## Saturday, July 11, 2009

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

Proposed Problem
Click the figure below to see the complete problem 319.

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Complete Problem 319
Level: High School, SAT Prep, College geometry

1. 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.
Is this proof enough, Antonio?
Ajit: ajitathle@gmail.com

2. Incontrovertible proof!

3. Problem 319
http://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