Proposed Problem

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Complete Problem 319

Level: High School, SAT Prep, College geometry

## Saturday, July 11, 2009

### Problem 319: Triangle, Altitudes, Perpendiculars, Collinear points

Labels:
altitude,
collinear,
perpendicular,
triangle

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H is on the circumcircle of Tr. ABD which has AB as diameter. HF is perpendicular to AB,HG to AD & HN to BD extended. So by the Wallace-Simson Theorem F,G & N are collinear. Now consider Tr. CEB with H on its circumcircle and three perpediculars to its sides. Once again their feet N,M & F are collinear. This makes F,G,M & N all belong to the same line which is te common Simson Line of the two triangles mentioned above.

ReplyDeleteIs this proof enough, Antonio?

Ajit: ajitathle@gmail.com

Incontrovertible proof!

ReplyDeleteProblem 319

ReplyDeletehttp://img686.imageshack.us/img686/1417/probem319.png

Connect ED

1. Note that CM/CE=CH/CA=CN/CD ( AE//HM and AD//HN)

So MN //ED

Similarly we can show that FG//ED

2. We have BH^2=BN.BC=BF.BA ( Relations in right triangles BHC and BHA)

So BN/BA=BF/BC and Tri. FBN similar to Tri. CBA ( case SAS)

And m(BNF)=m(A)

3. AEDC is a concyclic quar. So BE.BA=BD.BC

So BD/BA=BE/BC and Tri. EBD similar to Tri. CBA ( case SAS)

And m(BDA)=m(A)

4. m(BNF)=m(BDE) so FN// ED

MN, FG and FN all parallel to ED so F,G,M and N are collinear

Peter Tran