<s>[INST] 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 $\measuredangle MGN=120^o$
 [/INST] Since $$BM+GN=BN+MG,$$ in the standard notation we obtain:
$$\frac{c}{2}+\frac{1}{6}\sqrt{2b^2+2c^2-a^2}=\frac{a}{2}+\frac{1}{6}\sqrt{2a^2+2b^2-c^2}$$ or
$$3(a-c)+\sqrt{2a^2+2b^2-c^2}-\sqrt{2b^2+2c^2-a^2}=0$$ or
$$(a-c)\left(1+\frac{a+c}{\sqrt{2a^2+2b^2-c^2}+\sqrt{2b^2+2c^2-a^2}}\right)=0,$$ which gives $a=c$ and our triangle is an equilateral triangle.
Thus, $$AC=2BN=2(EN+BE)=2(\sqrt3+3).$$
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