Thursday, November 12, 2015

Geometry Problem 1164: Triangle, Angle Bisector, Cevian

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1164: Triangle, Angle Bisector, Cevian.

Posted by Antonio Gutierrez at 12:51 PM
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7 comments:

  1. Peter TranNovember 12, 2015 at 4:35 PM

    Let AD=x so DE=2x, EC=6x and AB=3x
    calculate BD and BC using cosine formula in triangles ABD and ABC .we get
    BD^2=10x^2-6x^2.cos(A)
    BC^2=90x^2-54x^2.cos(A)
    (BD/BC)^2= 1/9 => BD/BC=1/3= ED/EC
    so BE bisect angle DBC

    ReplyDelete
  2. Sumith PeirisNovember 12, 2015 at 11:26 PM

    Let AD = p so that DE = 2p, CE = 6p and AB = 3p

    Now AB^2 = 9p^2 = AD.AC so AB is tangential to DBC at B

    From similar Tr.s AD/AB = BD/BC = 1/3 = 2p/6p = DE/EC

    Hence BE bisects < DBC

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. c.t.e.oNovember 13, 2015 at 10:11 AM

    Tr BAE isoceles => ang ABE=ang AEB => ABD + DBE = EBC + ECB
    => DBE = EBC ( ABD = ECB )

    ReplyDelete
    Replies
    1. Sumith PeirisNovember 13, 2015 at 6:07 PM

      Excellent simple solution!

      Delete
    2. Ana de la FuenteJanuary 31, 2016 at 1:25 PM

      Yes, but why ¿ABD=ECB?, because:
      The triangle ▲ABD is proportional to the triangle ▲BCA
      BAC = BAD (They have in common the angle A)
      Its sides are proportional:AC/AB=AB/AD=3

      Delete
  4. c.t.e.oJanuary 31, 2016 at 3:07 PM

    See relations between tang and secant
    Ang ABD = 1/2 arc BD, Ang ACB = 1/2 arc BD
    or Tr ABD similar to Tr ABC

    ReplyDelete
  5. Benjamin LeisNovember 18, 2017 at 9:08 PM

    Use Stewarts Law with triangle ABC and the cevian BD:

    AB = 3x, AD = x, DE = 2x, CE = 6x

    (3x)^2*8x + BC^2*x = 9x(BD^2 + 2x^2)
    That simplifies to BC^2 = 9BD^2 or BC = 3BD

    Since BC:BD = 3:1 and CE:DE = 3:1 also BE is the angle bisector by the angle bisector theorem.
















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