We have $\angle ADE = \angle B$ and $\angle ADF = \angle C$. So $AFDE$ is a cyclic

quadrilateral and
$$\angle AFE = \angle ADE = \angle B \text{ and } \angle AEF = \angle ADF = \angle C,$$
Hence, $EF$ is parallel to $BC$ and $DE = DF$, so the triangle $FDE$ is isosceles.

Let $M$ be the midpoint of $EF$, $L = AD \cap EF$, and $H = DG \cap EF$. By Thales Theorem, we obtain
$$\frac{LE}{LF} = \frac{DC}{BD}$$
Also, from $\triangle FGH \sim \triangle CGD$ it follows that
$$\frac{HF}{DC} = \frac{GH}{GD}$$
From $\triangle EGH \sim \triangle BGD$, we get
$$\frac{HE}{BD} = \frac{GH}{GD}$$
and from this
$$\frac{HF}{HE} = \frac{DC}{DB}$$
then $H$ and $L$ are symmetric with center $M$, as we will see later and $\angle LDM = \angle HDM = \alpha$ with
$$\alpha = 90^\circ - \angle ADC = \frac{\angle C - \angle B}{2} \text{ if } C > B,$$
and
$$\alpha = 90^\circ - \angle ADB = \frac{\angle B - \angle C}{2} \text{ if } B > C.$$
Finally,
$$\angle GDE = \angle ADE - \alpha = \angle ADF \text{ if } B > C,$$
and
$$\angle GDE = \angle ADE + \alpha = \angle ADF \text{ if } B < C.$$
It remains to prove that $L$ and $H$ are symmetric with center $M$. To do so denote
$$LF + LE = HF + HE = s, \text{ and then}$$
$$\frac{LE}{LF} = \frac{HF}{HE}$$
hence
$$\frac{s}{LF} = \frac{s}{HE} \Rightarrow LF = HE \text{ and } LE = HF$$
that is, $HM = ML$, as we wanted to prove.