Prolongamos $DG$, mas alla de $G$, hasta el punto $X$. Si $\angle DGE + \angle DGF=180$ entonces $\angle DGE=\angle FGX$ (*) $FC\parallel EB$ entonces $\angle FCD=\angle EBD= \alpha$ $DCFG$ es ciclico, luego $\angle FGX=\alpha$ (1) $P=EB\cap AC$, entonces $AB=AP$ y $DB=DP$, luego $\angle DPE=\alpha$. $DEGP$ es ciclico, como $\angle PDE=90-\alpha$ entonces $\angle PGE=90+\alpha$. Luego $\angle DGE=(90+\alpha)-90=\alpha$ (2) (1)=(2)=(*)

Let the internal bisector of $\angle BAC$ intersect $BC$ at $T$. Let the rays $DG$ and $EB$ intersect at $S$. Now $\angle CAT=\dfrac{\angle A}{2}, \angle CAF=\dfrac{180^{\circ}-\angle A}{2}=90^{\circ}-\dfrac{\angle A}{2}$. So, $\angle TAF=90^{\circ}=\angle TAE$. Since $\angle CAT=\angle BAT$, we have that $\angle EAB=\angle FAC$. Now since $\angle AES+\angle AGS=180^{\circ}$ we have that $E,A,G,S$ are concyclic. So, $\angle BSD=\angle ESG=180^{\circ}-\angle EAG=\angle FAC=\angle EAB=180^{\circ}-\angle BAD$. So $B,A,D,S$ are cyclic as well. Now, $\EGD=180^{\circ}-\angle EGS=180^{\circ}-\angle EAS=\angle SAD=\angle SBD=180^{\circ}-EBD=180^{\circ}-\angle FCD$. But $\angle CFD+\angle CGD=180^{\circ}$ means that $C,F,D,G$ are concyclic. So, $\angle FCD=\angle FGD$. Thus $\angle EGD=180^{\circ}-\angle FGD$, as desired.

Let the internal angle bisector of $\angle{BAC}$ meets $BC$at $D'$ then the ratio $(B, C ; H, D)$ is harmonic and because $F, E, A, D$ are the perpendicular projections from $B, C, D', D$ respectively, then the ratio $(F, E; A, D)$ is also harmonic. Since $\angle{AGD} = 90^{\circ}$ and $(F, E; A, D) = -1$ then $GA$ bisects $\angle{EGF}$. Notice that $\angle{CED} + \angle{CGD} = 90^{\circ} + 90^{\circ} =180^{\circ}$, we have $CGDE$ is cyclic, hence if $GF$ meets $BD$ at $P$, we get $\angle{PGA} = \angle{CGE} = \angle{PDA}$, so $ADGP$ is cyclic hence $\angle{APD} = 90^{\circ}$. Thus, $APBF$ is cyclic, so $\angle{DGE} + \angle{DGF} = \angle{DCE} + \angle{PAF} = \angle{DBF} + \angle{PAF} = 180^{\circ}$.

Nothing complicated is required. $EAG$ is similar to $BAD$ then $\angle DGE=90+D$ and $\angle DGF=90-D$.

Another solution: Let X be the projection of A on to BC.Now,we have two cyclic quadritedrals,GDAX and BXAE.Now,from cyclic quadritedral GDFC we have <DGF=<DCF=90-(<C-<B)/2.Also,<EXC=90+GXD+<EXA=90+<XAD+<XDA=180 => G,X and E are collinear => <EGA=<XGA=<XDA=(<C-<B)/2,and now the conclusion easily follows...

here is one line solution : DI pedencular AB then DGE=DIE=DBE=DCF=180-DGF ( symetric , AD bisector)

Purely Projective! Let $X$ be the intersection of $A-$angle bisector with $BC$, then, $$-1=(D,X;B,C) \overset{\infty_{AX}}{=} (D,A;E,F) \implies \angle EGA=\angle FGA \implies \angle DGF+\angle DGE=180^{\circ}$$

Solution without projective geometry (Same logic though ) First step is the same. Let the internal bisector of $\angle A$ intersects $BC$ at $X$. I think this is the most important step.

Furthermore we know that from similarity ( $EB // AX // FC$ ) and properties of bisectors $$\frac{EA}{AF} = \frac{BX}{XC} = \frac{DB}{DC} = \frac{DE}{DF}$$We can easily see that (from the properties of bisectors, again) $GA$ is the bisector of $\angle EGF$. WLOG $\angle EGA = \alpha$. Then we have that $\angle DGF = 90+\alpha \; \; \angle DGE = 90- \alpha \Rightarrow \angle DGF + \angle DGE= 180$.

Mine is very brute force solution. Note that it is equivalent to show $\angle DGE = \angle EBD$. By the ratio lemma, we get $\frac{DE}{DF} = \frac{GE\cdot \sin{\angle DGE}}{GF\cdot \sin{(\angle C + \angle{\frac{A}{2}})}} = \frac{DB}{DC} = \frac{AB}{AC} = \frac{\sin{\angle C}}{\sin{\angle B}}$. We also know $\frac{GE}{GF} = \frac{\sin{\angle C}}{\sin{\angle GED}}$ So, $\frac{\sin{\angle DGE}}{\sin{\angle GED}} = \frac{\sin{(\angle C + \angle{\frac{A}{2}})}}{\sin{\angle B}}$ But we also know that $\angle DGE + \angle GED = 180 - \angle{\frac{A}{2}} = (\angle C + \angle{\frac{A}{2}}) + \angle B$. So, by a well known lemma, we get that $\angle DGE = \angle C + \angle{\frac{A}{2}} = \angle EBD$. Done!