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

Labels:
center,
diagonal,
parallel,
parallelogram

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http://img29.imageshack.us/img29/8348/problem621.png

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