Monday, May 19, 2008

Geometry Problem 107



See complete Problem 107
Triangle, Angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 10:23 PM
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7 comments:

  1. AnonymousMay 27, 2008 at 4:40 AM

    Construct E on AC such that angles ABC=BEC, then angles CBD=EBD and we have DE/BE=DC/CB=AB/BC so triangles ABC~DEB giving triangle BAD isosceles thus x=pi-4a.

    ReplyDelete
  2. AnonymousMay 14, 2009 at 6:14 PM

    http://geometri-problemleri.blogspot.com/2009/05/problem-24-ve-cozumu_14.html

    ReplyDelete
  3. Nilton LapaMay 7, 2012 at 9:14 PM

    Solution to problem 107.
    Take E on BC, such that AE is bisector of ang(BAC). Notice that ang(BAE) = ang(DBE) = alpha and both subtend the same segment DE. So ABED is cyclic, ang(BDE) = alpha, tr. BDE is isosceles and BE = DE.
    In triangle ABD, ang(BDC) = x + 2.alpha, so
    ang(CDE) = x + alpha.
    As AB = CD, BE = DE and
    ang(ABE) = ang(CDE) = x + alpha,
    triangles ABE and CDE are congruent, with
    ang(BAE) = ang(DCE) = alpha.
    Thus, in triangle ABC,
    (x + alpha) + 2.alpha + alpha = 180
    and finally x = 180 – 4.alpha.

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  4. Sumith PeirisJanuary 25, 2016 at 4:12 AM

    Let AE be the bisector of < BAC where E is on BC

    Tr. s ABE and CDE are congruent SAS so < EDC = < ABE implying that ABED is cyclic and < DBC = @

    So x = 180-4@

    Sumith Peiris
    Moratuwa
    Sri. Lanka

    ReplyDelete
  5. Sumith PeirisJanuary 25, 2016 at 4:18 AM

    Let E be on BC such that AE bisects < BAC

    Then ABED is cyclic and so Tr.s ABE and DEC are congruent SAS

    Hence AE = EC and so < ACE = @

    So x = 180-4@

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. APOSTOLIS MANOLOUDISApril 27, 2016 at 11:57 AM

    Bring AE=//DC (AB=DC) then AECD is parallelogram so <BEA=<BAE=<CAE=a then AE is bisector angle BAC.Is <ECB=<DBC=a=<BAE so BECA is isosceles trapezoid or concyclic.
    Therefore <BCA=<BAE=a and <BDA=2a,so x=180-4a.

    ReplyDelete
  7. MarcoDecember 27, 2022 at 12:57 AM

    [For easy typing, I use "a" instead of alpha]
    <BCD=180-x-3a
    Consider triangle BCD
    sin(180-x-3a)/BD=sina/CD
    BD/CD=sin(x+3a)/sina---------(1)

    Consider triangle ABD
    sin(180-x-2a)/AB=sin2a/BD
    BD/AB=sin2a/sin(x+2a)----------(2)

    Since AB=CD, by equating (1) & (2)
    sin(x+3a)/sina=sin2a/sin(x+2a)
    sin(x+3a)sin(x+2a)=sinasin2a
    cosa-cos(2x+5a)=cosa-cos3a
    cos(2x+5a)=cos3a
    2x+5a=3a or 360-3a
    x=-a (rej) or 2x=360-8a
    x=180-4a

    ReplyDelete

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