Given a triangle ABC. point M lies on the side BC it is known that AB = BM and AM = MC

Given a triangle ABC point M lies on the side BC it is known that AB = BM and AM = MC, the angle B = 120 degrees. find the remaining angles of the triangle ABC.

Consider a triangle ABM, it is isosceles (AB = BM by condition). The angle B at the apex of the triangle is 120 °, we find the angles at the base:
∠ BAM = ∠ BMA = (180 ° – 120 °) / 2 = 60 ° / 2 = 30 °.
Consider an isosceles triangle AMC (AM = MC by condition). The angle M at the apex of the triangle is adjacent to the angle BMA, which means that it is equal to:
∠ AMC = 180 ° – ∠ BMA = 180 ° – 30 ° = 150 °.
Find the angles at the base:
∠ МАС = ∠ МСА = (180 ° – 150 °) / 2 = 30 ° / 2 = 15 °.
We find the angle A of the triangle ABC:
∠ А = ∠ BAM + ∠ МАС = 30 ° + 15 ° = 45 °.
Answer: angle A is 45 °, angle C is 15 °.