tag:blogger.com,1999:blog-6933544261975483399.post5941604078004338590..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1129: Isosceles Triangle, Circumcenter, Incenter, Parallel Lines, Perpendicular LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-51616733392640111842016-08-23T02:25:44.860-07:002016-08-23T02:25:44.860-07:00"Quad CDOI is, therefore, concyclic"
How..."Quad CDOI is, therefore, concyclic"<br />How does it follow from your previous statements?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55240080022068254702015-08-27T01:04:52.079-07:002015-08-27T01:04:52.079-07:00< IOC = < B from Tr. BOC = < IDC since ID...< IOC = < B from Tr. BOC = < IDC since ID //AB. So IODC is cyclic and so < MOI = < ICD = < A/2 = < ICN. <br /><br />Hence MOCN is cyclic and so < OMC = < ONC = 90.<br /><br />Note - M is where DO meets CI and N is the mid point of AC (note that B, O, I, N are all collinear since Tr. ABC is isoceles) <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27272836912102156822015-07-10T00:52:10.056-07:002015-07-10T00:52:10.056-07:00Let DO & CI, both extended, meet in X. /_AOC=2...Let DO & CI, both extended, meet in X. /_AOC=2B since O is the circumcentre and thus /_COI=/_B. Since DI//BA, /_CDI=/_B. Quad CDOI is, therefore, concyclic and thus, /_IOX=/_DCI=/_C/2. <br />We note that /_OIX=90° - /_C/2 which, in turn, means that /_DXI =180° - (90-C/2) - C/2 = 90°. QED.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com