MathDB
AD/BC=AF/FB=DG/GC

Source: Bosnia and Herzegovina 2011

May 16, 2011
trigonometrygeometrygeometric transformationhomothetygeometry proposed

Problem Statement

In quadrilateral ABCDABCD sides ADAD and BCBC aren't parallel. Diagonals ACAC and BDBD intersect in EE. FF and GG are points on sides ABAB and DCDC such AFFB=DGGC=ADBC\frac{AF}{FB}=\frac{DG}{GC}=\frac{AD}{BC} Prove that if E,F,GE, F, G are collinear then ABCDABCD is cyclic.
Back to ProblemsView on AoPS