See complete Problem 177 at:
www.gogeometry.com/problem/p177_parallelogram_area_midpoint.htm
Parallelogram with Midpoints, Areas. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Friday, September 12, 2008
Elearn Geometry Problem 177
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BH and FD cut AG at I and J, FD cuts EC at K, L is a point on BH such that LJ//AD
ReplyDeleteDJ=JK, S_1=[JAD], so are the others
S_1=S/5 is obvious
Very good solution
DeleteNICE SOLUTION:
ReplyDeleteCLICK HERE: http://forogeometras.com/index.php?topic=957.0
...(>_<)...
Name P,Q,R,S vertex of S1. So S(BFDH) = 2,5 S1 (FQ middleline => FQ=1/2(BP)
ReplyDelete=> 2,5 S1 = 1/2(S) => S1 = S/5
Graphical proof that there are 5 identical parallelograms in the figure
ReplyDeleteQED
Let the 4 points of S1 be W, X, Y and Z, naming clockwisely with W being the upper point on BH
ReplyDeleteLet [BEW]=[DYG]=x and [AHZ]=[CXF]=y
By similar triangle, [CWB]=4y so [BFXW]=3y
Since BE=EA and EW//AZ, by mid-pt. theorem, BW=BZ
Join BX, [BXF]=[CXF]=y
[ZWX]=[BWX]=2y
So, S1=4y
On the other hand, [BAH]=4x+y so [BHDF]=8x+2y
S1=[BHDF]-[BWXF]-[HZYD]=8x-4y
So 4y=8x-4y, which means x=y
So S=20y and S1=4y=S/5