Wednesday, February 18, 2009

Problem 11: Right triangle, Cevian, Angles

Proposed Problem
Problem: Right triangle, Cevian, Angles.

See complete Problem 11 at:
http://gogeometry.com/problem/problem011.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

11 comments:

  1. Triangles DAB & ACB are similar. Hence, BD/AB = AB/BC or AB^2 = BD*BC = 3BD^2. In triangle DAB, we've AD^2 = AB^2+BD^2 = 3BD^2+BD^2 = 4BD^2. Thus, BD = AD/2. In other words, DAB is 30-60-90 with angle DAB = x = 30 deg.
    Ajit: ajitathle@gmail.com

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  2. Solucion:
    http://img14.imageshack.us/img14/6471/rsolmarzo6.jpg

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

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  4. This comment has been removed by the author.

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  5. Let AB=c and DC=2a=2BD
    From triangle ABD and ABC:tanx=a/c=c/3a
    so 3a^2=c^2 or a/c=(sqrt3)/3 So tanx=(sqrt3)/3
    and x=30 deg

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  6. This comment has been removed by the author.

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  7. My solution

    http://lectiimatematice.blogspot.com/2012/02/var-docstocdocid112425849var.html

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  8. The solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJZmw0MVhmWHFFLVE

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  9. in triangle ABC, tanx= AB/BC....In triangle ABD, tanx =BD/AB.....Multipying both we get x=30

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  10. https://www.youtube.com/watch?v=ofDdrHRObOI

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  11. Extend CB to E such that BD = BE.

    Then AB is the perpendicular bisector of DE and < EAB = < BAD = x = < ACE.
    Hence < EAC must be a right angle and so AD=DE=DC
    Therefore < DAC = x, hence 3x = 9and x =30

    Sumith Peiris
    Moratuwa
    Sri Lanka

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