tag:blogger.com,1999:blog-6933544261975483399.post1005766133052474622..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1180: Quadrilateral, 120 Degrees, Diagonals, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-37369062233644949922016-01-18T04:02:07.322-08:002016-01-18T04:02:07.322-08:00Interestingly Tr. BGC is equilateral from which th...Interestingly Tr. BGC is equilateral from which the result EG = 2 FE is obvious.Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3674262279248813702016-01-17T07:18:39.952-08:002016-01-17T07:18:39.952-08:00OK to Sumith
From congr in tr BGC, GF median and B...OK to Sumith<br />From congr in tr BGC, GF median and BM median (M on GC)<br />=>EG = 2FEc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34102275570866126152016-01-17T05:28:49.138-08:002016-01-17T05:28:49.138-08:00ASA. ED common. <GDE = <CDE and < GED = &...ASA. ED common. <GDE = <CDE and < GED = < CED = 60 which I'm sure u can figure out why.Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26790791872432675042016-01-17T03:08:55.999-08:002016-01-17T03:08:55.999-08:00To Sumith & Peter
Why Tr CED congrc to EGDTo Sumith & Peter <br />Why Tr CED congrc to EGD<br />c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79070844104816483312016-01-16T15:31:35.083-08:002016-01-16T15:31:35.083-08:00Just a typo replace < BGE = < DGE = 30 with ...Just a typo replace < BGE = < DGE = 30 with < BHE = < DHE = 30Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60888401319741114252016-01-16T11:46:53.709-08:002016-01-16T11:46:53.709-08:00I am not sure from "E is the incentre of Tr. ...I am not sure from "E is the incentre of Tr. AHD" , you get " < BGE = < DGE = 30.<br />please explain .<br /><br />Peter<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27128497181731323712016-01-16T11:13:55.303-08:002016-01-16T11:13:55.303-08:00http://s1.postimg.org/kgkafchkf/pro_1180.png
BA m...http://s1.postimg.org/kgkafchkf/pro_1180.png<br /><br />BA meet CD at M<br />Since ∠(A)+ ∠(D)= 120 =>∠(M)=60 <br />Since EA and ED bisect ∠(A) and ∠(D) => E is in center if trị. MAD and ME bisect ∠((M) and ∠( AED)=120<br />MBEC is cyclic and chord BE=chord EC => trị BED is isosceles <br />Triangle EFC is 30-60-90 triangle => EC= 2. EF<br />Triangles ECD and EDG are congruence…. ( case ASA) => EG=EC=2. EF <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39284691061810259262016-01-16T10:09:38.296-08:002016-01-16T10:09:38.296-08:00I could not complete the last line
Hence EG = 2FE...I could not complete the last line<br /><br />Hence EG = 2FE <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50288727426315754742016-01-16T10:07:51.821-08:002016-01-16T10:07:51.821-08:00Let AB, DC meet at H so that E is the incentre of ...Let AB, DC meet at H so that E is the incentre of Tr. AHD and < BGE = < DGE = 30.<br /><br />Also easily < ACD = 60 + A/2 = < DBG ( A + 60 - A/2) making BECH concylic<br /><br />Hence BE = CE = 2FE since Tr . ECF is 30-60-90.<br /><br />Further Tr. s GED is congruent to Tr. CED case ASA making EG = EC Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com