Thursday, February 24, 2011

Problem 583: Quadrilateral, Diagonals, Circumcenters, Parallelogram, Octagon, Areas

Geometry Problem
Click the figure below to see the complete problem 583.

Problem 583: Quadrilateral, Diagonals, Circumcenters, Parallelogram, Octagon, Areas.
Zoom at: Geometry Problem 583

Posted by Antonio Gutierrez at 5:38 AM
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1 comment:

  1. Peter TranFebruary 24, 2011 at 6:19 AM

    1. GO=GD and HD=HO so HG is the perpendicular bisector of OD and HG perpen. to BD.
    Similarly we also have EF perpen. BD , EG perpen. AC and EH perpen. AC
    So EF//HG , FG//EH and EFGH is a parallelogram.
    2. Note that triangle HDG congruence to OHG (case SSS)
    similarly We also have 3 other pair of triangles congruence. and area(AEBFCGDH) =2 area(EFGH)

    Peter Tran

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