Monday, January 20, 2014

Geometry Problem 963: Right Triangle, 30-60-90 Degrees, Angle Bisectors, Metric Relations

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

This entry contributed by Mr Petre Ciobanu, Math Teacher from Sibiu, Romania.
Click the figure below to see the complete problem 963.

Online Geometry Problem 963: Right Triangle, 30-60-90 Degrees, Angle Bisectors, Metric Relations

Posted by Antonio Gutierrez at 6:38 PM
Labels: , , ,

8 comments:

  1. Peter TranJanuary 20, 2014 at 7:32 PM

    http://imagizer.imageshack.us/v2/800x600q90/203/r3th.png
    Make AH ⊥BC and HF=HD (see sketch)
    Triangle CAH is 30-60-90 triangle
    Triangle AFD is isoceles
    Triangles BEC similar to AFC .. ( case AA) => AF/BE=AC/BC= ½
    Bu AF= AD => AD= ½ BE

    ReplyDelete
  2. nakaJanuary 20, 2014 at 8:54 PM

    Since it is a 30-60-90 triangle, the length ratio would be 1:sqrt3:2
    the remaining can be done by angle bisector length theorem

    ReplyDelete
  3. Ivan BazarovJanuary 23, 2014 at 10:04 PM

    Midpoint of EB is P. <DIB=60 where I is incenter, so CEPD is cyclic, so then since <ECI=<DCI=30, EI=ID. But <IAP=<IPA=30, so AI=IP. AI+ID=EI+IP=AD=EP=EB/2

    ReplyDelete
  4. Sumith PeirisJune 28, 2015 at 6:25 AM

    2 ways of doing it.

    Draw altitude AH. Tr.s AHD and ABE are simiar from which the result follows

    Or draw median AO of Right Tr. ABC. Tr.s AOD and BEC are similar from which the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Arindam BanerjeeJuly 31, 2015 at 8:35 PM

    F is a point on BC such that CA=CF. Then F is the mid point of BC. Triangles BCE and ADF are equiangular and similar as well. So, BE/AD=BC/AF=2/1. Therefore, AD=(1/2)BE.

    ReplyDelete
    Replies
    1. APOSTOLIS MANOLOUDISFebruary 21, 2016 at 8:00 AM

      APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE
      Solution 1
      Even M is midpoint of BE.Bring EN perpedikular in BC.Then AM=NM=1/2.BE .Is AENB concyclic .Is <DAN=30, <AND=75 and <DNA=75, therefore AN=AD. Is <NMA=60 then triangle NMA is equilateral.
      Therefore AD=AN=AM=1/2.BE.
      Solution 2
      EB intersects AD to I is incenter to triangle ABC .Then <DIB=60 =ACB therefore CEID is concyclic,
      <BED=30=<ICD=<ICA=<ADE. The ED intersects AB to F. Is <EFA=15=<EBA , then AF=AB. Bring from B
      Parallel to AD where intersects the ED to N. Is <BNE=<ADE=30=<BEN .So BE=BN. Is AD=1/2.BN=1/2.BE.

      Delete
  6. MarcoApril 1, 2024 at 12:05 AM

    In triangle ADB
    sin30/AD=sin105/AB
    AD/AB=sin30/sin105=sin30/sin75--------(1)

    In triangle ABC
    cos15=AB/BE---------(2)

    (1)*(2)
    AD/BE=sin30=1/2
    AD=BE/2

    ReplyDelete
  7. Sumith PeirisApril 1, 2024 at 7:44 PM

    Pure Geometry Solution 3

    Let BE, AD intersect at N. Let M be midpoint of BE

    < BAM = 15, < AME = 30, < MAD = 30, < BND = 60

    Hence CEND is concyclic and since CN bisects < BAC, < NDE = < NCE = 30
    Hence < AME = < ADE = 30 and AMDE is concyclic, so < EMD = < EAD = 45

    So < AMD = 75 and < ADM = 180 -75 -30 = 75
    Therefore AD = AM = BE / 2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

Subscribe to: Post Comments (Atom)