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Lines AP, AK must be isogonal

Source: Sharygin Correspondence Round 2024 P10

March 6, 2024
geometry

Problem Statement

Let ω\omega be the circumcircle of triangle ABCABC. A point TT on the line BCBC is such that ATAT touches ω\omega. The bisector of angle BACBAC meets BCBC and ω\omega at points LL and A0A_0 respectively. The line TA0TA_0 meets ω\omega at point PP. The point KK lies on the segment BCBC in such a way that BL=CKBL = CK. Prove that ∠BAP=∠CAK\angle BAP = \angle CAK.
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