euclidean-geometry
$E$ 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.
Well, when there is no idea then coordinate system comes at handy. And actualy it is an easy problem with c.s.
Let $B=(2b,0)$, $C= (2c,0)$, $A= (0,2a)$ and $A'= (2t,0)$, for some fixed $a,b,c$ and variable $t$. The midpoint of $A'B$ is $N = (b+t,0)$. Since $B'$ is on a line $$AC:\;\;\;{x\over 2b}+{y\over 2c}=1$$ we have $$B' = (b+t,{a(b-t)\over b}) $$ and analougly we get $$C' = (c+t,{a(c-t)\over c}) $$
Now the slope of segment $B'C'$ is $$k= {at\over bc}$$ so the slope of $d$ is $$k' = -{1\over k} = -{bc\over at}$$
So the line $d$ has equation $$ y= {bc\over at}x +{2bc\over a}$$
which means that this line goes always through the point $F=(0,{2bc\over a})$.
Ignoring vertex $A$, this becomes a problem on the inscriptable $\square BCNM$. Let $L$ be the midpoint of $\overline{BC}$. Also, let $M'$ (instead of $P$), $N'$ (instead of $Q$), $H'$, $K'$ be the projections of $M$, $N$, $H$, $K$ onto $\overline{BC}$. Let the tangent segments from $B$ and $C$ to the incircle have length $d$; let the tangent segments from $M$ and $N$ have length $m$ and $n$. Finally, define $m' := |MM'|$, $n':=|NN'|$, $m'':=|M'L|$, $n'':=|N'L|$. Without loss of generality, $m\leq n$ so that $m'\leq n'$.
Certainly, if $m=n$, then $\overleftrightarrow{HK}$ meets $\overline{BC}$ at $L$. For $m \neq n$, we'll prove that $L$ is on $\overleftrightarrow{HK}$ by showing $\triangle HH'L\sim \triangle KK'L$ via $$|HH'||K'L|=|KK'||H'L| \tag{$\star$}$$
Parallelism and proportionality rules tell us that $$\frac{|M'H'|}{|M'N'|}=\frac{|MH|}{|MN|}=\frac{m}{m+n} \qquad |HH'|=m'+\frac{m}{m+n}(n'-m')=\frac{m'n+mn'}{m+n} \tag{1}$$ The Crossed Ladders Theorem tells us that $$\frac{1}{|KK'|}=\frac{1}{m'}+\frac{1}{n'} \quad\to\quad |KK'| = \frac{m'n'}{m'+n'} \tag{2}$$ (and, in fact, $K$ is the midpoint of the extension of $\overline{KK'}$ that meets $\overline{MN}$), whereupon some proportional thinking then yields $|M'K'|:|K'N'|=m':n'$, so that we have $$\frac{|M'K'|}{|M'N'|}=\frac{m'}{m'+n'} \tag{3}$$ Therefore, $$\begin{align} |H'L|&=|M'L|-|M'H'| = m'' - \,\frac{m}{m+n}(m''+n'') \;= \frac{m''n-mn''}{m+n} \\[6pt] |K'L|&=|M'L|-|M'K'| = m'' - \frac{m'}{m'+n'}(m''+n'')=\frac{m''n'-m'n''}{m'+n'} \end{align} \tag{4}$$ Substituting in $(\star)$, and clearing denominators, we need only verify that $$(m'n+mn')(m''n'-m'n'') = m'n'(m''n-mn'') \tag{5} $$ That is, $$\frac{m}{n}\cdot\frac{m''}{n''} = \left(\frac{m'}{n'}\right)^2 \tag{6}$$
It seems like there's a geometric mean argument to be made, but I'm not seeing it. So, writing $\theta$ for the common angle at $B$ and $C$, we have $$\frac{m}{n}\cdot\frac{d-(m+d)\cos\theta}{d-(n+d)\cos\theta} = \left(\frac{m+d}{n+d}\right)^2 \quad\to\quad (d+m)(d+n)\cos\theta = d^2 - m n \tag{7}$$ This same relation results (for $\theta \neq 0$) from the observation that there's a right triangle with hypotenuse $|MN|$ and legs $|m'-n'|$ and $m''+n''$. $$(m+n)^2 = (m'-n')^2 + (m''+n'')^2 \qquad\to\qquad (7) \tag{8}$$ This equality establishes $(\star)$ and completes the proof. $\square$
I believe there's a cleaner way to link $(6)$ and $(8)$ (or to demonstrate $(6)$ some other way) without having to show equality through $(7)$. Again, I'm not seeing it. Perhaps I'll return to this question.
Best Answer
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!