Nice problem! Let BK = CN = AM = x; AB = 2a. MH perpendicular to AB. \[AH = MH = \frac{x}{sin{45^{\circ}}}\]\[ \frac{MH}{DH}\ = \frac{x}{x + a\cdot\sin{45^{\circ}}} = \frac{CN}{CK} \]$\angle NCK = \angle MHD = \angle 90^{\circ}$ so we have $MHD \sim NCK$ Let NK intersection AB at L. Then $\angle NKC = \angle MDH$ and $\angle LBK = \angle DAM$ so $MAD \sim LBK$ $\angle ({KN}, {DM}) = \angle ({KN}, {AB}) + \angle ({AB}, {DM}) = \angle ({DM}, {AC}) + \angle ({AB}, {DM}) = \angle ({AB}, {AC}) = \angle 45^{\circ}$ Answer: $\angle 45^{\circ}$