Thursday, December 19, 2013

Geometry Problem 946: Triangle, Quadrilateral, Area, Diagonal, Midpoint, Parallel

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 946.

Online Geometry Problem 946: Triangle, Quadrilateral, Area, Diagonal, Midpoint, Parallel

2 comments:

  1. let's take the informal pragmatic approach: if the problem statement does not depend on A, B, C, D position but only on quadrilateral area, it holds for a particular case where, say, B=C=E. In this degenerate configuration, it's easy to see that area(CEFH) = area(ABCD)/4 = 7. Hence this must hold for *any* ABCD ;-)

    bleaug

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  2. Let S(ABC) denote the area of ABC, and so on.

    Since EF//BD//GH, so S(CEHF) = S(CEGF).

    Consider quadrilateral CEGF.
    CE = 1/2 CB
    EG = 1/2 BA
    GF = 1/2 DA
    FC = 1/2 DC
    So CEGF~CBAD.

    Hence, S(CEHF) = S(CEGF) = 1/4 S(CBAD) = 7

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