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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 865.
OM//FD => ΔHMO~ΔHFD => HM/HF = HO/HD OE//GD => ΔHOE~ΔHDG => HE/HG = HO/HDSo HM/HF = HE/HG, ΔHME~ΔHFG and hence ME//FG
http://img521.imageshack.us/img521/741/58400192.pngBS//GC and angle SBE=angle ECG ,angle SEB=angleCEG,BE=EC ,so ,triangles SBE,ECG are equal,therefore ,SB=CG.QS//EF,so,HB/HO=BS/OE=BQ/OF. But, OF=OE, so ,SB=BQ=CGBQ/QA=NB/AF, CG/GD=CT/FD .But, BQ/QA= CG/GD .So, NB/AS= CT/FD. Because AS=FD,is ,NB=CT,therefore, NB+BE=EC+CT , namely,NE=ET and E ,is,midpoint of NT.Because,O midpoint of EF and OM//AD//BC, M,is midpoint of FN.So ,from the triangle FNT ,is ME//FT .
Since OE//DG => HE/HG= HO/HDSince OM//FD => HO/HD=HM/HFFrom above relations we have HE/HG=HM/HF => ME//FG