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.

Click the diagram below for more details.

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

6 comments:

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

    ReplyDelete
    Replies
    1. 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. 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