Wednesday, September 26, 2018

Geometry Problem 1389: Triangle, 40-120-20 degree, Angle, Congruence

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

Details: Click on the figure below.

Geometry Problem 1389: Triangle, 40-120-20 degree, Angle, Congruence.

Posted by Antonio Gutierrez at 4:03 PM
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11 comments:

  1. Sumith PeirisSeptember 26, 2018 at 7:48 PM

    Find E on AC such that AE = BE. Construct equilateral Tr. BEF, F within Tr. BCE

    Since BC = AD = CE, AE = BE = EF = CD

    So Tr.s CEF & BCD are congruent SAS

    Hence < FCE = < CBD = 10 and so
    x = 30

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Dinesh Nayak H.February 21, 2019 at 12:39 PM

      Sumith Sir,

      You are truly an expert at geometry problems. You always provide very elegant solutions.

      Delete
    2. Sumith PeirisFebruary 21, 2019 at 11:27 PM

      Thank u for your kind words Dinesh. I thank God for whatever small talent I may have

      Delete
    3. AnonymousJanuary 31, 2024 at 2:29 PM

      thanks for you solution. How do you know that CD = EF?

      Delete
  2. APOSTOLIS MANOLOUDISSeptember 27, 2018 at 2:16 AM

    Let point E between A and D such as AE=BE.So <DBC=10.

    ReplyDelete
  3. AnonymousSeptember 27, 2018 at 6:51 AM

    Luăm punctul E pe prelungirea lui BC astfel încât CE = CA; triunghiul AEC este isoscel (80-20-80).
    Ducem BF paralel cu EA și atunci se arată ușor că EB=BF=FA=CD(triunghiul ABF isoscel 40-100-40).
    Construim un triunghi isoscel CDP cu unghiul DPC = 20.
    Deoarece CD = BF, implică triunghiul CBF = CDP și de aici CP = CB. De asemeni unghiul BCP = 60, ceea ce face triunghiul BPC echilateral(60-60-60) și PB =PC=PD. În triunghiul BDP isoscel BP = DP și
    unghiul BPD =40. De aici unghiul PDB = 70, adică unghiul BDC = 150 și deci BDA =30.
    Florin Popa , Comanesti, Romania.

    ReplyDelete
  4. Sailendra TSeptember 27, 2018 at 3:49 PM

    Considering usual triangle notations,
    let BC=AD=a and DC=x
    Mark a point E on AC such that CE=a => AE=x ------------(1)
    Observe that BEC is an 80-80-20 isosceles and ABE is a 40-40-100 isosceles with AE=BE=x -----------(2)
    Now extend the internal angle bisector of m(ABC) to a point P such that BP=a and this forms an equilateral triangle BPC
    From this construction it can be observed that C is the circumcenter of EBP
    => m(BEP) = (360-m(BCP))/2 = 360-60/2 = 150 degrees----------(3)
    Another observation is that EB=DC=x, m(EBP)=m(DCB)=20 & BP=BC=a => EBP is congruent to DCB (SAS)
    =>m(CDB)=m(BEP)=150
    =>m(BDA)=x=30 Q.E.D

    ReplyDelete
  5. Stan FULGEROctober 3, 2018 at 5:01 AM

    Extend (AB to E so that triangle BCE be equilateral, take F on AC so that EF=EC, we get <EFC=<FCE=80, and <BEF=40, thus AF=EF and F=D, hence <AFB=<BEC/2=30.

    ReplyDelete
  6. AmiltonMarch 5, 2019 at 1:48 PM

    Hello !

    Construct a triangle PCD outside the triangle BDC , DPC=20° , PD=PC=BC.
    PCB=60° and PBC is equilateral. So, PB=PD and BPD=40°. We conclude that DBP=BDP=70°. As we know, CBP=60°, so DBC=10°.
    Finally, x=DBC+DCB=10°+20°=30°.

    ReplyDelete
  7. AnonymousApril 5, 2020 at 10:59 AM

    More general form: A is twice of C.
    The many solution for 10-20.
    https://artofproblemsolving.com/community/c6h487183p2729760

    ReplyDelete
  8. MarcoJune 29, 2024 at 9:41 PM

    In triangle ABC
    sin20/AB=sin40/BC
    AB/BC=sin20/sin40
    AB/BC=1/2cos20--------(1)

    In triangle ABD
    <ABD=140-x
    sin(140-x)/AD=sinx/AB
    AB/AD=sinx/sin(140-x)----------(2)

    Since AD=BC, (1)=(2)
    sinx/sin(140-x)=1/2cos20
    sin(140-x)=2sinxcos20
    sin(140-x)=sin(x+20)+sin(x-20)
    sin(140-x)-sin(x-20)=sin(x+20)
    2cos60sin(80-x)=sin(x+20)
    sin(80-x)=sin(x+20)
    80-x=x+20
    x=30

    ReplyDelete

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