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A special incenter configuration

Source: 2022 China TST, Test 1, P4

March 24, 2022
geometryincenter

Problem Statement

Let ABCABC be an acute triangle with ∠ACB>2∠ABC\angle ACB>2 \angle ABC. Let II be the incenter of ABCABC, KK is the reflection of II in line BCBC. Let line BABA and KCKC intersect at DD. The line through BB parallel to CICI intersects the minor arc BCBC on the circumcircle of ABCABC at E(E≠B)E(E \neq B). The line through AA parallel to BCBC intersects the line BEBE at FF. Prove that if BF=CEBF=CE, then FK=ADFK=AD.
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