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:

Then I added a few extra lines, like this:

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.