<Tri-2025-2243>, hint

Let line AM cut BC at D and line CM cut AB at D

By Ceva's (BD/DC) x (CN/NA) x(AE/EB)=1 --> 

(AB sin20/AC sin60) x(CN/NA)x(AC sin20/BC sin10)=1   ....<1> 

In triangle  AM,  CN/NA=(CM sin NMC/AM sin NMA)=(sin60/sin20)x (sinNMC/sinNMA)..<2>

AB/BC=sin30/sin80=sin30/cos10,  sin20=2 sin10 cos 10...<3>

From <1>, <2>,<3> sinNMC/sinNMA=1, <NMC=<NMA.