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 $\measuredangle NMC$.
 A: Let $\Delta MNP$ be such that $P$ placed inside $\Delta ABC$, $BP=CP$ and $\measuredangle PCB=\measuredangle PBC=30^{\circ}$.
Also, let $BP\cap MC=\{Q\}$.
Thus, since $CN=NB$, $\measuredangle CNP=\frac{1}{2}\measuredangle CNB=40^{\circ}$.
From here $\measuredangle CPN=180^{\circ}-20^{\circ}-40^{\circ}=120^{\circ}$ and since $\measuredangle QPC=60^{\circ}$, we see that
$CQ$ and $PQ$ are bisectors of $\Delta CNP$.
Thus, $NQ$ is a bisector of $\Delta CNP$, which gives $\measuredangle CNQ=20^{\circ}$ 
and from here $\measuredangle NQM=10^{\circ}+20^{\circ}=30^{\circ}$.
But $NB\perp QM$ and $BN$ is a bisector of $\angle QBM$.
Thus, $BN\cap QM$ is a midpoint of $QM$, which says that $NQ=NM$ and
$$\measuredangle NMC=\measuredangle NQM=30^{\circ}$$
and we are done!
A: 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,
$$\angle \, CMN = \frac{1}{2} \, \angle \, CKN = \frac{1}{2} \, 60^{\circ} = 30^{\circ}$$
