ABC is an isosceles triangle in which the equal sides are AB and A.If AD is an altitude and |BC|=1/2|AB|.Prove that |AD|^2=15|BC|^2

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asked by Pamela
  1. I feel like this question might be missing some vital information. Is this a 3D trigonometry question? If so, I'm not sure it can be solved without being given an additional length or some bearings/elevation angles.

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    posted by Chloe
  2. The altitude in an isosceles triangle right-bisects the base, so BD = DC

    let BD = x
    then BC = 2x
    AB = 4x

    in the right-triangle ABD
    AD^2 + x^2 = (4x)^2 = 16x^2
    AD^2 = 15x^2

    AD^2 = 15BD^2 , not 15 BC^2 as you stated.

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    posted by Reiny

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