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ABCD convex, BE=BD/3, AF=AC/3 => parallelogram

Source: Romanian District Olympiad 2006, Grade 9, Problem 3

March 11, 2006
geometryparallelogramgeometry proposed

Problem Statement

Let ABCDABCD be a convex quadrilateral, MM the midpoint of ABAB, NN the midpoint of BCBC, EE the intersection of the segments ANAN and BDBD, FF the intersection of the segments DMDM and ACAC. Prove that if BE=13BDBE = \frac 13 BD and AF=13ACAF = \frac 13 AC, then ABCDABCD is a parallelogram.
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