Saturday, June 11, 2011

Online Geometry Problem 621: Two Parallelograms, Diagonals, Centers, Parallel Lines

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

Click the figure below to see the complete problem 621.

 Online Geometry Problem 621: Two Parallelograms, Diagonals, Centers, Parallel Lines.

2 comments:

  1. http://img29.imageshack.us/img29/8348/problem621.png
    Let KJ cut AD at M
    Since L and H are midpoints of AC and GC so We have LH//AG
    AD=EJ, EK=GD and angle( KEJ)= angle (GDA)
    So triangle AGD congruence with triangle JKE , case SAS
    So angle (DAG)= angle (EJK)= angle (JDM)
    And AG//KJ//LH

    Peter Tran

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  2. LH ∥ AG (∵ LH joins midpoints of CA, CG)
    Diagonal ED is common to parallelograms
    EADJ, EKDG.
    So ED,AJ,KG bisect each other.
    Hence AKJG is a parallelogram and AG ∥ KJ
    Follows LH ∥ KJ

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