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.
Sunday, January 18, 2009
Elearn Geometry Problem 226: Triangle, Centroid, Perpendiculars
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ReplyDeleteThe median MN of the trapezoid ADFC.
Triangle NMG similar to triangle BEG.
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
ReplyDeleteThanks for the hint, AG!
Ajit: ajitathle@gmail.com
See the drawing
ReplyDelete- 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