# Why do rhombus diagonals intersect at right angles?

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.

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You've looked all over?? I typed "rhombus" into google, and the first hit was the wikipedia article on rhombus, which suggests a proof using high school geometry: the diagonals divide the rhombus into four triangles. The top two triangles are congruent by the "side-side-side" criterion and their congruence shows that the bottom angle in the left triangle has the same measure as the bottom angle in the right triangle. But these are supplementary angles, so they must both be right angles. – Pete L. Clark Mar 6 '11 at 19:41

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By the way, here's a suggestion to show that two numbers $X$ and $Y$ are negative reciprocals: show that $XY = -1$. In your case, the identity $a^2-b^2 = (a+b)(a-b)$ will be useful. – Pete L. Clark Mar 6 '11 at 19:42
– Pete L. Clark Mar 6 '11 at 19:46
@Pete L. Clark I meant to say that I couldn't find the algebraic proof. I had found the geometric proof, but I was having trouble translating it into an algebraic form. – quanticle Mar 6 '11 at 20:21
Thanks for editing. I have removed my downvote. – Pete L. Clark Mar 6 '11 at 23:08

## 4 Answers

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

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

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Thanks! That's exactly what I was looking for. My problem was that I was looking for something that looked like a reciprocal without going ahead and multiplying out my two slopes. – quanticle Mar 6 '11 at 20:11
@quanticle: When you're looking for things that look like reciprocals and you've got radicals, doing some algebra may make the connection clearer.  For example: $$\frac{\sqrt{a^2-b^2}}{a+b}\cdot\frac{\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}=\frac{a^2‌​-b^2}{(a+b)\sqrt{a^2-b^2}}=\frac{(a+b)(a-b)}{(a+b)\sqrt{a^2-b^2}}=\frac{a-b}{\sqr‌​t{a^2-b^2}}$$ – Isaac Mar 6 '11 at 20:17
n.b. in both my answer and my immediately-previous comment, $a>b$, so $$(\sqrt{a^2-b^2})^2=\left|a^2-b^2\right|=a^2-b^2.$$ – Isaac Mar 6 '11 at 20:18

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.

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Hint: Multiply the slopes together and simplify.

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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.

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