Thursday, September 22, 2011

Problem 671: Triangle, Cevian, Angles, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 671.

Online Geometry Problem 671: Triangle, Cevian, Angles, Congruence
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5 comments:

  1. Peter TranSeptember 22, 2011 at 10:57 AM

    http://img838.imageshack.us/img838/3824/problem671.png
    Draw triangle BDE as image of triangle BDA ( see picture)
    Note that BE=CD and (DCB)=(EBC)=4. alpha
    And DBCE is an isosceles trapezoid
    (BDE)=(BDA)= 180-7.alpha
    In triangle ABD , x+180-7.alpha+3.alpha= 180 >> x=4. alpha
    In triangle ABC , 7.alpha+4.alpha+4.alpha=180 >> alpha=10 and x=40
    Peter Tran

    ReplyDelete
  2. AlejandroSeptember 22, 2011 at 7:44 PM

    Okay, I have a really simple solution.

    At first the problem seems to be awful but it can be taken down faster.

    Let P be a point on BC so that DB = DP, hence <CDP = 3a and then ∆CDP = ∆ABD and x = 4a, hence a = 10° whereat x = 40°.

    ReplyDelete
  3. Sumith PeirisOctober 23, 2016 at 7:03 AM

    Find E on DC such that < CBE = 3@
    Let BD and BE extended meet the circle ABC at X and Y.

    AX = CY so ACYX is an isoceles trapezoid.

    Now < AEB = 7@ = < ABE

    Hence AB = c = AE = CD
    So AD = CE = b-c
    Further < XAC = YCA
    So tr.s AXD and CYE are congruent SAS.

    < BDC = < BEA = 7@

    Hence 7@ + 7@ + 4@ = 180 and therefore

    @ = 10 and x = 40

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. MarcoJanuary 3, 2024 at 11:18 PM

    [For easy typing, I use a instead of alpha]
    sin(180-14a)/BD=sin11a/AB
    AB/BD=sin11a/sin14a--------(1)
    sin7a/CD=sin4a/BD
    CD/BD=sin7a/sin4a---------(2)
    Since AB=CD, (1)=(2)
    sin7a/sin4a=sin11a/sin14a
    sin4asin11a=sin7asin14a
    cos7a-cos15a=cos7a-cos21a
    cos21a-cos15a=0
    -2sin18asin3a=0
    sin18a=0 or sin3a=0 (rej)
    18a=180 [180 is the only suitable value]
    a=10
    x=180-14a=40

    ReplyDelete

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