Geometry Problem
Click the figure below to see the complete problem 603.
Sunday, May 22, 2011
Geometry Problem 603: Parallelogram, Diagonals, Congruence, Concurrency
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Geometry Problem
Click the figure below to see the complete problem 603.
http://img405.imageshack.us/img405/8158/problem603.png
ReplyDelete1. Since EHGF is a parallelogram so EF=GH and
Angle (BHG)= Angle (BMF)= Angle (MFD)
and Angle (MEA)= Angle (FPC)= Angle (PGN)
and Triangle(BHG) congruence to Triangle (DFE) Case ASA
AH=BH-AB =DF-DC=CF similarly we also have AE= CG
2. Let O is the intersection of AC and BD
we have Triangle (AEO) congruence to Triangle (CGO) case SAS
so Angle(COG)= Angle (AOE) and E,O,G are collinear
Similarly H,O,F are collinear.
3. We have Triangle ( AQH) congruence to Triangle (CPF) caseASA
and Triangle (HQO) congruence to Triangle (FPO) case SAS
Angle(HOQ)= Angle (FOP) so P,O, Q are collinear
Similarly M,O,N are collinear
Peter Tran
AH // and = CF imply BHDF is a parallelogram and so FH, BD bisect each other at O.
ReplyDeleteSimilarly BEDG is a parallelogram implies that
EG and BD bisect each other at O.