tag:blogger.com,1999:blog-6933544261975483399.post796996822198351842..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 479: Triangle, Cevians, Concurrency, Transversal, ProportionAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-61780493598806582132010-07-27T14:10:07.191-07:002010-07-27T14:10:07.191-07:00Below is the proving of 2nd part of Problem 479
1....Below is the proving of 2nd part of Problem 479<br />1. Apply Ceva’s theorem to triangle DBE and point of concurrence O . We have:<br />(HD/HE)*(CE/CB)*(AB/AD)=1 so HD/HE= (CB/CE)*(AD/AB)<br />2. Apply Menelaus’s theorem for triangle DBE and secant ACG . We have:<br />(GE/GD)*(AD/AB)*(CB/CE)=1 so GD/GE=(AD/AB)*(CB/CE)<br />3. From the result of steps 1 and 2 we get HD/HE=GD/GE<br /><br />Peter Tran<br />vstran@yahoo.comPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76491144788459149162010-07-26T12:23:32.051-07:002010-07-26T12:23:32.051-07:00join B to G
in ▲ABG from ceva's theorem
BD...join B to G <br />in ▲ABG from ceva's theorem <br />BD/DA∙AG/CG = BE/EC => AG/CG = BE/EC∙DA/BD (1)<br />in ▲ABC from ceva's theorem<br />BD/DA∙AF/FC∙EC/BE = 1 (2)<br />multiply (1) to (2)<br />AG/CG = (BE/EC∙DA/BD)∙(BD/DA∙AF/FC∙EC/BE)<br /><br />AG/CG = AF/FCc .t . e. onoreply@blogger.com