Circumcentre has another name if you play inside

$\textbf{Solution:}$

By the definition of the circumcentre, $GD$ , $GE$ and $GF$ are the $\perp$ bisectors of $BC$ , $AC$ and $AB$ respectively. In particular, $D$ , $E$ and $F$ are the midpoints of $BC$ , $AC$ and $AB$ respectively. By the midpoint theorem, joining the midpoints $D$ and $E$ with a line yields the segment $DE$ being parallel to $AB$ .

Let $FG$ produced meet $DE$ at $J$ . Since $DE$ is parallel to $AB$ , we have $\angle FJD = \angle AFG$ as a pair of alternate angles. Since $GF\perp AB$ , we have $\angle AFG = 90^{\circ}$ . Hence $\angle FJD = \angle AFG = 90^{\circ}$ . Therefore, $FJ$ is an altitude with respect to $DE$ in $\triangle DEF$ .

Similarly, by joining $DF$ and $EF$ , we will have $DG$ produced and $EG$ produced being perpendicular to $FE$ and $DF$ respectively.

In conclusion, $G$ is the orthocentre of $\triangle DEF$ . $\Box$