Thursday, December 29, 2011

Problem 709: Parallelogram, Midpoint, Diagonal, Metric Relations, Triangle

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

Click the figure below to see the complete problem 709.

Online Geometry Problem 709: Parallelogram, Midpoint, Diagonal, Metric Relations, Triangle.

2 comments:

  1. http://img607.imageshack.us/img607/5235/problem709.png

    Extend AF and BF to AD and BC( see picture)
    Let AD=BC=DN=CL=2b
    Let CD=2a, AF=FL=2c
    Note that triangle AHD similar to triangle EHL so
    AH/HL=AD/EL=HD/HE=2/3
    From this ratio we get HF=2c/5 and AH/HF=4
    Triangle BGE similar to triangle NGA so AG/GE=4b/b=4
    So GH//EF//BD
    Triangle HAD similar to triangle HLE
    So AG/GE=AD/El=2/3 and EH/ED=3/5=3/x
    So x=5

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  2. Trace EF. BD meet AE and AF at P and Q.
    ΔABQ ~ QFD --> BQ = 2QD
    ΔAPD ~ EPB --> PD = 2BP
    BD = BQ + QD = 3 QD --> QD = 1/3x
    BD = BP + PD = 3 BP --> BP = 1/3x
    By Thales’ theorem: EF // BD , EF = 1/2x.
    ΔBPG ~ ΔGEF --> PG/GE=BP/EF =(x/3)/(x/2)= 2/3
    ΔQHD ~ ΔHEF --> QH/HF=QD/EF =(x/3)/(x/2)= 2/3
    Then PG/GE = QH/HF = 2/3 ----- ( 1 )
    In the trapezoid EFQP, (1) --> GH // BD //EF or MH // BD //EF.
    ΔMEH ~ ΔBED --> MH/BD = GE/EP = 3/(3+2) = 3/5
    Then 3/x = 3/5 --> x=5.

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