Geometry Problem

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## Thursday, February 24, 2011

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

Labels:
area,
circle,
circumcenter,
diagonal,
octagon,
parallelogram,
quadrilateral

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1. GO=GD and HD=HO so HG is the perpendicular bisector of OD and HG perpen. to BD.

ReplyDeleteSimilarly 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