By Pascal on $(A_{1}A_{1}C_{1}B_{1}B_{1}K)$, we get that $C$, $N$ and $M$ are collinear. $C_{1}K$ being a diameter, implies that $\angle NB_{1}M$ and $\angle NA_{1}M$ are right. So, quadrilateral $NA_{1}B_{1}M$ is cyclic. Now, $\angle NMB_{1} = \angle C_{1}A_{1}B_{1} = \angle C_{1}B_{1}A = \angle CB_{1}M \implies C$ is the midpoint of $MN$. Finally, $MN=2*MC=2*CB_{1}=CB_{1}+CA_{1}=CA+CB-AB=b+a-c$. Note: You probably meant that $M$ is the intersection between $C_{1}B_{1}$ and $A_{1}K$.

$(A_{1}A_{1}C_{1}B_{1}B_{1}K)$ by packal teorem M C N are collinear.C mid point for MN