tag:blogger.com,1999:blog-6933544261975483399.post431119326573225996..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 365. Circular Sector of 60 degrees, Midpoints, Perpendicular, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-56429812441407426212015-07-31T11:42:22.766-07:002015-07-31T11:42:22.766-07:00OA=OC=OB=AB since Tr. OAB is equilateral
Then fr...OA=OC=OB=AB since Tr. OAB is equilateral <br /><br />Then from mid point theorem GDEF is a rhombus whose diagonals are perpendicular to each other<br /><br />Sumith Peiris<br />Moratuwa <br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19158005615288198012010-10-24T14:23:07.199-07:002010-10-24T14:23:07.199-07:00Anonymous
Thank you for your explanationAnonymous<br />Thank you for your explanationPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76177093924022695212010-10-23T11:48:36.740-07:002010-10-23T11:48:36.740-07:00Midpoints of a quadrilateral connected together al...Midpoints of a quadrilateral connected together always form a parallelogram. But the two diagonals of Quadrilateral ACBO are equal. (Connect OC, revealing OC=OA=OB. Connect AB, revealing Equilateral OAB, so OC=AB.) Quadrilateral ACBO is a rhombus.<br /> Therefore, DF is perpendicular to EG.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23402078169754331272010-09-10T15:07:59.238-07:002010-09-10T15:07:59.238-07:00It is OKIt is OKPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3531394133240478062010-09-08T12:42:50.197-07:002010-09-08T12:42:50.197-07:00To Peter
It is necessary to be rhombus
OAB equilat...To Peter<br />It is necessary to be rhombus<br />OAB equilateral => OC = AB => GD = DE = EF = GFc .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57183212337201830952010-07-14T23:36:40.577-07:002010-07-14T23:36:40.577-07:00FGDE is a parallelogram per C.t.e.o comment not rh...FGDE is a parallelogram per C.t.e.o comment not rhombus.Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69496058847922276422009-12-19T12:23:40.795-08:002009-12-19T12:23:40.795-08:00GF = 1/2 AB ( as middle line of OAB ) (1)
GF // AB...GF = 1/2 AB ( as middle line of OAB ) (1)<br />GF // AB<br /><br />DE = 1/2 AB ( as middle line of ACB ) (2)<br />DE // AB<br /><br />from (1) & (2)<br /><br />GF = De, GF // DE<br /><br />at the same way<br />GD = EF, GD // EF<br /><br />it easy FGDE rombhusc .t . e. onoreply@blogger.com