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Found problems: 1

2024 Yasinsky Geometry Olympiad, 3

Inside triangle \( ABC \), points \( D \) and \( E \) are chosen such that \( \angle ABD = \angle CBE \) and \( \angle ACD = \angle BCE \). Point \( F \) on side \( AB \) is such that \( DF \parallel AC \), and point \( G \) on side \( AC \) is such that \( EG \parallel AB \). Prove that \( \angle BFG = \angle BDC \). [i]Proposed by Anton Trygub[/i]