It’s a very nice problem. I’ll add my solution later.

The important step is noticing $EGAC$ is cyclic and this can be proved by angle chasing only. After this it is easy to see external angle bisector of $AGD$ is line $EF$

Lemma. Let $AD$ be a segment and $H_1$ and $H_2$ be two points on the same plane such that $\measuredangle AH_1D + \measuredangle AH_2D = \alpha$. Then the homothety center of the circumcircles of $\Delta AH_1B$ and $\Delta AH_2B$ is a fixed point depending only on $\alpha$. Idea of proof. Let $T$ be the point such that $AT=DT$ and $\measuredangle ATB = \alpha$. Then if line $AT$ additionally intersects the two circles at points $U$ and $V$, a simple angle chase yields $TU \cdot TV = TA^2$. Let $X$ and $Y$ be points on lines $AB$ and $CD$ respectively such that $ABCD \sim XADY$. The lemma implies that there exists a point $W$ different from $G$ (as long as $AB \neq CD$, which is necessary for the problem to make sense) serving as the homothety center for two pairs of circles $(ABD), (ADY)$ and $(ACD), (XAD)$. By Monge applied to $(ABC)$, $(ACD)$ and $(XAD)$, point $E$, $G$ and $W$ are collinear. Similarly we show that $F$, $G$ and $W$ are also collinear, which completes the proof.

The main Claim is the following. Claim: Quadrilaterals $EGAC$ and $FGDB$ are both cyclic. Proof: We will only prove that the first one is cyclic, as the proof for the second follows in a similar manner. Let $EA,EC$ intersect cycle $(ABC)$ at points $P,Q$, respectively and circle $(ADC)$ at points $P',Q'$ respectively. Note that $\angle AEC=\angle AQC-\angle EAQ=(180^\circ-\angle ABC)-\dfrac{\overset\frown{PQ}}{2}=$ $=180^\circ-\angle ABC-\dfrac{\overset\frown{P'Q'}}{2}=180^\circ-\angle ABC-(180^\circ-\angle ADC)=\angle AGD=\angle AGC,$ and so quadrilateral $EGAC$ is cyclic, as desired $\blacksquare$ Back to the problem, we obviously have $EA=EC$, and so $\angle EGC=\angle EAC=\angle ECA=180^\circ-\angle EGA,$ that is $GE$ is the exterior angle bisector of $\angle AGC$. Similarly, the same holds for $GF$, hence points $E,F,G$ are collinear, as desired.

I just tried 2017 G7 today and bruh. My in-contest solution is very similar to it lmaooo. If only I had done 2017 G7 a few weeks earlier.. then this problem probably wouldn’t have had taken me 2 HOURS. Anyway, does anyone know if I can disclose the paper I submitted while I took the exam?

Actually, this problem becomes obvious if you consider Poincare's model of hyperbolic geometry on a half-plane formed by the external bisector of AGD. Let us consider circles (ABC) and (ACD). Obviously, there exists inversion at G such that its composition with reflection over bisector of AGD changes points A and D, B and C, therefore it changes (ABC) into (ACD). All inversions at G are movements in Poincare's model. Reflection over the bisector of AGD is also a movement. Therefore, circles (ABC) and (ACD) are equal. Therefore, there exists a symmetry (on hyperbolic plane) such that the reflection of (ABC) is (ACD). However, symmetry in Poincare's model is inversion with center on the external bisector of AGD in Euclidian geometry. Therefore, common tangents to (ABC), (ACD) intersect on the external bisector.

Nice problem! Here is my solution Let $O_d$ be a center of $(ABC)$, $O_b$ be center of $(ACD)$. By law of sines we can see that $\frac{O_dA}{O_bD}=\frac{R_{(O_d)}}{R_{(O_b)}}=\frac{\frac{AB}{2\sin \angle ACB}}{\frac{CD}{2\sin \angle CAD}}=\frac{AB}{CD}=\frac{AG}{DG}$. Also, since $AD \parallel BC$ we have $\angle O_dAB=90^{\circ}-\angle ACB = 90^{\circ} - \angle CAD = \angle O_bDG$. So, triangles $GAO_d$ and $GDO_b$ are similar and we have that $\angle AGO_d = \angle DGO_b$ and $\frac{GO_d}{GO_b}=\frac{R_{(O_d) }}{R_{(O_b)}}=\frac{EO_d}{EO_b}$, so $GE$ is external bisector of angle $O_dGO_b$. Now, not hard to see that points $O_d, O_b$ both lies inside or both lies outside angle $\angle AGD$. So, we achieved that $GE$ is external bisector of angle $\angle AGD$! But by analogy, we have that $GF$ is external bisector of angle $\angle AGD$! So, $G, E, F$ are collinear! $\blacksquare$