<Tri-2024-2094>, hint

Let internal bisector of <A cuts BN at A', Since ACNB is a parallelogram

Triangles CDN and BAA' are isosceles  and <CDN=<CND= <BAA'=<BA'A=<ACB 

Since triangle A'AM is right triangle  (<A'AM=90),  BA'=BA=BM.

BM=CD and parallel, BCDM is a parallelogram , and so <DMN=<DNM 

Hence DM=DN. ( AA'//DN Triangles BAA' and CDN are congruent )