Friday, June 8, 2012

Problem 763: Parallelogram, Diagonal, Parallel, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 763.

Online Geometry Problem 763: Parallelogram, Diagonal, Parallel, Congruence.

6 comments:

  1. AG/GC = AE/DC = AE/AB = CF/BC = CH/AH
    1 + AG/GC = 1 + CH/AH
    AC/GC = AC/AH
    GC = AH
    GC - GH = AH - GH
    CH = AG

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  2. Since EF//AC, so let BE/EA = BF/FC = k.
    Then, FH/HD = EG/GD = 1/(k+1).

    Let AD=BC=a, AB=CD=b, then
    S(ADE) = 1/2*a*b*sin A*1/(k+1)
    S(CDF) = 1/2*a*b*sin C*1/(k+1)
    then S(ADE)=S(CDF) (notice that A=C).

    Therefore,
    S(ADG)
    = S(ADE)*DG/DE
    = S(ADE)*(k+1)/(k+2)
    = S(CDF)*(k+1)/(k+2)
    = S(CDF)*DH/DF
    = S(CDH)

    Since ADG and CDH have the same height,
    thus, AG = CH.

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  3. Denote the area of triangle PQR by S(PQR).

    From EF//AC,
    so BE/EA = BF/FC and EG/GD = FH/HD.

    AG/CH
    = S(ADG)/S(CDH)
    = S(ADE)/S(CDF)
    = S(ADB)/S(CDB)
    = 1

    Hence, AG=CH.

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  4. Using the vector approach to solve the problem.
    Denote vector DA = a, vector DC = c.
    Let AE:EB=k:1-k, then BF:FC=k:1-k (as triangle BEF ~ triangle BAC)
    By considering that AG is along AC , AG = rAC = -ra + rc
    By considering that AG divides DE into s:1-s, AG = -sa + k(1-s)c
    Hence r = s = k/(1+k)
    => AG = (k/(1+k))(c-a)
    Symmetrically, CH = -(k/(1+k))(c-a)
    Q.E.D.

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  5. Let BD intersect EF, AC, at L and O respectively.
    In ∆ABC,
    O is the midpoint of AC and EF ∥ AC imply L is the midpoint of EF.
    In ∆DEF,
    L is the midpoint of EF and GH ∥ EF imply O is the midpoint of GH.
    Hence
    AG = OA - OG = OC - OH = CH

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  6. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27378&p=133967

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