Let us show that $|FA| = |FG|$. We will proceed: from behind. On the extension of $CD$ we take the point $P$ such that $|DP| = |DA|$ and similarly on extension of $DC$ we take the point $Q$ such that $|CQ| = |CA|$. Then, using well known properties of the incircle, we have
$$\begin{aligned} |PE| &= |PD| + |DE| = |DA| + \frac{|BD| + |CD| - |BC|}{2} = \frac{|BD| + |CD| + |BC|}{2} \\ |QE| &= |QC| + |CE| = |AC| + \frac{|BD| + |CD| - |BC|}{2} = \frac{|BD| + |CD| + |BC|}{2} \end{aligned}$$
This means that the line $EF$ is the axis of the segment $PQ$. In particular, it means that the circumcenter $O$ of triangle $\triangle APQ$ lies on the line $EF$ as well as on the axes of segments $AP$ and $AQ$. This means that
$$\angle DAO = \angle OPD = \angle CQO = \angle OAC$$
So points $O$ and $F$ coincide. Moreover
$$\angle AOP = 2\angle AQP = \angle AQC + \angle CAQ = 180^{\circ} - \angle QCA = \angle ACP$$
Therefore the points $A$, $C$, $F$, $P$ are concyclic. Thus, necessarily $P = G$, and thus $|FP| = |FG| = |FA|$. We have proved what we need. $\Box$

We will show that $FA = FG$. Let $H$ be the center of the excircle of triangle $\triangle ACD$ opposite vertex $A$. Then $H$ lies on the angle bisector $AF$. Let $K$ be the point where this excircle touches $CD$. By a standard computation using equal tangents, we see that $CK = (AD+CD-AC)/2$. By a similar computation in triangle $\triangle BCD$, we see that $CE = (BC+CD-BD)/2 = CK$. Therefore $E = K$ and $F = H$.
Since $F$ is now known to be an excenter, we have that $FC$ is the external angle bisector of $\angle DCA = \angle GCA$. Therefore
$$\angle GAF = \angle GCF = 90^\circ - \frac{1}{2}\angle GCA = 90^\circ - \frac{1}{2}\angle GFA$$
We conclude that the triangle $\triangle GAF$ is isosceles with $FA = FG$, as desired. $\square$