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

4 comments:

  1. 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. 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]

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  3. 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

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

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