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Math Help - Perpendiculars

  1. #1
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    Perpendiculars

    (See Attachment)
    Given a Triangle ABC with circumcenter Z.
    And a line l outside the triangle ABC.
    There are four perpendiculars to L, starting in A, B, C and Z. (ending in respectively A',B',C' and Z')
    I want to proof:

    ZZ'= \frac{1}{3}(AA'+BB'+CC')

    Thanks for you help
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  2. #2
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    Quote Originally Posted by bramkierkels View Post
    (See Attachment)
    Given a Triangle ABC with circumcenter Z.
    And a line l outside the triangle ABC.
    There are four perpendiculars to L, starting in A, B, C and Z. (ending in respectively A',B',C' and Z')
    I want to proof:

    ZZ'= \frac{1}{3}(AA'+BB'+CC')

    Thanks for you help
    This is best done using vectors. Let P be a point on the line \ell, and let \mathbf{p} be the position vector of P. Let \mathbf{n} be a unit vector perpendicular to \ell. Given a point X, with position vector \mathbf{x}, let X' be the foot of the perpendicular from X to \ell. Then the position vector of X' is \mathbf{x} + d\mathbf{n}, where d is the distance XX'. Also, the line PX' is perpendicular to \mathbf{n}, and so (\mathbf{x}+d\mathbf{n} - \mathbf{p})\mathbf{.}\mathbf{n} = 0 (where the dot represents the inner product). Therefore d = (\mathbf{x} - \mathbf{np})\mathbf{.}\mathbf{n} (because \mathbf{n}\mathbf{.}\mathbf{n} = 1).

    If A, B, C, Z have position vectors \mathbf{a},\ \mathbf{b},\ \mathbf{c},\ \mathbf{z}, then \mathbf{z} = \frac13(\mathbf{a}+\mathbf{b}+ \mathbf{c}). Thus \frac13(AA'+BB'+CC') = \frac13\bigl((\mathbf{a} - \mathbf{p})\mathbf{.}\mathbf{n} + (\mathbf{b} - \mathbf{p})\mathbf{.}\mathbf{n} + (\mathbf{c} - \mathbf{p})\mathbf{.}\mathbf{n}\bigr), which simplifies to (\mathbf{z} - \mathbf{p})\mathbf{.}\mathbf{n} = ZZ'.
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  3. #3
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    Thanks Opalg, but actually I was looking for a solution only using Euclidean Geometry. Can you help me?
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