$ D $ and $ E $ are obviously the reflections of $ C $ and $ B $ over the lines $ BI $ and $ CI $ respectively and so $ F $ is the orthocenter of $ \triangle IBC $, or $ I $ is the orthocenter of $ \triangle FBC $. Denote by $ M_1,M_2 $ the midpoints of $ BD,CE $ respectively. By the Lemma here, we get $ IH \perp M_1M_2 $, and the diameters $ BD,CE $ are equal, so the radical axis $ IH $ also becomes the perpendicular bisector of $ M_1M_2 $. Now, $ BM_1=CM_2 $, $ BM=CM $, $ \angle ABM= \angle M_1BM=\angle ACM=\angle M_2CM $, so $ \triangle M_1BM, \triangle M_2CM $ are congruent, and $ MM_1=MM_2 $, or $ M $ is on the perpendicular bisector of $ M_1M_2 $, i.e $ IH $.

use from: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=575006&p=3388429#p3388429

My solution sketch: 1.) Draw circumcircles of triangle with diameter $BD$ and $CE$ . Note that radius is same. 2.) Using this, prove that $M$ lies on the radical axis of these two circles. 3.) Obviously H is on the radical axis(power of the point) and since I is orthocentre of triangle $FBC$ , I also lies on radical axis. So $M,I,H$ are collinear(Lies on radical axis) .

Let $BI,CI,HD,HE$ meet $AM$ at $I_B,I_C,K,L$, $HD,HE$ meet $CM,BM$ at $Q,P$, $BI,CI$ meet $AC,AB$ at $X,Y$. Note that $HD,HE$ are parallel to $CI,BI$ respectively. ($HD$ perpendicular to $EF$ perpendicular to $CI$ etc.) Hence $\angle MKQ=\angle AYC-\angle KAD=\frac{\angle B}{2}$ and similarly $\angle MLP=\frac{\angle C}{2}$ Together with $\angle AMB=\angle C$, we get that $\triangle KMQ$ (and similarly $\triangle LMP$) is isoceles. Now we want to show that $\triangle PHQ$ and $\triangle BIC$ are homothetic w.r.t. $M$. This would imply that $M,H,I$ collinear and is equivalent to showing $MQ=MP\Leftrightarrow MK=ML\Leftrightarrow KI_C=LI_B$. But $KI_C=DY\frac{sin(\angle A+\frac{\angle C}{2})}{sin(\frac{\angle B}{2})}=DC\frac{sin\frac{\angle A}{2}}{sin\frac{\angle B}{2}}=2BCsin\frac{\angle A}{2}$ and similarly $LI_B=2BCsin\frac{\angle A}{2}$ and hence we are done.

Here is a very ridiculous solution, although at heart I think it's equivalent to shivangjindal's solution, just using more arcane weaponry. [asy][asy] size(8cm); pointpen = black; defaultpen(fontsize(8pt)); pathpen = black; pair A = Drawing("A", dir(110), dir(110)); pair B = Drawing("B", dir(235), dir(235)); pair C = Drawing("C", dir(-55), dir(-55)); draw(A--B--C--cycle); pair I = Drawing("I", incenter(A,B,C), dir(-125)); pair D = Drawing("D", 2*foot(C,B,I)-C, dir(200)); pair E = Drawing("E", 2*foot(B,C,I)-B, dir(-20)); pair F = Drawing("F", extension(B,E,C,D), dir(200)); pair H = Drawing("H", orthocenter(D,E,F), dir(175)); pair M = Drawing("M", dir(90), dir(90)); draw(B--E--D--C); draw(circumcircle(A,D,E)); draw(unitcircle); pair X = Drawing("X", midpoint(B--D), dir(A+B)); pair Y = Drawing("Y", midpoint(C--E), dir(A+C)); pair P = Drawing("P", midpoint(B--C), dir(B+C)); draw(X--Y, black+1.5); draw(M--I, black+1.5); draw(M--X--I--Y--cycle, black+1.5); draw(I--P, dashed); [/asy][/asy] Denote by $X$ and $Y$ the midpoints of $BD$ and $CE$ and by $P$ the midpoint of $BC$. Observe that $IX=IP=IY$. Next, consider the second intersection $T$ of $(ABC)$ and $(ADE)$. It's the center of the spiral similarity $BD \to CE$, but then it must be an (oriented) congruence, so $TB = TC \implies T \equiv M$. Hence $MX = MY$; thus $MI$ is the perpendicular bisector of $XY$. Now $XY$ is the Gauss line of complete quadrilateral $DBEC$. Since $I$ is the orthocenter of triangle $FBC$, line $MI$ must be the Steiner line of $DBEC$ (since the Steiner and Gauss lines are always perpendicular), which by definition passes through $H$.

Preliminary: By $SAS$ we have congruent triangles $MEC,MDB$. Let $MD\cap CI=D', ME\cap BI=E'$. Lemma: $MD'=ME'$. Proof: Let $E''$ be a point on $ME$ such that $E$ is between $M,E''$ and $EE''=DD'$. Hence we can prove $MD'B, ME''C$ are congruent$\Rightarrow CE''=BD'=D'E=DE''$(here we used the fact that $DEE''D'$ is an isosceles triangle), this means $E''$ is on $BI$ $\Rightarrow E'\equiv E''$ and we are done. Main problem: Let pedals from $C,D,M$ onto $BE$ be $C_1,D_1,M_1$ and the pedals from $B,E,M$ onto $CD$ be $B{}_1{}_1,E_1,M_2$. Hence $M,H,I$ are collinear$\iff \frac {M{}_1D{}_1}{M{}_1C{}_1}=\frac {M{}_1E{}_1}{M{}_1B{}_1}\iff \frac {MD}{MD'}=\frac {ME}{ME'}$ which by the lemma is true and we are done.

Take $S\in EH\cap MB$ and $T\in DH\cap MC$. If we show that $ST\parallel BC$, then we are done because a homothety with center $M$ sends $\triangle HST$ into $\triangle IBC$. In order to prove this we just have to see that $\widehat{ESB}=360^\circ - 90^\circ - \widehat{BCF}-\widehat{MBC}$. But $\widehat{BCF}=90^\circ - B/2$ and $\widehat{MBC}=90^\circ - A/2$ therefore $\widehat{ESB}=180^\circ - C/2$, so $S\in \odot (C,CB)$ and similarly $T\in \odot (B,BC)$. Now, since $\triangle MBC$ is isosceles, we get that $BS=CT$ and the conclusion follows.

Dear Mathlinkers, this is just an idea... can we make a link with a mixtilinear incircle? Sincerely Jean-Louis

For those who are less skilled in geometry, here's a solution that can be finished in at most 1 hour (even if you are past your MATHCOUNTS prime like me). This is more bashy that most other problems though. COMPLEX NUMBERS!!! Work on the unit circle and set the vertices as $a^2, b^2, c^2$ and the minor arc midpoints $M_c, M_a, M_b$ as $-ab, -bc, -ac$. Therefore $m = bc$, and $i = -ab-bc-ac$. We show that the perpendiculars from $D$ to $BE$, $E$ to $CD$ and $MI$ are concurrent by showing that the intersection $MI$ with the perpendicular from $E$ to $CD$ (let's call this point $K$) is symmetric in $b, c$. Note that $E$ is the reflection of $B$ over $CM_c$. Therefore, the $e = c^2-ab+abc^2\bar{b}^2 = c^2-ab+\frac{ac^2}{b}$. Note that the perpendicular from $E$ to $CD$ is parallel to $BM_b$, so $EK \| BM_b$. Therefore, $\frac{k-e}{\bar{k}-\bar{e}} = \frac{b^2+ac}{\bar{b}^2+\overline{ac}} = ab^2c \implies k-ab^2c\bar{k} = e-ab^2c\bar{e}$. (1) Now, $K$ is collinear with $M, I$, so $\frac{k-bc}{\bar{k}-1/bc} = \frac{bc+(ab+bc+ac)}{\overline{bc+(ab+bc+ac)}} = abc \cdot \frac{ab+ac+2bc}{b+c+2a} \implies$ \[ k - abc \cdot \frac{ab+ac+2bc}{b+c+2a} \bar{k} = bc - \frac{1}{bc} \cdot abc \cdot \frac{ab+ac+2bc}{b+c+2a} = \frac{(b+c)(bc-a^2)}{b+c+2a} \text{(2)} \] Now we can have some more fun solving (1)+(2). The best way to solve these is probably $\frac{ab+ac+2bc}{b+c+2a}$ times (1) minus $b$ times (2). This gives $\left(\frac{ab+ac+2bc}{b+c+2a}-b \right)k = \frac{ab+ac+2bc}{b+c+2a} \cdot (e-ab^2c\bar{e}) - \frac{b(b+c)(bc-a^2)}{b+c+2a}$ $(c-b)k = \frac{1}{a+b} \cdot \left((ab+ac+2bc) \cdot (e-ab^2c\bar{e}) - b(b+c)(bc-a^2) \right)$. Note that $e-ab^2c\bar{e} = c^2-ab+\frac{ac^2}{b}-\frac{ab^2}{c}+bc-\frac{b^3}{c} = \frac{(a+b)(c^3-b^3)+b^2c(c-a)}{bc}$. Plugging this into above, and reversing the roles of $b, c$ we need that (since we have $(c-b)k$ on the LHS) $\frac{1}{a+b} \left((ab+ac+2bc) \cdot \left(\frac{(a+b)(c^3-b^3)+b^2c(c-a)}{bc} \right) - b(b+c)(bc-a^2) \right) + \frac{1}{a+c} \left((ab+ac+2bc) \cdot \left(\frac{(a+c)(b^3-c^3)+bc^2(b-a)}{bc} \right) - c(b+c)(bc-a^2) \right) = 0$ Multiplying through by $(a+b)(a+c)$, we get $(ab+ac+2bc) \cdot \left(\frac{(a+b)(a+c)(c^3-b^3)+b^2c(c^2-a^2)}{bc} \right) + $ $(ab+ac+2bc) \cdot \left(\frac{(a+b)(a+c)(b^3-c^3)+bc^2(b^2-a^2)}{bc} \right)$ $ - (ab+bc)(b+c)(bc-a^2) - (ac+bc)(b+c)(bc-a^2) = $ $(ab+ac+2bc)\left( b(c^2-a^2)+c(b^2-a^2)-(b+c)(bc-a^2) \right) = 0$, so we're finished.

v_Enhance wrote: Here is a very ridiculous solution, although at heart I think it's equivalent to shivangjindal's solution, just using more arcane weaponry. [asy][asy] size(8cm); pointpen = black; defaultpen(fontsize(8pt)); pathpen = black; pair A = Drawing("A", dir(110), dir(110)); pair B = Drawing("B", dir(235), dir(235)); pair C = Drawing("C", dir(-55), dir(-55)); draw(A--B--C--cycle); pair I = Drawing("I", incenter(A,B,C), dir(-125)); pair D = Drawing("D", 2*foot(C,B,I)-C, dir(200)); pair E = Drawing("E", 2*foot(B,C,I)-B, dir(-20)); pair F = Drawing("F", extension(B,E,C,D), dir(200)); pair H = Drawing("H", orthocenter(D,E,F), dir(175)); pair M = Drawing("M", dir(90), dir(90)); draw(B--E--D--C); draw(circumcircle(A,D,E)); draw(unitcircle); pair X = Drawing("X", midpoint(B--D), dir(A+B)); pair Y = Drawing("Y", midpoint(C--E), dir(A+C)); pair P = Drawing("P", midpoint(B--C), dir(B+C)); draw(X--Y, black+1.5); draw(M--I, black+1.5); draw(M--X--I--Y--cycle, black+1.5); draw(I--P, dashed); [/asy][/asy] Denote by $X$ and $Y$ the midpoints of $BD$ and $CE$ and by $P$ the midpoint of $BC$. Observe that $IX=IP=IY$. Next, consider the second intersection $T$ of $(ABC)$ and $(ADE)$. It's the center of the spiral similarity $BD \to CE$, but then it must be an (oriented) congruence, so $TB = TC \implies T \equiv M$. Hence $MX = MY$; thus $MI$ is the perpendicular bisector of $XY$. Now $XY$ is the Gauss line of complete quadrilateral $DBEC$. Since $I$ is the orthocenter of triangle $FBC$, line $MI$ must be the Steiner line of $DBEC$ (since the Steiner and Gauss lines are always perpendicular), which by definition passes through $H$. Nah, not ridiculous, ours are practically the same, too: If $\odot ADE \cap \odot ABC = M'$, the midpoints points of $DB, EC$ be $D_M, E_M$, and their isotomic points be $D', E'$, then note $M'A$ is then tangent to $AD'E'$ due to spiral similarity, so by easy angle chasing $\angle M'AC = 90-A/2 \implies M=M'$. The transformation preserves lengths, so $MD_M = ME_M \implies M$ is on the perpendicular bisector of $D_ME_M$ (I used angles instead of easy lengths ). The circles with diametres $DB, EC$ have centres $D_M, E_M$, and obviously the two orthocentres lie on the radical axis of them (which proves perpendicularity of the steiner line with gauss line), which is the perpendicular bisector of $D_ME_M$.

Glad to see I'm not totally insane then It seems like the main difficulty of the problem is figuring out how to use the mysterious orthocenter $H$. As far as I can tell the two main ways are (1) using a homothety after some synthetic work, or (2) nuking the problem with Steiner lines (or, replicating the proof by constructing the circles and using radical axis). I'm of the opinion that (1) is actually hard to reliably pull off in-contest. In fact, I didn't initially realize it was possible, which is why I called my solution ridiculous: I don't think most contestants have seen (2) before, and I never ever thought that I would have to resort to Steiner lines on an Olympiad. By "resort" I mean I couldn't see any other way to do the problem after finding the Steiner line solution. But maybe that's because I was rocking out to sk8er boi in an empty MIT classroom while doing this problem.

v_Enhance wrote: Glad to see I'm not totally insane then It seems like the main difficulty of the problem is figuring out how to use the mysterious orthocenter $H$. As far as I can tell the two main ways are (1) using a homothety after some synthetic work, or (2) nuking the problem with Steiner lines (or, replicating the proof by constructing the circles and using radical axis). I'm of the opinion that (1) is actually hard to reliably pull off in-contest. In fact, I didn't initially realize it was possible, which is why I called my solution ridiculous: I don't think most contestants have seen (2) before, and I never ever thought that I would have to resort to Steiner lines on an Olympiad. By "resort" I mean I couldn't see any other way to do the problem after finding the Steiner line solution. But maybe that's because I was rocking out to sk8er boi in an empty MIT classroom while doing this problem. Hahahah! I neither saw the homothety solution... Naturally when I see conditions, like this one, that is kind of hard to deal with, I go in defensive mode and pull out any theorem possible... In this case, took their isotomic points to make a parallel lines then quickly angle chased. Interestingly enough, both our proofs bring up the following lemma: Let $ABC$ be a triangle, $M$ be the midpoint of arc $BAC$ (can be generalised further, ofc), then the circles through $MA$ intersect the sides of the triangle at points $DE$ s.t. $BD=CE$. The converse also... This actually means any problem involving equal lengths on different sides etc... we just use this and we immediately have made progress on the problem!

That's my problem. I hope you had nice time solving it. Here is another result influenced by Jean-Louis's idea (i noticed this fact in geogebra and i don't know a proof for it): let the mitilinear incircle touch arc $BC$ at $T$ (in fact it is the intersection point of $MI$ and circumcircle of $ABC$ ) and let $A'$ be the symmetric point to $A$ with respect to the perpendicular bisector of $BC$. Then points $T, F, H, A'$ are concyclic.

Let segments $MD$ and $CI$ intersect at point $T$, (It is known that $AEDM$ is syclic), then it is enough to prove that: $\frac{IT}{HD}=\frac{TK}{DK}$. We prove it using "sinus's" theorem...!!

McItran wrote: That's my problem. I hope you had nice time solving it. Here is another result influenced by Jean-Louis's idea (i noticed this fact in geogebra and i don't know a proof for it): let the mitilinear incircle touch arc $BC$ at $T$ (in fact it is the intersection point of $MI$ and circumcircle of $ABC$ ) and let $A'$ be the symmetric point to $A$ with respect to the perpendicular bisector of $BC$. Then points $T, F, H, A'$ are concyclic. Absolutely fantastic problem after a long time Here's my synthetic solution. Lemma 1: $ FH \parallel OI $ Proof: We have $ FH \perp DE $ and $ OI \perp DE $ (the last is well known, has appeared many times on AoPS), hence the result. Lemma 2: $ M,H,I,T $ are collinear. Proof: In the previous posts. So we get $ \angle FHT=\angle FHI=\angle HIO=\angle MIO $. Note that $ F $, the orthocenter of $ \triangle IBC $ is the reflection of $ I_a $, the $ A- $ excenter in the midpoint of $ BC $. Therefore, symmetric of $ F $ in the perpendicular bisector of $ BC $ is just the reflection of $ I_a $ in line $ BC $, name it $ L $. Let symmetric of $ T $ wrt the perpendicular bisector of $ BC $ be $ T' $. Note that $ T' \in (O) $, $ TT' \parallel BC $, so $ AT,AT' $ are isogonal lines wrt $ \angle A $ and so $ AT' $ is the $ A- $ nagel cevian, passing through the touchpoint $ S $ of the $ A- $ excircle with $ BC $. We hence get, $ \angle FA'T=\angle LAS $. Hence, it suffices to prove $ \angle MIO=\angle LAS $. Line $ OI $ meets line $ \overline{I_aSL} $ at the Bevan point $ V $. Now, another lemma. Lemma 3: In $ \triangle ABC $, with orthocenter $ H $ and circumcenter $ O $, and corresponding orthic $ \triangle DEF $, $ A' $ is the reflection of $ A $ in $ EF $. Then, $ HDA'O $ is a cyclic quadrilateral. Proof: Obviously, $ A' $ is on $ AO $ and $ \angle OBA=\angle OAB=\angle FA'A=\angle FA'O $ and so $ FBA'O $ is a cyclic quadrilateral, hence $ AO.AA'=AF.AB=AH.AD $, which makes $ HOA'D $ cyclic too. Applying the lemma in our problem to the excentral triangle, we get $ AIVL $ is cyclic, and so $ \angle LAI_a= \angle IVS= 180^{\circ}-\angle IOM $. Now, some computations. By the sine rule, \[ \frac{\sin \angle LAI_a}{\sin \angle ALS} = \frac{2r_a}{AI_a} \] \[ \frac{ \sin \angle LAS}{\sin \angle ALS}=\frac{r_a}{AS} \] \[ \implies \frac{ \sin \angle LAI}{\sin \angle LAS}=\frac{2AS}{AI_a} \] We also have in $ \triangle MIO $, \[ \frac{\sin \angle MOI}{\sin \angle MIO}=\frac{MI}{R} \] Equating $ \sin \angle MOI $ and $ \sin \angle LAI $ we obtain \[ \frac{\sin \angle MIO}{\sin \angle LAS}=\frac{2AS.R}{AI_a.IM} \] Since $ \triangle ABT \sim \triangle ASC $, we have $ AS=\frac{AB.AC}{AT} $ and its easy to see $ AI.AI_a=AB.AC $. The equation reduces to \[ \frac{\sin \angle MIO}{\sin \angle LAS}= \frac{AI.(2R)}{IM.AT} \] However, $ \angle MAI $ is right and $ \frac{AI}{MI}=\sin \angle AMT $, so we get \[ \frac{\sin \angle MIO}{\sin \angle LAS}= \frac{2R. \sin \angle AMT}{AT}=\frac{AT}{AT}=1 \implies \angle MIO=\angle LAS \] ending the proof.

McItran wrote: Here is another result influenced by Jean-Louis's idea (i noticed this fact in geogebra and i don't know a proof for it): let the mitilinear incircle touch arc $BC$ at $T$ (in fact it is the intersection point of $MI$ and circumcircle of $ABC$ ) and let $A'$ be the symmetric point to $A$ with respect to the perpendicular bisector of $BC$. Then points $T, F, H, A'$ are concyclic. We can generalize this result: $\circ$ Let $\triangle ABC$ be a triangle and $X, Y$ points on $AB, AC$ such that $BX = CY$. If $F$ is the intersection of $BY$ and $CX$ and $A'$ is the reflection of $A$ on the perpendicular bisector of $BC$, then the angle between $FA'$ and $BC$ is fixed, and equal to $90- \theta$, where $2\theta = \angle B - \angle C$ (WLOG $\angle B > \angle C$). Indeed, after some angle-chasing, this implies McItran's statement for the case when $X \equiv D, Y \equiv E$. Proof: Let $(M)$ denote the circle centered at the midpoint $M$ of arc $BAC$ that goes through $A, A'$, and let $Z$ be the second intersection of $(M)$ with $(AXY)$. Since $M \in (AXY)$, $MX = MY$ and $MA = MZ$, we have $AZ \parallel XY$ and the angle between $FZ$ and $XY$ is equal to $\angle FZA$, which is easily seen to be $90 - \theta$. Therefore, it remains to prove that $Z \in FA'.$ Note that $\varphi : X \mapsto Y$ is a projectivity with $\varphi(B) = C$ and $\varphi(X) = Y \Longrightarrow F$ lies on its axis $\tau$. Taking $X$ and $Y$ to coincide with $A$, we see that $\tau$ goes through the intersections $X_0, Y_0$ of $(M)$ with $AB, AC$ respectively, i.e. $\tau \equiv X_0Y_0$. Since $AX_0 = AY_0 = AC-AB$, $\tau$ is parallel to the tangent to $(M)$ at $A$, i.e. the $A$-bisector. Let $(AXY)$ cut $CX$ again at $V$. Note that $\angle X_0 A Z + \angle XAZ = 180 \Longrightarrow \angle X_0 Y_0 Z = \angle XVZ$ and $V \in (FY_0 Z)$. Further, $\angle MVX = \angle MAX_0 = \angle MY_0 A$, i.e. $V \in (MY_0 C)$ (its locus). Finally, $\angle Y_0 ZF = \angle Y_0 V F = \angle Y_0 MC = \angle Y_0 A A'$, i.e. $\angle Y_0 ZF + \angle Y_0 ZA' = 180$ and $Z \in FA'$, as desired.

Sorry for the revive, but I think I have a nice solution for this one. It is well known that $AMED$ is cyclic, so $\displaystyle \angle MDE=\angle MED=90-\frac{A}{2}.$ Simple angle chasing implies that $\displaystyle \angle DFE=90-\frac{A}{2}$. This means that $M$ is the intersection point of the tangets of the circle $(F,D,E)$ at $D,E.$ Let $K,L$ be the intersections of $DH,EH$ with $(F,D,E)$. $BIEA,CIDA$ are cyclic, so $I$ is the Miquel point of $BECDFA\Rightarrow BIFD,CIFE$ cyclic. This implies that $\angle FEI=\angle FDI=\angle \displaystyle \frac{A}{2}=90-\angle DFE$, so $I$ is the intersection of $DL,EK$. The conclusion follows using Pascal's theorem at $DDKEEL.$

Because $BD=CE$, $M$ is the center of spiral similarity which takes $DE$ to $BC$, therefore $\widehat{MDE}=\widehat{MED}=90^\circ-\hat{A}=\widehat{DFE}$, hence $MD,ME$ are tangent to circle $(DEF)$. As $IFDB$ and $IFEC$ are cyclic, we have that $\widehat{IDF}=\widehat{IEF}=\dfrac{\hat{A}}{2}$. Let $\{X\}=EH\cap DI,\ \{Y\}=DH\cap EI$. By angle chasing, we have that $D,X,F,Y,E$ are concyclic. Let $\{T\}=XY\cap DE$. The point $T$ is on $DE$ which is the polar of $M$ wrt $(DEF)$, hence $M$ is on the polar of $T$. But by Brokard's theorem, $\triangle{THI}$ is self-polar, so the polar of $T$ is actually $HI$, thereby proving the claim.

Dear Mathlinkers, I have this link http://egmo2014.tubitak.gov.tr/sites/default/files/solutions-day1.pdf Sincerely Jean-Louis

Notice $\triangle IBD\cong\triangle IBC\cong\triangle IEC$ and $\triangle MDB\cong\triangle MEC$ so $I$ and $M$ are the centers of the spiral similarities $\overline{BD}\mapsto\overline{EC}$ and $\overline{BD}\mapsto\overline{CE},$ respectively. Hence, $I$ and $M$ are the Miquel points of $ECDB$ and $BCED,$ respectively. Consider the circles $\omega=(BD)$ and $\gamma=(CE).$ We see $I$ is the orthocenter of $\triangle FBC$ as $BD=BC$ and $\overline{BI}$ bisects $\angle ABC,$ so by the Gauss-Bodenmiller Theorem, the radical axis of $\omega$ and $\gamma$ is $\overline{HI}.$ Let $D_1$ and $E_1$ be the feet from $D$ and $E$ to $\overline{MB}$ and $\overline{MC},$ respectively. Notice $D_1$ lies on $\omega$ and $E_1$ lies on $\gamma.$ Since $\triangle MDB\cong\triangle MEC,$ we see $MD_1\cdot MB=ME_1\cdot MC$ so $M$ lies on the radical axis of $\omega$ and $\gamma.$ $\square$ Remarks. The fact that $M$ is the Miquel point of $BCED$ is not necessary in my proof...

Is this new way? Let $l_B,l_C$ be the external angle bisectors of $B,C$ respectively. Define $l_B\cap EH=S, l_C\cap DH=T$ and $U,V$ be the feet of perpendicular from $M$ to $l_B,l_C$ respectively. Then $HS\perp l_B,HT\perp l_C$. By using $\triangle IBE\sim\triangle ICD$ and $\angle EBS=\angle DCT$, we get $\displaystyle\frac{CT}{BS}=\frac{CI}{BI}$. Also, $MB=MC$ and $\angle IBC=\angle BMU, \angle ICB=\angle CMV$ so it holds that $$\displaystyle\frac{CU}{BV}=\frac{CI}{BI}=\frac{CT}{BS},$$implies that $I,H,M$ are collinear by affinity.

Let $P,N$ be midpoint's of $BD,CE$. Claim $: MP = MN$. Proof $:$ Note that $MB = MC$ and $PB = NC$ and $\angle MBP = \angle MBA = \angle MCA = \angle MCN$. so $MCN$ and $MBP$ are congruent so $MP = MN$. So $M$ lies on Radical Axis of circles with centers $P,N$ and radius $PD,NE$. Claim $:IP = IN$. Proof $:$ Let $Q$ be midpoint of $BC$ note that $IP = IQ$ and $IN = IQ$ so $IN = IP$. so $I$ lies on Radical Axis of circles with centers $P,N$ and radius $PD,NE$. Let $EH$ meet $DF$ at $S$ and $DH$ meet $EF$ at $K$. Note that $K,S$ lie on circles so $H$ lies on Radical Axis. so now $M,H,I$ are collinear on Radical Axis of circles with centers $P,N$ and radius $PD,NE$.

SnowPanda wrote: \[\begin{vmatrix} u & v & w \\ -b & a & 0 \\ 1 & 1 & 1 \end{vmatrix} = ua + vb - w(a + b) = 0,\] Could you explain where that matrix comes from please, from some colinearity or some formula?

Note $BI$ is perpendicular bisector of $CD$, hence $BI \perp CF$ and similarly $CI \perp BF$. Hence $I$ is orthocentre of $FBC$, which implies $ \angle BFC= 90 - A/2$. Note we have $CI \parallel DH$ as both are perpendicular to $BE$, and similarly $BI \parallel EH$. Define $X = DH \cap CM$ and $Y = EH \cap BM$. Claim : $XY \parallel BC$. Proof : Note $\angle CDX = \angle FDH = 90 - \angle DFE = A/2$. Also $\angle MXD = \angle MCI = B/2$. By sine rule in triangles $CDX$ and $BDC$, we get $\frac{XC}{\sin A/2} = \frac{DC}{\sin B/2} = 2BC$. Similarly, we get $\frac{YB}{\sin A/2} = 2BC$. Hence, $ XC = YB$, which implies $XY \parallel BC$. $\square$ To finish, note triangles $HXY$ and $ICB$ are homothetic, hence $HI$ passes through $M$, and we are done!

there's no way what Let $O_B$ and $O_C$ be the midpoints of $BD$ and $CE$ and therefore centers of $(BD)$ and $(CE)$. Notice that $I$ is the orthocenter of $\triangle FBC$. By drawing altitudes we can use PoP to get $HI$ is the radical axis of $(BD)$ and $(CE)$. Now since $MO_B=MO_C$ clearly $M$ also lies on the radical axis so $MHI$ collinear. Done! motivation: steiner config also basically identical to 1995 G8

The main pitfall here is to forget that $F$ is a thing. If you do so, then you have no alternative but to bash, which would be an unfavorable one. Claim: [$F$ is a thing] $I$ is also the orthocenter of $\triangle BFC$. Proof. Realize that $\triangle BDC$ is isosceles with $BD=BC$. Thus $\overline{BI}$ is perpendicular to $\overline{CD}=\overline{CF}$ and $\overline{CI}$ is perpendicular to $\overline{BE}=\overline{BF}$, so $I$ is the orthocenter of $\triangle BFC$ as desired. Now due to 2020 HMMT T8 vibes, we realize that $M$ is the center of the spiral mapping that sends $\overline{BD}$ to $\overline{CE}$. This means that $MD=ME$, and recalling that $MB=MC$, we have that $M$ has an equal distance from both midpoints of $BD$ and $CE$. In particular, $M$ has equal power with respect to the circle $\omega_B$ with diameter $BD$ and the circle $\omega_C$ with diameter $CE$. It remains to show that $\overline{IH}$ is the radical axis of $\omega_B$ and $\omega_C$, since that would imply that $M$ lies on $\overline{IH}$. We will only show that $H$ lies on the radical axis, as showing $I$ lies on it would be a symmetric argument. Let $H_B$ be the foot of the altitude from $H$ onto $\overline{BE}$ and similarly define $H_C$. Note that \[ \text{Pow}(H, \omega_B) = HH_B \cdot HD = HH_C \cdot HE = \text{Pow}(H, \omega_C) \]by power of a point on $H$ with respect to the circle with diameter $DE$, and we are done.

Note that $M$ is the center of a rotation from $DB$ to $EC$ while $I$ is the center of a rotation from $BD$ to $EC$, so line $IM$ is the locus of points $P$ satisfying $PB^2 - PC^2 = PE^2 - PD^2$. Let $X$ and $Y$ be the feet from $D$ to $EF$ and $E$ to $DF$. Then we have $$HB^2 - HC^2 = (BD^2 - DX^2 + HX^2) - (CE^2 - EY^2 + HY^2) = (EY^2 - DX^2) + (HX^2 - HY^2) = (EY^2 - DX^2) + (FY^2 - FX^2) = FE^2 - FD^2 = HE^2 - HD^2$$ as desired.

THIS IS SO GOOD Let $(BDF)$ and $(CEF)$ intersect at $I'$. First, easy angle chasing gives $\angle BI'C=90+\angle A/2$. However, $I'$ is the center of a spiral similarity taking $BD$ to $EC$. As $BD=CE$ this spiral sim is a rotation so we find $BI'=I'E$, and as $ABI'E$ is cyclic by miquel we find $I'$ is on the angle bisector of $\angle BAC$. Thus $I\equiv I'$. Easy angle chasing then implies $I$ is also the orthocenter of $BFC$. To finish, there is a spiral similarity at $M$ taking $DB$ to $EC$ which is a rotation. Thus $M$ has equal power with respect to $(BD)$ and $(CE)$ and we finish by Steiner line.

Solved with dolphinday and Shreyasharma.Three heads are better than one lol. Beautiful problem! A pretty different solution. This is the promised homothety solution which is entirely synthetic (and elementary as well). We start off by proving the following. Claim : We have the equal line segments, $BI=EI$ and $CI=DI$. Proof : We note that, in triangles $\triangle DBI$ and $\triangle CBI$, $BC=BD$ and $BI$ is a common side. Also, since $BI$ is an angle bisector, $\measuredangle DBI = \measuredangle IBC$, from which it follows that $\triangle CBI \cong \triangle DBI$. Thus, $CI=DI$ as claimed. We can similarly show that $BI=EI$, proving the claim. Claim : Quadrilateral $AMED$ is cyclic. Proof : Note that due to the condition that $CE=BC=DB$, we have that in triangles $\triangle MEC$ and $\triangle MDB$, $CE=DB$ and $MC=MB$. Further, clearly, \[\measuredangle ECM = \measuredangle ACM = \measuredangle ABM = \measuredangle DBM\]from which it follows that $\triangle$ $MEC \cong \triangle MDB$. Thus, we clearly have that \[\measuredangle AEM = \measuredangle CEM = \measuredangle BDM = \measuredangle ADM\]from which the claim follows Claim : Lines $EH$ and $BI$ and lines $DH$ and $CI$ are parallel. Proof : Simply note that, \[2\measuredangle BEH = 2\measuredangle FEH = \measuredangle EMD = \measuredangle EAD = \measuredangle CAB = \measuredangle ACB + \measuredangle CBA = 2\measuredangle EBI\]Thus, $\measuredangle BEH = \measuredangle EBI$, from which it follows that $EH \parallel BI$. Similarly, we can show that $DH \parallel CI$, proving the claim. Now, we let $P$ and $Q$ be the reflections of $I$ across lines $DF$ and $EF$ respectively. We now locate $P$ and $Q$, with the following claims. Claim : $P$ and $Q$ lie on the lines $DH$ and $BI$ and $EH$ and $CI$ respectively. Proof : It is clear that $Q$ lies on $CI$ since $\triangle CEB$ is isosceles, and $CI$ is its angle/perpendicular bisector. We then wish to show that $Q$ lies on $\overline{EH}$. This is clear since, \[\measuredangle CEQ = \measuredangle IEC = \measuredangle ECI = \measuredangle CEH\]due to the previously proved claims. Thus, $Q$ lies on $\overline{EH}$. The other is entirely similar. Note that now it trivially follows that $HPIQ$ is a parallelogram. We can first prove the following important equality of angles. Note that, \[2 \measuredangle FBI = 2(\measuredangle EBC + \measuredangle CBI ) = \measuredangle ECB + \measuredangle CBA = \measuredangle ACB + \measuredangle CBA = \measuredangle CAB \]we can similarly show that \[2\measuredangle ICF = \measuredangle CAB\]Thus, $\measuredangle FBI = \measuredangle ICF$. This will be extremely useful in what follows. Claim : Points $B$ , $C$ , $P$ , $Q$ and $F$ are concyclic. Proof : Note that, \[\measuredangle QBF = \measuredangle FBI = \measuredangle ICF = \measuredangle QCF\]from which it is clear that $QBCF$ is cyclic. We can similarly show that $PBCF$ is also cyclic, which proves the claim. Now, we are almost there, note that if $L$ and $N$ are the midpoints of arcs $AC$ and $AB$ not including $B$ and $C$, respectively, it is clear that, \[\measuredangle BLN = \measuredangle BCN = \measuredangle BCQ = \measuredangle BPQ \]from which it is clear that $LN \parallel PQ$. Now, note that the homothety centered at $I$ mapping $\triangle IPQ$ to $\triangle ILN$, must map $H$ to $M$ (since $M$ is well known to be the point such that $LINM$ is a parallelogram). Thus, we must have that points $M$ , $H$ and $I$ are collinear as was required.

Since $BI$ is the angle bisector of $\angle DBC$, and $DB = DC$ we have that $BI$ is also an altitude of $\triangle DBC$, and similarly for $CI$, so $I$ is the orthocenter of $\triangle FBC$. Then note that the Steiner line of self-intersecting quadrilateral $BEDC$ is the radical axis of $(BD)$ and $(CE)$, which also passes through the orthocenter of $\triangle DEF$ and $I$. Let $P$ and $Q$ be the midpoints of $BD$ and $CE$. Since $(BD)$ is congruent to $(CE)$, it suffices to show that $M$ lies on the radical axis of $(BD)$ and $(CE)$ or on the perpendicular bisector of $PQ$. Since $M \in (ADE)$ and $(ABC)$, $M$ is the Miquel point of $BDEC$. Then $M$ is the center of the spiral similarity sending $DE$ to $BC$ and by the Gliding Lemma, it sends $PQ$ to $BC$. Then $\triangle MPQ \sim \triangle MBC \implies MP = MQ$ as desired.

Claim: $I$ is the orthocenter of $\triangle BFC$. We have $\angle EBC=90-\angle C/2$ but $\angle BCI=\angle C/2$, so $CI\perp BF$, and similarly $BI\perp CF$. By Gauss-Bodenmiller, this then means that $IH$ is the radical axis of $(BD)$ and $(CE)$. However, since $BD=CE$, there is a rotation at $M$ taking $DB$ to $EC$. This also means that $M$ is equidistant to the midpoints of $DB$ and $EC$. Since $(DB)$ and $(EC)$ also have the same radius, $M$ has equal power with respect to both of them, done.

Nice problem! Hopefully this solution is new and I’ll add some diagrams tomorrow. Define $P = MD \cap CI$ and $Q = ME \cap BI$. Note that $M$ is the center of similarity sending $BD$ to $CE$ which means that $MD = ME$. Now we have the main claim. Claim 1: We have $MP = MQ$ Proof: By angle chasing, we have \[\angle DEM = 90 - \frac{\angle A}{2} = \angle DIQ\]hence $D, E, Q, I$ are concyclic. Similarly, $D, E, P, I$ are cyclic as well so $DEQIP$ is cyclic. But since $MD = ME$ we must have $MP = MQ$ as desired. Now triangles $PIQ$ and $DHE$ are homothetic, so $IH$ passes through $M$ as desired.

Inverted problem wrote: In $\triangle ABC$, let points $E$ and $D$ lie on rays $BA$ and $CA$, respectively, such that $DB = BC = CE$. Let $\omega_B$ be the circle passing through $I$ and $E$ tangent to line $BI$, and let $\omega_C$ be the circle passing through $I$ and $D$ tangent to line $CI$. Prove that $\omega_B$ and $\omega_C$ intersect again on line $IK$, where $K$ is the midpoint of $\overline{BC}$. (Circles $\omega_1$ and $\omega_2$ are the images of the altitudes from $D$ and $E$ in $\triangle DEF$.) Let $X$ and $Y$ be the antipodes of $I$ WRT $\omega_B$ and $\omega_C$ and let $P = IX \cap AB$ and $Q = IY \cap AC$. Let $A'$ be the antipode of $A$ in $(ABC)$. We have $IEPX \sim IDQY$ from an angle chase left as an excercise to the reader, so it suffices to show that $\overline{A'I} \perp \overline{PQ} \parallel \overline{XY}$. By definition, $APIQ$ is a parallelogram; from $\triangle BIP \sim \triangle CIQ$, we have that $APIQ$ is in fact similar to $IBA'C$ by a $90^{\circ}$ rotation. It thus follows that $\overline{A'I} \perp \overline{PQ}$, as desired.