
2 Attachment(s)
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:
Attachment 21025
Then I added a few extra lines, like this:
Attachment 21026
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.

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 triangleAGK.(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.