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
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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