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Geometry Angles Problem

  1. Aug 6, 2014 #1
    1. The problem statement, all variables and given/known data
    ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Then prove that ∠D=90°-(∠A/2)


    2. Relevant equations



    3. The attempt at a solution

    I guess I want to prove an untrue thing because by drawing a diagram and working out I get ∠D=∠A/2. I think I have drawn it wrong by mis-understanding the question. Or the question is an error? Or did I work out wrong?
     

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  3. Aug 6, 2014 #2

    Mentallic

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    I get the same result you did and I also oriented the corners of the triangle such that ABC is labelled clockwise instead (C at the top, A at the left), so the intersection point is underneath the triangle and the answer is still the same.
     
  4. Aug 6, 2014 #3

    Simon Bridge

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    You can do a quick reality check by drawing an arbitrary triangle and constructing the bisectors and measuring the angles. Try it for several triangles.
     
  5. Aug 6, 2014 #4
    It seems a difficult work. Are there any thing to do with Theoretical geometry.
     
  6. Aug 6, 2014 #5
    Even I tried what Simon said with a simulator called GEOGEBRA. I get the same result.
     
  7. Aug 6, 2014 #6

    olivermsun

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    I get the same result as you using a + b + c = 180° = (180 - b)/2 + b + a + c/2.
     
  8. Aug 6, 2014 #7
    I guess there must be a printing error in the question like may be the angles changed or like that
     
  9. Aug 7, 2014 #8

    Simon Bridge

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  10. Aug 7, 2014 #9
    Yes I know just lazy to do that because when I miss 1° accuracy, I may get very strange results.
     
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