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prove that FOG = 180 - 2BAC

Source: Sharygin Finals 2023 10.3

August 2, 2023
geometrySharygin Geometry OlympiadSharygin 2023

Problem Statement

Let ω\omega be the circumcircle of triangle ABCABC, OO be its center, A′A' be the point of ω\omega opposite to AA, and DD be a point on a minor arc BCBC of ω\omega. A point D′D' is the reflection of DD about BCBC. The line A′D′A'D' meets for the second time at point EE. The perpendicular bisector to D′ED'E meets ABAB and ACAC at points FF and GG respectively. Prove that ∠FOG=180∘−2∠BAC\angle FOG = 180^\circ - 2\angle BAC.
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