Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 621.
Saturday, June 11, 2011
Online Geometry Problem 621: Two Parallelograms, Diagonals, Centers, Parallel Lines
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ReplyDeleteLet 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
LH ∥ AG (∵ LH joins midpoints of CA, CG)
ReplyDeleteDiagonal 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