[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.3) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.092188843899757, xmax = 4.9919338579071235, ymin = -4.012066949297928, ymax = 5.554852787620956; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-7.923937999999994,4.688918000000004)--(-9.2,-2.38), linewidth(1) + wrwrwr); draw((-9.2,-2.38)--(0.2369062699875711,-2.3653412974589467), linewidth(1) + wrwrwr); draw((0.2369062699875711,-2.3653412974589467)--(-7.923937999999994,4.688918000000004), linewidth(1) + wrwrwr); draw((-7.923937999999994,4.688918000000004)--(-4.481546865006214,-2.372670648729473), linewidth(1) + wrwrwr); draw(circle((-5.629010576670807,-0.018807765819647927), 2.6186529821890563), linewidth(1) + wrwrwr); draw((-6.7764742883354,2.335055117090177)--(-6.769156080615715,-2.3762240784300985), linewidth(1) + wrwrwr); /* dots and labels */ dot((-7.923937999999994,4.688918000000004),dotstyle); label("$A$", (-7.885808539059035,4.792423594613228), NE * labelscalefactor); dot((-9.2,-2.38),dotstyle); label("$B$", (-9.9600052725788,-2.9783238528693963), NE * labelscalefactor); dot((0.2369062699875711,-2.3653412974589467),dotstyle); label("$C$", (0.28158363809225156,-2.957435381828089), NE * labelscalefactor); dot((-4.481546865006214,-2.372670648729473),linewidth(2pt) + dotstyle); label("$M$", (-4.439210817243274,-2.95876808839005), NE * labelscalefactor); dot((-5.629010576670807,-0.018807765819647927),linewidth(2pt) + dotstyle); label("$G$", (-5.5880767245151945,0.06118490375705552), NE * labelscalefactor); dot((-6.7764742883354,2.335055117090177),linewidth(2pt) + dotstyle); label("$K$", (-6.928942631787114,2.501582131424815), NE * labelscalefactor); dot((-6.769156080615715,-2.3762240784300985),linewidth(2pt) + dotstyle); label("$N$", (-7.246498396266461,-2.95876808839005), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $K$ be the second intersection of $AM$ and $\Omega$. Clearly $KN \perp BC$. Then consider a homothety of scale $2$ centered at $A$, which takes $N$ to $S$ and $K$ to $G$. Thus $SG \perp BC$ as desired.

instead of using a homothety one could use the median line in $\triangle{AGS}$

You can think me a dumb, but what does symmetric point mean?

@bove you can understand symmetric point as a reflection of a point in a point ,ie, $S$ is reflection of $A$ in $N$ As for the question: As in rocketscience's diagram, $\implies$ $AK=KG$, $AN=NS$ and $\angle KNM=90^{\circ}$ $\implies$ $GS \perp BC$

Here is my solution for this problem Solution Let $K$ be midpoint of $AG$ then: $K$ $\in$ ($G$; $GM$) or $KN$ $\perp$ $BC$ But: $KN$ $\parallel$ $GS$ so $GS$ $\perp$ $BC$

circumcircle of $\triangle{NGS}$ , cut AG at R. so $\triangle{RSA} \sim \triangle{NGA}$ => RS = SN $\angle NSR$ = 360 - 2$\times$$\angle NMR$ it means that S is circumcenter of $\triangle{NMR}$ SN=SM , GM=GN GS Is perpendicular to BC

Let $ABC$ be an acute-angeled triangle , non-isosceles and with barycentre $G$ (which is , in fact , the intersection of the medians).Let $M$ be the midpoint of $BC$ , and let $\Omega$ be the circle with centre $G$ and radius $GM$ , and let $N$ be the point of intersection between $\Omega$ and $BC$ that is distinct from $M$.Let $S$ be the symmetric point of $A$ with respect to $N$ , that is , the point on the line $AN$ such that $AN=NS$. Prove that $GS$ is perpendicular to $BC$.