Angle Chasing Problem - Moscow 1952 | A Nice Geometry Challenge 

Did it through trigonometry and algebra, alternative way. After assignin angles properly everywhere, and using AC=BC, realizing MC=MB and BN = AB, and then using Sine theorems for triangles MAB, and MNB, I got Sin(Theta+20) = 2 Sin(Theta)Cos 40.. Further trig. expansion yields tan (Theta).. There is a tricky step at this very end with tan (Theta) = sin 20 / (2Cos 40 - Cos 20) result obtained. Only by considering Cos 40 = Cos (60-20) and then expanding, substituting there resulted exact result of tan(Theta)=1/sqrt(3) and therefore Theta should be 30 degrees.

Wow verry nice !!!

Interesting geometrical problem :)

It was also possible to do so. Let the point L on the BC side be such that the angle BAL is 60 degrees and the lines AL and BM intersect at point K. Then the triangles AKB and MKL are equilateral. Given now that the angles of BAN and BNA are 50 degrees each, we get the equations BK=BA=BN. And since the angle of KBN is 20 degrees, the angles of BKN and BNK are 80 degrees each. Further, it is easy to calculate that the angles of NKL and NLK are equal to 40 degrees each. Thus, we finally have: KM=ML, KN=NL. Therefore, the KML and KNL triangles are equal on three sides. Therefore, MN is the bisector of the angle KML equal to 60 degrees. Therefore, the desired angle BMN is 30 degrees.

what can go wrong? once the distance ab is chosen, xe=lab/2 and ye can be calculated and for the calculation of angles just calculate the scalar product of coordinates in intersections

Wow

Too easy

using trig: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Qaj8aOOu2P4.html

Plz can you tell me a soln involving SOT and algebra.... Thinking of where to construct is an art which I think will take quite a bit to master!

what a joke w=180-50-60-2*20

То, что рисует этот нерусский остолоп, не является очевидным, так как геометрия это наука о красоте линий и форм, а не каракули младенца на стенах!