Hint: take the Carnot circle $\odot BCH$ ($H$ - the orthocenter) and see that it passes through the reflection of $A$ in $M$, hence $X$ belongs to this circle as well. Subsequently, by symmetry about $BC$, we get $HZYX$ is cyclic, hence the two lines are parallel. Best regards, sunken rock

HXBC is cyclic, this is a very traditional problem, if not noticed that , maybe the Menelause theorem works, but the calculation must be tough

Because $BDHF,MDHX$ are both cyclic, so $AF \cdot AB = AH \cdot AD =AX \cdot AM$, and $BMXF$ is cyclic, hence $\angle ZXY = \angle BXM = \angle BFM = \angle FBM = \angle DHC = \angle ZHY$, so $XYZH$ is cyclic, and we have $\angle XYZ = \angle AHX = \angle XMD$, so $YZ\parallel BC$.

Attachments:

Let $P = AM \cap BE$, and note that \[{YZ \parallel BC \iff Y(Z,M;B,C) = -1 \iff B(Z, A; YB\cap AD, H) = - 1 \iff (X, A; Y, P) = -1,}\] which is true iff $P$ on the polar $\rho$ of $Y$ w.r.t. $(AEF).$ But the polar of $M$ is $EF$, hence $AA \cap EF \in \rho.$ But also $AH \cap XF \in \rho$, hence by Pascal's theorem on $AAHEFX$ we get $P \in \rho$, as desired.

Dear Mathlinkers, 1. T the second point of intersection of BX with the circle with diameter AH 2. the tangent to this circle at F goes through M 3. by using two time the Pascal theorem in two degenerated case, we are done. Sincerely Jean-Louis

sunken rock wrote: Hint: take the Carnot circle $\odot BCH$ ($H$ - the orthocenter) and see that it passes through the reflection of $A$ in $M$, hence $X$ belongs to this circle as well. Subsequently, by symmetry about $BC$, we get $HZYX$ is cyclic, hence the two lines are parallel. Best regards, sunken rock Is there any motivation to take the symmetric of $A$ in $M$?

Here's my solution: Let $H$ be the orthocenter of $\triangle ABC$, and let $BX \cap \odot (AEF)=K$. Then, by Pascal on $AHFKXA$, we get that the given statement is equivalent to proving that $FK$ is parallel to $BC$. But this directly follows from $$\angle AFK=\angle MXB=\angle ABC \Rightarrow FK \parallel BC$$where we use the fact that $\odot (AXB)$ is tangent to $BC$ (as $X$ is the $A$-Humpty point). Hence, done. $\blacksquare$

$\text {Lemma}$:$FM$ and $EM$ is tangent to the $\odot (AEF)$. $\text {proove}$; Just from the nine point circle theorem, the midpoint of $AH$ and $M$ are diametrically opposite points and hence prof the lemma. Then, we can easily observe that $BFMD$ and $HXDM$ are conclic and so $\angle AFX=\angle 180-BFX =\angle XMD$ and hence, $BXFM$ is conclic. But $MFC=90-\angle B$ and $BFC=90$ and so $BFM=\angle B$ and $BXM=\angle B$ and $DHX=180-(\angle A+\angle C$ =$\angle B$ . Hence, $HXZY$ is conclic and so $ZY \parallel DM$.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 51... Sincerely Jean-Louis

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Very nice problem and solitions!

See my solution on my Youtube channel here: https://www.youtube.com/watch?v=CZkZxd-fC2o

Mehhhhhhhh Let $H$ be the orthocenter, just remembering: if you know Humpty Dumpty stories, you're done. Opinion:

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Hint 1:

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Hint 2:

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Hint 3:

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All of those hints are well known facts, note that $YZ || BC \Leftrightarrow YXHZ$ cyclic, and this is true iff $\angle XZH=\angle HYX$, and hence we just need to prove $(ABX)$ tangent to $BC$, by easy angle chase ($\angle XMD=\angle XHA$, and $\angle BAM=90-\angle HYX$, which is equal to $\angle 90-DZB$ iff the tangency is true). Now just $ME^2=MB^2=MX.MA$ finishes the problem.

half line sol with proj: let $K = BX \cap CF$. We know that $(A,X;E,F) \stackrel{H}{=} (B,K;X,Z) \stackrel{Y}{=} (B,C;M, YZ \cap BC) = -1 \Rightarrow YZ // BC$.