Geometry Problem

Click the figure below to see the complete problem 603.

## Sunday, May 22, 2011

Subscribe to:
Post Comments (Atom)

skip to main |
skip to sidebar
## Sunday, May 22, 2011

###
Geometry Problem 603: Parallelogram, Diagonals, Congruence, Concurrency

Online Geometry theorems, problems, solutions, and related topics.

Labels:
concurrent,
congruence,
diagonal,
parallelogram

Subscribe to:
Post Comments (Atom)

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.