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Part of 2014 Iran Geometry Olympiad (senior)

Problems(1)

Iranian geometry olympiad 2014

Source: Iranian geometry olympiad 2014

2/22/2015
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
geometrygeometry proposed