Since $\angle{ADH}=\angle{AEH}=90$, $E,D$ are the feet of the $B, C$ altitudes. Additionally, it is well known that $MD, ME$ are tangents to $(AH)$. Then as the center of $(AH)$ lies on $AH$, $Q=DD \cap HH$, and Pascal's on $DDAHHE$ gives the desired collinearity.

Here goes my solution , notice some super well known facts such as $D,E$ being feet of the altitudes and $MD$ being tangent to $(AH)$ at $D$. Then just to make the finish different than the previous solution , observe that by La Hire $Q\in{BP}\hspace{3 mm}\textbf{Q.E.D.}$

2020 Korea National Olympiad P2 wrote: $H$ is the orthocenter of an acute triangle $ABC$, and let $M$ be the midpoint of $BC$. Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$, and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$. Prove that $P,Q,B$ are collinear. Great problem! Though trivialized by projective geometry. This is extremely detailed solution.

RIP to my Polars bash.

MNJ2357 wrote: $H$ is the orthocenter of an acute triangle $ABC$, and let $M$ be the midpoint of $BC$. Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$, and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$. Prove that $P,Q,B$ are collinear. Solution similar to above few solutions. By three tangents lemma , $MD$ and $ME$ are tangent to $(ADHE)$ and by pascal's theorem on $DDAHHE$ we get $P,Q,B$ are collinear.

The orthocenter $H$ is unnecessary. The problem can be generalized as follows: Generalization: Let $D,E$ be arbitrary points on $AB$ and $AC$. Let $H = CD \cap BE$, $P=AH \cap DE$. The line through $H$ parallel to $BC$ meets $DM$ at $Q$. Prove that $P,Q,B$ are collinear.

any solution for people who do not know about pascal's theorem or la hire's theorem?

Tragic Problem. All poles and polars will be taken wrt $(AH)$. By Brokard on $ADHE$ we see that $BP$ is the polar of $C$, and since $C \in DH$ is the polar of $Q$, by La Hire $Q$ lies on the polar of $C$, hence $B, Q, P$ collinear as desired.

Let $\omega$ be the circle through $(ADEH)$. Recall that $M \in \overline{DD}$ by Three Tangents lemma. As a result, $Q = \overline{DD} \cap \overline{HH}$. Now, Pascal's on $DDAHHE$ gives us $Q$, $B = \overline{DA} \cap \overline{HE}$, and $P = \overline{AH} \cap \overline{DE}$ are collinear, as desired.

MNJ2357 wrote: $H$ is the orthocenter of an acute triangle $ABC$, and let $M$ be the midpoint of $BC$. Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$, and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$. Prove that $P,Q,B$ are collinear. Even if is a bit known, is cute :3. (Love projections :3). It is know that $D,E$ are feets of altitude. We will take polars w.r.t. $(AH)$ Let $BP \cap DH=S$ and $BP \cap AE=T$ and let $CP \cap EH=U$ and $CP \cap AD=V$ $$(D, S; H, C) \overset{B}{=} (A, T; E, C) \overset{P}{=} (H, S; D, C) \implies -1=(D, S; H, C) \overset{B}{=} (A, T; E, C) \implies B \in \mathcal P_P$$$$(E, U; H, B) \overset{C}{=} (A, V; D, B) \overset{P}{=} (H, U; E, B) \implies -1=(E, U: H, B) \overset{C}{=} (A, V; D, B) \implies C \in \mathcal P_P$$And both of them implies that $\mathcal P_P=BC$. We also have: $$-1=(D, S; H, C) \overset{B}{=} (A, T; E, C) \implies T \in \mathcal P_C$$$$-1=(E, U; H, B) \overset{C}{=} (A, V; D, B) \implies V \in \mathcal P_B$$But otherside by the first ratio chasing $P \in \mathcal P_B$ and $P \in \mathcal P_C$ and then $\mathcal P_B=PV$ and $\mathcal P_C=PT$ We know that $\mathcal P_Q=DH$ and $\mathcal P_P=BC$. So proving that $B,P,Q$ are colinear is equivalent to prove that its polars are concurrent but: $\mathcal P_B \cap \mathcal P_Q \cap \mathcal P_P=VC \cap DH \cap BC=C$ Thus we are done

Same as #9 - posting for storage. Notice that both $QH$ and $QD$ are tangent to the circle with diameter $AH$, then $DH$ is the polar of $Q$ with respect to the aforementioned circle which contains $C$ and consequently by La Hire, we get that the polar of $C$ which by Brokard is $BP$ and contains $Q$, as desired. $\blacksquare$

Bruh, absolutely trivial. Note that $QH,QD$ are tangent to $(AH)$ by trivial angle chase. Hence, by Pascal's theorem on $DDEHHA$, we get that $P,Q,B$ are collinear.

We present two solutions, the second written on 6/6/23 in Korea. Solution One: Three Tangents Lemma and Pascal. $\square$ Solution Two: Let $Q'=\overline{BP}\cap\ell$ where $\ell$ is the line through $H$ perpendicular to $\overline{AH}$. Then, letting $M'=\overline{BC}\cap\overline{DQ}$, it suffices to show $M'=M$. Indeed, letting $P_{\infty}$ be the point at infinity along $\ell$, we have \[(B,C;M,P_{\infty})\stackrel{Q}=(\overline{BP}\cap\overline{CD},C;D,H)\stackrel{P}=(B,\overline{AB}\cap\overline{CP};D,A)=-1\]$\square$

Bruh, $MD$ is tangent to $(MH)$. Now $Q$ is pole of $DH$,and it lies on $PB$ ,the polar of $C$ by Brocard.

Note that in $\triangle PBC$ : $P \equiv c ; B \equiv c$ By chasing angles: $MD ; ME$ are tangents to $(AH)$ $\implies GD ; GH$ are tangents to $(AH)$ $\implies G \equiv c$ $\implies P,Q,B$ are collinear.

very easy $D-H-C$ and $E-H-B$ are obvious $MD$ and $ME$ are tangent to $(ADHE)$ Thus Pascal yields $DDAHHED$ and we are done

oops. Rename $D$ to $F$ for the rest of this solution since this problem has strange naming conventions. Furthermore, let $D$ then be the foot from $A$ to $BC$. Employ barycentric coodinates with reference triangle $\triangle DEF$, and let $EF=a,DF=b,DE=c$. Our strategy will be to let $Q'$ denote the intersection of $BP$ and $MF$, and show that $HQ'$ is parallel to $BC$, which would solve the problem. Note that $$A=(-a:b:c),B=(a:-b:c),C=(a:b:-c).$$We will first compute $M$. Multiply $B$ by $a+b-c$ and $C$ by $a-b+c$ to make them the same sum to get $$B=(a^2+ab-ac:-ab-b^2+bc:ac+bc-c^2)$$and $$C=(a^2-ab+ac:ab-b^2+bc:-ac+bc-c^2).$$Averaging these (this is allowed now since $B$ and $C$ have the same sum) gives $$M=(a^2:-b^2+bc:-c^2+bc).$$For now, we also know that $\overrightarrow{BC}$ is a scalar multiple of $$B-C=(2ab-2ac:-2ab:2ac)$$so it is also a multiple of $$(b-c:-b:c),$$which we will use later. Then, cevian $MF$ is described by $$(a^2:-b^2+bc:t).$$We can find that the equation of line $BP$ is $$2bcx+acy-abz=0,$$so if $Q'=(a^2:-b^2+bc:t)$ then we can solve to get $$t=2ac-bc+c^2,$$which means that $$Q'=(a^2:-b^2+bc:2ac-bc+c^2).$$The sum of this is $$a^2+2ac+c^2-b^2=(a+c)^2-b^2=(a+b+c)(a-b+c),$$so we multiply $H=(a:b:c)$ by $a-b+c$ to get $$H=(a^2-ab+ac:ab-b^2+bc:ac-bc+c^2),$$so $\overrightarrow{Q'H}$ is a scalar multiple of $$Q'-H=(ab-ac:-ab:ac)$$so it is also a multiple of $$(b-c:-b:c).$$Since this is also true of $\overrightarrow{BC}$, they are parallel, so we are done.

First of all, it is well known that $D$ and $E$ are the feet of the altitudes from $C$ and $B$ respectively. Further, let $F= \overline{AH} \cap \overline{BC}$ and $H' = \overline{HM} \cap (AH)$. Also, let $R= \overline{BE} \cap \overline{DM}$. We observe that $QH$ is tangent to $(AH)$ (since the center of $(AH)$ lies on the segment $AH$). It is well known that $MD$ and $ME$ are tangent to $(AH)$. Thus, \[-1 = (DE;HH') \overset{H}{=}(DR;QM)\] Now, let $X_C = \overline{EF} \cap \overline{AB}$ and $Q'= \overline{BP} \cap \overline{DM}$. Then, \[-1=(AH;PF) \overset{B}{=}(DR;Q'M)\] But this means that $Q'=Q$ and indeed $B-Q-P$ as desired.

Has anybody done it by trig bash?

By the Three Tangents Lemma, $ME$, $MD$ are both tangents to $(ADHE)$. Then $HQ$ is also a tangent since $\angle QHA = \angle HEA$. So Pascal's on $DDAHHE$ finishes.

cute I claim that the three points lie on the polar of $B$ with respect to $(AH)$ under renamed points. This is since $B$ is the pole on line $AC$ by Brokard Since $MF$ and the line through $A$ parallel to $BC$ are tangents to $(AH)$, we find $Q$ is the pole of line $AFB$.

Note that $D$ and $E$ are feet of the altitudes from $B$ and $C$ respectively. It is well-known that QH and MD are tangent to $(AEHF)$. Pascal on $DDAHHE$ yields the desired result.

Let $T=AB \cap HQ$ Since $HT$ is parallel to $BC$ $\triangle DQT \sim \triangle DMB \implies \frac{DQ}{DM}=\frac{QT}{MB}$ And $\triangle DHQ \sim \triangle DCM \implies \frac{DQ}{DM}=\frac{HQ}{MC}$ Hence $\frac{HQ}{MC}=\frac{QT}{MB}$ and since $M$ is the midpoint of $BC$ we get $HQ=QT$ $\star$ Denote $\angle CAB=\alpha$ $\angle ABC =\beta$ and $\angle ACB=\gamma$ By Law of sine In triangle $PHD$ we get $\frac{\sin(\angle PDH)}{PH}=\frac{\sin(\angle HPD)}{HD}$ It is well Known that $EDBC$ is cyclic hence $\angle PEC=180-\beta$ Thus $180-\angle HPD =\angle EPH =360-(\angle PEC + \gamma +90 )= 360-(180-\beta +90+\gamma)$ $=90+\beta -\gamma$ Thus $\sin(\angle HPD)=\sin(90+\beta-\gamma)$ And also we have $90-\gamma=\angle EBC =\angle EDC=\angle PDH$ so $\sin(\angle PDH)=\sin(90-\gamma)=\cos(\gamma)$ Hence $PH=\frac{HD\cos(\gamma)}{\sin(90+\beta+\gamma)}$ Similarly in triangle $APD$ by law of sine we get $\frac{\sin(\angle APD)}{AD}=\frac{\sin(\angle ADP)}{PA}$ And we have that $\angle APD =\angle EPH=90+\beta-\gamma$ (got it before) And also by cyclic quad $EDBC$ $\gamma =\angle EDA=\angle PDA$ Hence $PA=\frac{AD\sin(\gamma)}{\sin(90+\beta-\gamma)}$ $\frac{PH}{PA}=\frac{HD\cot(\gamma)}{AD}$ and since $\frac{HD}{AD}=\cot(\beta)$ we get $\frac{PH}{PA}=\cot(\beta)\cot(\gamma)$ $\star$ We know that $BT=AB-AT$ In right Triangle $AHT$ $\frac{AH}{AT}=\sin(\angle ATH)=\sin(\beta)$ hence $AT=\frac{AH}{\sin(\beta)}$ So $\frac{BT}{BA}=1-\frac{AH}{AB\sin(\beta)}$ And by Law of sine in triangle $AHB$ $\frac{\sin(\angle HBA)}{AH}=\frac{\sin(AHB)}{AB} \implies \frac{AH}{AB}=\frac{\sin(\angle HBA)}{\sin(AHB)}$ by angle chasing we get that $\angle AHB =180-\gamma$ and $\angle HBA=90-\alpha$ Hence $\frac{AH}{AB}=\frac{\cos(\alpha)}{\sin(\gamma)}$ So $\frac{BT}{BA}=1-\frac{\cos(\alpha)}{\sin(\beta)\sin(\gamma)}$ $=\frac{\sin(\beta)\sin(\gamma)-cos(180-(\beta+\gamma))}{\sin(\beta)\sin(\gamma)}$ $=\frac{\sin(\beta)\sin(\gamma)+cos(\beta)\cos(\gamma)-\sin(\beta)\sin(\gamma)}{\sin(\beta)\sin(\gamma)}$ $=\cot(\beta)\cot(\gamma)$ Thus $\frac{BA}{BT}=\tan(\beta)\tan(\gamma)$ $\star$ Finally Using the 3 $\star$ By Menelaus Theorem $\frac{PH}{PA}$×$\frac{BA}{BT}$×$\frac{QT}{QH}$$=\cot(\beta)\cot(\gamma)$×$\tan(\beta)\tan(\gamma)$×$\frac{QT}{QT}=1$ $\iff$ $P,Q,B$ collinear

Cute yet easy, here is a solution that doesn't involve projective geo: Let $O_1$ be the midpoint of $AH$ and let $E$, $D$ be the feet of the altitudes from $B, C$ respectively. Define $Q'$ to be the intersection of $BP$ with the line through $O_1$, parallel to $AB$ and let $R'$ be the intersection of $PC$ with the line through $O_1$, parallel to $AC$ (in other words, let $Q', R'$ be the images of $B, C$ of a homothety centered at $P$ that sends $A$ to $O_1$. It is well-known (and follows directly from PoP) that $\frac{PO_1}{PA} = \frac{PH}{PH_a}$, so $H_a$ is sent to $H$ with this homothety. Hence $Q',R',H$ are collinear and $Q'R'\perp AH_a$ and $Q'R'\parallel BC$. Note that by trivial angle-chase $\angle O_1DH = \beta = \angle ABC = \angle O_1Q'R'$, so $DO_1HQ'$ is cyclic, so $\angle Q'DO_1 = 90^{\circ}$. However, as mentioned in pervious posts and also by angle-chase, $\angle MDO_1 = 90^{\circ}$, so $D, Q', M$ are collinear, hence $Q'\equiv Q$. $\square$

By Brokard's, we see that $PB$ is the polar of $C$ with respect to $(AH)$ so if $Q$ lies on this, La Hire's Implies that $C$ lies on the polar of $Q$. Now it is well known that $MD$ is tangent to $(AH)$ and obviously $QH$ is tangent to $(AH)$ so this implies that $Q$ is pole of $DH$ or that $DH$ is polar of $Q$ but $C$ obviously lies on $DH$.

Let $F=AH\cap BC$, $G=DM\cap BH$ and $Q'=BP\cap DM$. Then\begin{align*}\{G,D;M,Q\} & \underset{H}{=}\{B,C;M,\infty \} \\ & =-1 \\ & =\{H,A;F,P\} \\ & \underset{B}{=}\{G,D;M,Q'\}, \end{align*}so $Q=Q'$, and therefore $P,Q,B$ are collinear.