Problem

Source:

Tags: geometry, angle bisector



In the triangle $ABC$ given that $\angle ABC = 120^\circ .$ The bisector of $\angle B$ meet $AC$ at $M$ and external bisector of $\angle BCA$ meet $AB$ at $P.$ Segments $MP$ and $BC$ intersects at $K$. Prove that $\angle AKM = \angle KPC .$