sketch: $G$ is the incenter of $DFC$. We just need to prove that $DE=DF$ or $DF\cdot DC=DB\cdot DA$ which is true in any triangle($\sqrt{bc}$ inversion easily shows this).

$\angle BAE = \angle ECB = \angle B - \angle C = \angle FCD = \angle FAD$. $\angle AFC = \angle ABC - \angle FAD = \angle C, \angle AEC = \angle ABC = \angle B, AC=AE=AF$. $A$ is the circumcenter of $CEF$. Let $G'$ be the arc midpoint of $FE$ in this circle. Then $AG$ passes through $G'$ by isosceles triangle, and $GC$ passes through $G'$ by HAT. So $G = G'$, and we are done.

Angle chase to show $\triangle AFC$ is isosceles with $AF=AC$ so $A$ is the midpoint of arc $FC$ in $(AFDC)$ and $G$ lies on $AD$ and angle bisector of $\angle DCB$ so $D$ is the incenter of $\triangle DFC \implies AF=AG=AC$ (Fact 5). $$\angle AEC = \angle ABC = \angle BFA + \angle BAF = \angle BCA + \angle DCB = \angle BCA + \angle BCE = \angle ACE$$So $AE = AC \implies F,G,E,C$ lie on the circle centered at $A$ with radius $AC$.

$AE=AC$ by construction. Inversion $(A;AC)$ swaps $CE\mapsto (ABC)$, $B\mapsto D$ , $F\mapsto BC\cap (ACD)=F$. Then $G\in (CEF)$ by Apolonian circle or by Fact 5.

Claim: $\triangle BED$ and $\triangle BFD$ are congruent $\angle BED = \angle BAC = \angle DFB$ and $\angle CBA = \angle A \text{ wrt A-tangent} = \angle ACE = \angle ABE$. Proving the claim. $\square$. Thus, $G$ is the intersection of the angle bisector of $\angle DCF$ and the perpendicular bisector of $EF$. Notice that $G$ is unique and the midpoint of arc $EF$ satisfies the same conditions. Thus, $G$ is the midpoint of arc $EF$. $\blacksquare$

Unexpectedly nice. Let $\odot{EFC} \cap AB=G’$. We claim that $G’ \equiv G$. First, angle chase to get $AF=AC=AE$, so the center of this circle if $A$. Next, we compute $\angle{CG’B}$ in two different ways $$\angle{CG’B}=\angle{G’CD}+\angle{G’DC}=\angle{G’CD}+C$$And $$AG’=AC \implies \angle{AG’C}=\angle{ACG’}=C+\angle{BCG’}$$ These two imply $\angle{BCG’}=\angle{G’CD}$, as desired $\blacksquare$

Idk about the level of CentroAmerican but I think it is safe to say that this was easy, maybe because I was hunting the circumcenter in my diagram (drawn using paper and pencil and not geogebra , which was beneficial because you could move your compass around to see if $A$ is the circumcenter) which turned out to be the correct approach finishing the problem in like 5 minutes Let $T$ be a point on the tangent at $A$ on the same side of $A$ as $B$ Key Claim: $A$ is the centre of $\odot(EGFC)$ (Which actually finishes the problem ) It is obvious that $\triangle ACE$ is isosceles (because of tangents and parallels or just by a one line angle chase) Now notice that $$\measuredangle AFC = \measuredangle ADC = \measuredangle DAT = \measuredangle FCA \implies AC=AF$$Finally $$\measuredangle AGC =\measuredangle DGC = \measuredangle ADC + \measuredangle DCG = \measuredangle FCA +\measuredangle GCF= \measuredangle BCA + \measuredangle GCB = \measuredangle GCA \implies AC=AG$$which implies $A$ is the circumcenter of $\odot(EGFC)$ which means that $E,G,F,C$ lie on a circle centred at $A$ proving the claim and the problem $\blacksquare$

Let $P$ be an arbitrary point on the tangent to $(ABC)$ at $A$. We have $\measuredangle FBA = \measuredangle CBA = \measuredangle CAP = \measuredangle ACE = \measuredangle ABE$ and $\measuredangle FAB = \measuredangle FCD = \measuredangle CBA + \measuredangle BDC = \measuredangle CEA + \measuredangle DAP = \measuredangle ACE + \measuredangle BCA = \measuredangle BCE = \measuredangle BAE$ so $\triangle AEB\cong \triangle AFB$ and $AB$ is the perpendicular bisector of $EF$. But then $G$ is the intersection of the perpendicular bisector of $EF$ and the angle bisector of $\angle ECF$, which is the midpoint of $\overarc{EF}$ of $(CEF)$, as desired.