MathDB
Iranian geometry olympiad 2014

Source: Iranian geometry olympiad 2014

February 22, 2015
geometrygeometry proposed

Problem Statement

In the Quadrilateral ABCDABCD we have ∡B=∡D=60∘ \measuredangle B=\measuredangle D = 60^\circ .MM is midpoint of side ADAD.The line through MM parallel to CDCD meets BCBC at PP.Point XX lying on CDCD such that BX=MXBX=MX.Prove that AB=BPAB=BP if and only if ∡MXB=60∘\measuredangle MXB=60^\circ.
Author: Davoud Vakili, Iran
Back to ProblemsView on AoPS