Lemma 1: As $X$ varies on ray $BC$, $F$ lie on a line passing through $G$. Proof: Set up barycentric co-ordinate system with $A=(1,0,0),B=(0,1,0)$ and $C=(0,0,1)$. Let $X=(0,x,1-x),Y=(1-y,0,y)$ and $Z=(z,1-z,0)$. Then $BX^2=CY^2=AZ^2\implies a^2(x-1)^2=b^2(y-1)^2=c^2(z-1)^2\implies y,z$ are of the form $ax+b$ for some $a,b\in\mathbb{R}$. $\therefore$ Centroid of $\triangle XYZ$ lie on a line as $X$ varies on ray $BC$. For $X=B$, $F=G$. This proves lemma 1. Lemma 2:Let WLOG, $BC$ be smallest side of$\triangle ABC$. Let $D,E$ lie on segments $CA$ and $AB$ respectively s.t. $BC=CD=BE$. Then $OI\perp DE$ Proof: Let $M$ be the second intersection of $BI$ and $(O)$. Then $CDE\sim MIO$ because $\angle DCE = \angle IMO$ and \[ \frac{CD}{CE} = \frac{CD}{CB} = \frac{MA}{MO} = \frac{MI}{MO}. \]Furthermore, $CD \perp MI$ and $CE \perp MO$, so $DE \perp IO$ as well. This proves lemma 2. (Note: Proof of lemma 2 is due to huajd) Lemma 3: Let $X=C$. Then $Y=D$. Let $Z=K$. Then line joining centroid $F$ of $\triangle CDK$ and $G$ is parallel to $DE$,ie $FG$ is parallel to $DE$. Proof: Set up the barycentric co-ordinate system with $A=(1,0,0),B=(0,1,0)$ and $C=(0,0,1)$. Let $BC=a,CA=b$ and $AB=c$. Let $\frac{1}{a}=u,\frac{1}{b}=v$ and $\frac{1}{c}=w$. Then $D=(\frac{a}{b},0,\frac{b-a}{b})$, $E=(\frac{a}{c},\frac{c-a}{c},0)$. The equation of line $DE$ is $\frac{x}{a}+\frac{ya}{a-c}+\frac{za}{a-c}=0$, ie $x+\frac{yw}{w-u}+\frac{zv}{v-u}=0$. Also we obtain $Z=K=(\frac{c-a}{c},\frac{a}{c},0)$. So centroid $F$ of $XYZ=(\frac{c-a}{c}+\frac{a}{b}:\frac{a}{c}:2-\frac{a}{b})$. Also $G=(1:1:1)$. Equation of line $FG$ becomes $(2v+2w-u)x+(2w+2u-v)y+(2u+2v-w)z=0$. $$ \begin{vmatrix} 2(u+v+w) & 2(u+v+w) & 2(u+v+w) \\ 1 & \frac{w}{w-u} & \frac{v}{v-u} \\ 1 & 1 & 1 \end{vmatrix}=0$$ $$ \begin{vmatrix} -3u & -3v & -3w \\ 1 & \frac{w}{w-u} & \frac{v}{v-u} \\ 1 & 1 & 1 \end{vmatrix}=0$$ $\therefore$ $$ \begin{vmatrix} 2v+2w-u & 2w+2u-v & 2u+2v-w \\ 1 & \frac{w}{w-u} & \frac{v}{v-u} \\ 1 & 1 & 1 \end{vmatrix}=0$$ $\therefore FG\parallel DE$ as desired. Now let $X,Y,Z$ be the on ray $BC,CA,AB$ respectively s.t. $BX=CY=AZ$. Then centroid of $\triangle XYZ$ lies on line joining centroid of $\triangle CDK$ and $G$. This line is perpendicular to $OI$ from lemma 1 and lemma 2. This completes the proof. Q.E.D.

$BS=CT=BC$ S lies on AB, T lies on AC, ST and IO are perpendicular. (Pythagorean) And prove FG and ST are parellel by vector (it's trivial)

An extension?! In $\triangle ABC$, we select three points $X,Y,Z$ on the sides $BC,CA,AB$ respectively such that $BX=CY=AZ$. Define the points $S,T$ in the lines $AB,AC$ such that $BS=BC=CT$, the centroid of $\triangle ABC$ is $G$, the centroid of $\triangle XYZ$ is $G_1$. Prove that the lines $SG$ and $TG_1$ meets on the $B$-median of $\triangle BZC$.

Let $D$ and $R$ lie on $AC$ and $E$ and $S$ lie on $AB$ such that $D$ and $E$ are intouch points and $SB = BC = CR$. Lemma: $RS \perp OI$. Proof: If $AC = AB$, the result is trivial. Without loss of generality assume $AC > AB$. Then, \begin{align*} OS^2 - OR^2 & = (OS^2-OA^2)+(OA^2-OR^2) \\ & = -SA\cdot SB + RA \cdot RC = a(RA - SA) = a(b - c). \end{align*}We also find \begin{align*} IS^2 - IR^2 & = ES^2 - DR^2 \\ & = ((a - s + b)^2 - (a - s + c)^2) = a(b - c). \end{align*}Hence, by Perpendicularity Lemma we obtain the desired. $\square$ Let $BX = CY = AZ = k$. We can determine the direction of $FG$: \begin{align*} \overrightarrow{FG} & = \frac{A + B + C}{3} - \frac{X + Y + Z}{3} = \frac{1}{3}(\overrightarrow{ZA} + \overrightarrow{XB} + \overrightarrow{YC}) \\ & = \frac{k}{3}(\vec u_{BA} + \vec u_{CB} + \vec u_{AC}) = \frac{k}{3a}(\overrightarrow{BS} + \overrightarrow{CB} + \overrightarrow{RC}) = \frac{k}{3a}\overrightarrow{RS}. \end{align*}We conclude that $FG \parallel RS$. Hence $FG \perp OI$.

Too easy with linpop. Let $BX = CY = AZ = s$ and $BC = y+z, CA = z+x, AB = x+y$. Let $f(\bullet) = \text{Pow}(\bullet, \omega) - \text{Pow}(\bullet, (ABC))$ where $\omega$ is the incircle, then \[ f(F) = \frac 13 \sum_{\text{cyc}} f(X) = \frac 13 \sum_{\text{cyc}} ((y-s)^2 + s(y+z-s)) = \frac 13 \sum_{\text{cyc}} y^2 \]which does not depend on $s$, so $F$ moves along a line perpendicular to $OI$. Note that $F=G$ when $s=0$, hence proved.