Why do rhombus diagonals intersect at right angles?

Question

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.

Answer 1

Another way to say that the slopes are opposite reciprocals is to say that their product is $-1$.

$$\begin{align} \frac{\sqrt{a^2-b^2}}{a+b}\cdot\frac{-\sqrt{a^2-b^2}}{a-b} &=\frac{-(\sqrt{a^2-b^2})^2}{(a+b)(a-b)} \\ &=\frac{-(a^2-b^2)}{a^2-b^2} \\ &=-1 \end{align}$$

Answer 2

Hint: Multiply the slopes together and simplify.

Answer 3

You don't have to work through square roots if you use the properties of the vector dot product and the parallelogram law to construct the rhombus.

I.e one of the rhombus's diagonals can be identified with $\mathbf{a + b}$ where $\mathbf{b}$ is a vector added head-to-tail to vector $\mathbf{a}$, according to the parallelogram law. Similarly the other diagonal is given by $\mathbf{b - a}$. The constraint for a rhombus is $\lVert \mathbf{a} \rVert^2 = \lVert \mathbf{b} \rVert^2$. Two vectors $\mathbf{u, v}$ are perpendicular iff $\mathbf{u} \cdot \mathbf{v} = 0$, and as $\mathbf{(a + b)} \, \mathbf{\cdot} \, \mathbf{(b - a)}$ = $ \lVert \mathbf{b} \rVert^2 - \lVert \mathbf{a} \rVert^2 = 0$, the two diagonals are perpendicular.

Answer 4

One more way of looking at it is by factoring $a^2-b^2$ while it's inside the square root. Then

$$ \begin{align} \frac{\sqrt{a^2-b^2}}{a+b} &= \frac{\sqrt{(a-b)(a+b)}}{a+b} = \frac{\sqrt{a-b}}{\sqrt{a+b}} \\ - \frac{\sqrt{a^2-b^2}}{a-b} &= \frac{\sqrt{(a-b)(a+b)}}{a-b} = -\frac{\sqrt{a+b}}{\sqrt{a-b}} \end{align} $$ From here, it's easier to see that they have the correct relationship.