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

6 comments:

  1. 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. 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. 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. 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. 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 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