My solution : Let the tangent of $ \odot (ABC) $ passing through $ B, C $ meets each other at $ T $ . Let $ S $ be the isogonal conjugate of $ T $ WRT $ \triangle ABC $ ($ SB \parallel CA, SC \parallel AB $) . From $ PM \perp AB, TM \perp BC \Longrightarrow \angle TMP=\angle CBA=\angle TCP $ , so $ C, P, T, M $ lie on a circle with diameter $ CT $ $ \Longrightarrow $ $ P $ is the projection of $ T $ on $ CA $ . Similarly, $ Q $ is the projection of $ T $ on $ AB $ $ \Longrightarrow \odot (PMQ) $ is the pedal circle of $ T $ WRT $ \triangle ABC $ , so $ X $ is the projection of $ S $ on $ BC $ due to $ T, S $ share the same pedal circle (WRT $ \triangle ABC $) $ \Longrightarrow BH=CX $ . Q.E.D

Let $U \equiv QM \cap AC,$ $V \equiv PM \cap AB.$ $T$ is the midpoint of $PQ$ and $AM$ cuts $\odot(APQ)$ again at $N$ (reflection of orthocenter $M$ of $\triangle APQ$ on $PQ$). By symmetry, reflections $Y$ and $Z$ of $X$ on $PQ$ and $T$ lie on $\odot(APQ)$ and $PQ \parallel YZ$ $\Longrightarrow$ $\angle (XZ,XM)=\angle (ZN,ZX)=\angle ANZ=\angle AYZ$ $\Longrightarrow$ $AHY \perp BC.$ But by Butterfly theorem for the cyclic $PQVU,$ it follows that $TM \perp BC$ $\Longrightarrow$ $TM$ is X-midline of $\triangle XHY$ $\Longrightarrow$ $MX=MH$ or $BH=CX.$

Let $I$ be the midpoint of $PQ$. Since $M$ is the orthocenter of triangle $APQ$ and $MB=MC$ then applying butterfly theorem for the circle with diameter $PQ$, $IM\perp BC$. Let $J$ be the reflection of $A$ wrt $M, O$ be the circumcenter of triangle $APQ, N$ is the center of $(MPQ)$, $L$ be the antipode of $M$ wrt $(MPQ)$, $Y$ be the midpoint of $LJ.$ We have $NY\parallel =\frac{1}{2}MJ\parallel =\frac{1}{2}AM\parallel =IN$ hence $N$ is the midpoint of $IY$. This means $IY\parallel =MJ$ or $JY\parallel IM$. Therefore $LJ\perp BC$ at $X$. But $A$ and $J$ are symmetric wrt $M$ hence $MH=MX$ or $BH=CX.$

This question is a little bit silly but how your guys can thinking like that ? ANY SECRET

Let $PM\cap AB=D, QM\cap AP=E$ performing inversion wrt $M$ and ratio $\sqrt{-MD.MP}$ it amounts to showing that $H'=DE\cap BC$ & $X'=PQ\cap BC$ are equidistant from $M$. But by converse of butterfly thm on $DQPE$ with $DQ$ and $EP$ as 'wings', it follows that $OM\perp BC$($O$ is centre of $\odot DQPE$. Now we see that the above thing that we wanted to prove is nothing but butterfly on $DQPE$ with $DE$ and $QP$ as wings.

Here is my solution. Reflecting $P,Q$ in $M$ the problem is equivalent to the following: In triangle $ABC$ the feet from $A$ onto $BC$ is $H$ and $M$ is the midpoint of $BC$. Points $K,N$ are on $AB,AC$ such that $\angle AKM=\angle ANM=90^{\circ}$. Let rays $MK,MN$ meet the lines through $B,C$ parallel to $AC,AB$ respectively at $P,Q$. Prove that $P,Q,H,M$ are concyclic. For this newer statement, the proof is as follows: It suffices to prove that $H$ is the centre of a spiral similarity sending $KP$ to $NQ$ since we already know that $A,H,M,K,N$ are concyclic. Now, for this it suffices that $\frac{HK}{HN}=\frac{KP}{NQ}$. Consider the following equalities: \begin{align*} \frac{HK}{HN}=\frac{\sin HNK}{\sin HKN}=\frac{\sin HAK}{\sin HAN}=\frac{\cos B}{\cos C} \end{align*} And observe that $BM\cdot \cos B=BK$ and so $KP=BK\cdot \tan A$. Similarly, we observe that $CM\cdot \cos C=CN$ and $NQ=CN\cdot \tan A$. Therefore, we conclude that \begin{align*} \frac{HK}{HN}=\frac{\cos B}{\cos C}=\frac{BM\cdot \tan A}{CM\cdot \tan A}\cdot \frac{\cos B}{\cos C}=\frac{KP}{NQ} \end{align*} Thus, our claim holds. Now, it follows that points $P,Q,H,M$ are concyclic.

Consider triangle $APQ$, which has orthocenter $M$. Extend line $BC$ to intersect the circle with diameter $PQ$ at $Z$, $Y$. Then $YZ$ also has midpoint $M$ by Butterfly theorem. By inversion at $M$, Butterfly theorem then implies $MH = MX$ too.

Let $N$ be the midpoint of $PQ$. By the converse of butterfly theorem, $MN \perp BC$. Also $M$ is the orthocenter of $APQ$. Now let the intersection of $(AM),(APQ)$ be $L \ne A$. Then we claim that $AHML$ is a rectangle. Proof: Make $APQ$ our "reference triangle". Then the antipode of $A$ in $(APQ)$ is reflection of $M$ over $N$, and so as $LA \perp LM$ , $A', N' M; L$ are collinear. So the conclusion follows. So $HM = AL$. Consider the translation with vector $AM$.It sends $AQP)$ to $(MQP$ as these circles are congruent and are reflections over $PQ$ and their radical axis is perpendicular to $AM$. As $MX \parallel AL$, and $X \in (MQP)$, $L$ is mapped to $M$, so $MX = AL = HM$ and so $BH = CX$.

Dear Mathlinkers, according a converse of the butterfly theorem and https://www.artofproblemsolving.com/community/c6h1387927_a_nice_lemma we can have a nice proof of the problem... Who want to start in this adventure? Sincerely Jean-Louis

Dear Mathlinkers, see http://jl.ayme.pagesperso-orange.fr/Docs/Simplicity%201.pdf p. 19. Sincerely Jean-Louis

No need for summoning butterflies silouan wrote: In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$ Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$. The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$ M. Didin Draw parallelogram $ABDC$ and let $T$ be the intersection of tangents to $\odot(ABC)$ at $B$ and $C$. Notice that $$\angle MQB=90^{\circ}-A=\angle MTB$$so $BMTQ$ is cyclic. Likewise, $CMTP$ is cyclic. It follows that $\odot(PMQ)$ is the pedal circle of point $T$ wrt $\triangle ABC$. Since $T$ and $D$ are isogonal conjugates in $\triangle ABC$, and $\overline{DX} \perp \overline{BC}$, we see that $X \in \odot(PMQ)$, as desired. $\blacksquare$

Solved with Christopher Qiu, Elliott Liu, Isaac Zhu, Jeffrey Chen, Kevin Wu, Rey Li, and Ryan Yang. We present two comedies. First comedy (inversion) Invert at \(M\); butterfly. Second comedy (wild goose chase) Let \(APA'Q\) be a parallelogram. By Reim's theorem, \(\overline{A'Q}\) is tangent to \((QME)\) and \(\overline{A'P}\) is tangent to \((PMF)\). Now \(\angle A'QM=\angle A'PM=90^\circ\), so it will suffice to show \(\angle A'XM=90^\circ\). In other words, if \(O\) is the midpoint of \(\overline{PQ}\), it will suffice to show \(\overline{OM}\perp\overline{BC}\). Let \(B'\) and \(C'\) be the reflections of \(B\) and \(C\) over \(Q\) and \(P\), and let \(N\) be the midpoint of \(\overline{B'C'}\). By homothety \((B,2)\), we have \(\overline{B'C}\perp\overline{AC'}\), and analogously \(\overline{C'B}\perp\overline{AB'}\), so points \(B\), \(C\), \(B'\), \(C'\) lie on a circle centered at \(N\). Now \(NQMP\) is a Varignon parallelogram, and since \(NB=NC\), we have \(\overline{NOM}\perp\overline{BC}\). The end.

Let $U$ be the reflection of $A$ over $MQ$. Note that $\measuredangle QUP = \measuredangle PAQ = \measuredangle QMP$. Hence $P, Q, M, U$ are concyclic. Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$. Since $QM$ is the perpendicular bisector of $BF$, $AF = BU$. And note that $\triangle BPM \sim \triangle HAC$, hence $$ BU \times BP = AF \times (CH \times \frac{BM}{AH}) = BM \times (CH \times \frac{AF}{AH}) = BM \times CD = BM \times BX $$Hence $M, X, U, P$ are concyclic, which means $X \in \odot(MPQ)$, as desired.

Define $MP\cap AB=R,MQ\cap AC=S.$ Since $H\in \odot (ARMS),$ inversion wrt $M,$ which fixes $\odot (PQRS),$ maps $H,X$ onto points $QP\cap BC, RS\cap BC$ respectively. These points are symmetric wrt $M$ by DIT, thus the conclusion.

synthetic skill issue. Rename $H$ to $D$; let $D'$ be the reflection of $D$ over $M$ and $A'$ be the reflection of $A$ over the midpoint of $\overline{PQ}$. I claim that $\overline{A'D'} \perp \overline{BC}$. This is by the sophisticated method of brainlessly coordinate bashing. Let $A=(a,b),B=(-1,0),C=(1,0)$, so $M=(0,0)$ and $D=(a,0),D'=(-a,0)$. Then $\overline{MQ}$ has equation $y=-\tfrac{a-1}{b}x$ by computing the slope of $\overline{AC}$, and $\overline{AB}$ has equation $y=\tfrac{b}{a+1}(x+1)$, so their intersection $Q$ has $x$-coordinate $-\tfrac{b^2}{a^2+b^2+1}$. Likewise, $\overline{MP}$ has equation $y=-\tfrac{a+1}{b}x$ and $\overline{AC}$ has equation $y=\tfrac{b}{a-1}(x-1)$, so their intersection $P$ has $x$-coordinate $\tfrac{b^2}{a^2+b^2+1}$, so the midpoint of $\overline{PQ}$ lies on the $y$-axis and hence $A'$ has $x$-coordinate $-a$. To finish, note that this means $\angle A'PM=\angle A'QM=\angle AD'M=90^\circ$, hence $A'PQMD'$ is cyclic which finishes. $\blacksquare$ Remark: This finish seems much cleaner than all the other solutions that prove the perpendicularity synthetically (imagine butterfly)

This is very nice problem and it is my favorite problem. I would like to introduce an extension to this problem. Given a cyclic quadrilateral $ABCD$ and $ABCD$ is not a trapezoid. Let $M$ be the midpoint of $AD$. Points $P$ and $Q$ lie on lines $CD$ and $AB$ respectively such that $MP \parallel BD$ and $MQ \parallel AC$. The circumcircle of triangle $MPQ$ intersects $AD$ again at $N$ different from $M$. Let $AC$ intersect $BD$ at $E$. $F$ is the projection of $E$ onto $AD$. a) Prove that $AF = ND$. b) Take points $I$ and $J$ on $MQ$ and $MP$ respectively such that $AI \parallel CD$ and $DJ \parallel AB$. Prove that $M, N, J, I$ lie on the same circle.

Another generalization of this beautiful problem Given an acute triangle $ABC$. Two points $M$ and $N$ lie on side $BC$ such that $M$ and $N$ are different from $B$ and $C$, and $BM = CN$. The line through $M$ perpendicular to $CA$ intersects line $AB$ at $F$. The line through $N$ perpendicular to $AB$ intersects line $AC$ at $E$. $MF$ intersects $NE$ at $P$. On the circumcircle of triangle $PEF$, take $Q$ such that $PQ \parallel BC$. Let $R$ be the point symmetric to $Q$ with respect to the midpoint of $BC$. Prove that $AR \perp BC$.

And now this is the combination of both two extesions above Given a cyclic quadrilateral $ABCD$ that is not a trapezoid. On $CD$, take points $M$ and $N$ that are symmetric with respect to the midpoint of $CD$. On lines $BC$ and $AD$, take points $E$ and $F$ respectively such that $ME \parallel AC$ and $NF \parallel BD$. Let $P$ be the intersection of $ME$ and $NF$. On the circle $(PEF)$, take $Q$ such that $PQ \parallel CD$. Let $R$ be the point symmetric to $Q$ with respect to the midpoint of $CD$. Prove that $IR \perp CD$ where $I$ is the intersection of $AC$ and $BD$.