Let NE be perpendicular to BC and intersect BC at point e,

Let CN=t, then BM=2t.

Because ∠ C = 60, Ne = 4/5t and Ce = 3/5t.

ME = BC-BM-CE = 10-2t-3/5t = 10- 13/5t

When MN∨AB, ∠ NMC = 45, NE=ME= 10- 13/5t.

10- 13/5t=4/5t

t=50/ 17

When △MNC is an isosceles triangle, there are three situations, namely, MN=NC or MN=MC or MC=CN. Because ∠ C = 60 and △MNC are equilateral triangles, the above three situations are combined into one.

Let CN=t, then BM=2t and MC= 10-2t.

t= 10-2t

t= 10/3