tag:blogger.com,1999:blog-6933544261975483399.post2055867431697805573..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Triangle with the bisectors of the exterior angles.Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-56427922227060048692009-08-13T21:15:51.622-07:002009-08-13T21:15:51.622-07:00AD is the external bisector of angle A. Hence, BD/...AD is the external bisector of angle A. Hence, BD/DC = - AB/AC. Likewise, CE/AE = -BC/AB & AF/BF=-AC/BC. Now (AF/BF)*(BD/DC)*(CE/AE)=(-AC/BC)*(-AB/AC)*(-BC/AB)= -1. Therefore, by the converse of Menelaus's Theorem D, E & F are collinear. Would that be correct, Antonio?<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com