Solution with skipiano and wu2481632 Let $D,E,F$ be the feet of the altitudes to $BC, AC, AB$ respectively. Let $EF$ and $BC$ intersect at $T$. Lemma 1. $H$ is the orthocenter of $ATM.$ Proof. This is the result of a USA TST problem; clearly, $M$ is the circumcenter of $(BFEC),$ so by Brokard, $M$ is the orthocenter of $ATH$ and $MH\perp AT$. Since also $AD\perp TM,$ $H$ is the orthocenter of $ATM$ as desired. Lemma 2. Triangles $HYM$ and $AQM$ are similar. Proof. This is bashable with a variety of techniques (coordinate, trig, bary). We spent a very long time looking for a synthetic solution but we couldn't find it Main problem. By Lemma 1, we know that $TH\perp AM.$ Thus, using phantom points, the condition we're asked to prove is equivalent to showing that $J,$ the intersection of $TH$ and $QM,$ is the projection of $Y$ onto $QM.$ In other words, it suffices to show that $HYJM$ is cyclic. Since $AQ\perp AM,$ we have $AQ\parallel TH,$ from which we deduce that $\angle AQM=\angle HJM.$ But by Lemma 2, $\angle HYM=\angle AQM,$ so $\angle HYM=\angle HJM,$ implying the desired cyclicy.

Any synthetic proof for Lemma 2 above?

Here is a synthetic proof of Lemma 2 from post #2 . By simple angle chasing we can get $\angle HAY = \angle MBH$ and $\angle AHY = \angle BMH$, hence $\triangle AHY \sim \triangle BMH$. Similarly, we have $\angle AHQ = \angle MCA$ and $\angle AQH = \angle MAC$, so $\triangle AQH \sim \triangle MAC$. From these similarities we obtain $\frac{HY}{HM} = \frac{AH}{BM}$ and $\frac{AQ}{AM} = \frac{AH}{MC}$. Since $BM = MC$, it follows that $\frac{HY}{HM} = \frac{AQ}{AM}$, which combined with the fact that $\angle MAQ = \angle MHY = 90^\circ$ implies $\triangle AQM \sim \triangle HYM$.

Let $E$ be the foot from $B$ to $AC$. The key claim is Claim : $\Delta AQM\sim\Delta HYM$. Proof : Let $U, V$ be midpoints of $CE, BH$. Since $\angle QAM=90^{\circ}$, $AM$ is tangent to $\odot(QAE)$ which implies $\measuredangle AQE=\measuredangle UAM$ therefore $\Delta AUM\sim \Delta QEA$. Proceeding similarly on other side, we get that $\measuredangle HYE=\measuredangle VHM$. Therefore $\Delta HVM\sim\Delta YEH$. Since $\Delta MUV$ is the midpoint triangle of $\Delta EBC$ and $\angle EBC=\angle EAH$, therefore $\Delta MUV\sim\Delta EBC\sim\Delta EAH$. Therefore $$\frac{AQ}{AM}=\frac{AE}{MU} =\frac{HE}{MV}=\frac{HY}{HM}$$implying the desired similarity. Back to the main problem, note that $\measuredangle HJM=\measuredangle HYM = \measuredangle AQM$. Therefore $JH\parallel AQ\perp AM$ as desired.

This is a very beautiful problem! I'll provide one more synthetic solution of the fact $\triangle AQM\sim\triangle HYM$: Let $HY\cap AQ=L$. Note that it is enough to prove that $\angle{HMY}=\angle{AMQ}$. Let $HY$ intersect $AB$ at $X$, by butterfly we have that $\triangle{MHX} \equiv \triangle{MHY}$, thus it is enough to show that $\angle{XMH}=\angle{AMQ}$. Now applying Isogonality Lemma on $\{Q,X\}$ wrt $\angle{AMH}$, we have that it is enough to show that $\angle{AML}=\angle{BMH}$, but because $ALMH$ is cyclic, we have the following relations: $$\angle{AML}=\angle{AHL}=90^{\circ}-\angle{MAH}-\angle{ALH}=90^{\circ}-\angle{MAH}-\angle{AMH}=\angle{BMH} \ \ \ \blacksquare$$

It's pretty tricky with synthetic techniques. I'll post the trig solution which many posters rightly claim to be feasible. Also, @above, nice solution! Ferid.---. wrote: Let $ABC$ be an acute triangle. Denote by $H$ and $M$ the orthocenter of $ABC$ and the midpoint of side $BC,$ respectively. Let $Y$ be a point on $AC$ such that $YH$ is perpendicular to $MH$ and let $Q$ be a point on $BH$ such that $QA$ is perpendicular to $AM.$ Let $J$ be the second point of intersection of $MQ$ and the circle with diameter $MY.$ Prove that $HJ$ is perpendicular to $AM.$ (Steve Dinh) Clearly, we want $HJ \parallel AQ \iff \angle MYH=\angle MJH=\angle MQA \iff \triangle MHY \sim \triangle MAQ$. Equivalently, we'll show that $\tan \angle MYH=\tan \angle MQA$. Observe that $$\frac{AQ}{\cos A}=\frac{AB}{\sin CAM}\iff AQ=(2R\cos A) \left(\frac{\sin ACM}{\sin CAM}\right)=\frac{2AM \cdot AH}{BC}.$$Hence $\tfrac{AQ}{AM}=2\cot A$. Let $BB_1$ be the altitude in $\triangle ABC$ and $N$ be the midpoint of $CB_1$. Then $HMNY$ is cyclic, so $$\frac{HY}{HM}=\tan \angle B_1NH=\frac{2HB_1}{CB_1}=2\cot A$$proving $\tan \angle MYH=\tfrac{1}{2}\tan A=\tan \angle MQA$ as desired. $\blacksquare$

Possibly insightful solution that looks at the bigger side of what's going on. Let $E,F$ be the feet of $B,C$ on the opposite sides. $X=YH \cap AB$, $P=AQ\cap CH$. From butterfly theorem on $BCEF$ twice, once with $A$ and then with $H$, we have $HXM \cong HYM$ and $APM \cong AQM$. Letting the midpoints of $BF,CE,BE,CF$ to be $R,S,T,U$ we have that since $ABE \cup \{ T \} \sim HBF \cup \{ R \}$ so $\angle MQA = 90^\circ - \angle AMQ 90^\circ - \angle ATQ = 90^\circ - \angle ATE = 90^\circ -\angle HRF = \angle HRM = \angle HXM = \angle HYM$, so $HJ \parallel AQ$ so $HJ \perp AM$. $\blacksquare$ NB: The angle chase setup we used can also be used to give another proof of the Butterfly theorem.

Clearly, \(\Delta AQH \sim \Delta MAC \Rightarrow \dfrac{AH}{MC} = \dfrac{AQ}{AM} -(i)\) Let \(YH \cap AB = Z\), by butterfly theorem, \(HY = HZ\), also we have: \(\Delta AHZ \sim \Delta CMH \Rightarrow \dfrac{AH}{CM} = \dfrac{HZ}{HM} =\dfrac{HY}{HM} - (ii) \) From \((i),(ii): \dfrac{AQ}{AM} =\dfrac{HY}{CM}\) also, \(\angle MAQ = \angle MHY =90^{\circ}\) \(\Rightarrow \Delta MAQ \sim \Delta MHY \Rightarrow \angle AQM = \angle HYM = \angle HJM \Rightarrow HJ || AM \Longrightarrow HJ \perp AM.\) I see @above now, essentially the same.

Here’s my solution (read sketch) which I guess is not yet posted Let $X,Z$ be the reflection of $A,H$ over $M$. Angle chasing yields $HYZC$ and $ABQX$ to be cyclic. Further angle chasing yields $\angle{YZH} = \angle{QAX}$. Also, we know that $\angle{YHZ} = \angle{PAX} = 90^{\circ}$ So, $\Delta AXQ \sim \Delta HRY$ It is not difficult to see that $\Delta AQM \sim \Delta HYM$. Note that the above condition is equivalent to $HJ \parallel AQ$ Thus, $AM \perp HJ$. $\blacksquare$

RC. wrote: Clearly, \(\Delta AQH \sim \Delta MAC \Rightarrow \dfrac{AH}{MC} = \dfrac{AQ}{AM} -(i)\) Let \(YH \cap AB = Z\), by butterfly theorem, \(HY = HZ\), also we have: \(\Delta AHZ \sim \Delta CMH \Rightarrow \dfrac{AH}{CM} = \dfrac{HZ}{HM} =\dfrac{HY}{CM} - (ii) \) From \((i),(ii): \dfrac{AQ}{AM} =\dfrac{HY}{CM}\) also, \(\angle MAQ = \angle MHY =90^{\circ}\) \(\Rightarrow \Delta MAQ \sim \Delta MHY \Rightarrow \angle AQM = \angle HYM = \angle HJM \Rightarrow HJ || AM \Longrightarrow HJ \perp AM.\) I see @above now, essentially the same. excuse me, may I ask to which cyclic quadrilateral do you apply Butterfly theorem?

@above. To a circle with diameter $BC$. Also there's a typo: $\dfrac{HY}{CM}$ needs to be $\dfrac{HY}{HM}$

First of all let's $D$ and $E$ be foot of perpendicular from $A$ and $B$,respectively. $HJ\perp AM$ $\iff$ $HJ||AQ$ $\iff$ $\angle AQM=\angle HJM$ $\iff$ $\angle AQM=\angle HYM$ $\iff$ $\triangle AQM\sim \triangle HYM$ $\iff$ $\frac{AQ}{AM}=\frac{HY}{HM}$. Now from angle chasing you can easily obsserve that $\angle HAY=\angle HBM$ and $\angle AYH=\angle BHM$. So $\triangle AYH\sim \triangle BHM$. So $\frac{HY}{HM}=\frac{AH}{BM}$. Since $\angle AHQ=\angle ACM$ and $\angle AQH=\angle MAC$ w eget $\triangle AHQ\sim \triangle MCA$. So $\frac{AQ}{AM}=\frac{AH}{MC}=\frac{AH}{BM}$ $\implies$ $\frac{AQ}{AM}=\frac{AH}{BM}=\frac{HY}{HM}$. So we are done!

A problem filled with similarities ... Let MH meet AQ at S. we have to prove HJ || AQ or in fact ∠MHJ = ∠MSQ. Let's go and prove some similarities first'. step1 : HAY and MBH are similar. ∠HAY = ∠MBH and ∠AYH = 90 - ∠YHQ = ∠QHS = ∠BHM. step2 : AQH and MAC are similar. ∠AHQ = ∠ACM and ∠AMC = 180 - ∠AMB = 180 - ∠HAS = ∠HAQ. step3 : MHY and MAQ are similar. ∠MHY = 90 = ∠MAQ and MH/HY = MB/AH = MC/AH = AM/AQ. Now that we have MHY and MAQ are similar we have AHM and QYM are similar as well. now let's get back to our angle chasing. ∠MHJ = 90 - ∠JHY = 90 - ∠JMY = 90 - ∠AMH = ∠MSA so HJ || AQ. we're Done.