Sunday, May 18, 2008

Elearn Geometry Problem 7

Geometry Problem, Triangle, Angles

See complete Problem 7 at:
www.gogeometry.com/problem/problem007.htm

Triangle, Cevian, Angles, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

13 comments:

  1. trazar la ceviana exterior BE (que corta a la prolongacion de AC en E)
    Si el angulo EBC=x
    entonces el angulo BEC=3x...(1)
    Ademas los triangulos ABC y BED son congruentes
    y de (1) se tiene que: el angulo ABC=3x
    Finalmente en el triangulo ABC
    3x+4x+3x=180º
    10x=180º
    x=18º

    atte:tino

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  2. http://geometri-problemleri.blogspot.com/2009/10/problem-34-ve-cozumu.html

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

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  4. Given:
    ABC=A+B+C
    ABC=180 A=3x B=x+2x C=4x
    Solution:
    ABC=A+B+C
    180=3x+x+2x+4x
    180=10x
    180/10=10x/10
    18=x

    answer:
    A=3x B=x+2x C=4x
    A=3(18) B=18+2(18) C=4(18)
    A=54 B=18+36 C=72
    B=54

    180=54+54+72
    180=180

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  5. To Jonathan (Problem 7): Why B=x+2x?

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  6. Video-solution to the problem by me:

    http://www.youtube.com/watch?v=cMkCjlKlgRk

    Greetings.

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  7. The solution is uploaded to the follwing link

    https://docs.google.com/open?id=0B6XXCq92fLJJLWMyd3E5THpTTTJKZHdOaVU2UjRzZw

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    Replies
    1. Triangles in the first step can not be congruent by AAS
      there is no criteria of AAS for congruence

      Delete
    2. There is no rule of AAS congruence, side common between angles must be same...................

      Delete
  8. Sea E el punto en la prolongación de A-C tal que CE = AD
    Sea F tal que F-E es paralela a D-B y además EF=DB
    Con esto se arma un rombo DBFE
    La diagonal B-E corta a C-F en un punto G
    Esta diagonal también es bisectriz por lo que ang(GEC)=3α, lo que hace que
    △CGE y △BGF sean isósceles.
    Como BFEC es un paralelogramo recto se deduce que:
    ang(ECB) = ang(FEC)
    Pero esto implica:
    6α=180°-4α
    α=18° Q.E.D.

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  9. Construct isoceles Tr. ABE where E is on AC. Then Tr. BDE is isoceles and so DE = BD= AC

    Hence Tr.s ABC & BED are congruent SAS from which it follows that Tr. BDC is isoceles with internal angles adding upto 10x and so x=18

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  10. Construct point E such that BA=BE (hence triangle ABE is formed and it is an isosceles triangle, where <BAE=<BEA=3x). Additionally, E lies on AC produced.
    Consider triangle ABC and EBD.
    <ABD=180-9x
    <ABE=180-6x
    Hence <EBC=(180-6x)-2x-(180-9x)
    <EBC=x
    Hence <DBE=3x
    Since <DBE=<DEB=3x, DB=DE
    Since AC=DB (given), AC=DE

    1.BA=BE (from above)
    2.<BAC=<BED=3x (proven)
    3.AC=ED (proven)
    Hence triangle ABC is congruent to EBD (SAS)
    Hence, BD=BC
    <BDC=4x
    In triangle BCD,
    2x+4x+4x=180
    x=18

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  11. By the law of sines,
    sin(∠BDC):sin(∠C) = BC:BD = BC:AC = sin(∠A):sin(∠ABC).
    That is,
    sin(180°-6x):sin(4x) = sin(3x):sin(180°-7x),
    thus,
    sin(6x):sin(4x) = sin(3x):sin(7x).
    This implies
    sin(7x)sin(6x) = sin(4x)sin(3x)
    By the product-to-sum identities,
    -(1/2)(cos(13x) - cos(x)) = -(1/2)(cos(7x) - cos(x)),
    which can be reduced to
    cos(13x) - cos(7x) = 0 (since cos(x) = 0 is not valid).
    Then by the sum-to-product identity,
    -2sin(10x)sin(3x) = 0
    which suggests
    sin(10x) = 0 or sin(3x) = 0.
    This means x is either a multiple of 18° or a multiple of 60°.
    Since ∠A = 3x, ∠C = 4x, and ∠ABC = 180°-7x must all be in (0°, 180°), we must have x = 18°.

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