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Math Help - a difficult geometry proof

  1. #1
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    a difficult geometry proof

    Here is the problem.

    Consider an isoceles triangle ABC.

    Please see the diagram below and IGNORE the dots which are used to show the positions of the points. Point D is a point on side AB.

    ......................A


    .................D

    ..............B__________C

    suppose that side AB=side AC. angle BAC=20 degrees and angle BDC= 30 degrees.

    Prove that side AD=side BC.

    Please help. thanks
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  2. #2
    MHF Contributor

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    Quote Originally Posted by davedave View Post
    Here is the problem.

    Consider an isoceles triangle ABC.

    Please see the diagram below and IGNORE the dots which are used to show the positions of the points. Point D is a point on side AB.

    ......................A


    .................D

    ..............B__________C

    suppose that side AB=side AC. angle BAC=20 degrees and angle BDC= 30 degrees.

    Prove that side AD=side BC.

    Please help. thanks
    using law of sines we have \frac{\sin (20)}{DC}=\frac{\sin (10)}{AD} and \frac{\sin (80)}{DC}=\frac{\sin (30)}{BC}. dividing these two relations gives us: \frac{AD}{BC}=\frac{\sin (80) \sin (10)}{\sin(20) \sin (30)}=\frac{2 \cos(10) \sin(10)}{\sin (20)}=1.
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  3. #3
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    Joined
    May 2009
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    3

    non-trigonometric proof

    I guess I forgot to state a requirement for this problem.

    We should prove it without using trigonometry.

    sorry about my oversight
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