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Math Help - Triangles

  1. #1
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    Triangles

    In a Triangle ABC, |_A = 90 degree and D is the mid point of AC.The value of BC^2-BD^2 = AD^2.Prove.
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  2. #2
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    Quote Originally Posted by devi View Post
    In a Triangle ABC, |_A = 90 degree and D is the mid point of AC.The value of BC^2-BD^2 = AD^2.Prove.
    If D is the midpoint of AC, then AD = DC....and AC = 2(AD)

    In right triangle ABC,
    (BC)^2 = (AB)^2 +(2(AD))^2
    (BC)^2 = (AB)^2 +4(AD)^2 -----------(i)

    In right triangle ABD,
    (BD)^2 = (AB)^2 +(AD)^2 ----------(ii)

    Eq,(i) minus Eq.(ii),
    (BC)^2 -(BD)^2 = 3(AD)^2

    Oops...it is not proven.
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  3. #3
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    Hello, devi!

    The problem has strange wording ... and the statement is not true.


    In \Delta ABC,\;\angle A = 90^o and D is the midpoint of AC.
    The value of: . BC^2-BD^2 \:= \:AD^2 .
    . . . "the value of..." ?
    Prove.
    Consider a 3-4-5 right triangle.
    Code:
           B*
            | *  *
            |   *     *    5
           3|     *        *
            |       *           *
            |         *              *
            * - - - - - - * - - - - - - - *
            A      2      D       2       C

    We have: . BC = 5,\;BD = \sqrt{13},\;AD = 2

    . . But: . (5)^2 -(\sqrt{13})^2 \;\;{\bf{\color{red}\neq}}\;\;(2)^2

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  4. #4
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    Triangle BCD is not right-angled so you cannot apply Pythagoras Theorem to it.
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