Let the rhombus $ADKL \equiv ABCD$ and let $P \in (DK)$ such that $BM=DP.$ Obviously, $\Delta ABM \equiv \Delta ADP.$ Result $AM=AP, \angle BAM \equiv \angle DAP.$ Follows $\Delta MAN \equiv \Delta PAN,$ so $MN=PN.$ From $MC+CN+MN=2$ and $BM=DP$ we obtain $MN=BM+DN=DP+DN.$ But $MN=PN$ So $PN=DP+DN.$ Result that points $N, D, P$ are collinear, so $ABCD$ is a square.

Remark. Let $w$ be the exincircle w.r.t. the vertex $C$ for the triangle $CMN$. From the relation $CM+MN+NC=2$ results $B,D\in w$ and $MN=BM+ND$, a.s.o.