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Math Help - isosceles triangle

  1. #1
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    isosceles triangle

    In an isosceles triangle ABC , AB=AC and angle BAC =120 degree. Points D and E lie on BC such that BD=DE=EC . Prove that ADE is an equilateral triangle .
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    Using Sine Law, AD and AE can be proved to have equally divided angle BAC into 3 smaller ones, 40 degrees each. However, angles ADE and AED are 70 degrees (exterior angle of triangles), so triangle DAE should be isoscles, not equilateral.
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  3. #3
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    Quote Originally Posted by ukorov View Post
    Using Sine Law, AD and AE can be proved to have equally divided angle BAC into 3 smaller ones, 40 degrees each. However, angles ADE and AED are 70 degrees (exterior angle of triangles), so triangle DAE should be isoscles, not equilateral.
    Thanks , so the problem is with the question . Are you really sure ??
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  4. #4
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    In an isosceles triangle ABC , AB=AC and angle BAC =120 degree. Points D and E lie on BC such that BD=DE=EC . Prove that ADE is an equilateral triangle .
    Quote Originally Posted by ukorov View Post
    Using Sine Law, AD and AE can be proved to have equally divided angle BAC into 3 smaller ones, 40 degrees each. However, angles ADE and AED are 70 degrees (exterior angle of triangles), so triangle DAE should be isoscles, not equilateral.
    Quote Originally Posted by thereddevils View Post
    Thanks , so the problem is with the question . Are you really sure ??
    There's nothing wrong with the question. Just use some simple trigonometry and Pythagoras' Theorem.

    Let AB= d, and then if M is the mid-point of BC:
    BD = \tfrac13BC = \tfrac23BM =\tfrac23d\cos30^o = \frac{d}{\sqrt3}

    \Rightarrow DM = \tfrac12BD = \frac{d}{2\sqrt3}
    Also
    AM = d\sin30^o=\frac{d}{2}
    So using Pythagoras
    AD^2 = DM^2+AM^2 =\frac{d^2}{12}+\frac{d^2}{4}=\frac{d^2}{3}

    \Rightarrow AD = \frac{d}{\sqrt3}= BD = DE = AE (by symmetry)
    and hence \triangle ADE is equilateral.

    Grandad
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