Geometry Problem
Click the figure below to see the complete problem 613.
Wednesday, June 1, 2011
Problem 613: Altitude of a Triangle, Perpendicular, Collinear Points
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Geometry Problem
Click the figure below to see the complete problem 613.
Triangle Transversal By Menelau
ReplyDeleteABC NEF => AN.CF.BE = NC.FB.EA
CGB DMF => CM.GD.BF = MG.DB.FC
BGA EHD => GH.AE.BD = HA.EB.DG
Multiply out respective sides of the equations, we get
AN.CF.BE.CM.GD.BF.GH.AE.BD
= NC.FB.EA.MG.DB.FC.HA.EB.DG
Cancelling common factors:
AN.CM.GH = NC.MG.HA
Hence again by Menelau:
N, H, M are collinear
Refer to my solution of Problem 613:
ReplyDeleteLast but one sentence should read as:
"Hence again by Menelau applied to
triangle AGC"
Remark:
I haven't used perpendicularity conditions.Please point out flaw, if any, in my solution
To Pravin:
ReplyDeleteThanks for your comments, in effect, the perpendicularity conditions is not necessary.
By Menelaus in triangle ABC with transversal NEF, NA/NC*FC/BF*BE/EA=1.
ReplyDeleteFC/BF=[DFC]/[DBF] and BE/EA=[DBE]/[DEA].
However [DBE]/[DBF]=[DGE]/[DGF].
Plugging these into original equation yields NA/NC*[DFC]/[DGF]*[DGE]/[DEA]=1.
[DFC]/[DGF]=MC/GM and [DGE]/[DEA]=GH/HA, meaning that NA/NC*MC/GM*GH/HA=1 which is converse of Menelaus in triangle AGC and points N, H, and M.