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 $\measuredangle MGN=120^o$

Since $BNGM$ has an inscribed circle, it must follow pitot theorem i.e. the opposite sides must sum to the same value. Let $x,y,d$ be the lengths of $BM, BN, BG$ respectively.
By law of cosines, we have
$$MG=\sqrt{x^2+d^2-xd\sqrt{3}}$$
$$NG=\sqrt{y^2+d^2-yd\sqrt{3}}$$
By pitot theorem,
$$x+\sqrt{y^2+d^2-yd\sqrt{3}}=y+\sqrt{x^2+d^2-xd\sqrt{3}}$$
$$x-\sqrt{x^2+d^2-xd\sqrt{3}}=y-\sqrt{y^2+d^2-yd\sqrt{3}}$$
Consider the function $f(x)=x-\sqrt{x^2+a^2-xa\sqrt{3}}$ for some real parameter $a$. We have that
$$f'(x)=1-\frac{2x-a\sqrt{3}}{2\sqrt{x^2+a^2-xa\sqrt{3}}}$$
If $u=\frac{x}{a}$, this is equivalent to
$$1-\frac{2u-\sqrt{3}}{2\sqrt{u^2-u\sqrt{3}+1}}$$
$$=1-\sqrt{\frac{4u^2-4u\sqrt{3}u+3}{4u^2-4u\sqrt{3}+4}}$$
$$=1-\sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}$$
Since $u\in\mathbb{R}$, we have $4u^2-4u\sqrt{3}+4\in [1,\infty)$, which implies
$$\frac{1}{4u^2-4u\sqrt{3}+4}\in (0,1]$$
$$\implies 1-\frac{1}{4u^2-4u\sqrt{3}+4}\in [0,1)$$
$$\implies \sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}\in [0,1)$$
$$\implies 1-\sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}\in (0,1]$$
Hence, $f'(x)$ is positive$~\forall x\in\mathbb{R}$, which means $f(x)$ is monotonically increasing.
This means that the only solution to
$$f(x)=f(y)$$
is when $x=y$. So $BM=BN$. From there it follows that the original triangle is equilateral. The rest of the answer follows.
