$G$ is just the point on $(ABC)$ such that $AI \perp IG$, and similarly, $H$ is just the point on $(ABC)$ such that $AJ \perp JH$ (here $J$ is clearly the $A-excenter$ The proof of this is to consider phantom points, let $G$ be the point on $(ABC)$ such that $AI \perp IG$, and let $AG \cap BC = E$, it suffices to prove that $EI \perp AI$, but now consider two circles $C_1$ is the circle with diameter as $AI$ and $C_2$ is circle with center $D$ and radius $DI$, now note clearly $IE$ is the radical axis of these two circles (As power of $I$ is $0$ for both and as $AG \cap BC = E, EG \cdot EA=EB \cdot EC$ so power of $E$ is also equal for both circles), but we know radical axis is perpendicular to line joining centers but in this case the line joining centers is precisely $AI$... Proving this result for $H$ is also similar, just we will have to consider cirlce with diameter $AJ$ instead of $AI$ Now, I will prove $\Delta BGC \cong \Delta CHB$ which will clearly prove the result (the motivation for this is that if $BG=CH$, by symmetry $BH$ must be $CG$ also which is equivalent to congruency.) Let $A'$ be the $A$ antipode in $(ABC)$. It is well known and very easy to prove that $A'-I-G$ collinear and $J-A'-H$ collinear. So now, $\angle BGC= \angle CHB$ due to cyclic quads, so it suffices to prove $\angle BCG=\angle CBH$ but $\angle BCG= \angle BA'I$ and $\angle CBH = 180 - \angle CA'J$, so it suffices to prove $\sin \angle BA'I = \sin \angle CA'J$ Also, note $A'I=A'J$ (as $A'D \perp IJ$ and $DI=DJ$). So now by sine rule in $\Delta BA'I$ noting $\angle A'BI= 90 - \frac{B}{2}$ and sine rule in $\Delta JCA'$ and noting $\angle JCA'= \angle JCA - 90= \frac{C}{2}$, It suffices to prove $BI \cdot \cos \frac{B}{2} = CJ \cdot \sin \frac{C}{2}$ but both are just $s-b$ (where $s$ is semiperimeter of $\Delta ABC$)

Let $X, Y$ the contact point of the incirle and the $A$-excircle with $BC$. We know that $J$ is the $A$-excenter and $D$ is the center of $(BICJ)$. $\textbf{Claim 1:}$ $G,H\in (ABC)$ $EI$ is tangent to $(BICJ)$ at $I$, then $EB\cdot EC=EI^2 = EG\cdot EA$ meaning that $GACB$ is cyclic. Analogously for $H$. $\textbf{Claim 2:}$ $D,X,G$ are collinear as well as $D,Y,H$. This is well known!!! Let $\phi$ be the inversion centered at $D$ with radious $DI$. The circle $(EGIX)$ remains fixed so, $\phi(G)=\phi((ABC))\cap \phi((EGIX))= BC\cap (EGIX)=X$, this implies that $D,X,G$ are collinear. Analogously for $D,Y,H$. The problem concludes by noticing that $D$ es the midpoint of arc $BC$ in $(ABC)$ then $X,Y$ being symmetric w.r.t the perpendicular bisector of $BC$ implies that $G,H$ are also symmetric, so $GH\parallel BC$ and we are done.