Midpoint = 10 points. Take the midpoint $K$ of $AB$ and justify by quick angle chasing that $MSKH$ is cyclic. Thus $\angle MSH = \angle MKH = 180^{\circ} - \angle KHD = \angle ADB = 90^{\circ}$, so $SM \perp ST$; analogously $TN \perp ST$ and we are done.

Wow, here is an absolutely slick solution without any additional points! (Big gratitude to the Bosnian Deputy Leader!) Angle chasing shows that $\triangle AB_1D \sim \triangle HTC$ and since $TH$ and $TN$ are corresponding medians, we get $\angle HTN = \angle AB_1H = 90^{\circ}$. So $TN \perp HT$, analogously $MS \perp HT$, happy ending!

I present another solution which doesn't require introducing the midpoints. I will prove that $TN$ is perpendicular to $ST$, which is obviously sufficient. Since $\angle THC= \angle HBC$, we want triangles $HTN$ and $HBD$ to be similar, which is equivalent to $\frac {HB} {BD}=\frac {HN}{HT}= \frac {CH} {2HT}$. So we want $\frac {CH} {HT}=\frac {2BH}{BD}$. Note that due to the tangency, the former is $\frac {CA} {AH}$, so we want $CA.BD=2BH.AH=AD.BH \iff \frac {AD} {BD} =\frac {CA} {BH}$ but $BH=2Rcos(\beta)$, $AC=2Rsin(\beta)$ and the former is $tg(\beta)$ so we are done.

wrong post

Proposed by Jason Prodromidis, Greece

Pretty nice

Let $O,P$ be the circumcenter of $\triangle ABC$ and the midpoint of $BC$, respectively. Let $AO$ meet $ST$ at $Q$. Then $\displaystyle OP=\frac{1}{2}AH=\frac{1}{2}HD$. This implies $M,O,N$ are collinear and $PO\perp MN$. It’s easy to show that $\square BSTC$ is cyclic by angles chasing. Hence, $\triangle ATS \sim \triangle ABC$ and, consequently, $AO\perp ST$. Moreover $\triangle MPN\sim \triangle CHB$. Hence, $\displaystyle \frac{SQ}{QT}=\frac{CD}{DB}=\frac{MO}{ON}$, i.e. $SM\parallel TN$ as desired.

By trivial angle chase we get that both of the lines are perpendicular to ST thus they are parallel to each other.

From the given condition $AH=HD$ we have $2R \cos A=2R \cos B \cos C$ $$\cos (A)=\cos (\pi-B-C)=-\cos (B+C)=\cos B \cos C \implies 2\cos B \cos C=2 \cos A= \sin B \sin C$$ Since $SH$ is tangent to $(BHC)$; $\angle SHB= \angle HCB=90-B$ Now $\angle ASH=\angle SHB+\angle SBH=C$. Symmetrically $\angle ATH=B$ Let $E= AH \cap MN$ Claim: $A,S,E,T$ are concylic Proof: By sine rule $SH=\frac{2R \cos A \cos B}{\sin C}$ and $TH=\frac{2R \cos A \cos C}{\sin B}$ Clearly $SH \cdot TH=\frac{4R^2 \cos^2 A \cos B \cos C}{\sin B \sin C}=(2R^2 \cos A \cos B \cos C ) \times \left(\frac{2 \cos A}{\sin B \sin C}\right)=2R^2 \cos A \cos B \cos C =AH \cdot HE$ From this, we have $\angle HSE=\angle EAT=90-C \implies ES \perp AB$ Note that $MN || BC \implies AH \perp MN$ $$\angle SEM=90-\angle SEH=90-\angle ATS=90-B=\angle SHB \implies M \in \odot(SHE)$$ Hence $\angle MSH=90 \implies SM \perp ST$. Due to symmetry $TN \perp ST$

Define the midpoints of $AC, AB$ as $X,Y,$ respectively. Observe that \begin{align*}\angle{THC} = \angle{HPC} = 90 - c = \angle{HAC},\end{align*}so $\triangle{THC} \sim \triangle{HAC}.$ As a result, $\triangle{THN} \sim \triangle{HAX},$ since $N$ and $X$ are both midpoints of respective sides. So, $\angle{HTN} = \angle{AHX},$ and analogously $\angle{MSH} = \angle{AHY}.$ Then since $AH = HD,$ it follows that $X,H,Y$ are collinear, since a homothety of factor 2 about $A$ sends all $X,H,Y$ onto $BC.$ Hence, $\angle{AHX} + \angle{AHY} = 180,$ equivalently, \begin{align*}\angle{HTN} + \angle{MSH} &= 180 \\ \Leftrightarrow \angle{STN} + \angle{MST} &= 180, \end{align*}which implies the desired.

Easy question for g3

I enjoyed solving this problem, it was a fun one! 2022 JBMO Problem 2 wrote: Let $ABC$ be an acute triangle such that $AH = HD$, where $H$ is the orthocenter of $ABC$ and $D \in BC$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $BHC$. Let $S$ and $T$ be the intersection points of $\ell$ with $AB$ and $AC$, respectively. Denote the midpoints of $BH$ and $CH$ by $M$ and $N$, respectively. Prove that the lines $SM$ and $TN$ are parallel. Let $M_1$, and $M_2$ be midpoints of $AB$, $AC$ respectively. Taking a homothety centered at $A$ with factor $2\implies \overline{M_1-H-M_2}$. $$\angle SHB=\angle HBC=\angle HAB \implies \Delta AHB\cup\{M_1\}\sim\Delta HSB\cup\{M\}\implies \angle M_1HA=\angle MSH.$$Similarly, we can get that $\angle AHM_2=\angle NTH$. Now, $$SM\parallel TN\iff \angle MSH+\angle NTH=180^{\circ}\iff \angle M_1HA+\angle AHM_2=180^{\circ}.\blacksquare$$[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.713333333333319, xmax = 9.62, ymin = -5.106666666666666, ymax = 6.133333333333336; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen xfqqff = rgb(0.4980392156862745,0.,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen ffqqff = rgb(1.,0.,1.); /* draw figures */ draw((-3.14,4.48)--(-3.18,-2.3466666666666653), linewidth(0.8)); draw((-3.14,4.48)--(-6.280159884151869,-2.328501667345463), linewidth(0.8)); draw((-3.14,4.48)--(4.336286178660645,-2.3907074059947546), linewidth(0.8)); draw((4.336286178660645,-2.3907074059947546)--(-6.280159884151869,-2.328501667345463), linewidth(0.8)); draw(circle((-0.981936852745612,-4.0662712033367745), 5.5759313348303685), linewidth(0.8) + qqwuqq); draw((-5.092464197484706,0.24666277054954366)--(-0.6053697952168444,2.1506747035489453), linewidth(0.8) + red); draw((-3.16,1.0666666666666675)--(-6.280159884151869,-2.328501667345463), linewidth(0.8) + xfqqff); draw((-3.16,1.0666666666666675)--(4.336286178660645,-2.3907074059947546), linewidth(0.8) + xfqqff); draw((-4.710079942075935,1.0757491663272687)--(0.5981430893303223,1.044646297002623), linewidth(0.8) + ffxfqq); draw((-5.092464197484706,0.24666277054954366)--(-4.720079942075935,-0.6309175003393978), linewidth(0.8) + ffqqff); draw((-0.6053697952168444,2.1506747035489453)--(0.5881430893303223,-0.6620203696640435), linewidth(0.8) + ffqqff); /* dots and labels */ dot((-3.14,4.48),dotstyle); label("$A$", (-3.0866666666666496,4.613333333333337), NE * labelscalefactor); dot((-3.18,-2.3466666666666653),dotstyle); label("$D$", (-3.1266666666666496,-2.2133333333333316), NE * labelscalefactor); dot((-3.16,1.0666666666666675),linewidth(4.pt) + dotstyle); label("$H$", (-3.0466666666666495,1.2133333333333358), NE * labelscalefactor); dot((-6.280159884151869,-2.328501667345463),dotstyle); label("$B$", (-6.713333333333317,-2.8133333333333317), NE * labelscalefactor); dot((4.336286178660645,-2.3907074059947546),linewidth(4.pt) + dotstyle); label("$C$", (4.393333333333353,-2.28), NE * labelscalefactor); dot((-5.092464197484706,0.24666277054954366),linewidth(4.pt) + dotstyle); label("$S$", (-5.54,0.17333333333333564), NE * labelscalefactor); dot((-0.6053697952168444,2.1506747035489453),linewidth(4.pt) + dotstyle); label("$T$", (-0.5533333333333154,2.253333333333336), NE * labelscalefactor); dot((-4.710079942075935,1.0757491663272687),linewidth(4.pt) + dotstyle); label("$M_1$", (-5.4,1.2266666666666692), NE * labelscalefactor); dot((0.5981430893303223,1.044646297002623),linewidth(4.pt) + dotstyle); label("$M_2$", (0.6733333333333518,1.2133333333333358), NE * labelscalefactor); dot((-4.720079942075935,-0.6309175003393978),linewidth(4.pt) + dotstyle); label("$M$", (-4.713333333333317,-1.), NE * labelscalefactor); dot((0.5881430893303223,-0.6620203696640435),linewidth(4.pt) + dotstyle); label("$N$", (0,-1.0533333333333312), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

geometry6 wrote: $ \Delta AHB\cup\{M_1\}\sim\Delta HSB\cup\{M\}\implies \angle M_1HA=\angle MSH.$ What does this mean?

Let the midpoints of $AB$ and $AC$ be $P$ and $Q$ respectively. Because $AH = HD$, we know $H, P, Q$ are collinear. Now, observe that $$\angle BHS = \angle BCH = 90^{\circ} - \angle B = \angle BAH$$which clearly implies $BHS \sim BAH$. Similarly, we deduce $CHT \sim CAH$. Because $M$ is the midpoint of $BH$ and $P$ is the midpoint of $BA$, we know $M$ and $P$ are corresponding points of $BHS$ and $BAH$, yielding $BHSM \sim BAHP$. Now, since $MN \parallel BC$ follows from midlines, $$\angle SMN = \angle SMH + \angle HMN = \angle HPA + \angle HBC = \angle APQ + \angle HBC.$$An analogous process gives $\angle TNM = \angle AQP + \angle HCB$, so $$\angle SMN + \angle TNM = (\angle APQ + \angle HBC) + (\angle AQP + \angle HCB)$$$$= (\angle APQ + \angle AQP) + (\angle HBC + \angle HCB)$$$$= (180^{\circ} - \angle A) + (180^{\circ} - \angle BHC)$$$$= (180^{\circ} - \angle A) + (180^{\circ} - (180^{\circ} - \angle A)) = 180^{\circ}$$as required. $\blacksquare$ Remarks: Unlike most people, I used $MN$ as my transversal, not $ST$. This leads to a more elaborate angle chase, but it's still manageable.

Coordinate bash 1. Take D as Origin, $A(0,1), B(b,0), C(c,0) \implies H = (0,0.5) $ 2. H is also $(0, -bc) \implies bc = -0.5 $ 3. reflection of A in H lies on circumcircle of BHC $ \implies $ the y coordinate of circumcenter of BHC {O} is $ \frac{-1+0.5}{2} = \frac{-1}{4} $ 4. The x coordinate of circumcenter of BHC {O} is $ \frac{b+c}{2} $ 5. The slope of HT can be calculated as OH is perpendicular to HT. It comes out to be $\frac{2(b+c)}{3}$. 6. The coordinate T comes out to be $( x, 1-\frac{x}{c} )$ where $x = \frac{3}{4(b+c) + \frac{6}{c}}$. N is $(\frac{c}{2}, 0.25)$ 7. The slope TN after using $bc = -0.5$ comes out to be $\frac{-3}{2b + 2c}$, which is symmetric in b,c Hence TN || SM, also TN is perpendicular to ST

Let $BH,CH$ intersect sides $AC,AB$ at points $B',C'$ respectively.So we have $\displaystyle \frac{SH}{HM*2}=\frac{AH}{AB}= \frac{AD}{AB*2}=\frac{HB'}{HT*2}$ $\rightarrow$ $\displaystyle \frac{SH}{HM}=\frac{HB'}{HT}$ . So $SM \perp ST$ .Similarly $NT \perp ST$ $\rightarrow$ $SM||NT$.

$$\triangle {BSH}\sim \triangle{BHA}\Longrightarrow \frac{SH}{AH}=\frac{BH}{AB},\triangle {CTH}\sim\triangle {CHA}\Longrightarrow \frac{TH}{AH}=\frac{CH}{AC}$$Thus $\frac{SH\cdot TH}{AH^2}=\frac{BH\cdot HC}{AB\cdot AC}$ Its clear that $AREA(\triangle {ABC})=2AREA(\triangle BCH)$. Thus by sine area we get Last ratio is equals to $1/2$ So we get $AH^2=2TH\cdot SH=BH\cdot HE=2HM\cdot HE$ thus $HM\cdot HE=TH\cdot SH$ So $TSME$ is concylic thus $\angle{TSM}=90^\circ$ Similarly $\angle{STN}=90^\circ$ so $SM||TN$

$Q\in AB \to BQ=AQ$ $QSMH$ cyclic $\angle{MSH}=\angle{MQH}=180-\angle{QHD}=90$ $SM \perp ST$ and $TN \perp ST$ done