What is the value of side $AC$ in the triangle below? For reference:The angle B of a triangle ABC measures 60°. The AN and CM medians are plotted. The radius of the circle inscribed in the MBNG quadrilateral
(G is centroid(barycenter) of ABC) measures $\sqrt3$ . Calculate AC.
My progress

$\triangle BED: \\sen30 = \frac{\sqrt3}{BD}\therefore BD = 2\sqrt3\\
cos 30 = \frac{BE}{BD}\rightarrow \frac{\sqrt3}{2}=\frac{BE}{2\sqrt3}\therefore BE = 3\implies BN = 3+\sqrt3\\
\triangle BNG:cos 30 = \frac{BN}{BG}\rightarrow \frac{\sqrt3}{2}=\frac{3+\sqrt3}{BG}\rightarrow BG = \frac{6+2\sqrt3}{\sqrt3}=2\sqrt3+2\\
BG = \frac{2BP}{3}\rightarrow BP = 3\sqrt3 + 3\\
\triangle BPC: tg30 = \frac{PC}{BP}\rightarrow \frac{\sqrt3}{3} = \frac{PC}{3\sqrt3+3} \implies \boxed{ PC = 3+\sqrt3}\\
\therefore \boxed{\color{red}AC = 2(3+\sqrt3) = 6+2\sqrt3}$
My question...only the equilateral triangle meets the conditions? Why if the quadrilateral is indescribable $\measuredangle MGN=120^o$


Given $ \small MBNG$ is a tangential quadrilateral, it is easy to see that,
$ \small BN + MG = BM + GN$
[How? $ \small BE + EN = BF + NH, MK + KG = MF + GH$. In fact there is a theorem called Pitot Theorem that states the same.]
So if $ \small AN = d, CM = e$, we have $ \small MG = \frac{e}{3}, \small GN = \frac{d}{3}$ and we rewrite $\small BN + MG = BM + GN$ as,
$\frac{c}{2} + \frac{d}{3} = \frac{a}{2} + \frac{e}{3}$
$ d + \frac{3c}{2} = e + \frac{3a}{2} \tag1$
Now given the circle is also the incircle of $\triangle BCM$ and $\triangle ABN$, AND the area of both $\triangle ABN$ and $\triangle BCM$ are same, which is half the area of $\triangle ABC$, we conclude they must have the same perimeter.
$c + d + \frac{a}{2} = a + e + \frac{c}{2}$
$d + \frac{c}{2} = e + \frac{a}{2} \tag2$
Subtracting $(2)$ from $(1)$, $c = a$.
As $a = c ~$ and angle between the sides is $60^\circ$, $\triangle ABC$ must be an equilateral triangle.
So we easily find that,
$ \small AC = BC = 2 BN = 2 (BE + EN) =  2 (3 + \sqrt3)$
