Problem

Source: Iran Team selection test 2024 - P7

Tags: geometry



Let $\triangle ABC$ and $\triangle C'B'A$ be two congruent triangles ( with this order and orient. ). Define point $M$ as the midpoint of segment $AB$ and suppose that the extension of $CB'$ from $B'$ passes trough $M$ , if $F$ be a point on the smaller arc $MC$ of circumcircle of triangle $\triangle BMC$ such that $\angle FB'A=90$ and $\angle C'CB' \neq 90$ , then prove that $\angle B'C'C=\angle CAF$. Proposed by Alireza Dadgarnia