# Angle chasing geometry problem [duplicate]

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.

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 $$\angle BMN=\angle EMN = \angle FMN=\frac 12 \angle EMF = 30^\circ$$.