Sunday, July 27, 2008

Elearn Geometry Problem 146 Varignon s theorem and more

See complete Problem 146
Varignon's Theorem: Quadrilateral, Midpoints, Parallelogram, Area, Perimeter. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. well, to prove that EFGH is a parallelogram, one would take triangle CBD, and use the the midline theorum to get that FG is parallel to BD, then do the same for triangle ADB to get that EH is parallel to BD, and FG is parallel to BD, thus, EH and FG are parallel. Then do it again with triangles BAC and DCQA to get that HG, EF, and AC are all parallel. Now that EFGH has sides EF and GH parellel, and sides FG, and EH parallel, it is mow a parallelogram.

    How would the others be proven?

  2. 2. By the midline theorem: EF = GH = AC/2 and EH = FG = BD/2.

  3. iii
    area of trangle EBF = area of trangle FAE(as they have the same vertex and equal bases), area of trangle BAF = area of trangle FAC
    so EBF = 1/4 ABC Using the sam method it can be proved that GDH = 1/4 ADC so EBF + GDH = 1/4 ABCD

    Using the sam method it can be proved that EAH + FCG = 1/4 ABCD
    so EAH + FCG + EBF + GDH = 1/2 ABCD
    by subtracting the areas EFGH = 1/2 ABCD

  4. in triangle ABC, E and F are the midpoints of AB, AC respectively. thus by midpoint theorem we have EF=(1/2)*AC. similarly applying the theorem to triangle ADC, we get GH=(1/2)*AC. also by the same theorem we get that EF, AC and GH are parallel to each other. therefore EFGH is a parallelogram because a pair of opposite sides are equal and parallel.
    EF=GH=(1/2)*AC and EH=FG=(1/2)*BD
    these results are due to midpoint theorem.
    adding these results we get EF+FG+GH+HE=AC+BD.
    join AG. in triangleAGD, GH is the median as H is the midpoint of AD. so S4=(1/2)*area of triangleAGD. similarly in triangleACD, AG is the median and area of triangleAGD=(1/2)*area of triangleACD. so from the above relations we get, S4=(1/4)*area of triangleACD.
    similarly, S2=(1/4)*area of triangleABC
    S3=(1/4)*area of triangleBCD
    S1=(1/4)*area of triangleBAD
    S1+S2+S3+S4=(1/4)*(area of triangles ACD+ABC+BCD+BAD)=(1/4)*2*area of quadrilateral ABCD=(1/2)*area of qudrilateralABCD
    it is easy to see that the area of quadrilateralABCD=S1+S2+S3+S4+S5
    but S1+S2+S3+S4=(1/2)*area of quadrilateralABCD
    so S5=(1/2)*area of quadrilateralABCD.
    Q. E. D.

  5. (To Antonio Gutierrez)
    sir,I think there's a small mistake:
    EF=1/2AC=GH & EH=1/2BD=FG
    so,the perimeter of EFGH=2*1/2AC+2*1/2BD=AC+BD [Not 1/2(AC+BD)]

    1. To Aditya
      the figure of problem 146 has been updated. Thanks.