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Math Help - Proof: Right Triangle and a Perpendicular

  1. #1
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    Proof: Right Triangle and a Perpendicular

    [/URL]
    Hypothesis: MN Perpendicular to CB
    AM = MC
    Thesis: BN^2= CN^2 + AB^2

    I've thought this could be resolved using some similarity theorem. In fact ABC is similar to MNC because they've got three corresponding pairs of angles. So CM/BC = MN/AB = CN/AC = k. However I don't know what to do now!

    Thank you very much!
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Proof: Right Triangle and a Perpendicular

    Quote Originally Posted by goby View Post
    [/URL]
    Hypothesis: MN Perpendicular to CB
    AM = MC
    Thesis: BN^2= CN^2 + AB^2

    I've thought this could be resolved using some similarity theorem. In fact ABC is similar to MNC because they've got three corresponding pairs of angles. So CM/BC = MN/AB = CN/AC = k. However I don't know what to do now!

    Thank you very much!
    BC^2=AC^2+AB^2

    (BN+CN)^2=(2CM)^2+AB^2

    BN^2+CN^2+2BN*CN=4CM^2+AB^2 ----------------- (1)

    By similar triangles,

    \frac{CM}{BC}=\frac{CN}{AC}

    \frac{CM}{BN+CN}=\frac{CN}{2CM}

    2CM^2=BN*CN+CN^2

    Substituting this in (1),

    BN^2+CN^2+2BN*CN=2BN*CN+2CN^2+AB^2

    BN^2=CN^2+AB^2

    QED
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  3. #3
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    Re: Proof: Right Triangle and a Perpendicular

    Yeah, it was easier than I thought! Thank you again!
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