tag:blogger.com,1999:blog-6933544261975483399.post7098018974253040416..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1270 Triangle, Excircle, Excenter, Escribed Circle, Tangency Points, Perpendicular, 90 Degrees, Angle BisectorAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-69307354893285856002016-10-02T05:43:40.521-07:002016-10-02T05:43:40.521-07:00Further if AO meets DE at A1 and CA1 and C1A meet ...Further if AO meets DE at A1 and CA1 and C1A meet at X then O,X,F are collinear points. <br /><br />Further if CX cuts FD at Z and AX cuts FE at Y then XF and YZ bisect each other. (XYFZ is a parallelogram) Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20176670681149011422016-10-01T18:54:10.586-07:002016-10-01T18:54:10.586-07:00< EAO = < EC1O = 90-A/2
(since < C1CD = ...< EAO = < EC1O = 90-A/2 <br />(since < C1CD = 90-C/2 and < C1DC = 90-B/2)<br /><br />Hence EAC1O is a cyclic quadrilateral <br /><br />So < AC1C = < AEO = 90<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com