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Showing the parallelogram law with a simplistic knowledge of vectors

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the sum of the squared side lengths of a parallelogram is equal to the sum of its squared diagonals.

    2. Somewhat relevant thoughts

    I've decided to try to show the parallelogram law with vectors, since I already managed to find an Elements-inspired proof of it. However, I couldn't translate that into vectors, and I can't seem to figure this one out on my own. I found another proof that uses normed vector spaces that I didn't understand. My knowledge of vectors is pretty much limited to scalar products, vector products and the volume function.

    3. The attempt at a solution

    http://img525.imageshack.us/img525/1478/parai.png [Broken]

    Using the parallelogram representation in the figure above, it seems relevant to show that [tex]|\overrightarrow{AB}|^2 + |\overrightarrow{BC}|^2 + |\overrightarrow{DC}|^2 + |\overrightarrow{AD}|^2 = |\overrightarrow{AC}|^2 + |\overrightarrow{BD}|^2[/tex] (1)

    [tex]|\overrightarrow{AB}|^2 + |\overrightarrow{BC}|^2 + |\overrightarrow{DC}|^2 + |\overrightarrow{AD}|^2 = 2|\overrightarrow{AB}|^2 + 2|\overrightarrow{AD}|^2 = 2|\overrightarrow{AM} + \overrightarrow{MB}|^2 + 2|\overrightarrow{AM} + \overrightarrow{MD}|^2 = 2|.5 \overrightarrow{AC} + .5 \overrightarrow{DB}|^2 + 2|.5 \overrightarrow{AC} - .5 \overrightarrow{DB}|^2 =[/tex]
    [tex]= \frac{1}{2} (|\overrightarrow{AC} + \overrightarrow{DB}|^2 + |\overrightarrow{AC} - \overrightarrow{DB}|^2)[/tex]

    The latter step seems like a reasonably important geometric identity, but I'm unable to get anywhere with it. All further attempts I've made have either created more complexity and ultimately been dead-ends, or have been wrong. Any suggestions?

    I also noticed that the R.H.S. of (1) can be written as [tex]2|\overrightarrow{AM}|^2 + 2|\overrightarrow{MD}|^2[/tex], which is somewhat algebraically and geometrically similar to the [tex]2|\overrightarrow{AM} + \overrightarrow{MB}|^2 + 2|\overrightarrow{AM} + \overrightarrow{MD}|^2[/tex] step above.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 5, 2009 #2

    lanedance

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    Homework Helper

    scratch last bit....

    if i follow correctly, i think you might be close, how about expanding your last line by expressing the magnitude squared as a dot product of the vector inside with itself
     
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