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Math Help - Prove CD + AE = AC.

  1. #1
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    Prove CD + AE = AC.

    Given a triangle ABC with ∠ B = 60. The bisectors of angle A and C intersect BC and AB at D and E respectively. Prove that CD + AE = AC.

    Please help me with this.... I tried hours but still can't prove it.
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  2. #2
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    Re: Prove CD + AE = AC.

    Did you google "angle bisector theorem"?
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  3. #3
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    Re: Prove CD + AE = AC.

    Quote Originally Posted by MichaelLight View Post
    Given a triangle ABC with ∠ B = 60. The bisectors of angle A and C intersect BC and AB at D and E respectively. Prove that CD + AE = AC.

    Please help me with this.... I tried hours but still can't prove it.
    I retract my answer, sorry about that. Will make sure my answer is correct next time.
    Last edited by Vingar; September 9th 2011 at 08:15 AM. Reason: error
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  4. #4
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    Re: Prove CD + AE = AC.

    No Vingar: the line that bisects an ANGLE does not always cut opposite side in half.
    Draw a 30-60-90 triangle and bisect the 90degree angle...you'll "see"...
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  5. #5
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    Re: Prove CD + AE = AC.

    Using angle bisector theorem, i get (EB/AC)= (AE)/(AC) and (BD)/(DC) = (AB)/AC). So by combining the results, i get

    (AE + DC) / (AC) = [ (EB)/BC)] + [(BD)/(AB)]

    (AE + DC) = (AC) [(EB)/BC)] + [(BD)/(AB)]

    Am i in the correct path? How should i proceed?
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  6. #6
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    Re: Prove CD + AE = AC.

    Michael, is this homework...or what?
    Did you "kinda see what's going on" by drawing a representative triangle?
    Anyway, I'll give you this, then you're on your own:
    Code:
                 B(60)
     
                   c-e
          a-d             
                           E
        D
                 M            e
      d
     
    C(2u)          b                 A(120-2u)
    M is the point where the bisectors cross...OK?
    Let a = BC, b = AC, c = AB, d = CD and e = AE; then BD = a-d and BE = c-e.
    (Easier to work, as example, with "b" instead of "AC"...get my drift?)

    So the ratios are: a / (c+e) = b / e and c / (a-d) = b / d

    OK: let 2u = angleACB ; then angleACE = u.
    Also, angle BAC becomes 120-2u ; then angleCAD = 60-u

    Now using triangleAMC, angleAMC = 180-u-(60-u) = 120 : see that?
    This forces angleAME = 60 (also angleCMD)

    Carry on...
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  7. #7
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    Re: Prove CD + AE = AC.

    Its simple...
    Let the point of intersection of AD and CE be O
    draw the angle bisector of O meeting AC at M.
    Now, LCOD = 60 so, LCOM = 60 = LMOA = LAOE
    Thus, Triangles COD and COM && Triangles AOM and AOE are congruent..
    Thus, AE = AM and CD = CM or, CD + AE = AM + CM = AC. Done...!!
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  8. #8
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    Re: Prove CD + AE = AC.

    Hmmmm...well, true Burnit...but we usually do not give full solution here.
    I think Michael would have been able to wrap this up from my hints:
    he seems to know what he's doing...
    Showing him that the "trick" is to arrive at central angle = 120 should then
    make it evident to him that next step is to create these congruent triangles.

    But then who knows...he hasn't responded !
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  9. #9
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    Re: Prove CD + AE = AC.

    Got it guys, thanks a lot!
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