My solution (The name was deliberaltely given "Sharrygin" to distinguish it from the original, wrong pdfs i had created)

Switch $C,D$ in the problem. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -15.24, xmax = 15.56, ymin = -10.26, ymax = 10.08; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); draw((3.3468145877396585,6.870836624266258)--(3.3619212990982517,6.446841592135069)--(3.7859163312294406,6.461948303493663)--(3.7708096198708474,6.885943335624852)--cycle, linewidth(1.6)); /* draw figures */ draw((-6.5,6.52)--(1.92,6.82), linewidth(0.8) + dtsfsf); draw((-6.5,6.52)--(-8.18,1.66), linewidth(0.8) + dtsfsf); draw(circle((-2.133956533357085,2.2903800362222024), 6.078817367761839), linewidth(0.8) + sexdts); draw((1.92,6.82)--(3.941619239741696,2.091886671249705), linewidth(0.8) + dtsfsf); draw((-6.5,6.52)--(3.941619239741696,2.091886671249705), linewidth(0.8) + rvwvcq); draw((-8.18,1.66)--(3.7708096198708474,6.885943335624852), linewidth(0.8) + rvwvcq); draw((1.92,6.82)--(3.7708096198708474,6.885943335624852), linewidth(0.8) + wrwrwr); draw((3.7708096198708474,6.885943335624852)--(3.941619239741696,2.091886671249705), linewidth(0.8) + wvvxds); draw((-8.18,1.66)--(3.941619239741696,2.091886671249705), linewidth(0.8) + dtsfsf); /* dots and labels */ dot((-6.5,6.52),dotstyle); label("$A$", (-6.42,6.72), NE * labelscalefactor); dot((1.92,6.82),dotstyle); label("$B$", (2,7.02), NE * labelscalefactor); dot((-8.18,1.66),dotstyle); label("D", (-8.1,1.86), NE * labelscalefactor); dot((3.941619239741696,2.091886671249705),linewidth(4pt) + dotstyle); label("C", (4.02,2.26), NE * labelscalefactor); dot((3.7708096198708474,6.885943335624852),linewidth(4pt) + dotstyle); label("$F$", (3.86,7.04), NE * labelscalefactor); dot((-0.21279358675277127,5.1439622237499005),linewidth(4pt) + dotstyle); label("$G$", (-0.14,5.3), NE * labelscalefactor); label("$90^\circ$", (3.46,6.54), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Let $a,b,c,d,f,g$ denote the complex numbers for $A,B,C,D,F,G$ Let $\odot(ABCD)$ be the unit circle. So by egmo Lemma 6.12 part(b) page 102 we have that $g=\frac{1}{3}a+b+c$ and by Lemma 6.11 egmo page 101 we have that $f=\frac{1}{2}(a+b+c-ab\overline{c})=\frac{1}{2}(a+b+c-\frac{ab}{c})$ since $c$ lies on the unit circle.Now since $ABCD$ is iscoseles trapeziod $\implies AB \parallel CD$ or by the parallel condition for complex numbers we have $$\frac{a-b}{\frac{b-a}{ab}}=\frac{c-d}{\frac{d-c}{cd}}\implies ab=cd$$. Putting this value in our complex number for $f$ we get $$f=\frac{1}{2}(a+b+c-ab\overline{c})=\frac{1}{2}(a+b+c-\frac{ab}{c})=\frac{1}{2}(a+b+c-\frac{cd}{c})=\frac{1}{2}(a+b+c-d)$$.Now it suffices to prove $G-F-D$ collinear or by lemma 6.6 egmo page 100 we have to prove $$\frac{f-g}{\overline{(f-g)}}=\frac{f-d}{\overline{(f-d)}}$$evaluating the lhs by putting values of $f$ and $g$ we have $$LHS=\frac{\frac{a+b+c}{6}-\frac{d}{2}}{\overline{\frac{a+b+c}{6}-\frac{d}{2}}}=\frac{\frac{a+b+c-3d}{6}}{\frac{6}{a}+\frac{6}{b}+\frac{6}{c}-\frac{2}{d}}=\frac{a+b+c-3d}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{3d}}$$. Now we leave the $LHS$ in this form and work with the $RHS$ by putting values of $f$ and $d$. $$RHS=\frac{\frac{a+b+c}{2}-\frac{3d}{2}}{\overline{\frac{a+b+c}{2}-\frac{3d}{2}}}=\frac{\frac{a+b+c-3d}{2}}{\frac{2}{a}+\frac{2}{b}+\frac{2}{c}-\frac{2}{3d}}=\frac{a+b+c-3d}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{3d}}$$. And now we observe that $LHS=RHS$ and so we are done. $\blacksquare$

My solution We will denote the area of $\triangle XYZ$ by $[XYZ]$, wherever required. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.26, xmax = 16.82, ymin = -6.8, ymax = 9.64; /* image dimensions */ draw((-1,6)--(1,6)--(4,0)--(-4,0)--cycle, linewidth(0.4)); /* draw figures */ draw((-1,6)--(1,6), linewidth(0.4)); draw((1,6)--(4,0), linewidth(0.4)); draw((4,0)--(-4,0), linewidth(0.4)); draw((-4,0)--(-1,6), linewidth(0.4)); draw((-1,6)--(-1,0), linewidth(0.4) + linetype("4 4")); draw((1,6)--(1,0), linewidth(0.4) + linetype("4 4")); draw((-1,0)--(1,6), linewidth(0.4) + linetype("2 2")); draw((-1,6)--(0,0), linewidth(0.4) + linetype("2 2")); draw((-1,6)--(4,0), linewidth(0.4) + linetype("4 4")); draw((8,0)--(8,6), linewidth(0.4)); draw((10,0)--(10,6), linewidth(0.4)); draw((10,6)--(8,6), linewidth(0.4)); draw((10,0)--(8,0), linewidth(0.4)); draw((8,0)--(10,6), linewidth(0.4) + linetype("2 2")); draw((8,6)--(9,0), linewidth(0.4) + linetype("2 2")); draw((8,6)--(10,0), linewidth(0.4) + linetype("2 2")); draw((8.666666666666666,2)--(10,0), linewidth(0.4) + dotted); /* dots and labels */ dot((-4,0),linewidth(1pt) + dotstyle); label("$A$", (-4.46,-0.2), NE * labelscalefactor); dot((4,0),linewidth(1pt) + dotstyle); label("$B$", (4.2,-0.1), NE * labelscalefactor); dot((1,6),linewidth(1pt) + dotstyle); label("$C$", (1.18,6.02), NE * labelscalefactor); dot((-1,6),linewidth(1pt) + dotstyle); label("$D$", (-1.5,6.02), NE * labelscalefactor); dot((0,0),linewidth(1pt) + dotstyle); label("$M$", (-0.14,-0.4), NE * labelscalefactor); dot((-1,0),linewidth(1pt) + dotstyle); label("$F$", (-1.1,-0.4), NE * labelscalefactor); dot((1,0),linewidth(1pt) + dotstyle); label("$E$", (0.92,-0.42), NE * labelscalefactor); dot((8,0),linewidth(1pt) + dotstyle); label("$P$", (7.52,-0.24), NE * labelscalefactor); dot((10,0),linewidth(1pt) + dotstyle); label("$Q$", (10.24,-0.26), NE * labelscalefactor); dot((10,6),linewidth(1pt) + dotstyle); label("$R$", (10.18,5.98), NE * labelscalefactor); dot((8,6),linewidth(1pt) + dotstyle); label("$S$", (7.56,5.92), NE * labelscalefactor); dot((9,0),linewidth(1pt) + dotstyle); label("$N$", (8.9,-0.4), NE * labelscalefactor); dot((8.666666666666666,2),linewidth(1pt) + dotstyle); label("$G$", (8.16,1.92), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We first prove a well-known lemma. Refer to the figure on the right. $\textbf{Lemma.}$ Let $PQRS$ be a rectangle. Let the mid-point of $PQ$ be $N$. Let $SN$ meet the diagonal $PR$ at point $G$. Then, we may have $AG:GN = 2:1$. $\emph{Proof of lemma.}$ We first note that as $N$ is the mid-point of $PQ$, we have $[PGN]=[GNQ]$. Now, we know that the diagonals of a rectangle bisect each other and so, $[PGS]=[PGQ]$. But, $[PGQ]=[PGN]+[GNQ]=2[PGN]$, and hence, $[PGS]=2[PGN]$, which thus establishes that $AG:GN=2:1$. $\square$ Now, back to the problem. Let $E$ be the projection of $C$ onto $AB$, and $M$ be the mid-point of $AB$. Since, the trapezoid is isosceles, it follows that $\triangle ADF \cong \triangle BCE$, which implies that $AF=BE$. It follows that $M$ is also the mid-point of $EF$. Also, note that we have $FECD$ as a rectangle. Now, from the lemma it follows that the centroid of $\triangle ABD$ lies on $CF$, as desired. $\blacksquare$

Strange, they have put well known Lemmas too! Sharygin 2020 CR P4 wrote: Let $ABCD$ be an isosceles trapezoid with bases $AB$ and $CD$. Prove that the centroid of triangle $ABD$ lies on $CF$ where $F$ is the projection of $D$ to $AB$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -11.28, xmax = 11.62, ymin = -6.11, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((0.738304521499782,0.03985700379309175), 4.57585797006303), linewidth(1.2) + wrwrwr); draw((-3.2,-2.29)--(4.7,-2.25), linewidth(1.2) + wrwrwr); draw((-1.94,3.75)--(3.4098136002973725,3.7548960889295318), linewidth(1.2) + wrwrwr); draw((3.4098136002973725,3.7548960889295318)--(4.7,-2.25), linewidth(1.2) + wrwrwr); draw((-1.94,3.75)--(-3.2,-2.29), linewidth(1.2) + wrwrwr); draw((-1.94,3.75)--(4.7,-2.25), linewidth(1.2) + wrwrwr); draw(circle((-0.589152260749891,-0.414928501896546), 2.287928985031515), linewidth(1.2) + wrwrwr); draw((-1.94,3.75)--(-1.9094508072217353,-2.283465573707452), linewidth(1.2) + wrwrwr); draw((-1.94,3.75)--(0.75,-2.27), linewidth(0.4) + wrwrwr); draw((-3.2,-2.29)--(1.38,0.75), linewidth(0.4) + wrwrwr); draw((-2.57,0.73)--(4.7,-2.25), linewidth(0.4) + wrwrwr); draw((-1.9094508072217353,-2.283465573707452)--(3.4098136002973725,3.7548960889295318), linewidth(1.2) + wrwrwr); /* dots and labels */ dot((-1.94,3.75),dotstyle); label("$D$", (-2.16,3.99), NE * labelscalefactor); dot((-3.2,-2.29),dotstyle); label("$A$", (-3.6,-2.61), NE * labelscalefactor); dot((4.7,-2.25),dotstyle); label("$B$", (4.98,-2.63), NE * labelscalefactor); dot((3.4098136002973725,3.7548960889295318),dotstyle); label("$C$", (3.48,3.95), NE * labelscalefactor); dot((0.75,-2.27),linewidth(4pt) + dotstyle); label("$C'$", (0.98,-2.75), NE * labelscalefactor); dot((-2.57,0.73),linewidth(4pt) + dotstyle); label("$A'$", (-2.94,0.85), NE * labelscalefactor); dot((1.38,0.75),linewidth(4pt) + dotstyle); label("$B'$", (1.46,0.91), NE * labelscalefactor); dot((-1.9094508072217353,-2.283465573707452),linewidth(4pt) + dotstyle); label("$F$", (-2.08,-2.77), NE * labelscalefactor); dot((-0.1466666666666666,-0.26333333333333325),linewidth(4pt) + dotstyle); label("$G$", (-0.22,0.01), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] As $ABCD$ is an Isosceles Trapezoid, so $ABCD$ must be a cyclic quadilateral too. Let $G$ be the centroid of $\triangle ABD$ and $A'B'C'$ be the Medial Triangle of $\triangle BAD$. Now consider the Homothety ($\mathcal H)$ at $G$ with a scale factor of $-1/2$ which maps $\triangle ADB$ to $\triangle A'B'C'$. Note that $F\in\odot(A'B'C')$ as $\odot(A'B'C')$ is the Nine-Point Circle of $\triangle BAD$ and the line $D$ parallel to $AB$ which is $DC$ maps to $AB$. So, $\mathcal H:C\mapsto AB\cap \odot(A'B'C')\neq C'=F$. So, $\overline{C-G-F}$.

here a nice solution-

NJOY wrote: Let $PN$ meet the diagonal $QS$ at point $G$. Then, we may have $AG:GN = 2:1$. What is this?

Anyone has a coord bash proof for this one? Thanks @below

Set $D = (0,0) , C = (1,0)$ and $A = (x_1,y_1)$. We get that $F = (0,y_1)$ and $B = (z_1,y_1)$ $DA = CB \Longrightarrow x^2 = (z_1-1)^2 \Longrightarrow x = \pm(b-1)$ The centroid of $G$ of $\triangle ABD$ has coordinates $\left(\frac{x_1 + z_1}{3},\frac{2y_1}{2}\right)$ Thus, $G \in CF \iff \frac{-y_1}{3} \cdot \frac{3}{x+b} = -y_1 \iff x+b = 1$ If $x = z_1+1$ , then $z_1 = 1 \Longrightarrow DFBC$ is a rectangle, which is a absurd. Thus we must have $x = 1- z_1$ completing the proof.

This problem was proposed by Burii.

timon92 wrote: This problem was proposed by Burii. How can this well known Property be Proposed by a single person?

Imagine not coord bashing and absolutely incinerating this problem Let $F$ be the origin, $G$ be the centroid and $A=(x_1,0)$ $B=(x_2,0)$ $D=(0,y_0)$ Since $C$ is the reflection of $D$ about perp bisector of $AB$ (the equation for which is $x=\frac{x_1+x_2}{2}$) we must have $C=(x_1+x_2,y_0)$ We also know that $G=(\frac{x_1+x_2}{3},\frac{y_0}{3})$. Now notice that the slope of both $FG$ and $FC$ are the same, which means they are collinear, as desired $\blacksquare$ I haven't seen a shorter coord bash in a long time, this might be too short to even call it a "bash"

After using Phantom Points, we will solve this problem with the following labeling: 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.

[asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-71.31011,50.59116); pair B = (-23.61373,50.26502); pair C = (34.27022,-32.53019); pair D = (-130.32090,-31.40474); pair F = (-129.75749,50.99082); pair E = (-24.17714,-32.13054); pair Cp = (-82.62452,-31.73088); pair M = (-76.96732,9.43014); pair G = (-75.08158,23.15048); import graph; size(12.1cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw((-129.80059,44.68773)--(-123.49749,44.64463)--(-123.45439,50.94772)--F--cycle, linewidth(0.75) + blue); draw((-24.13404,-25.82744)--(-30.43714,-25.78434)--(-30.48024,-32.08744)--E--cycle, linewidth(0.75) + blue); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5) + blue); draw((6.26755,9.52410)--(4.38893,8.21071), linewidth(0.5) + blue); draw(A--D, linewidth(0.5) + blue); draw((-99.88526,8.92373)--(-101.74575,10.26268), linewidth(0.5) + blue); draw(B--D, linewidth(0.5)); draw(D--F, linewidth(0.5)); draw(F--A, linewidth(0.5) + red); draw((-100.97166,51.94011)--(-100.98733,49.64796), linewidth(0.5) + red); draw((-100.08027,51.93402)--(-100.09594,49.64187), linewidth(0.5) + red); draw(B--E, linewidth(0.5)); draw(F--C, linewidth(0.5)); draw(Cp--B, linewidth(0.5) + blue); draw((-54.04937,9.93654)--(-52.18888,8.59759), linewidth(0.5) + blue); draw(A--Cp, linewidth(0.5)); draw(F--E, linewidth(0.5)); draw(C--E, linewidth(0.5) + red); draw((5.48439,-33.47949)--(5.50007,-31.18734), linewidth(0.5) + red); draw((4.59300,-33.47339)--(4.60867,-31.18124), linewidth(0.5) + red); draw(E--Cp, linewidth(0.5) + red); draw((-52.96297,-33.07983)--(-52.94730,-30.78768), linewidth(0.5) + red); draw((-53.85436,-33.07374)--(-53.83869,-30.78159), linewidth(0.5) + red); draw(Cp--D, linewidth(0.5)); dot("$A$", A, N); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, SW); dot("$F$", F, NW); dot("$E$", E, SE); dot("$C'$", Cp, SW); dot("$M$", M, 2*dir(190)); dot("$G$", G, 2*dir(130)); [/asy][/asy] Let $E$ denote the foot of the perpendicular from $B$ onto $CD$. Also let $C'$ be the reflection of $C$ over $BE$, and $M$ be the midpoint of $BD$. Finally let $G=CF\cap AC'$. Clearly $BEDF$ is a rectangle, so we get that $\overline{E-M-F}$ are collinear. Now as $C'$ is the reflection of $C$ over $BE$, we get that $BC'=BC=AD$ which along with the fact that $AB\parallel CD\equiv C'D$ gives us that $ABC'D$ is a parallelogram. This further means that $\overline{A-M-C'}$ are collinear. So we now have that $\overline{G-F-C}$ and $\overline{G-A-C'}$ and that $AF\parallel CC'$, which on combining gives us that $\triangle GAF\sim\triangle GC'C$. Now note that $CC'=2C'E=2(DE-DC')=2(FB-AB)=2FA$. So $\triangle GAF\sim\triangle GC'C\implies\dfrac{GA}{GC'}=\dfrac{AF}{C'C}=\dfrac{1}{2}$. Thus we finally have \[\dfrac{GM}{AG}=\dfrac{AM}{AG}-1=\dfrac{1}{2}\cdot\dfrac{2AM}{AG}-1=\dfrac{AC'}{AG}-1=\dfrac{GC'}{AG}=2,\]which proves that $G$ is the centroid of $\triangle ABD$ as $AM$ is the $A$-median.