Solve for $x$ in the following figure, given that $AM=MC$ I need to find $x$ in the following figure

My progress: I tried to draw some auxiliary lines but I was unsuccessful. In geogebra I discovered an interesting relationship: $\triangle AFC$ is rectangle. An equation is still missing
 A: From triangle $\triangle AMB$:
$$\frac{AM}{BM}=\frac{\sin105^\circ}{\sin x}\tag{1}$$
From triangle $\triangle BMC$:
$$\frac{MC}{BM}=\frac{AM}{BM}=\frac{\sin15^\circ}{\sin \angle C}=\frac{\sin15^\circ}{\sin (60^\circ-x)}\tag{2}$$
From (1) and (2):
$$\frac{\sin105^\circ}{\sin x}=\frac{\sin15^\circ}{\sin (60^\circ-x)}$$
$$\frac{\cos15^\circ}{\sin x}=\frac{\sin15^\circ}{\sin (60^\circ-x)}$$
$$\sin (60^\circ-x)=\tan15^\circ\sin x$$
$$\frac{\sqrt{3}}2\cos x-\frac12\sin x=\tan15^\circ\sin x$$
$$\sqrt{3}=(1+2\tan15^\circ)\tan x$$
$$\tan x=\frac{\sqrt{3}}{1+2\tan15^\circ}$$
$$x=\arctan \frac{\sqrt{3}}{1+2\tan15^\circ}\approx 48.43^\circ$$
A: Here is a similar diagram to yours but instead of $BD$, I have $BN$ which is perpendicular to $AB$.

$\angle CBF = 60^0, \angle CBM = \angle NBM = 15^0$
Also I will use $BC = a, AC = b, AB = c$.
In $\triangle NBC$, by angle bisector theorem,
$\cfrac{BN}{a} = \cfrac{NM}{b/2} \tag1$
We also have $\triangle ABN \sim \triangle AFC$
$\cfrac{AN}{b} = \cfrac{BN}{CF} = \cfrac{c}{AF}\tag2$
Now $CF = \cfrac {\sqrt3 \ a}{2}, BF = \cfrac {a}{2}$
From $(2)$, $BN = \cfrac{\sqrt3 \ a}{2b} AN$ and from $(1)$, $NM = \cfrac {b}{2a} {BN} \ $
So, $NM = \cfrac{\sqrt3}{4} AN$
$AN + NM = \cfrac{b}{2} \implies AN = \cfrac{2 b}{4 + \sqrt3}$
From $(2)$, $\cfrac{2}{4+\sqrt3} = \cfrac{c}{c + a/2} \implies c = (2-\sqrt3) a$
So $AF  = c + \cfrac{a}{2} = \cfrac{5-2 \sqrt3}{2} a$
$\tan x = \cfrac{CF}{AF} = \cfrac{\sqrt3} {5-2 \sqrt3} \implies x \approx 48.43^0$
