Point K is marked on the base AC of an isosceles triangle ABC, and points M and P are marked on the sides AB and BC

Point K is marked on the base AC of an isosceles triangle ABC, and points M and P are marked on the sides AB and BC, respectively, with PK = MB, ﮮ KPC = 80 °, ﮮ C = 50 °. Prove that KMBP is a parallelogram.

Our triangle is isosceles, and in an isosceles triangle, the two sides are equal.
Since AB = BC, and MV = RK, then RK = BP. On the other hand, the segments of points M and P are released to one common point K, and since we have an isosceles triangle, these two segments are equal, that is, RK = MK. Hence MK = PK = MB = BP. That is, the CMWR figure is a rhombus. And any rhombus is a parallelogram.