Let $N$ on $AC$ extended be such that $EN \parallel BC$. Join $M$ to $N$.

Triangles $\triangle ABC$ and $\triangle AEN$ are similar, hence $|AE| = |AN|$ and so $|BE| = |CN|$. Alternatively, we may define $N$ to be the point on $AG$ extended such that $|BE| = |CN|$ and $C$ is between $A$ and $N$.

Since $|BM| = |CM|$, $|BE| = |CN|$ and $\angle EBM = \angle CGM = 120^\circ$, triangles $\triangle MBE$ and $\triangle MGN$ are congruent by SAS. This implies $|MN| = |ME| = |MD|$.

This means that $M$ is the centre of the circle through $D$, $E$ and $N$. The Central Angle Theorem implies now $\angle DME = 2 \cdot \angle DNE = 2 \cdot 60^\circ = 120^\circ$.

Alternatively, we may observe that the congruence of $\triangle MBE$ and $\triangle MGN$ and the equality $|MN| = |MD|$ imply that $\angle AEM = \angle ANM = \angle ADM$. This implies that the four points $E$, $M$, $A$, $D$ are concyclic. Therefore, the two inscribed angles $\angle DME$ and $\angle DAE = 120^\circ$ are equal. This can also be seen from $\angle AEM = \angle ADM$ by considering the two triangles $\triangle DAF$ and $\triangle EMF$.

Triangles $\triangle MFE$ and $\triangle BME$ are similar, because $\angle MBE = 120^\circ = \angle DME = \angle FME$ and both triangles share an angle at $E$. Therefore, $\angle BFM = \angle MFE = \angle BME$.

First note that $\cos(\theta)$ determines $\theta$ if it is known that $0 \le \theta \le \pi$. Also note that if $n_1$ and $n_2$ are unit vectors, then the angle between them is given by $\cos(\theta) = n_1 \cdot n_2$. We solve the problem by calculating the relevant unit vectors and showing that a certain two dot products are the same.

We place $A$ at $(-\sqrt{3}, 0)$, $B$ at $(0, 1)$ and $C$ at $(0, -1)$ in a Cartesian coordinate system. Then $M$ is at $(0, 0)$. Now $\overrightarrow{MD} = \overrightarrow{MA} + s \cdot \overrightarrow{CA} = (-\sqrt{3} - s\sqrt{3}, s)$ for some $s > 0$. Also $\overrightarrow{ME} = \overrightarrow{MB} + t \cdot \overrightarrow{AB} = (t\sqrt{3}, t+1)$ for some $t > 0$.

Since $|MD|^2 = |ME|^2$, we get $3(s+1)^2 + s^2 = 3t^2 + (t+1)^2$. That is $4s^2 + 6s + 3 = 4t^2 + 2t + 1$. So $(2s + \frac{3}{2})^2 + \frac{3}{4} = (2t + \frac{1}{2})^2 + \frac{3}{4}$. Since $s$ and $t$ are positive, we must have $2s + \frac{3}{2} = 2t + \frac{1}{2}$. That is $t = s + \frac{1}{2}$.

Now
$$\cos(\angle EMB) = \frac{\overrightarrow{MB} \cdot \overrightarrow{ME}}{|MB| \cdot |ME|} = \frac{(0,1) \cdot (t\sqrt{3}, t+1)}{|ME|} = \frac{t+1}{|ME|}$$
Also
$$\begin{aligned} \cos(\angle MFE) &= \frac{\overrightarrow{FB} \cdot \overrightarrow{FM}}{|FB| \cdot |FM|} = \frac{\overrightarrow{AB} \cdot \overrightarrow{DM}}{|AB| \cdot |DM|} \\ &= \frac{(\sqrt{3},1) \cdot (\sqrt{3}+s\sqrt{3},-s)}{2 \cdot |MD|} = \frac{3+3s-s}{2 \cdot |MD|} = \frac{2s+3}{2 \cdot |MD|} \end{aligned}$$
But $|ME| = |MD|$ and $t+1 = s + \frac{3}{2} = \frac{2s+3}{2}$, so $\cos(\angle EMB) = \cos(\angle MFE)$ and hence $\angle EMB = \angle MFE$.