I've looked all over and I can't find a good proof of why the diagonals of a rhombus should intersect at right angles. I can intuitively see its true, just by drawing rhombuses, but I'm trying to prove that the slopes of the diagonals are negative reciprocals and its not working out.
I'm defining my rhombus as follows: $[(0,0), (a, 0), (b, c), (a+b, c)]$
I've managed to figure out that $c = \sqrt{a^2-b^2}$ and that the slopes of the diagonals are $\frac{\sqrt{a^2-b^2}}{a+b}$ and $\frac{-\sqrt{a^2-b^2}}{a-b}$
What I can't figure out is how they can be negative reciprocals of one another.
EDIT: I mean to say that I could not find the algebraic proof. I've seen and understand the geometric proof, but I needed help translating it into coordinate form.
Best Answer
Another way to say that the slopes are opposite reciprocals is to say that their product is $-1$.
$$\begin{align} \frac{\sqrt{a^2-b^2}}{a+b}\cdot\frac{-\sqrt{a^2-b^2}}{a-b} &=\frac{-(\sqrt{a^2-b^2})^2}{(a+b)(a-b)} \\ &=\frac{-(a^2-b^2)}{a^2-b^2} \\ &=-1 \end{align}$$