By Stewart's Theorem on the angle bisector, $$C X^{2}=A C \cdot B C\left(1-\frac{A B}{A C+B C}^{2}\right)$$ Thus, $$C X^{2}=4 \cdot 6\left(1-\frac{5}{10}^{2}\right)=18$$ Since $\overline{X M} \| \overline{A C}$ and $\overline{X N} \| \overline{B C}$, we produce equal angles. So, by similar triangles, $X M=X N=\frac{4 \cdot 6}{10}=\frac{12}{5}$. Moreover, triangles $M C X$ and $N C X$ are congruent isosceles triangles with vertices $M$ and $N$, respectively. Since $C X$ is an angle bisector, then $C X$ and $M N$ are perpendicular bisectors of each other. Therefore, $$M N^{2}=4\left(X N^{2}-(C X / 2)^{2}\right)=4 \cdot\left(\frac{12}{5}\right)^{2}-18=\frac{126}{25}$$ and $$M N=\frac{3 \sqrt{14}}{5}$$