It is well-known that's $PA \parallel BC$. And if the second intersection of $HM$ and $\omega$ is $T$ then $TM=MH$. By easy angle chasing (or because $HD = DK$ implies that $K$ lies on $\omega$ and $\angle AKT = \angle HDM = 90^o$) we have that $AT$ is a diameter of $\omega$. So $\angle MQT = \angle DAM$(since $AD$ and $QT$ are simmetric about the perpendicular bisector of $BC$) and $\angle APM = 90^o = \angle ADM$ implies $APDM$ cyclic and $\angle DPM = \angle DAM$ so we are done.

Let $S$ $\equiv$ $QM$ $\cap$ $\omega$; $I$ $\equiv$ $AP$ $\cap$ $BC$ We have: $\dfrac{BS}{CS} = \dfrac{BM}{CM} . \dfrac{QC}{QB} = \dfrac{AB}{AC}$ Then: $\dfrac{BP}{CP} = \dfrac{IB}{IC} . \dfrac{AC}{AB} = \dfrac{BD}{CD} . \dfrac{CS}{BS}$ or $D$, $P$, $S$ are collinear So: intersection of $DP$ and $QM$ lies on $\omega$

BMO Shortlist 2017 G5 wrote: Let $ABC$ be an acute angled triangle with orthocenter $H$. centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $AH$ and $AG$ with side $BC$. Rays $MH$ and $DG$ interect $ \omega$ again at $P$ and $Q$ respectively. Prove that $PD$ and $QM$ intersect on $\omega$. Notice that a Homothety $\mathcal H$ at $G$ with a scale factor of $-\frac{1}{2}$ sends $\odot(X_5)$ to $\odot(ABC)$, so $\mathcal H:D\leftrightarrow Q$. Hence, we get that $QA\|BC$. Now introduce the orthic triangle $DEF$ of $\triangle ABC$. Note that $P$ is the $A-\text{Queue Point}$ of $\triangle ABC$, hence, $\angle APH=90^\circ$. Now Radical Axis Theorem on $\odot(ABC),\odot(BC)$ and $\odot(AH)$ we get that $AP,EF,BC$ are concurrent at the $A-\text{Expoint} (X_A)$ of $\triangle ABC$. Let $PD\cap\odot(ABC)=T$ and $QM\cap\odot(ABC)=T'$. Now notice that $$-1=(X_AD;BC)\overset{P}{=}(AT;BC)\implies ABTC\text{ is a Harmonic Quadrilateral.}$$Similarly we get that $$-1=(BC;M\infty_{BC})\overset{Q}{=}(BC;T'A)\implies ABT'C\text{ is a Harmonic Quadrilateral.}$$So from this we can conclude that $\boxed{T'\equiv T}$. Hence, $PD\cap QM\in\odot(ABC). \blacksquare$

It is well - known that $P$ is $A$ Queue point and that $AQ \parallel BC$. Assume that $F$ intersects $\odot(ABC)$ at point $F$. Claim: $F$ is intersection of $A$ symmediaa and $\odot(ABC)$ Proof: Consider $X = PA \cap BC$. It is well - known that $X$ is $A$ Ex point. Therefore: $$ -1 = (X,D;B,C) \overset{P}{=} (A,F;B,C) $$This proves our claim: Now assume that $QM \cap \odot(ABC) = F'$. Note that: $$ -1 = (\infty_{BC},M;B,C) \overset {Q}{=} (A,F';B,C) $$This proves that $F = F'$. As a result that lines $PD \cap QM$ lies on $\odot(ABC)$ as desired.

We use the well known facts that $AQ \parallel BC$ and $HM$ passes through the antipode of $A$ on $\omega$, which we denote by $A'$. From here, we use complex numbers with $\omega$ as the unit circle. As usual, let lowercase letters denote the complex coordindates of their respective uppercase letter. Since $AQ \parallel BC$, we have $q = \frac{a}{bc}$. From the midpoint formula we have $m = \frac{b+c}{2}$. Since $D$ is the foot from $A$ to $BC$, \[ d = \frac{1}{2} \left(a + b + c - \frac{bc}{a} \right) \]Because $M$ lies on chord $PA'$, we have that \[ p = \frac{m + a}{1 + a \overline{m}} = \frac{ bc(2a + b +c)} {2bc + a(b+c)} \] Let $X$ be the second intersection of $QM$ with $\omega$. Then $X$ is on chord $QM$, so \[ x = \frac{m + q}{1 + q \overline{m}} = \frac{a(b+c) + 2bc}{a - b - c} .\] We want to check that $D$ is on chord $PX$. This is equivalent to \[ p + x = d + px \overline{d} \]Checking this is a straightforward computation.

We consider a projective transformation fixing $\omega$ and sending $M$ to the center of $\omega$. Now the problem is much simplified. Let $\omega$ be a circle with diameter $BC$ and $A$ be a point on it.Let $D$ be the feet of $A$ on $BC$ and $G$ be the centroid of $A,B,C$ . Let $Q$ be the intersection of $DG$ and $\omega$.Prove that $AD$ and $QM$ intersect on $\omega$. Easily,$AQBC$ is an isosceles trap. Now, $AD \cap \omega$ be $E$,thus $\angle EAQ=90$ and so $EQ$ is a diameter and we are done. (Just wanted to add that the points P and Q are well known and a complex bash would finish it but this is just shorter.)

Let $PD$ meet $\omega$ again at $K$. It's well-known, i.e. by 2011 G4, that $AQ \parallel BC$. Thus, symmetry and isogonality implies $QM$ and the $A$-Symmedian meet on $\omega$. In addition, the Orthocenter Reflection Lemma yields $P$ as the $A$-Queue point. Let $BH$ meet $CA$ at $E$ and $CH$ meet $AB$ at $F$. By Radical Axes, we know $AP, BC, EF$ are concurrent at some point $T$, so Ceva-Menelaus gives $$-1 = (T, D; B, C) \overset{P}{=} (A, K; B, C)$$which clearly finishes. $\blacksquare$ Remark: APMO 2012/4 directly implies $-1 = (A, K; B, C)$, but I included a proof for the sake of completeness.

Claim1 : $AQ || BC$. Proof : Let $K$ be a point on $\omega$ such that $AK || BC$. we have $AK = 2DM$,$AG = 2GM$ and $\angle AKG = \angle MDG$ so $AGK$ and $MDG$ are similar so $\angle AGK = \angle MGD$ so $K,G,D$ are collinear so $K$ is $Q$. Claim2 : $APDM$ is cyclic. Proof : It's well-known that reflection of $H$ across $M$ and $D$ lies on $\omega$ so If $S$ is reflection of $H$ across $M$ then $AS$ is diameter of $\omega$ so $\angle APM = \angle APS = \angle 90 = \angle ADM$. we will prove $\angle MQA + \angle DPA = \angle 180$. $\angle MQA = \angle MAQ = \angle AMB$ and $\angle DPA = \angle AMC$ so we're Done.

Wow! I am surprised to not see a really short and simple solution involving the butterfly theorem (I got to know this method by seeing a part of the solution to ISL 2016 G2, after that, this was pretty easy) It is well known (EGMO Lemma in chapter 3) that $AQ \parallel BC$ and let $HM \cap \omega = X$ and let $XQ \omega = Y$ Notice that by orthocenter reflections lemma and reflection symmetry we get $XQ \perp BC$, this means $MD = DY$ now, the butterfly theorem finishes $\blacksquare$ Came here from The problem proposer’s YT channel!

Is it just me or does it seem hella familiar with some problem, maybe i did this one before o wot? Let $Q' = AA \cap \odot ABC$, $X = A-$symmedian $\cap \odot ABC$. It is well known that $AQ \parallel BC$, that $AMXQ'$ is cyclic by the $A-$Apollonius circle wrt $BC$ and that $P$ lies on $\odot AH$. $\measuredangle APM = \measuredangle APH = 90^{\circ} = \measuredangle ADM \implies APDM$ is cyclic. Now perform an Inversion with center $A$ and with radius $\sqrt{AB \cdot AC}$ followed by a reflection wrt the Angle Bisector of $\angle BAC$. \begin{align*} Q \xleftrightarrow{} Q'\\ M \xleftrightarrow{} X\\ \end{align*}Thus $\overline{X - M - Q}$ are collinear. Now, $$\measuredangle APD = \measuredangle AMD = \measuredangle AMC = \measuredangle ABX = \measuredangle APX \implies \overline{P - D - X}\text{ are collinear.}$$Thus $X$ is our desired concurrency point.

BVKRB- wrote: Wow! I am surprised to not see a really short and simple solution involving the butterfly theorem (I got to know this method by seeing a part of the solution to ISL 2016 G2, after that, this was pretty easy) It is well known (EGMO Lemma in chapter 3) that $AQ \parallel BC$ and let $HM \cap \omega = X$ and let $XQ \omega = Y$ Notice that by orthocenter reflections lemma and reflection symmetry we get $XQ \perp BC$, this means $MD = DY$ now, the butterfly theorem finishes $\blacksquare$ Came here from The problem proposer’s YT channel! Sorry but can you tell me which lemma is that? I tried finding in EGMO, but didn't get it

MathsinMaths wrote: BVKRB- wrote: Wow! I am surprised to not see a really short and simple solution involving the butterfly theorem (I got to know this method by seeing a part of the solution to ISL 2016 G2, after that, this was pretty easy) It is well known (EGMO Lemma in chapter 3) that $AQ \parallel BC$ and let $HM \cap \omega = X$ and let $XQ \omega = Y$ Notice that by orthocenter reflections lemma and reflection symmetry we get $XQ \perp BC$, this means $MD = DY$ now, the butterfly theorem finishes $\blacksquare$ Came here from The problem proposer’s YT channel! Sorry but can you tell me which lemma is that? I tried finding in EGMO, but didn't get it Sorry for 1 month late reply, I didn't see the thread It is EGMO Problem 3.24.

Note that $P$ is the $A$-Queue point, so it's well known that if $T = \overline{AP} \cap \overline{BC}$ then $(T, D; B, C) = -1$ (as $T$ is the $A$-Ex point) $G$ is the insimilicenter of the nine-point circle of $\triangle ABC$ and $\omega$, so $\overline{AQ} \parallel \overline{BC}$ Thus, we have \begin{align*} -1 &= (T, D; B, C) \overset{P}{=} (A, \overline{PD} \cap \omega; B, C), \\ -1 &= (M, \infty_{BC}; B, C) \overset{Q}{=} (\overline{QM} \cap \omega, A; B, C), \end{align*}which finishes.

Posting for storage. We know that $M-H-Q_a=P$ and $AQ \parallel BC$. Let $A-symmedian \cap (ABC) $ at $K$ and take pencil from $P$. So, $ABKC$ is harmonic.Also,$(B,C;M,W^{\infty} )=-1$. Take pencil from $Q$, $(QM \cap (ABC), A;B, C) =-1$ Thus, $QM \cap (ABC) \equiv K$ No need for diagram just lemmas

Recall that $P$ is the $A$-Queue point and that $Q$ is the reflection of $A$ across the perpendicular bisector of $BC$. Let $K$ be the point on $\omega$ such that $(B,C;A,K)=-1$. We claim that $P-D-K$ and $Q-M-K$. Let $X$ be the $A$-Ex point. Then $-1=(B,C;X,D) \stackrel{P}{=} (B,C;A,PD \cap \omega)$, so $P-D-K$. $AK$ is the $A$-symmedian, so $\angle BQK = \angle BAK = \angle CAM = \angle BQM$, which implies that $Q-M-K$. Hence proved.

It is well known that $Q$ is the point on $(ABC)$ such that $AQ|| BC$ so $(B,C;M,P_\infty)=-1$. Projecting this through $Q$ onto $(ABC)$ gives us that $QM$ meets the circle at a point $X$ such that $ABXC$ is harmonic. Let $AP \cap BC= T$ , which is well known to be $A$ expoint. Then the fact that $(B,C;D,T)=-1$ is also well known. Projecting this bundle onto the circumcircle, if $PD \cap (ABC) =X'$, then $ABX'C$ is also harmonic. Hence, $X \equiv X'$ which is what we needed to show.

Let $S$ be second intersection of $PD$ with $(ABC)$ and $J \equiv AP \cap BC$. We have $\dfrac{PB}{PC} \cdot \dfrac{SB}{SC} = \dfrac{DB}{DC} = \dfrac{JB}{JC} = \dfrac{PB}{PC} \cdot \dfrac{AB}{AC}$. Then $\dfrac{AB}{AC} = \dfrac{SB}{SC}$. But $G$ is center of homothety that transforms NPC of $\triangle ABC$ to $(ABC)$ so it's easy to see that $AQ \parallel BC$. Hence $\dfrac{AB}{AC} = \dfrac{QC}{QB},$ so $\dfrac{SB}{SC} \cdot \dfrac{QB}{QC} = 1 = \dfrac{MB}{MC}$ or $Q, M, S$ are collinear