"Common twist" = radical axis in this case I think.

I'm sorry I don't know the word. This is in this case for exemple the line of power of the two circles... Do you understand ? "corde commune" in French... "coarda comuna" in Rumania...

See PP12 from here

Could you explain why $AA_3$ is an angle bisector of $OAH$ and why $A', M,H$ are collinear?

Megus wrote: Could you explain why $AA_3$ is an angle bisector of $OAH$ and why $A',M,H$ are collinear? Lemma. Let $w=C(O)$ be a circumcircle of the acute triangle $ABC$, where $AB<AC$. Denote the point $A'\in w\cap (AO$. Define the points $X\in AB$, $Y\in AC$ so that $B\in (AX)\ ,\ AX=AC$ and $Y\in (AC)\ ,\ AY=AB$. Then $m(\widehat {A'BY})=m(\widehat {A'CX})=\frac A2\ .$ Standard notation. $\boxed {\ (XY\ }$ - the ray without the point $X$.

Thank you, but I don't see how to apply this lemma

Another application of spiral similarity.

You are a clever young because you are right, Megus ! See more above the my first reply where is the complete solution of this problem.

BogG wrote: Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$. First, I repost a solution given by Andre.L which was deleted for an unclear reason (to anyone who feels spoken to: according to Valentin, these things are not supposed to be deleted!). andre.l wrote: The common chord of the cirmcumcircles of $ABC$ and $ADE$ is their radical axis. Thus, it's perpendicular to the line that passes through their centers. Let $O$ and $O_1$ be the circumcenters of $ABC$ and $ADE$, respectively. Then, $HM$ is perpendicular to the common chord of the circumcircles of $ABC$ and $ADE$ iff $HM$ is parallel to $OO_1$. Let $Q=OO_1\cap AH$. Since $OM\bot BC$ and $AH\bot BC$, $OM\parallel AH$. Thus, $OO_1\parallel HM \Leftrightarrow OQ \parallel HM \Leftrightarrow OMHQ$ is a parallelogram. Let $S$ be the midpoint of $OH$. Then, $OMHQ$ is a parallelogram $\Leftrightarrow MS=SQ$. But the circle with center $S$ which passes through $M$ is the nine-point circle of $ABC$. Thus, it intersects $AH$ at it's midpoint. What we will prove is that $Q$ is the midpoint of $AH$. The point $O_1$ is on the perpendicular bisector of $DE$, which is also the angle bisector of $BAC$. Let $T$ be the intersection of $OH$ and the angle bisector of $BAC$. Now, apply Menelaus' theorem to the line $OO_1Q$ and the triangle $HTA$. We get that $\frac{OH}{OT} \frac{O_1T}{O_1A} \frac{QA}{QH}=1$. Thus, $QA=AH \Leftrightarrow \frac{OH}{OT} \frac{O_1T}{O_1A} =1$. With some calculation, we can prove that the above relation is equivalent to $\cos A=2\cdot\cos^2\left(\frac{A}{2}\right)-1$, which is true. Let me add my solution. Let B' and C' be the feet of the altitudes of triangle ABC from the vertices B and C, respectively. Keep in mind that these altitudes intersect at the orthocenter H of triangle ABC. Since < BB'C = 90° and < BC'C = 90°, the points B' and C' lie on the circle with diameter BC; the center of this circle is the midpoint M of the segment BC. So the circle with diameter BC is a circle with center M which passes through the points B and C and intersects the sides CA and AB again at the points B' and C', respectively. Hence, according to IMO 1985 problem 5, if R is the point of intersection of the circumcircles of triangles ABC and B'AC' (apart from A), then < MRA = 90°. Thus, the point M lies on the perpendicular to the line AR at the point R. On the other hand, since < AB'H = 90° and < AC'H = 90°, the points B' and C' lie on the circle with diameter AH. Consequently, the circumcircle of triangle B'AC' is the circle with diameter AH; therefore, since the point R lies on this circumcircle, it satisfies < ARH = 90°. Consequently, the point H lies on the perpendicular to the line AR at the point R. Now as we know that both points M and H lie on the perpendicular to the line AR at the point R, we obtain $HM\perp AR$. Now we will prove that the point R lies on the circumcircle of triangle ADE. Once this is shown, it will follow that the points A and R are the two common points of the circumcircles of triangles ABC and ADE; this will mean that the line AR is the common chord of these circumcircles, so that the relation $HM\perp AR$ will tell us that the line HM is perpendicular to the common chord of the circumcircles of triangles ABC and ADE. This will solve the problem. So it only remains to show that the point R lies on the circumcircle of triangle ADE. Since AE = AD, the triangle DAE is isosceles, so that its base angle is $\measuredangle ADE=\frac{180^{\circ}-\measuredangle DAE}{2}=90^{\circ}-\frac{\measuredangle DAE}{2}=90^{\circ}-\frac{A}{2}$. In other words, $\measuredangle C^{\prime}DH=90^{\circ}-\frac{A}{2}$. Similarly, $\measuredangle B^{\prime}EH=90^{\circ}-\frac{A}{2}$. Thus, < C'DH = < B'EH. Together with < DC'H = < EB'H (both of these angles are 90°), this yields that the triangles C'DH and B'EH are similar; hence, $\frac{DC^{\prime}}{C^{\prime}H}=\frac{EB^{\prime}}{B^{\prime}H}$. On the other hand, since < BC'H = < CB'H (both of these angles are 90°) and < BHC' = < CHB', the triangles BC'H and CB'H are similar, so that $\frac{C^{\prime}H}{BC^{\prime}}=\frac{B^{\prime}H}{CB^{\prime}}$. Thus, $\frac{DC^{\prime}}{BC^{\prime}}=\frac{DC^{\prime}}{C^{\prime}H}\cdot\frac{C^{\prime}H}{BC^{\prime}}=\frac{EB^{\prime}}{B^{\prime}H}\cdot\frac{B^{\prime}H}{CB^{\prime}}=\frac{EB^{\prime}}{CB^{\prime}}$. Now, since the point R lies on the circumcircle of triangle B'AC', we have < RC'A = < RB'A, so that < RC'B = 180° - < RC'A = 180° - < RB'A = < RB'C. On the other hand, since the point R lies on the circumcircle of triangle ABC, we have < RBA = < RCA, what becomes < RBC' = < RCB'. From < RC'B = < RB'C and < RBC' = < RCB', it follows that the triangles RBC' and RCB' are similar. The points D and E are corresponding points in these two similar triangles (since they lie on the respective sides BC' and CB' and divide them in the same ratio $\frac{DC^{\prime}}{BC^{\prime}}=\frac{EB^{\prime}}{CB^{\prime}}$); since corresponding points in similar triangles form equal angles, we thus have < RDC' = < REB', or, equivalently, < RDA = < REA. This yields that the point R lies on the circumcircle of triangle ADE, and the problem is solved. Darij

Virgil Nicula wrote:

Can someone explain how to get this ones

abdurashidjon wrote: Virgil Nicula wrote:

Can someone explain how to get this ones Ok .We have $AH=2OM=\sqrt{4R^2-BC^2}=\sqrt{4r^2-4R^2(sinA)^2}=2RcosA$ By the low of sines $\frac{AP}{sin(B+\frac{A}{2})}=2R$ so $AP=2Rsin(90+\frac{B-C}{2})$ so $AP=2R\cos \frac{B-C}{2}$ And $MP=OP-OM=R(1-cosA)$ BTW , mr Nicula there exist a not metrical solution to this lemma ?

BogG wrote: Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$. All the problems from the Swiss Imo Selection Team 2006 This is problem is difficult, not because it's lacking swift solutions but because it's so easy to go pursuing some typical ideas who do provide you with ways to solve it but I think, more complicated ones. Here is my proof 1) Name $O$ and $O'$ the circumcenter of $ABC$ and of the other triangle. Let $F$ be the point where $AO$ intersects again the circumcircle of $ABC$. First thing to notice is that if $MH$ is to be perpendicular to $AK$ then it will intersect it precisely at $K$. This is because $MH$ passes through $F$ and hence, if it's going to cut $AK$ perpendicularly it must do it in a point who belongs to the big circle, ergo $K$. 1a) Notice that line $OO'$ cuts $AK$ perpendicularly at the midpoint. Hence it is parallel to line $FK$. It suffices to prove therefore that $H$ is in line $FK$ because then $M$ is in line $FH$ and they are all together beautifully. 2) With all this in mind the problem is actually half done. Triangles $BDH$ and $CEH$ are similar for $\angle HCE = \angle HBD$ and $\angle HDB = \angle HEC$. Hence $\frac{BD}{CE} = \frac{DH}{HE}$ 3) $\angle DKE = \angle BAC = \angle BKC \Rightarrow \angle BKD = \angle CKE$ 4) $\angle KBD = \angle KCE$ hence, using 3, triangles $KBD$ and $KEC$ are similar hence $\frac{KD}{KE} = \frac{BD}{CE}$ 5) With 3 and 4 $KH$ actually bisects $\angle DKE$ and we're set because if $G$ is the point where it touches again the small circle the $G$ and $A$ are diametrically opposite and hence $\angle HKA$ is right which means $H$ is in line $FK$. Daniel

Here is a proof with the use of trignometry. It is quite long so I think that I had better hide it.

.

Let $ O$ be the circumcenter, $ R$ be the radius and $ k$ be the circumcircle of $ \triangle ABC$ and let $ r$ be the radius and $ k'$ be the circumcircle of $ \triangle ADE$. Denote by $ S$ the intersection of $ AO$ with $ k$, by $ N$ the intersection of the line $ OM$ with $ k$ and let $ T$ be the intersection of the line $ AN$ with $ k'$. Moreover, let $ X$ be the intersection of $ DE$ with $ AT$. We'll prove that $ M$, $ T$, $ H$ are collinear and that $ AO$ and $ MH$ intersect in $ k$. Then it will follow that the point of intersection of the line $ MH$ with $ k$ and with $ k'$ will be a common point, let's say $ P$ and that the angle $ \angle HPA = 90$, as desired. It's well-known that $ AX$ and $ OM$ intersect in $ N$. Moreover, $ \angle HAX = \angle XAO$. First, we'll prove that $ S \in MH$. Let $ H_1$ and $ H_2$ be the reflections of $ H$ on $ AB$ and $ AC$, respectively. We have $ HC \Vert BS$ and thus $ BH = BH_1 = SL$ and $ HB \Vert CS$ and thus $ CH = CH_2 = BS$. Therefore, $ HBSC$ is a parallelogramm and it follows that $ HS$ intersects $ BC$ at its middle, i.e. $ M$, as desired. It's left to prove that $ T \in MH$, i.e. $ T \in SM$. Applying Menelaos' theorem to the triangle $ \triangle NAO$, it's remains to prove that $ AN/NO \cdot OM/MN \cdot NT/TA = - 1$ or equivalently $ MN/2OM = TN/TA = AN/AT - 1$. It's well-known that $ 2OM = AH$, thus $ MN/2OM = R - OM/2OM = R/2OM - 1/2 = R/AH - 1/2$. Handling $ AN/AT$ is the more difficult part. We have $ AX = AH \cdot \cos\angle HAX$ and $ AX = AE \cdot \cos \alpha/2$. Thus, $ AE = AH \cdot \cos \angle HAR / \cos \alpha/2$. From the equality of the area of triangle $ \triangle ADE$ it follows that $ DE = AE^2 \cdot \sin \alpha / AR \Leftrightarrow DE = AH \cdot \sin \alpha \cos \angle HAX / \cos^2 \alpha/2$. But $ AT = DE/\sin \alpha$ and therefore $ AT = AH \cdot \cos \angle HAX / \cos^2 \alpha/2$. Moreover, $ AN = 2R \cdot \cos \angle XAO$. Thus, $ AN/AT = 2R \cdot \cos^2 \alpha/2 /AH$. It's left to prove that $ 1/2 = R/AH \cdot (2\cos^2 \alpha/2 - 1)$. We have $ AH/2R = OM/R = \cos \alpha$. Therefore it's left to prove that $ \cos \alpha + 1 = \cos^2 \alpha/2$, which follows from the addition theorems.

in dhernandez solution i cannot understand why M is in line FK,and the last part of the solution with point G.how he get it?

dhernandez wrote: BogG wrote: Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$. All the problems from the Swiss Imo Selection Team 2006 This is problem is difficult, not because it's lacking swift solutions but because it's so easy to go pursuing some typical ideas who do provide you with ways to solve it but I think, more complicated ones. Here is my proof 1) Name $O$ and $O'$ the circumcenter of $ABC$ and of the other triangle. Let $F$ be the point where $AO$ intersects again the circumcircle of $ABC$. First thing to notice is that if $MH$ is to be perpendicular to $AK$ then it will intersect it precisely at $K$. This is because $MH$ passes through $F$ and hence, if it's going to cut $AK$ perpendicularly it must do it in a point who belongs to the big circle, ergo $K$. 1a) Notice that line $OO'$ cuts $AK$ perpendicularly at the midpoint. Hence it is parallel to line $FK$. It suffices to prove therefore that $H$ is in line $FK$ because then $M$ is in line $FH$ and they are all together beautifully. 2) With all this in mind the problem is actually half done. Triangles $BDH$ and $CEH$ are similar for $\angle HCE = \angle HBD$ and $\angle HDB = \angle HEC$. Hence $\frac{BD}{CE} = \frac{DH}{HE}$ 3) $\angle DKE = \angle BAC = \angle BKC \Rightarrow \angle BKD = \angle CKE$ 4) $\angle KBD = \angle KCE$ hence, using 3, triangles $KBD$ and $KEC$ are similar hence $\frac{KD}{KE} = \frac{BD}{CE}$ 5) With 3 and 4 $KH$ actually bisects $\angle DKE$ and we're set because if $G$ is the point where it touches again the small circle the $G$ and $A$ are diametrically opposite and hence $\angle HKA$ is right which means $H$ is in line $FK$. Daniel It suffices to prove therefore that H is in line FK because then M is in line FH and they are all together beautifully. Can somebody please explain why when $H$ is in $FK$ , $M$ should be in $FH$???Please!

In this solution, $(XYZ)$ will denote the circumcircle of $\triangle XYZ$. Also, assume without loss of generality that $A$ is not closer to $C$ than it is to $B$. Let $A_1$ be the foot of the angle bisector from $A$ to $BC$, and let $N$ be the intersection of $AA_1$ and $(ABC)$. Let $O$ be the center of $(ABC)$, let $A'$ be the intersection of $AO$ and $(ABC)$, and let $A_2$ be the intersection of line $AO$ and segment $BC$. It is well-known that $H$, $A'$, and $M$ are collinear. Let $A'M$ intersect $(ABC)$ at $P$. Let $\omega$ be the circle through $AP$ centered on line $AA_1$, and let $D'$ and $E'$ be the points where $\omega$ intersects $AB$ and $AC$, respectively. Let $F$ be the midpoint of $D'E'$ (which lies on $AA_1$), let $X$ be the point distinct from $A$ where $AF$ intersects $(AD'E')$, let $H'$ be the point of intersection of lines $PX$ and $D'F$, let $A_3$ be the point where $AH'$ intersects $BC$, and let $(AD'E')$ intersect line $AO$ at point $G$. We first note that $\angle A'PA = \angle XPA = 90^{\circ}$, so $A'$ and $X$ both lie on the line through $P$ perpendicular to $AP$, that is, $A'$, $P$, and $X$ are collinear. By a well-known lemma, there exists a spiral similarity centered at $P$ that maps $(ABC)$ to $(AD'E')$. $\widehat{A'C} = 2 \angle A'AC = 2\angle GAE' = \widehat{GE'}$, so the spiral similarity maps $A'$ to $G$. In addition, it clearly sends $M$ to $F$. It follows that $AO$ and $PF$ concur on $(AD'E')$, that is, $P, F, G$ and are collinear. Therefore, \begin{align*} (B,C; A_1, A_3) &= A(B,C; A_1 A_3) \\ &= A(D',E'; F,H') \\ &= P(D',E'; F,H') \\ &= P(D',E'; G,X) \\ &= A(D',E'; G,X) \\ &= A(B, C; A_2, A_1). \end{align*} Hence, $\frac{A_1 B \cdot A_3 C}{A_1 C \cdot A_3 B}= \frac{A_2 B \cdot A_1 C}{A_2 C \cdot A_1 B}$. By the angle bisector theorem, $\frac{A_1 B}{A_1 C} = \frac{AB}{AC}$, so $\left(\frac{AB}{AC} \right)^2 = \frac{A_3 B}{A_3 C} \cdot \frac{A_2 B}{A_2 C}$. It follows that lines $AA_2$ and $AA_3$ are isogonal. But line $AA_2$ is the line through the circumcenter; hence, $AA_3$ is an altitude. Since $P$, $H'$, and $X$ are collinear, $H'$ is the intersection of the altitude from $A$ to $BC$ and line $PM$. But $H$ is also the intersection of the altitude from $A$ to $BC$ and line $PM$, so $H = H'$. It follows that $D = D'$ and $E = E'$. Since $M$, $H$, and $P$ are collinear, the angle formed by the lines $MH$ and $AP$, the common chord of $(ABC)$ and $(ADE)$ is angle $XPA$, which is 90 degrees. [As a note, $O$ does not necessarily lie on line $DE$; the diagram is misleading in this case.]

Here's another solution by The QuattoMaster 6000. Let $X$ and $Y$ be the feet of the altitudes from $C$ to $AB$ and $B$ to $AC$, respectively. $\angle ADH = 90^{\circ} - \frac{A}{2}$, so $\angle XHD = \frac{A}{2}$. $\angle XHB = \angle YHC = 90^{\circ} - (\angle YCH) = \angle A$, so $\angle DHB = \angle XHD = \frac{A}{2}$. Hence, by the angle bisector theorem, $\frac{XD}{DB} = \frac{XH}{HB}$. In a similar way, we may deduce that $\frac{YE}{EC} = \frac{YH}{HC}$. But $XYCB$ is cyclic, so $HX \cdot HC = HY \cdot HB$, so $\frac{XD}{DB} = \frac{XH}{HB} = \frac{YH}{HC} = \frac{YE}{EC}$. Let $P$ be the Miquel point of complete quadrilateral $XYCB$. Then $PAYX$ and $PACB$ are cyclic. Since $\frac{XD}{DB} = \frac{YE}{EC}$, the spiral similarity centered at $P$ that maps $XB$ to $YC$ also maps $DB$ to $EC$, so $P$ is also the Miquel point of complete quadrilateral $DECB$. Hence, $PAED$ is cyclic. Let $O$ be the circumcenter of $\triangle ABC$, and let $A'$ be the point diametrically opposite to $A$ on the circumcircle of $\triangle ABC$. It is well-known that $A'$, $P$, and $H$ are collinear. $\angle HPA = \angle HXA = 90^{\circ}$, since $APXH$ is cyclic, and $\angle APA' = 90^{\circ}$, since $PACB$ is cyclic. The result follows trivially.

Dear Mathlinkers, several and interesting proof has been presented on this file all with calculation or with a function. An enterely pure and synthetic using solely the shape can be found. I think so. I am on this way and not so far from the result. Why not a challenge? Sincerely Jean-Louis

Let the reflection of $H$ about $M$ be $H'$. It is well known by orthocenter reflections that $H'$ lies on $(ABC)$ and $AH'$ is the diameter of $(ABC)$. Let line $HM$ intersect $(ABC)$ at $F \neq H'$. Since $\angle{AFH} = 90$, it follows that $F$ also lies on the circle with diameter $AH$. Let the circle with diameter $AH$ be $\omega_1$ and $(ABC)$ be $\omega_2$. It suffices to prove that $AFDE$ is cyclic and coaxial to $\omega_1$ and $\omega_2$ with common chord $AF$. By the Coaxility lemma, it suffices to prove $$\frac{\text{Pow}(D,\omega_1)}{\text{Pow}(D,\omega_2)} = \frac{\text{Pow}(E,\omega_1)}{\text{Pow}(E,\omega_2)}$$ Let $\omega_1$ intersect $DE$ at $X$. It follows that since $AX \perp DE$ and $AD = AE$, we know $X$ is the midpoint of $DE$. So, our power equation above simplifies to $$\frac{DH \cdot DX}{AD \cdot BD} = \frac{EX \cdot EH}{AE \cdot CE}$$which means that it suffices to prove $\frac{DH}{BD} = \frac{EH}{CE}$. This can be done with Law of Sines. Using the fact that $AD = AE$ and that $H$ is the orthocenter, we know that $\angle{DHB} = 180-\angle{DBH}-\angle{BDH} = 180-(90-A)-\left(90+\frac{A}{2}\right) = \frac{A}{2}$ and $\angle{EHC} = 180-\angle{ECH}-\angle{HEC} = 180-(90-A)-\left(90+\frac{A}{2}\right) = \frac{A}{2}$. Since $$\frac{DH}{AD} = \frac{\sin{DBH}}{\sin{DHB}} = \frac{\sin{(90-A)}}{\sin{\frac{A}{2}}} = \frac{\sin{ECH}}{\sin{EHC}} = \frac{EH}{CE}$$we're done.

Solved with hints Let the circumcircle of $\triangle ADE$ intersect $(ABC)$ at $Q$. Thus, $Q$ is the Miquel point of $DECB$. Let $B'$ and $C'$ denote the feet of the altitudes from $B$ and $C$. Consider the spiral similarity centered at $X$ that sends $DB$ to $E$. We claim that this also sends $C'$ to $B'$. This is just showing that $$\frac{BD}{C'D}=\frac{CE}{B'E}.$$Let $\angle BAC=2\alpha$. Then, $$\angle ADE=\angle AED=90-\alpha,$$so $$\angle C'HD=\angle B'HE=\alpha.$$Hence, we also have $$\angle BHD=\angle CHE=\alpha.$$Thus, by Angle Bisector Theorem, $$\frac{BD}{C'D}=\frac{BH}{HC'}=\frac{CH}{HB'}=\frac{CE}{B'E},$$where the second equality is from PoP on $BCB'C'$. Thus, this spiral similarity also sends $C'$ to $B'$. Hence, $Q$ lies on $(AC'B')$, so $\angle AQH=90$. This means that $Q,H,M$ are collinear (since they all lie on the line from $Q$ to $H'$, where $H'$ is the reflection of $H$ across the midpoint of $BC,$ which we call $M$). Therefore, $(ABC),(AH),(ADE)$ are coaxial as they all have $A$ and $Q$. Hence, their centers are collinear, which means that $P$, the circumcenter of $\triangle AEF$, lies on $ON$, where $N$ is the midpoint of $AH$. Since $NO\parallel HM\perp AQ,$ we are then done.

I think slightly misplaced for a G5. From a good diagram that helps us gain intuition, by letting F be the intersection of ray MH with circle ABC, we know that the intersection of ray HM with circle ABC is the diametrically opposite point of A, whence AFM is 90 degrees, and it also lies on the circle AIHJ, where I and J are the feet of altitude from B and C, respectively. Then, we can rephrase the problem as follows: Quote: Prove that the circle ADE is coaxial with the other two circles. This is equivalent to proving that $$\frac{DB}{DJ}=\frac{DA*DB}{DJ*DA}=\frac{EA*EC}{EI*EA}=\frac{EC}{EI}.$$First we chase some angles: $$JHD=90-(90-A/2)=\frac{A}{2}=\frac{JHB}{2},$$so HD is an angle bisector. Now the problem is evident, because $$\frac{DB}{DJ}=\frac{HB}{HJ}=\frac{HC}{HI}=\frac{EC}{EI}$$from similarity of triangles BHJ and CHI by AA (JBH=ICH by cyclic quad BCIJ), as desired. $\blacksquare$

In fact, I claim that $X$ is the $A$-queue point, which finishes by master Miquel. To do this, let the reflection of $H$ over $BC$ be $H'$ - by Brazil 2011/5 it suffices to show that $XBH'C$ is harmonic. However, noting that $\triangle HBD \sim \triangle HCE$, we have \[\frac{XB}{XC} = \frac{BD}{CE} = \frac{HB}{HC} = \frac{H'B}{H'C}\]and we are done. (a lot of the difficulty of this problem comes from the characterization of $X$ - once you successfully figure this out, the problem isnt very hard. although tbh, that might be a bit surprising to be true, which is why I took 30 min to think of that)

Let $\overline{BB_1}$, $\overline{CC_1}$ be altitudes of $\triangle ABC$ and $Q = (ABC) \cap (AB_1HC_1)$ be the $A$-Queue point of $\triangle ABC$. It's well-known that $Q$, $H$, and $M$ are collinear, and $\overline{AQ} \perp \overline{QHM}$, so it suffices to show that $AQDE$ is cyclic. By the (forgotten) Coaxiality lemma, it's equivalent to show that $$\frac{\text{Pow}_{(ABC)}(D)}{\text{Pow}_{(AB_1HC_1)}(D)} = \frac{\text{Pow}_{(ABC)}(E)}{\text{Pow}_{(AB_1HC_1)}(E)} \iff \frac{EB_1 \cdot EA}{EA \cdot EC} = \frac{DC_1 \cdot DA}{DA \cdot DB} \iff \frac{EB_1}{DC_1} = \frac{EC}{DB}.$$However, because $\measuredangle HDB = \measuredangle HDC_1 = - \measuredangle HEB_1 = -\measuredangle HEC,$ it's easy to see that $\triangle HDC_1 \overset{-}{\sim} \triangle HEB_1$ and $\triangle HDB \overset{-}{\sim} \triangle HEC,$ so $\tfrac{EB_1}{DC_1} = \tfrac{HE}{HD} = \tfrac{EC}{DB}$ and we are done.

Let the feet of the altitudes from $B$ and $C$ be $X$ and $Y$, respectively. It's easy to show that $\triangle HYB \sim \triangle HXC$ with corresponding angle bisectors $HD$ and $HE$, which tells us $YD: DB = XE: EC$. Thus there exists a spiral similarity sending $YDB$ to corresponding points $XEC$, which tells us circles $(AXY)$, $(ADE)$, and $(ABC)$ are coaxial, implying they share a common chord. Finally we note the common chord of $(AXY)$ and $(ABC)$ is well known to be $AQ$, where $Q$ is the $A$-Queue point, which is perpendicular to $HM$. $\blacksquare$ [asy][asy] size(200); pair A, B, C, H, D, E, X, Y; A = dir(110); B = dir(200); C = dir(340); H = A+B+C; X = foot(B, A, C); Y = foot(C, A, B); D = extension(A, B, H, incenter(H,B,Y)); E = extension(A, C, H, incenter(H,C,X)); draw(A--B--C--cycle^^D--E^^B--X^^C--Y); draw(circumcircle(A, D, E)^^circumcircle(A, B, C)); draw(circumcircle(A, X, Y), dashed); dot("$A$", A, dir(90)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$D$", D, dir(200)); dot("$E$", E, dir(30)); dot("$H$", H, dir(270)); dot("$X$", X, dir(0)); dot("$Y$", Y, dir(210)); [/asy][/asy]

Let $Q$ be the $A$-Queue point, $X$ be the midpoint of $DE$ so $AX \perp DE$, $A_1$ be the foot of the altitude from $A$, and $T$ be the point on $BC$ such that $(A_1,T;B,C)=-1$. It is well-known that $A,Q,T$ are collinear. Let $G$ be the intersection between $AT$ and $DE$, so $-1=(A_1,T;B,C) \stackrel{A}{=} (H,G;D,E)$. Now $GD \cdot GE = GH \cdot GX = GA \cdot GQ$, so $A,Q,D,E$ are concyclic. Since $AQ \perp HM$, we are done.

Define $X$, $Y$ as the feet of the altitudes from $B$, $C$. Firstly, we angle chase to find that $\triangle BHY \sim \triangle CHX$ and $HD$, $HE$ bisect $\angle BHY$, $\angle CHX$, so we have the ratio \[\frac{DY}{DB} = \frac{EX}{EC} \implies \frac{\operatorname{pow}(D,(AH))}{\operatorname{pow}(D,(ABC))} = \frac{\operatorname{pow}(E,(AH))}{\operatorname{pow}(E,(ABC))}.\] Thus $(AH)$, $(ADE)$, $(ABC)$ are coaxial through Coaxiality Lemma, with the radical axis being $AQ$, where $Q$ is the $A$-queue point, which is perpendicular to $HM$. $\blacksquare$

It suffices to prove that $(AH)\cap (ABC)$ lies on $(ADE)$ and let that intersection be $K$, this is because we have that $MH$ passes through $(AH)\cap(ABC)$ and is perpendicular to their radical axis. Let $X$ and $Y$ be the feet of the altitudes from $B$ and $C$. Clearly we have $(AHXY)$. Note that there is a spiral similarity sending $CX$ to $BY$ centred at $K$. Proving $DE$ is the angle bisector of $\angle BHY$ suffices as this implies $\frac{EX}{XC}=\frac{EH}{HC}=\frac{DH}{DB}=\frac{DY}{YB}$ which implies the spiral similarity centred at $K$ also sends $EX$ to $DY$ which suffices. $DE$ is the angle bisector as we have $\angle XHE=90-\angle AEH = \frac{\angle BAC}{2}$, and we have that $\angle XHC=90-\angle HCX=\angle ABC$ so this suffices.

The problem condition is equivalent to $(ADE)$ Passing through the $A$ queue point of $\Delta ABC$ which we call $Q$. Since $Q$ is the center of spiral similarity sending $XY$ to $BC$, where $X$ and $Y$ are feet of $B$ and $C$ altitudes respectively, it follows that must also send $DE$ to $BC$ or $BD$ to $CE$ for the problem condition to be true, but that follows from some ratio chasing so we are done.

Let $E$ and $F$ be the feet of altitudes from $B$ and $C$ respectively $\triangle BHF \sim \triangle CHE$ $\angle FMH = \angle ENH \implies \frac{BM}{MF} = \frac{CN}{NE}$ Let $Q$ be the A-queue point Consider the spiral similarity centered at $Q$ sending $\overline{BF}$ to $\overline{CE}$ Clearly, $M \longrightarrow N$ Thus, $Q$ is also center of spiral similarity sending $\overline{BM}$ to $\overline{CN}$ $\implies Q \in (AMN)$ It is now well known that $Q-H-D$ and $HD \perp AQ$ $\blacksquare$

Let $X$ and $Y$ be the feet of the altitudes from $B$ and $C$, respectively. Note that \[\angle YHD=90^{\circ}-\angle ADH=\frac12\angle A.\]Similarly, $\angle XHE=\angle YHD$. Also, since $\angle YHB=\angle XHC$, we also get $\angle DHB=\angle EHC$. Since $\triangle YBH\sim\triangle XCH$, we get that $\frac{YD}{DB}=\frac{XE}{EC}$. Let $P$ be the A-queue point; then it is on $(AXY)$ and $(ABC)$. We want to show that it is also on $(ADE)$, but since $D$ maps to $E$ in the same spiral similarity as $BY\to CX$ (by our ratio equivalence earlier), we get \[\angle DPE=\angle BPC=\angle BAC=\angle DAE,\]so indeed $P$ lies on $(ADE)$. Finally, since $AP\perp HM$ (since $\angle APH=90^{\circ}$ and the $A$-antipode lies on $HM$), we are done. $\blacksquare$

Denote by $Q$, $A-$queue point which lies on $MH$. If $H_b,H_c$ are the altitudes from $B,C$ then $\measuredangle BHE=\frac{\measuredangle A}{2}=\measuredangle EHH_C$ and similarily $HF$ bisects $\measuredangle H_BHC$. \[\frac{Pow(E,(AHQ))}{Pow(E,(ABCQ))}=\frac{EH_C.EA}{EA.EB}=\frac{HH_C}{HB}=\frac{HH_B}{HC}=\frac{Pow(F,(AHQ))}{Pow(E,(ABCQ))}\]Now, coaxiality lemma finishes it as desired.$\blacksquare$

See that due to the fact that the $A$-queue point lies on the line $\overline{HMA'}$ where $A'$ is the $A$-antipode, it suffices to prove that $(ABC)$, $(ADE)$, $(AH)$ are coaxial. Let $Y$ and $Z$ be foot of altitudes of $B$ and $C$. So due to the orz coaxiality lemma we just need to prove that \[\frac{\text{Pow}(D,(AH))}{\text{Pow}(D,(ABC))}=\frac{\text{Pow}(E,(AH))}{\text{Pow}(E,(ABC))} \iff \frac{DZ}{DB}=\frac{EY}{EC}\]But see that this is trivial because $\triangle HZB \cup D \overset{-}{\sim} \triangle HYC \cup E$ and done.