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Problem: In a triangle ABC, AC=BC and $ \angle C =20°$. There are points M and N on AC and BC respectively such that $\angle BAN=50°$ and $\angle MBA =60°$.Find the angle BMN?

I tried angle chasing and realised that something additional must be required to solve this problem like a construction somewhere. Can anyone tell me what that construction is? Also try to add your motivation behind the construction.

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Obviously $\angle BAC=\angle ABC=80^\circ$. It also means that $\angle NAC=30^\circ$ and $\angle MBC=20^\circ$

Construct point $E\in BM$ such that triangle ABE is equilateral. Extend $AE$ until it interesects $BC$ at point $F$.

It's easy to see that triangle $EFM$ is also equilateral and $MF\parallel AB$. It is also easy to claculate that $\angle NAE=10^\circ$, $\angle EAM=20^\circ$, $\angle EBF=20^\circ$.

From triangle $ABN$ you can calculate $\angle ANB=50^\circ=\angle NAB$. It means that triangle $ANB$ is isosceles and therefore $BN=BE=BE$. So you can draw a (pink) circle with center $B$ passing through points $N,E,A$. Inscribed angle $\angle ANE$ is one half of the central angle $\angle ABE=60^\circ$ i.e. $\angle ANE=30^\circ$.

Now that you know $\angle NAE=10^\circ$ and $\angle ANE=30^\circ$, from triangle $ANE$ you get $\angle NAF=10^\circ+30^\circ=40^\circ$ i.e. $\angle NEM=100^\circ$

On the other side, from triangle ABF it is obvious that $\angle BFA=40^\circ$ and $\angle NFM=100^\circ$.

Take a look at triangles $MEN$ and $MFN$. We know that: $ME=MF$ (from the equilateral triangle $MEF$), $MN=MN$ and that angles opposite to the longest sides are equal, $\angle MEN=\angle MFN=100^\circ$.

By SSU, triangles $MEN$ and $MFN$ are equivalent which means that $\angle BMN=\angle EMN = \angle FMN=\frac 12 \angle EMF = 30^\circ$.

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