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Math Help - Prove the following equation

  1. #1
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    Poland
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    Prove the following equation

    Hi. How to prove that equation:

    |AB|^{2} = |QA| * |AE| + |NB| * |BD|

    Additional informations:


    |BN| = |BC|
    |AQ| = |AC|

    |AD| and |BD| are heights of acute-angled triangle ABC.

    Illustration: http://i.imgur.com/9O8Bm.jpg

    I would really appreciate any help.
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  2. #2
    Super Member
    Joined
    Jun 2009
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    Re: Prove the following equation

    First point is that most of the diagram is unnecessary.
    If AQ=AC you don't need Q or P.
    If BN=BC you don't need N or M.
    Further, L and K serve no purpose.
    All that you need is the triangle ABC and the perpendiculars AD and BE.

    Using Pythagoras twice, with the triangles ADB and ADC,

    AB^{2}=AD^{2}+BD^{2}=(AC^{2}-CD^{2})+BD^{2}=AC^{2}-(BC-BD)^{2}+BD^{2}=AC^{2}-(BC^{2}+BD^{2}-2BC.BD)+BD^{2}=AC^{2}-BC^{2}+2BC.BD.

    Now repeat that sequence using the triangles ABE and BEC.

    Having done that, add the resulting equation to the one that's just been derived.
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  3. #3
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    Poland
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    Re: Prove the following equation

    Thank you very much. I've finally done that

    Here are full calculations:

    AB^{2} = AD^{2} + BD^{2} = (AC^{2} - CD^{2}) + BD^{2} = AC^{2} - (BC - BD)^{2} + BD^{2} = AC^{2} - (BC^{2} + BD^{2} - 2 * BC * BD) + BD^{2} = AC^{2} - BC^{2} + 2 * BC * BD

    AB^{2} = BE^{2} + AE^{2} = (BC^{2} - CE^{2}) + AE^{2} = BC^{2} - (AC - AE)^{2} + AE^{2} = BC^{2} - (AC^{2} + AE^{2} - 2 * AC * AE) + AE^{2} = BC^{2} - AC^{2} + 2 * AC * AE

    2AB^{2} = AC^{2} - BC^{2} + 2 * BC * BD + BC^{2} - AC^{2} + 2 * AC * AE
    2AB^{2} = 2 * BC * BD + 2 * AC * AE
    AB^{2} = BC * BD + AC * AE
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  4. #4
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