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Thread: Proof verification

  1. #1
    Junior Member
    Feb 2011

    Proof verification

    I was recently presented with this problem:

    $\displaystyle \mbox{Let ABCD be a parallelogram, where the diagonal BD is equal to the side AB. Show that}$

    $\displaystyle \displaystyle{\frac{\mid{AD}\mid}{\mid{AC}\mid} < \frac{2}{3}{$

    $\displaystyle \mbox{where}\mid{AD}\mid\mbox{and}\mid{AC}\mid\mbo x{are the lengths of the sides.}$

    I solved this problem by first drawing up the parallelogram, like this:
    Proof verification-parallellogram.gif

    Then I added a few extra lines, like this:
    Proof verification-parallellogram2.gif
    adding the height h and a point P.

    After that I set a condition that $\displaystyle h>0$ and stated that $\displaystyle \mid{BD}\mid=\mid{AB}\mid=\mid{DC}\mid$ and also the fact that
    $\displaystyle \mid{DP}\mid=\frac{\mid{AD}\mid}{2}$, which I'm too lazy to prove right now, as I'm still struggling with latex.

    After this I used the pythagorean theorem to show that $\displaystyle {\mid{AC}\mid}^2=(1.5{\mid{AD}\mid})^2+h^2$ which means that $\displaystyle \mid{AC}\mid=\sqrt{(1.5{\mid{AD}\mid})^2+h^2}$

    As an effect, this means that $\displaystyle \mid{AC}\mid>\sqrt{(1.5{\mid{AD}\mid})^2}$, as $\displaystyle \sqrt{(1.5{\mid{AD}\mid})^2+h^2}$ clearly is bigger than $\displaystyle \sqrt{(1.5{\mid{AD}\mid})^2}$

    Then, I just made myself an inequality.

    $\displaystyle \mid{AC}\mid>1.5{\mid{AD}\mid}$

    $\displaystyle \displaystyle{\frac{\mid{AC}\mid}{1.5\mid{AD}\mid} >1}$

    $\displaystyle \displaystyle{\frac{1}{1.5\mid{AD}\mid}>\frac{1}{\ mid{AC}\mid}}$

    $\displaystyle \displaystyle{\frac{\mid{AD}\mid}{1.5\mid{AD}\mid} >\frac{\mid{AD}\mid}{\mid{AC}\mid}}$

    And then finally

    $\displaystyle \displaystyle{\frac{\mid{AD}\mid}{\mid{AC}\mid}<\f rac{2}{3}}$, which I was supposed to show.

    My question is if I've done everything right, as the book in which I found the problem solves it with trigonometry, and the section for the problem was a section about trigonometry. If my solution is incorrect I need to know why.
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  2. #2
    Senior Member abhishekkgp's Avatar
    Jan 2011
    It's correct.
    but there is a geometrical proof i got.
    Drop perpendicular from B to AD and call the foot of the perpendicular as K. Let AC meet BD at L.
    then |AK|=|KD| and |AL|=|LC|.
    Let AL and BK meet at G. G is the centroid since AL and BK are medians.
    To prove: |AD|/|AC| < 2/3
    that is, |AK|/|AL| < 2/3.
    Now |AK|<|AG| since AG is the hypotenuse of triangle-AGK.(AGK is a right angled triangle).
    so |AK|/|AL| < |AG|/|AL|.
    we know that |AG|/|AL|=2/3 from elementary geometry because G is the centroid. which proves the result.
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