Sunday, January 29, 2012

Problem 723: Squares, Circumscribed Circles, Collinearity

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 723 details.

Online Geometry Problem 723: Squares, Circumscribed Circles, Collinearity

Posted by Antonio Gutierrez at 3:36 PM
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5 comments:

  1. PravinJanuary 29, 2012 at 6:54 PM

    First we observe that diagonal AOC is a diameter of circle (O) since <ADC is a rt angle.
    <AHC = <ADC = 90 deg
    Similarly <EHG = < EFG = 90 deg
    By Problem 722,
    A, E, H are collinear and so <AHG is the same as <EHG = 90 deg
    Hence <AHC + <AHG = 180 deg and
    C, H, G are collinear.

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  2. PravinJanuary 29, 2012 at 7:19 PM

    Solution 2:
    <DHG = <DEG = 45 deg
    <DHC = supplement of <DBC = 180deg - 45deg
    Sum = 180 deg
    So G, H, C are collinear.
    [Note: DB passes thro'O and bisects the rt <ABC]

    ReplyDelete
  3. Peter TranJanuary 29, 2012 at 9:56 PM

    We can prove this problem using result of problem 722 or
    Note that DF and DB are diameters of circles O and O’
    So (BHD)=(DHF)=90 >> B,H, F are collinear
    (GHF)=(CHB)=45 ----- (angles face 90 arc)
    So (CHG)=(BHF)+45-45= 180 >> C, H, G are collinear

    ReplyDelete
  4. Sumith PeirisSeptember 19, 2015 at 9:45 AM

    Reference 722 < AHC = 90 = < EHG. Hence CHG are collinear

    ReplyDelete
  5. MarcoJanuary 12, 2024 at 10:26 PM

    Let the interecting point of CG & EF be Y
    Let <ECY=x
    <EYC=90-x=<FYG
    <CYF=180-<EYC=90+x
    <CYF+<FYG=90+x+90-x=180
    CHG is a st. line

    ReplyDelete

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