Solution:

First method (angle chasing)
By the properties of the inscribed angle (also in the limiting case of the tangent) the claim is equivalent to proving that $\angle EGA = \angle EFG$.
Let us denote by $\theta$ the measure of $\angle GBC$. Observe that the points $B, E, G$ belong to the same circle centered at $A$ and therefore, since a central angle is twice a corresponding inscribed angle, we deduce that $\angle EAG = 2 \angle EBG = 2\theta$. Since triangle $AEG$ is isosceles on base $EG$, we conclude that
$$\angle EGA = \frac{1}{2}\left(180^\circ - \angle EAG\right) = \frac{1}{2}\left(180^\circ - 2\theta\right) = 90^\circ - \theta.$$
Let us now consider triangle $ABG$. It is an isosceles triangle on base $BG$, and the line $AD$ is the bisector of the angle at the vertex. It follows that the line $AD$ is also an altitude, and in particular is perpendicular to the line $BG$. Denoting by $K$ the intersection point of the lines $AD$ and $BG$, observe that $K$ is the midpoint of $BG$ (in an isosceles triangle the foot of the altitude is the midpoint of the base), while $D$ is the midpoint of $BF$ (by construction). It then follows from Thales' theorem that the line $GF$ is parallel to the line $AD$, and therefore perpendicular to the line $BG$. In particular we deduce that triangle $BGF$ is right-angled at $G$, from which
$$\angle EFG = 90^\circ - \angle GBC = 90^\circ - \theta.$$
Comparing (1) and (2) we obtain the required equality.

Variant
We could alternatively prove that $\angle FGC = \angle FEG$. To compute $\angle FGC$ we exploit the parallelism between $AD$ and $GF$, obtaining that
$$\angle FGC = \angle DAC = \frac{\alpha}{2},$$
where $\alpha$ denotes the measure of the angle at $A$ of triangle $ABC$.
To compute $\angle FEG$, consider the point $G'$, the reflection of $G$ with respect to $A$, and observe that the quadrilateral $BEGG'$ is cyclic and has the angle at $G'$ equal to $\alpha/2$, because the corresponding central angle (at $A$) equals $\alpha$. From the properties of cyclic quadrilaterals it then follows that
$$\angle FEG = 180^\circ - \angle BEG = \angle BG'G = \alpha/2,$$
as desired.

Second method (metric concyclicity)
By the tangent-secant theorem, the claim is equivalent to proving that
$$CG^2 = CF \cdot CE.$$
To do this we compute the lengths of the three segments in terms of the lengths of the three sides, for which we use the standard notations $a = BC$, $b = CA$, $c = AB$.
Since $AG = AB = c$ by construction, we immediately obtain that
$$CG = AC - AG = b - c.$$
By a well-known property of the bisector we know that $BD : DC = AB : AC$, from which
$$BD : (BD + DC) = AB : (AB + AC).$$
It follows that $BD = \frac{ac}{b + c}$, and therefore
$$CF = BC - BF = BC - 2BD = a - \frac{2ac}{b + c} = \frac{a(b - c)}{b + c}.$$
Finally, let us denote by $H$ the foot of the altitude of $ABC$ from vertex $A$. Observe that $AH$ is also the altitude of the isosceles triangle $ABE$, and therefore $H$ is also the midpoint of $BE$, from which
$$CE = BC - BE = BC - 2BH.$$
It remains only to compute the length of $BH$. The classical way to do this is to set $BH = x$ and $CH = y$, and observe that $x$ and $y$ satisfy the system
$$x + y = a, \quad c^2 - x^2 = b^2 - y^2,$$
in which the first equation follows from the fact that $BH + CH = BC$, and the second equation follows from having computed $AH^2$ in two ways, applying the Pythagorean theorem in the right triangles $ABH$ and $ACH$. Solving the system we find that
$$BH = \frac{a^2 + c^2 - b^2}{2a}$$
and in conclusion
$$CE = BC - 2BH = a - \frac{a^2 + c^2 - b^2}{a} = \frac{b^2 - c^2}{a}.$$
From equalities (4), (5) and (6), (3) immediately follows.

Remark
The length of $BH$ can also be obtained as
$$BH = BA \cdot \cos \beta = c \cos \beta,$$
where by $\beta$ we have denoted the angle $\angle ABC$. In turn, $\cos \beta$ can be derived in terms of the lengths of the sides using the law of cosines (sometimes referred to in Italy as Carnot's theorem), according to which
$$b^2 = a^2 + c^2 - 2ac \cos \beta.$$
The computation that we carried out solving the system in $x$ and $y$ is essentially the key step in the proof of the law of cosines.