Saturday, July 11, 2009

Problem 319: Triangle, Altitudes, Perpendiculars, Collinear points

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

Geometry Problem 319: Triangle, Altitudes, Perpendiculars, Collinear points<br />.
See more:
Complete Problem 319
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 1:55 PM
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6 comments:

  1. AjitJuly 12, 2009 at 4:44 AM

    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

    ReplyDelete
  2. Antonio GutierrezJuly 12, 2009 at 11:48 AM

    Incontrovertible proof!

    ReplyDelete
  3. Peter TranNovember 27, 2010 at 8:15 AM

    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

    ReplyDelete
  4. Peter TranJanuary 5, 2019 at 5:54 PM

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

    New 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

    ReplyDelete
  5. Sumith PeirisJanuary 6, 2019 at 10:56 PM

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

    Similarly for FGM

    Sumith Peiris
    Moratuwa
    Sri Lanka l

    ReplyDelete
  6. piet van KampenMay 3, 2020 at 3:23 AM

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

    ReplyDelete

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