tag:blogger.com,1999:blog-6933544261975483399.post4853108889884001280..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1269 Triangle, Incircle, Incenter, Inscribed Circle, Tangency Points, Perpendicular, 90 Degrees, Angle BisectorAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-55713776803026435462019-01-30T08:22:48.702-08:002019-01-30T08:22:48.702-08:00please explain why <BDA1 = <COA1
Peterplease explain why <BDA1 = <COA1<br />PeterPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39301824828919754952016-10-01T11:00:28.438-07:002016-10-01T11:00:28.438-07:00It follows that < OCA1 = B/2It follows that < OCA1 = B/2Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91951301567506142272016-10-01T10:54:05.599-07:002016-10-01T10:54:05.599-07:00< DEB = 90-B/2 and < A1AE = A/2 so
< EA1...< DEB = 90-B/2 and < A1AE = A/2 so <br />< EA1A = 90-B/2 - A/2 = C/2.<br /><br />But < OCD = C/2 hence OCDA1 is concyclic<br /><br />Therefore <OA1C = <ODC = 90<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52407884979118927642016-10-01T10:44:54.459-07:002016-10-01T10:44:54.459-07:00connect OC, <BDA1 = <COA1, so OCDA1 is on th...connect OC, <BDA1 = <COA1, so OCDA1 is on the same circle. <ODC = 90 = <OA1C =<AA1CAnonymousnoreply@blogger.com