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Math Help - Midpoint Theorum Help!!

  1. #1
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    Midpoint Theorum Help!!

    I need help in this sum.


    Q: In the adjoining figure, ABCD is a square and BCPQ is a parallelogram. If M is the midpoint of DC and PM produced meets AN at N, prove that AN=1/2 AD. If BD meets MN at R, prove that BQ=1/2 AC.

    I have no problem with the first part of the question but the second part confuses me. Please help, thanks.
    Attached Thumbnails Attached Thumbnails Midpoint Theorum Help!!-midpoint.jpg  
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  2. #2
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    Please check the statement.
    I think that it should be B{\color{red}R}=\frac{1}{2}AC.
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  3. #3
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    Quote Originally Posted by Insaeno View Post
    If BD meets MN at R, prove that BQ=1/2 AC.
    Quote Originally Posted by Plato View Post
    Please check the statement.
    I think that it should be B{\color{red}R}=\frac{1}{2}AC.
    I think it's more likely
    "If BD meets MN at Q, prove that BQ=1/2 AC."

    It amounts to the same thing, but at least Q is on the diagram.

    -Dan
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  4. #4
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    I'm pretty sure I copied the sum correctly. If it is so, then the question must be flawed. Thanks anyways!!!
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  5. #5
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    This diagram also fits the given. Clearly in this diagram R could not be Q.
    Letís suppose the R is the point of intersection of BD with NM (it may be Q or not). We know that NM is parallel to BC. In the triangle BCD, because M is the midpoint of CD then R is the midpoint of BD. The diagonals of a rectangle are the same length and bisect one another. Therefore, it follows that BR=\frac{1}{2}AC
    Attached Thumbnails Attached Thumbnails Midpoint Theorum Help!!-mid2.gif  
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