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KZ=KT implies XT \perp YZ

Source: Saint Petersburg olympiad 2024, 11.3

September 22, 2024
geometrycircumcircle

Problem Statement

In unequal triangle ABCABC bisector AKAK was drawn. Diameter XYXY of its circumcircle is perpendicular to AKAK (order of points on circumcircle is B−X−A−Y−CB-X-A-Y-C). A circle, passing on points XX and YY, intersect segments BKBK and CKCK in points TT and ZZ respectively. Prove that if KZ=KTKZ=KT, then XT⊥YZXT \perp YZ.
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