Geometry Problem

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

Click the figure below to see the complete problem 677.

## Saturday, October 15, 2011

### Problem 677: Parallelogram, Midpoint, Diagonal, Metric Relations

Labels:
diagonal,
metric relations,
midpoint,
parallelogram

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http://img542.imageshack.us/img542/4607/problem677.png

ReplyDeleteExtend FE to cut BC at H ( see picture)

Let AD=BC=a and DF=2a

Triangle EMC similar to tri. EDA ( case AA)

So EM/MD=MC/AD=1/2

Triangle EMH similar to tri. EDF ( case AA)

So MH/DF=ME/ED=1/2 and MH=a , CH=1.5a

Triangle GCH similar to tri. GDF

so CH/DF=GC/GD=1.5a/2a = 3/x

x=4

Peter Tran

Draw DH parallel FE (E on AC). Then

ReplyDeletex/3

= HE/EC

= (HE/AE)(AE/EC)

= (DF/AF)(AD/MC)

= (2/3)(2)

= 4/3

x = 4

Problem 677

ReplyDeleteAre triangles MEC and AED similar then CE/EA=MC/AD=1/2. In triangle ACD apply the of Menelaus’ theorem intersected by EGF. CE/EA.AF/FD.DG/GC=1 or 1/2.3/2.x/3=1 or x=4.

STOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE