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

https://photos.app.goo.gl/FKeFycxQbyovPigo7

ReplyDeleteNew solution from Peter Tran

Note that quadrilaterals AFGH, HGPM, HMNC, AEPH, HPDC,AEDC and BEPD are cyclic

HGPM is cyclic => ∠(MGH)=∠(MPH)=w

AEPH is cyclic => ∠(MPH)=∠(HAE)=w

so ∠(MGH)=∠(HAE)= w => F, G, M are collinear

similarly N, M, G are collinear => F, G, M, N are collinear

< HBC =90-C = < GHB = <EMG = < CHN = < CMN so GMN are collinear

ReplyDeleteSimilarly for FGM

Sumith Peiris

Moratuwa

Sri Lanka l

Consider ∠HFG and ∠HFN when these angles are equal than F,G,N are collinear. Similar for M , N and F.

ReplyDeleteAFGH is cyclic -> ∠HFG =∠HAD----------------------------------------(1)

HFBN is cyclic ; BH diameter -> HC is tangient to the circle-> ∠CHN=∠HFN

∠CHN=∠HAD (parallel line AD and HN) -> ∠HAD=∠HFN---------(2)

From (1) and (2) ∠HFG=∠HFN -> F,G and N collinear.

Piet van Kampen

Bergen Nh

The Netherlands