Sunday, April 4, 2010

Problem 433: Quadrilateral, Triangle, Area, Proportion, Measurement, Similarity

Proposed Problem
Click the figure below to see the complete problem 433 about Quadrilateral, Triangle, Area, Proportion, Measurement, Similarity.

Problem 433: Quadrilateral, Triangle, Area, Proportion, Measurement, Similarity.
See also:
Complete Problem 433

Level: High School, SAT Prep, College geometry

3 comments:

  1. draw BGH//CD, G on altitude from E, H on alt from A to CD
    name h1 altitude from B to CD, name h alt from E to CD
    h2 alt from A to CD
    name EG = x, AH = y
    =>
    h = h1 + x, h2 = h1 + y
    have to prove
    CF∙h1 + FD∙h2 = CD∙h ??? (1)

    CF∙h1 + FD∙(h1+y) = CD∙(h1+x)
    CF∙h1 + FD∙h1 + FD∙y = CD∙h1 + CD∙x
    (CF + FD)∙h1 + FD∙y = CD∙h1 + CD∙x
    FD∙y = CD∙x
    x/y = FD/CD ??? (2)

    BE/AB = x/y (3) from thales theorem
    AE/BE = CF/FD is given (P431)
    =>
    AB/BE = CD/FD
    =>
    BE/AB = FD/CD (4)
    from (3) and (4)
    x/y = FD/CD ( 2 is proved) =>

    (1) is proved

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  2. [ECA]/[BCA] = AE/AB = CF/CD = [ACF]/[ACD] = ([ECA]+[ACF])/([BCA]+[ACD]) = [ECFA]/[ABCD]
    similarly, EB/AB = [BFDE]/[ABCD]

    add them up we get

    AE/AB+EB/AB = [ECFA]/[ABCD]+[BFDE]/[ABCD] <=>
    [ABCD] = [ECFA]+[BFDE] = [BFA]+[CDE] <=>
    [ABCD]-[BFA] = [CDE] <=>
    S1+S2 = S

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  3. This quadrilateral can be transformed to a triangle with other conditions unchanged.
    Extending DC so that C is a vertex of tr(ADC) and the ratio CF:FD is unchanged,
    the problem becomes like this
    "The figure shows a triangle ADC with AE*DF=BE*CF. If S, S1, and S2 are the areas of triangles CED, BCF, and ADF, prove that S = S1 + S2."

    let AE=1, BE=a, DF=b, CF=b/a, AB=d,
    then, tr(AED)=tr(AFB)=tr(ADC)/d
    .....
    QED

    ReplyDelete