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Isotomic points intercepted by common inner tangents

Source: Sharygin Finals 2017, Problem 9.8

August 3, 2017
geometryhomothety

Problem Statement

Let AKAK and BLBL be the altitudes of an acute-angled triangle ABCABC, and let ω\omega be the excircle of ABCABC touching side ABAB. The common internal tangents to circles CKLCKL and ω\omega meet ABAB at points PP and QQ. Prove that AP=BQAP =BQ.
Proposed by I.Frolov
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