Friday, January 14, 2011

Problem 566: Quadrilateral, Diagonals, Parallel

Geometry Problem
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Problem 566: Quadrilateral, Diagonals, Parallel.
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Posted by Antonio Gutierrez at 7:14 AM
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2 comments:

  1. c .t . e. oJanuary 14, 2011 at 9:49 AM

    ▲EOB ~ ▲OCD => EO/OC = OB/OD
    ▲AOB ~ ▲OCF => AO/OC = OB/OF
    =>
    AO/EO = OF/OD
    from similarity or Thales theorem
    EF//AD

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  2. UnknownJanuary 14, 2011 at 11:13 AM

    Let I is the intersection of AC and BD
    1. Triangle IAB similar to IFC ( Case AA)
    And IF/IB=IC/IA
    So IF/ID=IF/IB * IB/ID= IC/IA * IB/ID (1)
    2. Triangle IBE similar to IDC ( Case AA)
    And IE/IC=IB/ID
    So IE/IA=IE/IC *IC/IA= IB/ID *IC/IA (2)
    3. compare (1) to (2) we have IF/ID= IE/IA and triangle IEF will similar to IAD (case SAS)
    And EF// AD

    Peter Tran

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