IE$. We are done!
