Friday, September 12, 2008

Elearn Geometry Problem 175

Quadrilateral, Midpoint, Triangles, Area

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.
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2 comments:

  1. AnonymousSeptember 12, 2008 at 11:59 PM

    BD//EH, [EAH]+[EHD]=[EAH]+[EHB], so
    [BEDG]=[BED]+[BDG]=[EAD]+[BGC]=[BAH]+[FDC]=[BHD]+[BDF]=[BHF]+[FHD]=[FHC]+[FAH]=[FAHC]

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  2. Gewürtztraminer68February 20, 2019 at 7:15 AM

    S being the area of quadrilateral ABCD,
    [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.

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