Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 603.
http://img405.imageshack.us/img405/8158/problem603.png1. 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 ASAAH=BH-AB =DF-DC=CF similarly we also have AE= CG2. Let O is the intersection of AC and BDwe have Triangle (AEO) congruence to Triangle (CGO) case SASso Angle(COG)= Angle (AOE) and E,O,G are collinearSimilarly H,O,F are collinear.3. We have Triangle ( AQH) congruence to Triangle (CPF) caseASAand Triangle (HQO) congruence to Triangle (FPO) case SASAngle(HOQ)= Angle (FOP) so P,O, Q are collinearSimilarly M,O,N are collinearPeter Tran
AH // and = CF imply BHDF is a parallelogram and so FH, BD bisect each other at O.Similarly BEDG is a parallelogram implies that EG and BD bisect each other at O.