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Source: 2025 CTST P2

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#1 h Mar 5, 2025, 3:28 AM • 7 Y

Y by MS_asdfgzxcvb, Retemoeg, Resolut1on07, sami1618, Rounak_iitr, NO_SQUARES, Deadline

Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$ . Let $P$ be an arbitrary point on the nine point circle of $ABC$ . The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$ . The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$ . If $M$ is the midpoint of $AQ$ , show that $HX \perp DM$ .

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#2 h Mar 5, 2025, 4:19 AM • 1 Y

Y by MS_asdfgzxcvb

Animate $Q$ on line $BC$ . Let $E$ be the midpoint of $AH$ , $F$ be the intersection of NPC and $AD$ , and $G$ be the intersection of $AH$ and $BC$ . $R$ is the reflection of $Q$ wrt $D$ . $Y$ is the intersectionf of $AR$ and $(AH)$ . It is easy to see that $DM \parallel AR$ , so it suffices to prove that $H, Y, X$ are collinear. Redefine $X$ as the intersection of $EF$ and line passing through $A$ perpendicular to $AQ$ . We will prove two claims; $H, Y, X$ are collinear and $Q, P, X$ are collinear. The second is equivalent to proving $AP \perp PX$ .

First, we only consider three cases where $Q = D$ , $Q = \infty_{BC}$ , and $Q = B$ or $Q = C$ , and they are all easy to verify that the claims above always hold.

Claim 1. $H, Y ,X$ are collinear.
Proof. Since $A, G, Q, P$ are concyclic, it is easy to see that the map $Q \mapsto P$ is projective, from inversion with center $A$ . Also, $Q \mapsto AQ \mapsto AX \mapsto X$ is projective. Also, $Q \mapsto R \mapsto AR \mapsto Y \mapsto HY \mapsto HY \cap EF$ is projective. Hence, it is easy to see that $H, Y, X$ are collinear for any $Q$ on $BC$ if it is true for three cases, but we verified it above.

Claim 2. $AP \perp PX$ .
Proof. Now, we prove that $AP \perp PX$ , which is equivalent to $A, F, Y, P$ concyclic. From inversion with center $A$ , we see that $Q \mapsto (AFY) \cap (EFG)$ is projective, which means that we only need to consider three cases here, and it is already verified at the beginning.

Therefore, combining the claims, we can see that $HX \perp DM$ for any $Q$ on $BC$ .

This post has been edited 1 time. Last edited by Seungjun_Lee, Mar 5, 2025, 4:37 AM

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#3 h Mar 5, 2025, 5:12 AM • 2 Y

Y by Kingsbane2139, k2i_forever

Let $G$ be the perpendicular foot from $A$ to $BC$ . Invert the problem about $H$ with radius $\sqrt{-AH \times HG}$ , then about $G$ with radius $\sqrt{-BG \times GC}$ , then flip some points over $BC$ . We get

Inverted Problem wrote:

Suppose $\triangle ABC$ is given with orthocenter $H$ and $H'$ is the reflection of $H$ over $BC$ . Let the midpoint of $BC$ be $M$ . Suppose $P$ be a arbitary point on the circumcircle of $\triangle ABC$ . $PH'$ meets $BC$ with $X$ , and $Q$ is the perpendicular foot from $A$ to $HX$ . Prove that $MP = MQ$ .

We solve the inverted problem. Reflect $P$ and $X$ over the perpendicular bisector of $BC$ and let them $P'$ and $X'$ . We get $A,Q,P'$ collinear and $\angle AP'X'=90^\circ$ . Since $XM=MX'$ , we get $MQ=MP'=MP$ as desired.

This post has been edited 3 times. Last edited by mathlove_13520, Mar 5, 2025, 5:18 AM
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#4 h Mar 5, 2025, 5:23 AM • 3 Y

Y by MS_asdfgzxcvb, QNam1106, Resolut1on07

Uhhh...what (P.D. I have to study for a test tomorrow instead of doing math aaaa).
Let the perpendicular from $D$ to $AQ$ hit $AQ,QP$ at $S,R$ respectively and $AH \cap BC=H'$ , then:
$\measuredangle DRP=\measuredangle RAP=\measuredangle DH'P \implies DH'SP \; \text{cyclic}$ Now let $AD$ meet the Nine-Point circle again at $G$ then from Reim's theorem we have that $AGPX$ is cyclic but if you now let $AH'$ hit the Nine-Point circle again at $N$ then clearly $ND$ is diameter and therefore $N,G,X$ are colinear, now let $T$ a point on $BC$ such that $TA$ is tangent to $(ABC)$ then it is easy to see from midpoint of arcs that $ND$ is perpendicular to the antiparallel of $BC$ w.r.t. $\measuredangle BAC$ and therefore $ND \perp AT$ while also $AN \perp DT$ so $N$ is orthocenter of $\triangle ADT$ and this means that $T,N,G$ are colinear but then notice $TN$ is the radical axis of $A,(BC)$ therefore $X$ lies on such radical axis and this means $\text{Pow}_{(BC)}(X)=XA^2=XP \cdot XQ$ and this means $X$ lies on the radax of $(AQ), (BC)$ but clearly by PoP so does $H$ and therefore $HD$ must be their radax and since $(AQ)$ has center $M$ and $(BC)$ as center $D$ we have $HX \perp DM$ as desired thus we are done :cool:

:cool:

.

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#5 h Mar 5, 2025, 7:08 AM • 3 Y

Y by aquelitl, hectorleo123, Resolut1on07

Just a sketch.
Probably too easy for a CTST2.
Denote the midpoint of $AX,AH$ as $K,H_a$ .
By perpendicular conditions we have $AH_aKXP \sim QDMAP$ ,and $H_aK \perp DM$
By $H_aK//HX$ ,the problem is done

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#6 h Mar 5, 2025, 12:41 PM

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Let $N,W$ be the midpoints of $AH,AX$ . Apply $\sqrt{\frac{bc}{2}}$ inversion.
New Problem Statement: $ABC$ is a triangle with circumcenter $O$ . Let $P$ be an arbitrary point on $(BOC)$ . $Q$ is the foot of the altitude from $A$ to $OP$ and $M$ is the reflection of $A$ over $PQ$ . Let $AW\parallel PQ$ and $PA=PW$ . Also $AO$ meets $(BOC)$ at $N$ and $D$ is $A-$ dumpty. Prove that $(BOC),(ADM),(AWN)$ concur.
Let $(ADM)\cap (BOC)=K$ . Note that $M$ lies on $(ABC)$ .
$\measuredangle PKD=\measuredangle POD=\measuredangle OAD=\measuredangle MAD=180-\measuredangle DKM$ thus, $P,K,M$ are collinear. Also $PA=PM=PW$ and $MA\perp AW$ hence $M,P,W$ are collinear.
$\measuredangle ANK=\measuredangle ONK=\measuredangle OPK=\measuredangle AWK$ So $A,W,N,K$ are concyclic as desired. $\blacksquare$

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#8 h Mar 5, 2025, 6:54 PM • 2 Y

Y by A_Crabby_Crab, centslordm

Let $X'$ be the midpoint of $AX$ , and let $A'$ be the midpoint of $H'$ . Let the circumcircle of $APM$ intersect the nine-point circle again at $Y$ . As $AX$ is tangent to the circumcircle of $APEQ$ with center $M$ and $AX'=PX'$ , we get that $APMX'$ is cyclic. Since $\angle MYP=180^\circ-\angle MAP=\angle QEP=180^\circ-\angle DEP=\angle DYP$ , we get $Y$ lies on $DM$ . Since $\angle MYX'=180^\circ-\angle MAX'=90^\circ$ , $X'Y\perp DM$ and $\angle X'YD=90^\circ=\angle A'YD$ , so $DM\perp X'A'\parallel HX$ .

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#9 h Mar 5, 2025, 7:35 PM • 1 Y

Y by qdvjp5689

Call the nine-point circle as $\omega$ , by definition of nine-point circle, $AH\cap \omega=M$ is the midpoint of $AH$ , also call the extension of $AH$ and $\omega$ as $N$ .

We have $AN\bot BQ, APNQ$ lie on the same circle, so $\angle{PAN}=\angle{PQN}; \angle{AMP}=\angle{PDQ}$ which implies $\triangle{AMP}\sim \triangle{QDP}$

Thus we have $\frac{AM}{QD}=\frac{AP}{QP}=\tan(\angle{AQP})=\frac{AX}{AQ}=\frac{AH}{2QD}=\frac{AX}{2MQ}, \frac{AX}{MQ}=\frac{AH}{QD}$

Since $\angle{XAQ}=\angle{ANQ}=90^{\circ}, \angle{XAH}=\angle{MQD}, \triangle{XAH}\sim \triangle{MQD}$

Since $AX\bot QM; AH\bot QD, XH\bot MD$ by spiral similarity.

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#10 h Mar 6, 2025, 2:55 AM

Y by

Here's a funny solution to this. First, rename $M$ to $N$ , $D$ to $M$ . Let $DEF$ be the orthic triangle, $I=AP\cap (DEF)$ , $K$ be the midpoint of $AH$ , and $G=AM\cap (DEF)$ . It clearly suffices to show that $X$ is on the radical axis of $(APDQ)$ and $(BCEF)$ . Note also that $MG^2-GA^2=MG'\cdot MA=MB^2$ is known since $G'$ is the reflection of $A$ over $G$ , or just the Humpty point.

The key claim is that $X,K,G$ are collinear, or $XG\perp AM$ . Invert about $A$ fixing the incircle, and let $Q$ map to $Q'=IK\cap AQ$ . By Reim, $AQ\parallel MI$ so by perpendicularity, $AQ'\perp Q'I$ so $IX'\perp AX$ . Since $MI\perp AX'$ also we obtain that $AX'\perp MX'$ or inverting back, $AG\perp XG$ .

To finish, the power from $X$ with respect to $(BCEF)$ can be calculated as follows:
$XM^2-MB^2=XG^2+GM^2-MB^2=XA^2-GA^2+GM^2-MB^2=XA^2,$ which concludes as $XA$ is clearly tangent to $(APDQ)$ .

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#11 h Mar 6, 2025, 3:15 AM

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Let $G$ be altitude from $A$ to $BC$ and $E,F$ is intersection of $AD,AG$ with nine point circle respectively .
Define $X' \in EF$ such that $\angle QAX' = 90^{\circ}$
Known that $A,Q,G,P$ concyclic with center $M$ . So line $AX'$ is tangent to $(AQGP)$ and line $AQ$ is tangent to $(AEX')$
Claim 1 $X=X'$ collinear .
Since $\angle QAP = \angle PGD = \angle PED$ so $QA$ is tangent to $(AEP)$
Implies that $A,E,P,X'$ concyclic . $\rightarrow \angle APX' = 90^{\circ}$
Thus $P,Q,X'$ collinear . So $X=X'$

Claim 2 $H \in \text{rad((AQ),(BC))}$
$\text{Pow(H,(BC))}= HA \times HG = \text{Pow(H,(AQ))}$

Claim 3 $X \in \text{rad((AQ),(BC))}$
Known that $(AH)$ is orthogonal with $(BC)$ and $F$ is midpoint of $AH$ .
So $F \in \text{rad(A,(BC))}$ .
And $EF \perp AD$ implies that $EF = \text{rad(A,(BC))}$
So $X \in \text{rad(A,(BC))}$ .
Since $\text{Pow(X,A)} = XA^2 = \text{Pow(X,(AQ))}$ so $\text{Pow(X,(AQ))} = \text{Pow(X,(BC))}$
Thus $X \in \text{rad((AQ),(BC))}$ .

By Claim 2,3 Conclude that $HX = \text{rad((AQ),(BC))}$
So $HX \perp DM$ . Done $\square$

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#12 h Mar 6, 2025, 4:36 AM

Y by

YaoAOPS wrote:

Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$ . Let $P$ be an arbitrary point on the nine point circle of $ABC$ . The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$ . The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$ . If $M$ is the midpoint of $AQ$ , show that $HX \perp DM$ .

Solution

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#13 h Mar 6, 2025, 4:28 PM

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Let $AH$ meet $BC$ at $E$ , and let $N$ be the midpoint of $AH$ . Define $S$ as the center of the spiral similarity that sends $AH$ to $QD$ , i.e., the intersection of the circumcircles $(APEQ)$ and $(HED)$ .

Consider the mapping that takes any point $Z$ on the circle $(APEQ)$ to the second intersection of the circle $(EDZ)$ with line $EA$ . By the pencil lemma followed by inversion about $E$ , this defines a projective transformation. Consequently, we obtain the cross-ratio relation:
$(A, S; P, Q) = (A, H; N, P_{\infty}).$ This implies that quadrilateral $APSQ$ is harmonic. Thus, the tangents to the circle $(APQS)$ from $A$ and $S$ intersect along line $PQ$ , meaning that $XS$ is tangent to this circle.

Since $\triangle SMX \sim \triangle AMX \sim \triangle SQA \sim \triangle SDH$ , it follows that $S$ also maps $HX$ to $DM$ . Finally, because $AH \perp QD$ , we conclude that $HX \perp DM$ .

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#14 h Mar 7, 2025, 9:39 AM

Y by

One answer using orthotransversal of mine.
We have a well-known lemma.
$\textbf{Lemma}.$ Let $P$ be an arbitrary point on the Euler circle. Then, the orthotransversal of $P$ with respect to $\triangle ABC$ passes through $O$ , the circumcentre of triangle $ABC$ . Furthermore, it's the Steiner line of $P$ with respect to the median triangle of $\triangle ABC$ .

$\textbf{Proof.}$ Denote $B', C'$ be the intersection of the $P$ 's orthotransversal with $CA, AB$ . Let $H$ be the orthocentre of the triangle. It's trivial that $(APA'), (BPB'), (CPC'), (O)$ , and $HP$ are concurrent at one point, let it be $W$ , then let $WB', W'C$ cut $(O)$ again at $B_1, C_1$ , and using Pascal's theorem, we get $B', O, C'$ are collinear. Then let $P'$ be the projection of $A$ onto $B'C'$ , and with some angle chasing, we get $P'$ and $P$ are symmetric to each other with respect to the midline of $A$ of triangle $ABC$ .

Back to the main problem, it's sufficient to prove that $\triangle MDQ \sim \triangle XHA$ , it's equivalent to
$\frac{DQ}{HA} = \frac{MQ}{AX}$ which simplifies to
$\frac{DQ}{2DO} = \frac{AQ}{2AX}$ $\frac{AQ}{AX} = \frac{DQ}{DO}$
Therefore, we only need to prove that $\angle AQP = \angle OQD$ .
Let $P'$ be symmetric to $P$ with respect to the midline of $A$ of triangle $ABC$ . Then
$\angle AQP = \angle DQP' = \angle OQD,$ since $Q, O, P'$ are collinear.

Hence, the problem is proved.

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#15 h Mar 8, 2025, 8:34 PM • 1 Y

Y by MS_asdfgzxcvb

Let $N$ midpoint of $AH$ , as $P$ belongs to nine point circle, $\angle NPD=90^{\circ} \implies \triangle APN \sim \triangle QPD$ where $P$ is center of spiral similarity of them, sending $A \rightarrow Q$ and $N \rightarrow D$ , where the rotation is $90^{\circ}$ , so at this spiral similarity each pair of homologous lines are perpendicular, now realize $X \rightarrow A$ because they are clearly homologous at the similarity, so take $L$ midpoint of $AX \implies NL \perp MD$ (homologous), thus $HX \perp MD \square$

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#16 h Mar 8, 2025, 11:34 PM • 1 Y

Y by qdvjp5689

we will coordinate bash this.

First, dilate at $A$ with factor $2$ so that $D$ is sent to $A'$ and $M$ to $Q$ . Reverse reconstruct such that $X$ satisfies only $AQ\perp AX$ and $HX\perp QA'$ . We shall prove that the foot of $A$ onto $QX$ , $P$ lies on the nine point circle.

We first claim as we move $Q$ on $BC$ , that $X$ moves on a line. Indeed, WLOG let $Q=(k,0)$ and $A=(0,1)$ . Then, $AX$ has slope $k$ hence equation $y=kx+1$ . Let $A'=(a,-1)$ as $A'$ lies somewhere on the dilation of $BC$ by two from $A$ . Then, $QA'$ has slope $\frac{1}{k-a}$ m so $HX$ has slope $a-k$ . Suppose that $H=(0,h)$ as $H$ is on the perpendicular line from $A$ to $BC$ . Then, this gives $HX$ equation $y=(a-k)x+h$ . Equate this with $y=kx+1$ . Then, $2y=ax+h+1$ , by summing, which is the equation of a line.

Now suppose $X$ lies on line $y=mx+c$ that is fixed. Now, we still have $AX$ as $y=kx+1$ . Equating, $c-1=x(k-m)$ , so $x=\frac{c-1}{k-m}$ . We also see that $y=\frac{ck-m}{k-m}$ . Now we'll calculate $P$ .

The slope of $QX$ is $\frac{ck-m}{c-1-k(k-m)}$ hence has equation
$y=\frac{ck-m}{c-1-k^{2}+km}\left(x-k\right)$ Meanwhile, $AP$ has equation
$y=-\frac{\left(c-1-k^{2}+km\right)}{ck-m}x+1$ Equate the two, so
$\begin{align*} x\left(\frac{ck-m}{c-1-k^{2}+km}+\frac{\left(c-1-k^{2}+km\right)}{ck-m}\right)&=\frac{ck^{2}-km+c-1-k^{2}+km}{c-1-k^{2}+km}\\ &=\frac{\left(k^{2}+1\right)\left(c-1\right)}{c-1-k^{2}+km} \end{align*}$ Now expand the coefficient of $x$ .
$\begin{align*} &\frac{c^{2}k^{2}+m^{2}-2ckm+c^{2}+1+k^{4}+k^{2}m^{2}-2c-2ck^{2}+2ckm+2k^{2}-2km-2k^{3}m}{\left(c-1-k^{2}+km\right)\left(ck-m\right)}\\ &=\frac{c^{2}k^{2}+m^{2}+c^{2}+1+k^{4}+k^{2}m^{2}-2c-2ck^{2}+2k^{2}-2km-2k^{3}m}{\left(c-1-k^{2}+km\right)\left(ck-m\right)}\\ &=\frac{\left(c^{2}-2c+k^{2}-2km+m^{2}+1\right)\left(k^{2}+1\right)}{\left(c-1-k^{2}+km\right)\left(ck-m\right)} \end{align*}$ So we have
$x=\frac{\left(c-1\right)\left(ck-m\right)}{c^{2}-2c+k^{2}-2km+m^{2}+1}$ Now solving for $y$ ,
$\begin{align*} y&=\frac{-\left(c-1\right)\left(c-1-k^{2}+km\right)}{c^{2}-2c+k^{2}-2km+m^{2}+1}+1\\ &=\frac{-c^{2}+2c+ck^{2}-ckm-1-k^{2}+km+c^{2}-2c+k^{2}-2km+m^{2}+1}{c^{2}-2c+k^{2}-2km+m^{2}+1}\\ &=\frac{\left(ck^{2}-ckm-km+m^{2}\right)}{c^{2}-2c+k^{2}-2km+m^{2}+1}\\ &=\frac{\left(ck-m\right)\left(k-m\right)}{c^{2}-2c+k^{2}-2km+m^{2}+1} \end{align*}$ This means that $P$ is given by
$\left(\frac{\left(c-1\right)\left(ck-m\right)}{\left(c-1\right)^{2}+\left(k-m\right)^{2}},\frac{\left(ck-m\right)\left(k-m\right)}{\left(c-1\right)^{2}+\left(k-m\right)^{2}}\right)$ Note that
$\frac{k-m}{c-1}=\frac{y}{x}$ Subbing this into the denominator for $x$ , we get
$x=\frac{(c-1)(ck-m)}{(c-1)^2\left(1+\frac{y^2}{x^2}\right)}$ Use $ck=\frac{y}{x}c(c-1)+cm$ and upon rearranging,
$x(c - 1)\left(1 + \frac{y^2}{x^2}\right) = c\left(\frac{y(c - 1)}{x} + m\right) - m$ Multiply both sides by $x$ so
$(x^2+y^2)(c-1)=c(c-1)y+x(cm-m)$ cancelling the $c-1$ , we get
$x^2+y^2=cy+mx$ Which is obviously a circle's equation, in particular
$\left(x - \frac{m}{2}\right)^2 + \left(y - \frac{c}{2}\right)^2 = \frac{m^2 + c^2}{4}$ where the radius squared is indeed non-negative.

Now, we have shown that the locus is a circle. $3$ points uniquely determine a circle. Here, let $Q=B$ , $Q=C$ and $Q=D$ . In the first two cases, $P$ is just one of the feet of the altitudes, hence clearly lies on the nine point circle. Finally, note that $X$ is the intersection of the perpendicular to $AD$ through $A$ with the perpendicular to $AA'$ through $H$ . But $AA'$ is the same as $AD$ so this point is at infinity. Now, the perpendicular direction to $DX$ must be $AD$ , hence $P=D$ , which lies on the nine point circle.

Thus we are done.

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#17 h Mar 18, 2025, 11:35 AM

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oly geometers who can bash vs oly geometers who can't bash
TestX01 CAN'T bash
Scilyse CAN bash

Let the nine-point circle of $\triangle ABC$ intersect $BC$ again at $K \neq D$ , which is the foot of the altitude from $A$ to $BC$ , and $AK$ again at $L \neq K$ , which is the midpoint of $\overline{AH}$ . Further let $Q'$ be the reflection of $Q$ over $D$ ; now $DM \parallel AQ'$ . Hence, redefine $X$ to be the intersection of the line through $A$ perpendicular to $AQ$ and the line through $H$ perpendicular to $AQ'$ . We will prove that $P$ , $Q$ and $X$ are collinear.

Set $K = (0, 0)$ , $D = (d, 0)$ , $Q = (q, 0)$ , $L = (0, \ell)$ and $A = (0, 1)$ so that $Q' = (2d - q, 0)$ and $H = (0, 2\ell - 1)$ . Now the line through $A$ perpendicular to $AQ$ has equation $y = qx + 1$ and the line through $H$ perpendicular to $AQ'$ has equation $y = (2d - q) x + 2\ell - 1$ . Intersecting these two lines yields $X = \left(\frac{1 - \ell}{d - q}, \frac{d - \ell q}{d - q}\right)$ . Since $P = (KAQ) \cap (KLD) \neq K$ , $P$ is the centre of the spiral symmetry taking $\overline{AQ}$ to $\overline{LD}$ , so
$\begin{align*} P &= \frac{AD - QL}{A + D - Q - L} \\ &= \frac{(d - \ell q)i}{(d - q) + (1 - \ell)i} \\ &= \frac{d - \ell q}{(1 - \ell) - (d - q)i} \\ &= \frac{(d - \ell q)((1 - l) + (d - q)i)}{(1 - \ell)^2 + (d - q)^2} \\ &= \left(\frac{(d - \ell q)(1 - \ell)}{(1 - \ell)^2 + (d - q)^2}, \frac{(d - \ell q)(d - q)}{(1 - \ell)^2 + (d - q)^2}\right). \end{align*}$ Now
$\begin{align*} |QXP| &= \begin{vmatrix} q & 0 & 1 \\ \frac{1 - \ell}{d - q} & \frac{d - \ell q}{d - q} & 1 \\ \frac{(d - \ell q)(1 - \ell)}{(1 - \ell)^2 + (d - q)^2} & \frac{(d - \ell q)(d - q)}{(1 - \ell)^2 + (d - q)^2} & 1 \end{vmatrix} \\ &= \text{stuff} \cdot \begin{vmatrix} q & 0 & 1 \\ 1 - \ell & d - \ell q & d - q \\ (d - \ell q)(1 - \ell) & (d - \ell q)(d - q) & (1 - \ell)^2 + (d - q)^2 \end{vmatrix} \\ &= \text{stuff} \cdot (q((d - \ell q)((1 - \ell)^2 + (d - q)^2) - (d - q)^2 (d - \ell q)) + ((1 - \ell)(d - \ell q)(d - q) - (d - \ell q)^2 (1 - \ell))) \\ &= \text{stuff} \cdot (q(d - \ell q)(1 - \ell)^2 + (1 - \ell)(d - \ell q)(\ell q - q)) \\ &= \text{stuff} \cdot q((d - \ell q)(1 - \ell)^2 - (d - \ell q) (1 - \ell)^2) \\ &= 0, \end{align*}$ as desired.

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#18 h Apr 8, 2025, 8:44 AM

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Let $K$ be the projection of $A$ onto $BC$ .
$AP,AD,AK,PQ$ intersect the nine point circle of $\triangle ABC$ again at $P',D',K',T$ , respectively.
Since $AQ$ is the diameter of the circumcircle of $\triangle APK$ , we have $AQ\perp P'K'$ . $(1)$
Also, since $QA$ is also the diameter of the circumcircle of $\triangle QPK$ , it follows that $AQ\perp DT$ . $(2)$
Now, combining with $(1)$ and $(2)$ , and applying Pascal's theorem to six points $\begin{vmatrix}P & K' & D \\ D' & T & P'\end{vmatrix}$ we conclude that the lines $PT,D'K'$ and the line through $A$ perpendicular to $AQ$ are concurrent.
Therefore, $X\in D'K'\Rightarrow XK'\perp AD$ . Otherwise, by cross ratio $D(QA,M\infty) = -1 = X(\infty K', HA)$ and since $DQ\perp HA, DA\perp XK'$ , and $QA\perp XA$ , we have $DM\perp XH$ . (q.e.d)

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