Hint: Let $P\in\omega|AP\parallel BC$ and $Q$ the projection of $P$ onto $BC$ ($HAPQ$ is a rectangle) and $N$ midpoint of $BC$. $N$ is also midpoint of $HQ$, thus $H-M-P$ are collinear. Now it is easy to see $\angle BA'P=\angle BCP=\angle ABC$, done. Best regards, sunken rock

This is also India IMOTC 2016, Practice Test 1, Problem 1.

Follows from a homothety at the centroid with ratio -2 and angle chasing.

Dear Mathlinkers, 1. A" the second point of intersection of MH with (O) 2. AA'' is parallel to BC 3. according to theReim's theorem, we are done... Sincerely Jean-Louis

Let $A''$ be point on $BC$ such that $BH=CA''$. Let $A_1$ be midpoint of $BC$, then $A_1$ is also midpoint of $HA''$. As $AM : MA_1=2$ we have that $\Delta ABC$ and $\Delta AHA''$ share centroid $M$. Let $L$ be midpoint of $AA''$. Then obviously $L,H,A''$ are collinear. Let $O$ be circumcenter of $\odot (\Delta ABC)$. We have that $O,L,A_1$ are collinear so $\angle HLO=\angle A''LO$ and $OA'=OY$ and $OL=OL$ thus $\Delta A'LO=\Delta LYO$ thus $LA'=LY$ so $A'Y || BC$ (where $Y=AA'' \cap \omega$). Let $X=\omega \cap A'L$. Then because of symmetry we have $AX||A'Y||BC$ thus $AX||BC$ so $\angle ABC=\pi - \angle BAX=\angle BA'X$ and we are done

Dear Mathlinkers, this problem is also based on http://www.artofproblemsolving.com/community/c6h1161858_two_parallels Sincerely Jean-Louis

Let Ray$ HM$ intersects circumcircle at $A''$ Applying euler circle and a homothety at point M send euler circle to circumcircle .We get $AA'||BC$. So $\angle AA'A''=\angle ACB=\angle ABC$

Just consider a Homothety taking medial triangle to Triangle $ABC$ with $M$ as centre of homothety and Ratio $1/2$

Let $K \in \omega$ such that $AK \parallel BC$. Consider the homothety with ratio $-2$ at $M$. Under this homothety $H \to K$. Thus, $H, M, K$ are collinear. Note that $AKBC$ is an isosceles trapezium. \[\measuredangle BA'H = \measuredangle BA'K = \measuredangle BCK = \measuredangle ABC = \measuredangle ABH\quad\blacksquare\]

ARO Restated wrote: An acute-angled $\triangle ABC \ (AB<AC)$ is inscribed into a circle $\omega$. Let $G$ be the centroid of $\triangle ABC$, and let $\overline{AD}$ be an altitude of this triangle. Ray $\overline{GD}$ meets $\omega$ at $X$. Prove that the circumcircle of the triangle $XDB$ is tangent to $\overline{AB}$. Here's my solution: Let $P$ be the point such that $APCB$ is an isosceles trapezoid. Then, by an easy homothety argument, we can see that $P$ lies on line $\overline{GDX}$. Invert about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the angle bisector of $\angle BAC$. Then $D$ gets sent to the antipode of $A$ in $\omega$ (say $A'$), while $B$ gets swapped with $C$. Also, $P$ goes to a point $T \in \overline{BC}$ such that $\overline{AT}$ is tangent to $\omega$, and so $X$ gets sent to the point $X'=\odot (AA'T) \cap \overline{BC}$. As $\angle A'AT=90^{\circ}$, so we get that $\overline{A'X'} \perp \overline{BC}$. Then the center of $\odot (A'CX')$ lies on $\overline{A'C}$, and so $\odot (A'CX')$ is tangent to $\overline{AC}$. Inverting back, we get the desired result. Hence, done. $\blacksquare$

We'll restate the whole problem and solve: Let $H$ be the foot from $A$ to $\overline{BC}$ in $\Delta ABC$ with $AB<AC$. Let $G$ be centroid of $\Delta ABC$. Let $M$ be midpoint of $\overline{BC}$. Let $AM \cap \odot (ABC)$ $=$ $S$. Let $A'$ $\in$ $\odot (ABC)$, such, $AA'$ $||$ $BC$ and Let $A'H$ $\cap$ $\odot (ABC)$ $=$ $X$ $\implies$ $XHMS$ is cyclic. Let $H'$ be reflection of $H$ over $M$ $\implies$ $G$ is centroid WRT $\Delta AHH'$ $\implies$ $G$ $\in$ $\overline{XA'}$ $\implies$ $\angle BXH$ $=$ $\angle BCA'$ $=$ $\angle ABH$

Beautiful problem. Let $HM\cap\odot(ABC)=K$, and let $KO\cap\odot(ABC)=L$ and let $H'$ be the orthocenter of $\triangle ABC$. Draw the Euler line of $\triangle ABC$. Now it's easy to see that $M$ is the centroid of $\triangle H'LK$. So, $H'H=HL\implies L\in\odot(ABC)$. Hence $L$ is the $K-$ antipode of $\odot(ABC)$, so $\angle HAK=90^\circ\implies AK\|BC$. Now the rest part is easy, $\angle ABC=\angle BCK=\angle BA'K$. Hence, proved. $\blacksquare$.

This is also Romania JBMO TST 2018

http://www.artofproblemsolving.com/community/c6h1159504

Let $X$ be the second intersection of $HM$ with $(ABC)$. We want to show that $BX=AC$, or by angle chase we want to show that $AX \parallel BC$. Throw the configuration onto the complex plane and let the circumcircle be the unit circle, and denote with lowercase letters of the points their complex values. Place then such that $c = \frac{1}{b}$. Then we have that: $$h=\frac{1}{2}\left( a+b+\frac{1}{b}-\frac{1}{a} \right) $$$$m = \frac{1}{3}\left(a+b+\frac{1}{b}\right)$$ We want to show that $x= \frac{1}{a}$, but we see that this value satisfies: $$\frac{h-m}{\overline{h-m}}=\frac{h-\frac{1}{a}}{\overline{h-\frac{1}{a}}}$$ Thus we have that $AX \parallel BC$.

EulersTurban wrote: Let $X$ be the second intersection of $HM$ with $(ABC)$. We want to show that $BX=AC$, or by angle chase we want to show that $AX \parallel BC$. Throw the configuration onto the complex plane and let the circumcircle be the unit circle, and denote with lowercase letters of the points their complex values. Place then such that $c = \frac{1}{b}$. Then we have that: $$h=\frac{1}{2}\left( a+b+\frac{1}{b}-\frac{1}{a} \right) $$$$m = \frac{1}{3}\left(a+b+\frac{1}{b}\right)$$ We want to show that $x= \frac{1}{a}$, but we see that this value satisfies: $$\frac{h-m}{\overline{h-m}}=\frac{h-\frac{1}{a}}{\overline{h-\frac{1}{a}}}$$ Thus we have that $AX \parallel BC$. I just want to know that why putting $c= \frac{1}{b}$ doesn't alter the question condition ??

$\measuredangle$ represents angles taken modulo $180^{\circ}$ [asy][asy] import olympiad; size(230); pair A = (-1 , 2) , B = (-3 , -2) , C = (4,-2); draw(A--B--C--cycle , green); draw(circumcircle(A,B,C) , cyan); pair D = (0.5 , -2); pair E = (1.5 , 0); pair F = (-2 , 0); pair G = centroid(A,B,C); pair H = foot(A,B,C); pair A_1 = (-1 , -4.5); draw(circumcircle(A_1, H , B) , orange); draw(A--H); dot("$A$" , A , N); dot("$B$" , B , W); dot("$C$" , C , NE); dot("$D$" , D , NE); dot("$E$" , E , NE); dot("$F$" , F , W); dot("$G$" , G , SW); dot("$H$" , H , SW); dot("$A_1$" , A_1 , S); [/asy][/asy] Let $X \in \omega$ such that $AX \parallel BC$ If we consider a homothety at $M$ with a factor of $-2$. As a result we have $H \mapsto X$ and so we have $H-M-X$. Also we have that $AXBC$ is an isosceles trapezium. Thus we get the following $$\measuredangle BA^{\prime}H = \measuredangle BA^{\prime}X = \measuredangle BCX = \measuredangle ABC = \measuredangle ABH$$$\blacksquare$

Let homothety with coefficient $-2$ wrt $M$ maps $H\mapsto D;$ easy to conclude, that $ABCD$ is an isosceles trepezoid. But then $\measuredangle BA'H=\measuredangle BA'D=\measuredangle BCD=\measuredangle CBA=\measuredangle HBA$, so the conclusion follows.

One thing I have realised is that Russians love this config a lot Well know that $A'-H-M-A''$ where $A''$ is such that $ABCA''$ is a cyclic isosceles trapezoid $$\measuredangle BA'D = \measuredangle BA'A'' = \measuredangle BCA'' = \measuredangle CBA = \measuredangle DBA \ \blacksquare$$

BVKRB- wrote: One thing I have realised is that Russians love this config a lot Same feeling

If $A''$ is the other intersection of $\overline{MH}$ with $\omega$ it's not hard to see that $AA''CB$ is an isosceles trapezoid (for instance, use coordinates to prove the converse), so $\angle BA'H=\angle BA'A''=\angle ABC=\angle ABH$ as desired.

What. You’ve got to be serious. By the configuration in ISL 2011/G4, if $A”=\overline{A’HM}\cap(ABC)\neq A’$ then it’s well known that $AA’’BC$ is an isosceles trapezium, so from there $\measuredangle BA’H=\measuredangle BA’A’’=\measuredangle BAA’’=\measuredangle ABC=\measuredangle ABH$, thus done from alt seg.