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

Posted by Antonio Gutierrez at 6:29 PM
Labels: , , ,

3 comments:

  1. Peter TranOctober 15, 2011 at 7:40 PM

    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. PravinOctober 16, 2011 at 10:05 AM

    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
  3. APOSTOLIS MANOLOUDISDecember 25, 2016 at 1:08 PM

    Problem 677
    Are 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

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