Geometry Problem

Click the figure below to see the complete problem 613.

## Wednesday, June 1, 2011

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## Wednesday, June 1, 2011

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Problem 613: Altitude of a Triangle, Perpendicular, Collinear Points

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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.