Monday, May 19, 2008

Geometry Problem 108



See complete Problem 108
Triangle, Angles, Median, Congruence, Cyclic Quadrilateral. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 10:24 PM
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8 comments:

  1. Antonio GutierrezMay 20, 2008 at 7:30 AM

    Think about a cyclic quadrilateral.

    Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

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  2. AnonymousMay 23, 2008 at 10:28 PM

    Draw a perpendicular through D, wich intersects BC in E. Then triangle DEC is congruent to triangle DEA. That implies that DEA=90-alpha, wich shows that ABED is cyclic, so EBD=EAD. this is alpha=x.


    srry for my english :$

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    Replies
    1. rv.littlemanDecember 8, 2019 at 1:11 PM

      Maybe you meant:
      Draw a perpendicular through D, which intersects BC in E.
      Then the triangle DEB is congruent to the triangle DEA.
      That implies that DEA=90-alpha
      which shows that ACED is concyclic,
      so EAD=alpha=ECD=x.
      Therefore is alpha=x.

      Delete
  3. AnonymousFebruary 18, 2009 at 5:33 PM

    This problem x has two solutions
    x=Alpha
    x=90-Alpha

    ReplyDelete
  4. AnonymousJune 3, 2009 at 1:12 PM

    http://geometri-problemleri.blogspot.com/2009/06/problem-29-ve-cozumu.html

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  5. Sumith PeirisJanuary 25, 2016 at 3:55 AM

    Let the perpendicular bisector of AC meet AB extended at E. Then < EAC = 90-x and so < AED = x = < DEC = < DBC

    Hence BDCE is cyclic and ABC is a right triangle with D the centre and so x = @

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. rv.littlemanDecember 8, 2019 at 12:53 PM

    Draw a circle passing through A, B and C
    Define E as intersection of CD and this circle
    ang(CDA)= a+x => ang(DAC)= 90-x
    ang(ECB)= x => ang(BAE)= x (intercepts the same arc BE)
    Therefore ang(CAE)= 90 and the center O of the circle is on CE

    Let’s suppose that D is not the center of the circle but O
    We already know that O is on CE
    On a circle, if we take two points, A and B, then the center of the circle is on the perpendicular bisector of AB
    Therefore, the center of our circle is on the perpendicular bisector of AB
    The intersection of the perpendicular bisector of AB and CE is D

    Therefore D is the center of the circle and DC=DB, ΔCDB is isosceles
    Therefore x=a

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  7. MarcoDecember 27, 2022 at 1:12 AM

    [For easy typing, I use "a" instead of alpha]
    <BAC=90-x
    Consider triangle BDC
    sinx/CD=sina/BD
    CD/BD=sinx/sina------(1)

    Consider triangle BDA
    sin(90-a)/AD=sin(90-x)/BD
    AD/BD=sin(90-a)/sin(90-x)-------(2)

    Since AD=DC, by equating (1) & (2)
    sinx/sina=sin(90-a)/sin(90-x)
    sinx/sina=cosa/cosx
    sinacosa=sinxcosx
    sin2a=sin2x
    2a=2x
    a=x

    ReplyDelete

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