Homothety with center $G$

Here you go the asymptote

Solution

This is way too well-known for an olympiad. For applications of this, see for example ISL 2011 G4 & BMO SL 2017 G5.

It is pretty trivial by complex numbers, calculating sholace determinant of points $S,D,G$ gives 0 therefore they are colinear.

Iora wrote: Let $ABC$ be an acute triangle and $G$ be the intersection of the meadians of triangle $ABC$. Let $D $be the foot of the altitude drawn from $A$ to $BC$. Draw a parallel line such that it is parallel to $BC$ and one of the points of it is $A$.Donate the point $S$ as the intersection of the parallel line and excircle $ABC$. Prove that $S,G,D$ are co-linear [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair A = dir(50), S = dir(130), B = dir(200), C = dir(-20), G = (A+B+C)/3, D = foot(A, B, C); draw(A--B--C--cycle, black+linewidth(1)); draw(A--S^^A--D, magenta); draw(S--D, red+dashed); draw(circumcircle(A, B, C), heavymagenta); string[] names = {"$A$", "$B$", "$C$","$D$", "$G$","$S$"}; pair[] points = {A, B, C,D,G,S}; pair[] ll = {A, B, C,D, G,S}; int pt = names.length; for (int i=0; i<pt; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] HOMOTHETY is a good solution but there is a different solution also: Let AG and BC intersect at M and SG and BC intersect at D'.Since Triangle SGA ~triangle MGD' and AG=2GM we get that AS=2MD'.Also since angle ASC=angle SCB we get that BS=AC thus ACBS is a isosceles trapezoid.Let SU be perpendicular to BC.Then BU=CD so MU=MD.since USAD is a rectangle UD=2MD=AS=2MD' so we conclude that D'=D

Let $M$ be the midpoint of the side $BC$ and $A^\prime$ be the reflection of $A$ with respect to $D$. $A^\prime M \cap AS=S^\prime$. We have: $A^\prime D = DA$ and $DM \parallel AS^\prime \implies A^\prime M = MS^\prime$ we also have $BM=MC \implies BS^\prime CA^\prime$ is parallelogram. Therefore we have: $BS^\prime = A^\prime C =CA \implies BS^\prime AC$ is a isosceles trapezoid, which means that $BS^\prime AC$ is cyclic $\implies S^\prime = S$. $MD$ is the mid line of triangle $A^\prime SA \implies MD = \frac{1}{2}SA$. Let $AM \cap SD = G^\prime$, as $AS \parallel MD \implies \triangle ASG^\prime$ and $\triangle MDG^\prime$ are similar. Then: $\frac{AG^\prime}{G^\prime M}=\frac{AS}{MD}=2$. Point $G^\prime$ divides the median $AM$ with the ratio of $2:1 \implies G^\prime = G$ $\square$

Take the homothety centered in $G$ that sends the nine point circle to $(ABC)$. This homothety sends the midpont of $BC$(call it $M$) to $A$, and $D$ to a point $S'$ in $(ABC)$ s.t. $DM\parallel S'A$. But notice that $S'=S$. Thus we're done.

Quote: Donate the point $S$ as the intersection of the parallel line and excircle $ABC$. Do you mean circumcircle?

This was also Sharygin 2020 Correspondence Round Problem 4 After using Phantom Points, we will solve this problem: Quote: In $\triangle ABCS$, let $D$ be the projection of $A$ onto $BC$ and let $G$ be the centroid of $\triangle ABC$. Let the line through $A$ parallel to $BC$ and line $DG$ meet at $S$. Show that $S$ lies on the circumcircle of $\triangle ABC$. Use barycentric coordinates on reference triangle $\triangle ABC$. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then clearly $G=(1:1:1)$ and $D=(0:a^2+b^2-c^2:a^2-b^2+c^2)$. Line $BC$ has equation $x=0$ so the point at infinity along $BC$ has coordinates $P_\infty=(0:-1:1)$. Thus, points on cevian $AP_\infty$ can be parameterized by $(t:-1:1)$. Since $S$ lies on $AP_\infty$ and $S,G,D$ are colinear, we have $$\begin{vmatrix}t&-1&1\\ 1&1&1\\ 0&a^2+b^2-c^2&a^2-b^2+c^2\end{vmatrix}=0 \iff t=\frac{a^2}{b^2-c^2}$$so $S=(a^2:-(b^2-c^2):b^2-c^2))$. Now just note that $$\text{Pow}_{(ABC)} (S)=a^2[-(b^2-c^2)(b^2-c^2)]+b^2[(b^2-c^2)(a^2)]+c^2[(a^2)(b^2-c^2)]=a^2(b^2-c^2)[-b^2+c^2+b^2-c^2]=0$$and we are done.

Let circle $(ABCS)$ be the unit circle hence $|a|=|b|=|c|=|s|=1$. It is well known that $G=g=\frac{1}{3} (a+b+c),D=d= \frac{1}{2}(a+b+c-\overline{a}{bc})$ Now we will derive $s$ in terms of $a,b,c,$. Since $\angle ASC= \angle SCB$ $$I = \frac{a-s}{s-c} \div \frac{b-c}{c-s}= \frac{a-s}{c-b} \in \mathbb{R} \rightarrow I=\overline{I}$$$$ \overline{I} = \frac{ \overline{a}- \overline{s}}{ \overline{c}- \overline{b}}= \frac{\frac{1}{a}- \frac{1}{s}}{ \frac{1}{c}- \frac{1}{b}}= \frac{ \frac{s-a}{as}}{\frac{b-c}{bc}}= \frac{bc(s-a)}{as(b-c)}$$$$ \overline{I}=I: \frac{bc(s-a)}{as(b-c)}= \frac{s-a}{b-c}\rightarrow \frac{bc}{as}=1 \rightarrow s= \frac{bc}{a}$$Since now we know the points $s,g,d$, we just have to prove their sholace determinant equals to 0 Before calculating the determinant, note that $\overline{g}= \frac{1}{3}( \frac{ab+bc+ac}{abc}),\overline{d}= \frac{1}{2}(\frac{ab+bc+ac-a^2}{abc}), \overline{s} = \frac{a^2}{abc}$. Now we are ready to calculate determinant $$A=\begin{vmatrix} \frac{bc}{a} & \frac{a^2}{abc} & 1 \\ \frac{1}{3}(a+b+c) & \frac{1}{3}( \frac{ab+bc+ac}{abc}) & 1 \\ \frac{1}{2} (a+b+c- \frac{bc}{a})& \frac{1}{2}(\frac{ab+bc+ac-a^2}{abc}) & 1 \end{vmatrix}= \begin{vmatrix} \frac{bc}{a} & \frac{a^2}{abc} & 1 \\ (a+b+c) & ( \frac{ab+bc+ac}{abc}) & 3 \\ (a+b+c- \frac{bc}{a})& (\frac{ab+bc+ac-a^2}{abc}) & 2 \end{vmatrix}=\begin{vmatrix} \frac{bc}{a} & \frac{a^2}{abc} & 1 \\ \frac{bc}{a} & \frac{a^2}{abc} & 1 \\ (a+b+c- \frac{bc}{a})& (\frac{ab+bc+ac-a^2}{abc}) & 2 \end{vmatrix} =$$$$\begin{vmatrix}0 & 0 & 0 \\ \frac{bc}{a} & \frac{a^2}{abc} & 1 \\ (a+b+c- \frac{bc}{a})& (\frac{ab+bc+ac-a^2}{abc}) & 2 \end{vmatrix}=0$$. This means area between points $S,G,D$ is 0, hence they are colinear. Therefore we are done

Exactly same problem as Evan Chen's EGMO Book problem 3.24

Let $X$ be the mid point of $BC$ And $SK$ be the perpendicular from $S$ to $BC$ $AS \parallel BC$ And $ACBS$ cyclic, So $ACBS$ is an isosceles trapezium. $AC=SB$, $AD=SK$ $\implies CD=BK$, So, $X$ is the mid point of $KD$ $ASKD$ is a rectangle, $AS=KD$ So, $AS=2XD$ $\triangle AGS$ is similar to $\triangle GDX$ because, $\frac{AG}{GX}=2$ $\frac{AS}{XD}=2$ $\angle GXD=\angle GAS$ Therefore, $\angle DGX=\angle AGS$ So, $D,G,S$ are collinear Asymptote code below [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair A = dir(50), S = dir(130), B = dir(200), C = dir(-20), G = (A+B+C)/3, D = foot(A, B, C), K= foot (S,B,C), X=(B+C)/2 ; draw(A--B--C--cycle, black+linewidth(1)); draw(A--S^^A--D, magenta); draw(S--K^^A--X^^B--S, blue+linewidth(1)); draw(S--D, red+dashed); draw(circumcircle(A, B, C), heavymagenta); string[] names = {"$A$", "$B$", "$C$","$D$", "$G$","$S$", "$K$", "$X$"}; pair[] points = {A, B, C,D,G,S,K,X}; pair[] ll = {A, B, C,D, G,S, K,X}; int pt = names.length; for (int i=0; i<pt; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy]

Okay, draw median $AM$, so $\triangle GDM \sim GSA$, with ratio $1:2$, (because of median) so $AS=2MD$, now by some more calculations, $\angle ASC=\angle BCS$, so $ACBS$ is a isosceles trapezoid, let $SI$ be perpendicular to $BC$, then we have $CD=UB$ and $MU=DM$, so $USAD$ is a rectangle, and $UD=2MD=AS$, hence $S,G,D$ are co-linear.

Use complex numbers with $(ABC)=\mathbb{S}^1$, then $AS\parallel BC\implies s=\frac{bc}{a}$. We also have $g=\frac{a+b+c}{3}$ and $d=\frac{a+b+c-\frac{bc}{a}}{2}$, so $g-s=\frac{a+b+c-3\frac{bc}{a}}{3}=2(d-g)$, implying that $S,G,D$ are collinear.

Use barycentric coordinates with respect to $ABC$. From $AS\parallel BC$ we have $S=(t:1:-1)$. From $S\in (ABC)$ we have $a^2(1)(-1)+b^2(-1)(t)+c^2(t)(1)=0$. Hence $S=(a^2/(c^2-b^2):1:-1)=(a^2:c^2-b^2:b^2-c^2)$. We know $D=(0:a^2+b^2-c^2:a^2-b^2+c^2)$ and $G=(1:1:1)$. We need to prove$$\begin{vmatrix}a^2&c^2-b^2&b^2-c^2\\0&a^2+b^2-c^2&a^2-b^2+c^2\\1&1&1\end{vmatrix}=0.$$Adding the first row to the second row we have that the determinant equals$$\begin{vmatrix}a^2&c^2-b^2&b^2-c^2\\a^2&a^2&a^2\\1&1&1\end{vmatrix}=0,$$as desired.

I shall prove an equivalent problem . Quote: Let $ABC$ be an acute triangle and $G$ .Let $D $be the foot of the altitude drawn from $A$ to $BC$. Draw a parallel line such that it is parallel to $BC$ and one of the points of it is $A$.Denote the point $S$ as the intersection of the parallel line and circumcircle $ABC$. Let G be the intersection of DS and median from point A in triangle ABC. Prove that G is intersection of medians in triangle ABC. In other words: Define G as the intersection of A-median and line DS. We shall prove that G is the centroid of ABC. SACB is isosceles trapezoid (as SA// BC and every cyclic trapezoid is isosceles ) Let K be the projection of S on BC and M be the midpoint of BC The midpoint M of BC is also midpoint of KD (easy) So in triangle AKD , AM is one median SADK is rectangle (easy) so diagonals bisect each other, so DS passes through midpoint N of AK, So in triangle ADK, AN is another median So the intersection of medians AM and DN is it’s centroid, so G is centroid of ADK So AG =2 /3 AM and as AM is median also of ABC, We conclude that G is centroid of ABC

Set $(ABC)$ as the unit circle... $3g=a+b+c$ , $2d=a+b+c-bc\overline{a}$ ,$s=bc/a$ And their conjugate by theorem... $A_{SGD}=\frac{i}{4}\begin{vmatrix} s& \overline{s} &1 \\ g& \overline{g} &1 \\ d & \overline{d} &1 \end{vmatrix}=0 $

We complex bash. Choose $b = -c \in i\mathbb{R}$.Then we have $s = \overline{a}$, $d = \frac{a-s}{2}$, $g = \frac{a}{3}$. But now $2d + s = 3g \implies g = \frac 23 \times d + \frac 13 \times s$, which implies the result. $\square$