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Geometry ProblemClick the figure below to see the complete problem 576.Zoom at: Geometry Problem 576
draw BEG, BFH, G on AC, H on CDBE/BG = BF/BH = 2/3 => EF//GHGH = 1/2 AD, GH//AD => EF//GH//ADEF = 2/3 GH = 2/3∙1/2AD = 1/3 AD
http://img694.imageshack.us/img694/5306/problem576.pngLet H is the midpoint of BC. Connect AH and DH ( see attached picture)Since E is the centroid of tri. ABC so A, E and H are collinear and HE=1/2 AESince F is the centroid of tri. DBC so D, F and H are collinear and HF=1/2 DFSo EE//AD (Thales theorem) and EF/AD=HF/HD= 1/3Peter Tran
let G be the midpoint of BC. join AG and DG. these are the medians of the triangles ABC and BDC respectively. so the centroids E and F will also lie on AG and DG respectively. that is AEG and DFG are line segments. as E and F are centroids, we have AE/EG = FD/FG = 2/1 as the centroid divides the medians in the ratio 2:1 internally. so we have EG/AG = FG/DG = 3/1. so that by the converse of basic proportionality theorem we have EF parallel to AD. and consequently the triangles GEF and GAD are similar and AD/EF = GD/GF = AG/GE = 3/1. which implies that EF = AD/3.Q. E. D.
If P us the midpoint of BC, AEF and DFP are collinear and both results follow