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.