Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 865.

## Saturday, March 30, 2013

### Problem 865: Parallelogram, Diagonal, Midpoint, Side, Triangle, Parallel, Similarity

Labels:
midpoint,
parallel,
parallelogram,
similarity,
triangle

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OM//FD => ΔHMO~ΔHFD => HM/HF = HO/HD

ReplyDeleteOE//GD => ΔHOE~ΔHDG => HE/HG = HO/HD

So HM/HF = HE/HG,

ΔHME~ΔHFG

and hence ME//FG

http://img521.imageshack.us/img521/741/58400192.png

ReplyDeleteBS//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=CG

BQ/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/HD

ReplyDeleteSince OM//FD => HO/HD=HM/HF

From above relations we have HE/HG=HM/HF => ME//FG