MathDB
Macedonia National Olympiad 2011 - Problem 2

Source:

April 16, 2011
greatest common divisorgeometrycircumcircleincentertrigonometrysymmetryfunction

Problem Statement

Acute-angled  ~ ABC\triangle{ABC}  ~ is given. A line  ~ ll  ~ parallel to side  ~ ABAB  ~ passing through vertex  ~ CC  ~ is drawn. Let the angle bisectors of  ~ BAC\angle{BAC}  ~ and  ~ ABC\angle{ABC}  ~ intersect the sides  ~ BCBC and  ~ ACAC at points  ~ DD  ~ and  ~ FF, and line  ~ ll  ~ at points  ~ EE  ~ and  ~ GG  ~ respectively. Prove that if  ~ DE=GF\overline{DE}=\overline{GF}  ~ then  ~ AC=BC.\overline{AC}=\overline{BC}\, .