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Let $BN=x$ and $NA=y$. From the conditions, let's deduct some convenient conditions that seem sufficient to solve the problem. $M$ is the midpoint of side $AC$. This implies that $[ABX]=[CBX]$. Given that angle $ABX$ is $60$ degrees and angle $BXC$ is $120$ degrees, we can use the area formula to get $$\frac{1}{2}(x+y)x\frac{\sqrt{3}}{2}=\frac{1}{2}x\cdot CX\frac{\sqrt{3}}{2}$$ So, $x+y=CX$ .....(1) $CN$ is angle bisector. In the triangle $ABC$, one has $BC/AC=x/y$, therefore $BC=2x/y$.....(2) Furthermore, triangle $BCN$ is similar to triangle $MCX$, so $BC/CM=CN/CX$, therefore $BC=(CX+x)/CX=(2x+y)/(x+y)$....(3) By (2) and (3) and the subtraction law of ratios, we get $$BC=2x/y=(2x+y)/(y+x)=y/x$$ Therefore $2x^2=y^2$, or $y=\sqrt{2}x$. So $BC=2x/(\sqrt{2}x)=\sqrt{2}$. Finally, using the law of cosine for triangle $BCN$, we get $$2=B{C}^2=x^2+(2x+y{)}^2-x(2x+y)=3x^2+3xy+y^2=\left(5+3\sqrt{2}\right)x^2$$ $$x^2=\frac{2}{5+3\sqrt{2}}=\boxed{\frac{10-6\sqrt{2}}{7}}.$$