The question is from primary school Mathematics competitions. It looks easy, but so hard to find the angle using elementary school knowledge.
In the diagram below, ABC is a triangle with $\angle BAC=120$ and $\angle ABC=40$. M is a point on AC such that AM=AB and N is a point on the extension of BA such that $\angle AMN=40$. What is the angle measure, in degrees, of $\angle BNC$?
I have found the degrees of most of the other angles as follows.
If you write $MN$ and $MC$ in terms of $BM$ by applying law of sines in respective triangles, it is easy to see $MN = MC$.
For a synthetic solution, draw equilateral triangle $\triangle BMP$ on side $BM$ as shown in the diagram. As $BN$ is perp bisector of $MP$, $PN = NM$ and we know $\angle PMN = 10^\circ$. So, $\angle MPN = 10^\circ$.
As $\angle PBC = \angle BPN = 70^\circ$ and $MA$ is perp bisector of $BP$, $PN$ extend will meet $AM$ extend at $C$.
$ \therefore \angle MNC = 20^\circ$.