Sunday, January 18, 2009

Elearn Geometry Problem 226: Triangle, Centroid, Perpendiculars

Triangle, Altitude, Angles

See complete Problem 226 at:
gogeometry.com/problem/p226_triangle_centroid_perpendicular.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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3 comments:

  1. Antonio GutierrezFebruary 27, 2009 at 9:33 AM

    Hints:
    The median MN of the trapezoid ADFC.
    Triangle NMG similar to triangle BEG.

    ReplyDelete
  2. AjitMarch 26, 2009 at 8:39 PM

    If M & N are midpoints of AC & DF resply. then MN=(d+f)/2. And in similar triangles NMG & BEG we've BE/MN = BG/MG = 2. Thus BE = e = 2MN =d + f
    Thanks for the hint, AG!
    Ajit: ajitathle@gmail.com

    ReplyDelete
  3. rv.littlemanSeptember 11, 2020 at 9:03 AM

    See the drawing

    - Define H as the intersection of BG and AC
    - G centroid => H is in the middle of AC
    - Define I in DF with IH ⊥ DF
    - DA//FC//IH and H middle of AC => IH=(d+f)/2
    - GIH is similar to GEB (aa)
    - =>GB/GH=EB/IH
    - GH=BH/3 and GB=2 BH/3 => GB=2GH
    - Therefore and EB=2IH, e=2(d+f)/2
    - e=d+f

    ReplyDelete

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