Saturday, October 15, 2011

Problem 677: Parallelogram, Midpoint, Diagonal, Metric Relations

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

Click the figure below to see the complete problem 677.

Online Geometry Problem 677: Parallelogram, Midpoint, Diagonal, Metric Relations

2 comments:

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

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  2. Draw DH parallel FE (E on AC). Then
    x/3
    = HE/EC
    = (HE/AE)(AE/EC)
    = (DF/AF)(AD/MC)
    = (2/3)(2)
    = 4/3
    x = 4

    ReplyDelete