WLOG assume that $CD>CB.$ Let $U,V$ be the projections of $A$ on $CD,CB$ ($U$ lying on segment $CD$ and $V$ lying on ray $CB$ beyond $B$). Clearly $\triangle AVB \cong \triangle AUD$ are directly congruent $\Longrightarrow$ $BV=UD.$ Since $\angle MAN=\tfrac{1}{2}\angle BAD=90^{\circ}-\tfrac{1}{2}\angle MCN$ and $CA$ bisects $\angle MCN,$ we deduce that $A$ is the C-excenter of $\triangle CMN$ $\Longrightarrow$ $\odot(A,AU)$ is C-excircle of $\triangle CMN$ touching $MN$ at $X$ $\Longrightarrow$ $MX=MV$ and $NX=NU$ $\Longrightarrow$ $MN=MV+NU=BM+BV+ND-UD=BM+ND.$

My solution: Let $K$ be the point on the ray $AB$ beyond $B$, such that $BK=DN$. Then we have $\angle ABK = \angle ADN$ and $AB=AD$, so the triangles $\triangle AND$ and $\triangle AKB$ are congruent. In particular, $AK=AN$. Moreover, we have $\angle KAM = \angle BAM + \angle NAD = \angle BAD - \angle MAN = \angle MAN$, hence the triangles $\triangle AMN$ and $\triangle AKM$ are also congruent. From this we infer $MN=KM=BM+BK=BM+DN$.

Let's construct the triangle $AED$ such that is congruent to the triangle $AMB$ ($AB=AD,BM=DE$ and $MA=EA$). Since $MA=EA$, $\angle MAN=\angle EAN$ and $AN$ is common, $\triangle MAN$ is congruent to $\triangle EAN$. Then $MN=NE=ND+DE=ND+BM$, as desired.

Continiue segment $BC$ Through $B$ to make point $T$ such that $TB=DN$ it is easy to find that the triangle $TBA$is eqal to triangle $ADN$ so angles $TAB,DAN$are equal so $AT=AN$ and angles $TAM=MAN$ so $AM$ is the perpendicular bisector of $TN$ so $TM =MN$ So $DN +BM =MN$