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< KCE = < LCP , 4 circles related, easy version

Source: 2019 RMM Shortlist G4, version 1

June 18, 2020
geometryequal anglescircles

Problem Statement

Let Ω\Omega be the circumcircle of an acute-angled triangle ABCABC. Let DD be the midpoint of the minor arc ABAB of Ω\Omega. A circle ω\omega centered at DD is tangent to ABAB at EE. The tangents to ω\omega through CC meet the segment ABAB at KK and LL, where KK lies on the segment ALAL. A circle Ω1\Omega_1 is tangent to the segments AL,CLAL, CL, and also to Ω \Omega at point MM. Similarly, a circle Ω2\Omega_2 is tangent to the segments BK,CKBK, CK, and also to Ω\Omega at point NN. The lines LMLM and KNKN meet at PP. Prove that ∠KCE=∠LCP\angle KCE = \angle LCP.
Poland
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