Let Ω be the circumcircle of an acute-angled triangle ABC. A point D is chosen on the internal bisector of ∠ACB so that the points D and C are separated by AB. A circle ω centered at D is tangent to the segment AB at E. The tangents to ω through C meet the segment AB at K and L, where K lies on the segment AL. A circle Ω1 is tangent to the segments AL,CL, and also to Ω at point M. Similarly, a circle Ω2 is tangent to the segments BK,CK, and also to Ω at point N. The lines LM and KN meet at P. Prove that ∠KCE=∠LCP.Poland geometryequal anglescircles