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Source: Iran MO 2023 3rd round , geometry exam P1

August 23, 2023
geometryincentercircumcircleangle bisector

Problem Statement

In triangle △ABC\triangle ABC , II is the incenter and MM is the midpoint of arc (BC)(BC) in the circumcircle of (ABC)(ABC)not containing AA. Let XX be an arbitrary point on the external angle bisector of AA. Let BX∩(BIC)=TBX \cap (BIC) = T. YY lies on (AXC)(AXC) , different from AA , st MA=MYMA=MY . Prove that TC∣∣AYTC || AY (Assume that XX is not on (ABC)(ABC) or BCBC)
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