tag:blogger.com,1999:blog-6933544261975483399.post1135763429866677704..comments2024-03-19T00:02:30.728-07:00Comments on Go Geometry (Problem Solutions): Problem 426: Triangle, Circumradius, Circumcenter, concurrent CeviansAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-2851261465730732902010-03-02T05:29:07.352-08:002010-03-02T05:29:07.352-08:00This is a special case of Ceva:
For any point P in...This is a special case of Ceva:<br />For any point P in ABC, if we draw three cevians APD, BPE, and CPF, then<br />AP/AD + BP/BE + CP/CF = 2Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9065913296386590802010-02-16T03:54:17.551-08:002010-02-16T03:54:17.551-08:00from ceva's theorem
OD/AD + OE/BE + OF/FC = 1...from ceva's theorem<br /><br />OD/AD + OE/BE + OF/FC = 1<br /><br />(AD-R)/AD + (BE-R)/BE + (FC-R)Fc = 1<br /><br />1 - R/AD + 1 - R/BE + 1 - R/FC = 1<br /><br />2 = R/AD + R/BE + R/FC<br /><br />2 = R ( 1/AD + 1/BE + 1/FC )<br /><br />2/R = 1/AD + 1/BE + 1/FC<br />-----------------------------------------c .t . e. onoreply@blogger.com