Find the angle NMC In triangle $ABC$,  $\measuredangle B =  70^{\circ}$,  $\measuredangle C = 50^{\circ} $. On $AB$ and $AC$ take points $M$ and $N$ such that $\measuredangle MCB =  40^{\circ}$,  $\measuredangle NBC= 50^{\circ}$. Find $\measuredangle NMC$.

Another solutions: 
Observe that by angle chasing $BN = CN$ and $BC = MC$. Let point $D$ be chosen on the line $AB$ so that the points $B$ and $M$ lie on the segment $AD$ and $MA = BD$. Consequently, triangles $ACM$ and $DCB$ are congruent by construction and therefore the triangle $ACD$ is equilateral. Draw the line passing through point $N$ and parallel to $AD$ and denote its point of intersection with $CD$ by $K$. Then triangle $CKN$ is also equilateral. Hence 
$$BN = CN = KN = CK$$ However, triangles $BCN$ and $MCK$ are congruent because $$CN = CK, \,\,\, BC = MC, \,\,\, \angle \, BCN = \angle \, MCK = 50^{\circ}$$ so $KM = BN = KN = KC$. Consequently, the circle centered at point $K$ and of radius $KM = KN = KC$ passes through the three points $C, \, M$ and $N$. Therefore, since $\angle \, CMN$ is incsribed in the circumcircle of $CMN$ while $\angle \, CKN$ is a central angle,
$$\angle \, CMN = \frac{1}{2} \, \angle \, CKN = \frac{1}{2} \, 60^{\circ} = 30^{\circ}$$
