tag:blogger.com,1999:blog-6933544261975483399.post5935955139945198064..comments2024-05-21T05:01:49.873-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 603: Parallelogram, Diagonals, Congruence, ConcurrencyAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-60819954529700509142011-06-01T03:50:46.009-07:002011-06-01T03:50:46.009-07:00AH // and = CF imply BHDF is a parallelogram and s...AH // and = CF imply BHDF is a parallelogram and so FH, BD bisect each other at O.<br /><br />Similarly BEDG is a parallelogram implies that <br />EG and BD bisect each other at O.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51549781922044469112011-05-22T21:56:59.833-07:002011-05-22T21:56:59.833-07:00http://img405.imageshack.us/img405/8158/problem603...http://img405.imageshack.us/img405/8158/problem603.png<br />1. Since EHGF is a parallelogram so EF=GH and <br /> Angle (BHG)= Angle (BMF)= Angle (MFD)<br />and Angle (MEA)= Angle (FPC)= Angle (PGN)<br />and Triangle(BHG) congruence to Triangle (DFE) Case ASA<br />AH=BH-AB =DF-DC=CF similarly we also have AE= CG<br />2. Let O is the intersection of AC and BD<br />we have Triangle (AEO) congruence to Triangle (CGO) case SAS<br />so Angle(COG)= Angle (AOE) and E,O,G are collinear<br />Similarly H,O,F are collinear.<br />3. We have Triangle ( AQH) congruence to Triangle (CPF) caseASA<br />and Triangle (HQO) congruence to Triangle (FPO) case SAS<br />Angle(HOQ)= Angle (FOP) so P,O, Q are collinear<br />Similarly M,O,N are collinear<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com