Sunday, May 18, 2008

Elearn Geometry Problem 6

Geometry Problem, Triangle, Angles

See complete Problem 6 at:
www.gogeometry.com/problem/problem006.htm

Triangle, cevian, equal segments, and angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

20 comments:

  1. CD/BD = sin(6x)/sin(3x)= 2cos(3x)
    AB/BD = sin(3x)/sin(2x)
    Thus, 2cos(3x) = sin(3x)/sin(2x)
    or sin(2x) = (1/2)* sin(3x)/cos(3x)
    Now if x=15 then sin(2x)=1/2 & sin(3x)=cos(3x) which satisfies the equation.
    So x = 15 deg.
    Ajit: ajitathle@gmail.com

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  2. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

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  3. bae deok rak(bdr@korea.com)October 3, 2010 at 2:10 AM

    Let E be a point with triangles DBA and EDC are congruence(AB=CD,AD=CE,ang(BAD)=x=ang(DCE), ang(ACE)=4x). Then DB=DE and ang(EDC)=2x and hence
    ang(EBD)=x=ang(BED).
    Furthermore, since BA=CD, BE=EC=AD=CA and ang(EBA)=3x=ang(ACD), triangles BAE and CDA are congruence. we see that AE=AD and so triangle AEC is an equilateral. we get 4x=60 degree, that is, x=15 degree.

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  4. Comment on "Joe": Trial-error is not accepted in the Netherlands

    Comment on "bae deaok rak":1)Where did you put point E? 2)AD=CA is not true

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  5. Given:
    ABC=180
    A=2x
    B=x+90(90 because that is right angle)
    C=3x
    Solution:
    ABC=A+B+c
    180=2x+x+90+3x
    180=6x+90
    -6x=90-180
    -6x/-6=-90/-6
    x=15

    answer:
    x=15+90 =2x =3x
    =2(15) =3(15)
    =30 =45
    180=15+90+30+45
    180=180

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  6. To Jonathan (Problem 6): In your statement B=x+90.... 90 is not a data.

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  7. if you notice the DBC this is 90%. and why B=x+90 because that is seperated. i'm right??

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  8. Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 6. Thanks Eder and Cristian.

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  9. The solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJWEZGVVpLQ2pRRUdza0E4NUoxODZUZw

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  10. To Rengaraj PradheepKannah
    Greetings.
    In your solution of problem 6, there’s something I didn’t understand.
    The first circle is defined as having center at the midpoint G of AB (with diameter AB) and the second circle has center at the midpoint E of CD (with diameter CD).
    I can’t see why the first circle meets CD in its midpoint E. If we don’t assure that E lies on the first circle, it’s not clear that AG = GE.

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  11. To Nilton Lapa
    Greetings to you too.
    Did you notice DBC is an Isosceles triangle.
    Since E is the Mid point of DC BEC is 90 Deg.
    As AB is the diameter of the first circle according to the angle on semi circle is 90 Deg
    The first circle should go thriugh E.

    Thanks.
    PRadheepKannah
    ..............

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  12. To Rengaraj PradheepKannah
    Thanks. The explanation is absolutly clear. My doubt in problem 6 was enlighted.
    Nilton Lapa

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  13. E es el punto sobre D-C tal que AD=DE
    F es tal que B-F paralelo a A-C y BF=AD
    D-B es paralela a E-F
    G es intersección entre E-F y B-C formando EBFC cíclico
    ang(DBC) = 180°-6α
    △FCD es isósceles
    (1) ang(FCD) =ang(DFC)=90°-α
    △BGE es congruente a △FGC por la simetría
    general en los triángulos isósceles △BFE y △EGC
    y el paralelismo de BF con EC.
    ang(CEB) = 90°-α
    ang(EBF)=ang(BFC)=[360-2(90°-α)]/2=90°+α
    ang(FHB)=180°-(90°+α)-2α=90°-3α
    ang(EHF)=3α
    Entonces △ABD es semejante a △EFH, por lo tanto
    (2) ang(GEB)=ang(FCG)=2α
    Comparando resutado (1) con (2)
    ang(FCE)=90°-α=ang(FCG)+ang(BCA)=5α
    6α=90°; α=15° Q.E.D.

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  14. Let E be a point on the extension AB such that AE=AC.
    Bisect ∠A and let a point F on the angle bisector of ∠A such that m∠AEF=m∠ACF=x.
    Then ∆AFE≅∆AFC by ASA and at the same time they are both isosceles.
    Let a point H on BC such that the angle bisector of ∠A intersect BC at H.
    Then m∠CFH=m∠EFH=m∠FEH=m∠FCG=2x, also m∠EBC=5x and m∠EGC=6x. An exterior angle is equal to the sum of its remote interior angles.
    Again by ASA, ∆EHF≅∆CHF and they are also isosceles triangles, then EH=FH=CH.
    By perpendicular bisector concurrence theorem, CG⊥EF that makes ∠EGC a right angle.
    Therefore: 6x=90°,x=15°

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  15. See a synthetıc solutıon on my facebook page https://www.facebook.com/photo.php?fbid=789390357808539&set=a.784793394934902.1073741826.100002127454113&type=1&theater

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  16. See my proof, maybe similar to the video-one mentioned by the owner of the site on Sept, 4, 2011, but I am not able to see that.

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    Replies
    1. To Stan Fulger
      Spanish solution video at http://www.gogeometry.com/geometria/p006_solucion_problema_video_triangulo_angulo_ceviana_trazo_auxiliar.htm
      Thanks

      Delete
  17. https://drive.google.com/file/d/0B6oYedIeLTUKczBPY0dfdWdsN0k/view?usp=sharing

    I think this is different from the previous solutions.

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  18. https://www.youtube.com/watch?v=FLLNvVRnUYI

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