See complete Problem 175 at:
www.gogeometry.com/problem/p175_quadrilateral_area_midpoint.htm
Quadrilateral 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 175
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BD//EH, [EAH]+[EHD]=[EAH]+[EHB], so
ReplyDelete[BEDG]=[BED]+[BDG]=[EAD]+[BGC]=[BAH]+[FDC]=[BHD]+[BDF]=[BHF]+[FHD]=[FHC]+[FAH]=[FAHC]
S being the area of quadrilateral ABCD,
ReplyDelete[EDGB]=[DEB]+[BGD], with:[DEB]=[DAE] and [BGD]=[BCG, thus[EDGB]=S/2
In the same manner:
[FAHC]= [AFC]+[CAH], with [AFC]=[ABF] and [CAH]=[CAD, thus [FAHC]=S/2
both quadrilaterals [EDGB] and [FAHC] have the same area.
Applying the carpet theorem proves the proposition.