Monday, May 19, 2008

Elearn Geometry Problem 59



See complete Problem 59
Right and Equilateral Triangles, Midpoints. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

3 comments:

  1. We let BA & BC be the x-axis & y-axis resply. and use standard nomenclature for Yr. ABC by calling EF=p and AD=q. We, therefore, need to prove that 4p^2=q^2. We've A:(c,0),B:(0,0),C;(0,a) and angle ABD=30 deg. Hence, D:(V3a/2,a/2),
    F:(V3a/4,a/4) and E:(c/2,a/2) since BC=BD=a.
    4p^2 =4[(V3a/4-c/2)^2+(a/4-a/2)^2
    = (V3a/2-c)^2+(-a/2)^2
    =(V3a/2-c)^2+(a/2)^2 -----------(1)
    while q^2 = (V3a/2-c)^2 + (a/2)^2 ---(2)
    By (1) & (2), it is evident that 4q^2 = q^2 or 4EF^2=AD^2 or EF = AD/2
    Ajit: ajitathle@gmail.com

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  2. http://geometri-problemleri.blogspot.com/2009/06/problem-27-ve-cozumu.html

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  3. http://img402.imageshack.us/img402/4481/problem59.png

    Let G is the midpoint of CD
    Connect BE & EG ( see sketch)
    Since E is the midpoint of AC => ∆BEC is isosceles and BE=EC, ∠EBC=∠ECB= α
    ∠FBE=∠ECG=60- α
    Triangle BFE congruence to ∆CGE …. ( Case SAS)
    So EF=EG= x
    E,G are midpoints of AC & CD => EG= a/2= x

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