∠DMB+∠ENB =∠NBM

∠NBM =∠ NBC +∠CBP+∠PBM.

∠NBC=(45-∠PBC)/2

∠PBM=(∠ABC-∠PBC)/2

∠ABC = 90°

∠NBM =(45-∠PBC)/2+(90-∠PBC)/2+∠PBC = 67.5

3\ Extend AB to B and then to D .. Let the angles FBG = V 1, ECB = V2, ABF= = AFB = V3 (known isosceles), DAG= = Ade = V4 (equal internal dislocation angles).

V (DBC) = v4-v2, and v (GBC) = v (abg) = v3+v 1, then the plane angle agd = v3+v1+v3+v1+v4-v2 =180.