Prolly some angle chasing approach exists (which I can see now bruh) but whatever Throw the figure into complex plane where $(ABC)$ is the unit circle and $\overleftrightarrow{AO}$ is the $\textsc{Re}$ axis with $a=-1, c=1/b$. The $OE\parallel BC$ condition is equivalent to $$b^4+2b^3-2b^2+2b+1=0\hspace{2 mm} (\star)$$While $\delta(H, BC)=\delta(H, GB)$ is equivalent to $$\frac{(b^4 - 1)^2}{4 (b^6 - 6 b^4 + b^2)}=\left(\frac{(b-1)^2}{2b}\right)^2\leftrightarrow \frac{(b - 1)^2 (b^4 + 2 b^3 - 2 b^2 + 2 b + 1)}{b (b^2 - 2 b - 1) (b^2 + 2 b - 1)} = 0\leftarrow (\star).$$Analogously $\delta(H, BC)=\delta(H, GC)$ and we're done. $\blacksquare$

Let $M$ be the midpoint of side $BC$ and $N$ be the midpoint of $OH$ (and the nine-point center of $\triangle ABC$). Note that $A, G, O, N, H, M$ are collinear. Since $G$ and $M$ lie on the nine-point circle of $\triangle ABC$, $N$ must be the midpoint of $GM$. Thus, $MO = GH$. Now, let $D \in BC$ such that $ED \perp BC$. Since $OM \perp BC$ and $OE \parallel BC$, $DEOM$ is a rectangle. Since $\angle AEH = 90^{\circ}$ and $G$ is the midpoint of $AH$, we have $GA = GH = GE$. Then $ED = OM = GH = GE$. Together with $GH \parallel ED$ and $GH = ED$, we see that $DEGH$ is a rhombus and $DG \perp EH$. Thus, $\angle GEB = 90^{\circ} - \angle EGD = 90^{\circ} - \angle EDG = \angle DEB$, and $\triangle GEB \cong \triangle DEB$. So $\angle GBE = \angle DBE$, and $BH$ is the angle bisector of $\angle GBC$. Since $GH$ is the perpendicular bisector of $BC$, $GH$ is the angle bisector of $\angle BGC$. Therefore, $H$ is the incenter of $\triangle GBC$.

It is also 2021 China South East Mathematical Olympiad Grade11 P1 My solution during the contest: