How prove this $\measuredangle DGE+\measuredangle B=\pi$ in $\Delta ABC$,and the point $D$ on $AB$ such $AD=BD$,and $E\in BC,F\in AB,G\in CF$ such
$$\measuredangle BDE=2\angle A, \measuredangle ACF=\measuredangle A+\measuredangle B, BF=2EG$$
show that
$$\measuredangle DGE+\measuredangle B=\pi$$

 A: It's kind of an annoying problem that most likely nobody will appreciate, but just for the sake of feeding my own narcissistic tendencies, I will write you a proof :D (with most likely many typos, which I am not going to try to fix for now).
Constructions and notations. Let point $B^*$ be the symmetric mirror image of point $B$ with respect to line $AC$. Then triangle $ABB^*$ is isosceles with $AB = AB^*$. If $H = AC \cap BB^*$ then $H$ is the midpoint of $BB^*$ and $AH$ is orthogonal to $BB^*$. By assumption, point $E \in BC$ is such that $$\angle \, BDE = 2 \, \angle \, BAC = 2 \, \alpha = \angle \, BAB^*$$ so $DE$ is parallel to $AB^*$. As $D$ is the midpoint of $AB$, the line $DE$ must contain the midsegment of triangle $ABB^*$ parallel to $AB^*$,  so line $DE$ must pass through the midpoint $H$ of segment $BB^*$. Therefore $E \in DH$. Moreover, $DH$ is parallel to $AB^*$ and is one half of it, i.e.  $$DH = \frac{1}{2} \, AB^* =  \frac{1}{2} \, AB^* = AD = BD$$ 

Let $F^* = BC \cap AB^*$. Then, by assumption $$\angle \, ACF = \angle \, BAC + \angle \, ABC = \alpha + \beta = \angle \, ACF^*$$ which means that $AC$ is the angle bisector of angle $\angle \, FCF^*$ so the mirror image of $F$ with respect to $AC$ is $F^*$. As $DE$ is parallel to $AF^*$ and $D$ is the midpoint of $AB$, point $E$ is the midpoint of $BF^*$ (midpoint segment property). Therefore $EH$ is the midsegment of triangle $BB^*F^*$ and is hence parallel to $B^*F^*$ and half of it. However, $B^*F^* = BF$ by reflection symmetry, so $$EH = \frac{1}{2} \, BF$$ Choose point $R$ on $BB^*$ so that line $ER$ is parallel to $AB$ and let $P = ER \cap CF, \,\, S = ER \cap AC$ and $Q = ER \cap AB^*$. Since $E$ is the midpoint of $BF^*$ and $EQ$ is parallel to $AB$, 
the segment $EQ$ is midpoint segment of triangle $ABF^*$ and therefore $$EQ = \frac{1}{2} \, AB = AD = BD$$ Therefore, the quads $ADEQ$ and $BEQD$ are parallelograms. 
Since $EH$ is parallel to $AB^*$ and $ER$ is parallel to $AB$,
$$\angle \, ERH = \angle \, ABB^* = \angle \, AB^*B = \angle \, EHR$$ which means that triangle $EHR$ is isosceles with $EH = ER$. Furthermore, as $D$ is the midpoint of $AB$ and $SR$ is parallel to $AB$,
$$ES = ER = EH = \frac{1}{2} \, BF$$ (if you wish, there is a circle, call it $k$, centered at $E$ and with radius $EH$ which passes through the points $H, R, S$ but it is not crucial for the proof. One can use it later to formulate one of the lemmas in terms ov inversions, but once can also go without it) However, by assumption, $G \in CF$ is chosen so that $$EG = \frac{1}{2} \, BF = EH = ER = ES$$ (which by the way implies that $G$ lies on the circle $k$ too.)  
Lemma 1. $$\frac{EQ}{ES} = \frac{AB}{BF} \,\,\, \text{ and } \, \,\, \frac{SE}{EP} = \frac{AB}{BF}$$  
Proof: As already established $EQ = AD = \frac{1}{2} \, AB$ and $ES = \frac{1}{2} \, BF$ so clearly $$\frac{EQ}{ES} = \frac{\frac{1}{2} \, AB}{\frac{1}{2} \, BF} = \frac{AB}{BF} $$ Since $SE$ is parallel to $AB$ in triangle $ABC$ and $P = SE \cap CF$ by the intercept theorem $$\frac{SE}{EP} = \frac{AB}{BF}$$
Lemma 2. $\,\,\, EP \cdot EQ = EG^2$
Proof: By lemma 1, $$\frac{EQ}{ES} = \frac{AB}{BF} =  \frac{SE}{EP}$$ which, after cross-multiplying, is equivalent to $EP \cdot EQ = ES^2 = EG^2$ because $ES = EG = EH = ER$.
Let $L$ be the intersection point of line $CF$ with line $DE$. 
Lemma 3. The quadrilateral $DLPQ$ is inscribed in a circle $s$. 
Proof: Indeed, since $BEQD$ is a parallelogram ($EQ = BD$ and $EQ \, || \, BD$) $$\angle \, DQP = \angle \, DQE = \angle \, DBE = \beta$$ However, $DL$ is parallel to $AB^*$ so $$\angle \, DLF = \beta = \angle \, DQP$$ which proves the lemma.
Corollary 1. $ \,\,\,\, EL \cdot ED = EG^2$
Proof without inversion: Since the quad $SKPQ$ is inscribed in circle $s$ the  intersecting secants theorem implies that $EL \cdot ED = EP\cdot EQ$. However, by lemma 2 $\, EP \cdot EQ = EG^2$ so $EL \cdot ED = EG^2$. 
Proof with inversion: By lemma 3, the circle $s$ superscribed around $DLPQ$ is mapped to itself by the inversion with respect to circle $k$ because point $P$ is mapped to point $Q$ under the inversion according to lemma 2. Since points $D$ and  $L$ are the intersection points of $ED$ with circle $s$, point $L$ is mapped to point $D$ under the inversion with respect to circle $k$. This implies the corollary. 
Corollary 2. Triangles $EGL$ and $EDG$  are similar.
Proof: The identity  $EL \cdot ED = EG^2$ from corollary 1 is
equivalent to $$\frac{EL}{EG} = \frac{EG}{ED}$$ which combined
with the fact that $\angle \, GEL = \angle \, DEG$ yields that
$EGL$ and $EDG$  are similar triangles.
Concluding the proof. By corollary 2, the two triangles $EGL$
and $EDG$  are similar so $$\angle \, DGE = \angle \, GLE$$
However, $\angle \, DLG = \angle \, DLF = \beta$ so $$\angle \,
DGE = \angle \, GLE = \pi - \angle \, DLG = \pi - \beta = \pi -
\angle \, ABC$$ Hence $\angle \, DGE + \angle \, ABC = \pi$.
