tag:blogger.com,1999:blog-6933544261975483399.post4664966237764954766..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1232: Triangle, Altitudes, Orthic Triangle, Collinearity, Orthic AxisAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-24715944679887041652016-07-06T18:16:22.424-07:002016-07-06T18:16:22.424-07:00Easily A1A2, B1B2 and C1C2 are external angle bise...Easily A1A2, B1B2 and C1C2 are external angle bisectors of the Orthic Triangle A1B1C1<br /><br />These angle bisectors meet the opposite sides at A2, B2 and C2<br /><br />From the result of Problem 631 these points are collinear <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70060392554137629162016-07-06T17:34:07.508-07:002016-07-06T17:34:07.508-07:00https://goo.gl/photos/1o6iZtQfZnuheshYA
Consider ...https://goo.gl/photos/1o6iZtQfZnuheshYA<br /><br />Consider complete quadrilateral ABC C1A1B2<br />Diagonals AA1 and CC1 meet at H<br />So ( ACB1B2)= -1 or B1, B2 are harmonic conjugate points of A and C…..<br />Or B2A/B2C=B1A/B1C…. (1)<br />Similarly we also have A2C/A2B= A1C/A1B…… (2)<br />And C2B/C2A=C1B/C1A….. (1)<br />Multiply expressions (1), (2) and (3) side by side<br />The value of right hand side is 1 per Ceva theorem <br />So the left hand side B2A/B2C x A2C/A2B x C2B/C2A= 1<br />And A2, C2, B2 are collinear per inverse of Menelaus theorem<br /><br />Note: This property can be generalized for any lines AA1, BB1, CC1 as long as that these lines concur at a point <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com