tag:blogger.com,1999:blog-6933544261975483399.post8298889450239055343..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 631: Triangle, Exterior Angle Bisectors, Collinear PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-38988304965370105802020-08-25T08:38:53.076-07:002020-08-25T08:38:53.076-07:00This comment has been removed by the author.sarbuneanghttps://www.blogger.com/profile/00575429067764567682noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27911228197027897372011-07-06T18:22:17.333-07:002011-07-06T18:22:17.333-07:00Since BB’, AA’ and C’C are external angle bisector...Since BB’, AA’ and C’C are external angle bisectors so<br />We have B’C/B’A= a/c<br />C’A/C’B=b/a<br />And A’B/A’C=c/b<br />And (B’A).(C’A/C’B).(A’B/A’C)=a/c.b/a.c/b= 1<br />So C’, A’ and B’ are collinear per Menelaus theorem<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com