<Cir-2024-780>, hint

Let M' be the reflection of M in K.

Then triangles NDM' and NBM are congruent

*ND=NB, DM'=MC=MB, Let line AB cut MN in P, <ADN=<NBP, 

<ADM'=<ACM=<MBP hence <NDM'=<NBM. MN=MM'

Since triangle MNM' is isosceles and MK=M'k.  <MKN=90.