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Geometry deduction

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Given a triangle ABC with BC = 2AB. D and E are the midpoints of BC and BD respectively. Show that AD bisects angle CAE.

    2. Relevant equations

    3. The attempt at a solution

    Let AB= x, so DC= AB= x and ED = x/2. If AD bisects angle CAE,
    then AC/AE = DC/DE
    AC/AE = x/(x/2)
    AC: AE = 2:1

    That's how far i can do... I tried using sine rule to prove that AC:AE = 2:1 but it did not help. Can anyone guide me?
  2. jcsd
  3. Oct 28, 2011 #2
    Can anyone help me? =(
  4. Oct 28, 2011 #3

    I like Serena

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    Homework Helper

    Hi Michael_Light! :smile:

    You might try the cosine rule on triangles AED and ADC with respect to the angles you're interested in.
    Followed by the cosine rule on triangles ABE, ABD, and ABC with respect to the angle at B.

    Find the cosines of the 2 angles you're interested in and you should find they are the same...
  5. Oct 28, 2011 #4


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    Homework Helper

    If you know about dot products, this problem becomes easy. If not, you could do it with components of vectors and trigonometry, but not without some difficulty. But to do it like that, place A at (0,0), D at (4,0), and go from there.
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