# Diagonals of Rhombus Intersect at Right Angles

## Theorem

Let $ABCD$ be a rhombus.

The diagonals $AC$ and $BD$ of $ABCD$ intersect each other at right angles.

## Proof

Without loss of generality, let $ABCD$ be embedded in the complex plane so that vertex $A$ coincides with the origin $0 + 0 i$.

Let $AB$ and $AD$ be represented by the complex numbers $a$ and $b$ respectively, expressed as vectors $\mathbf a$ and $\mathbf b$ respectively.

By Geometrical Interpretation of Complex Addition, the diagonal $AC$ is represented by the complex number $a + b$, expressed as a vector $\mathbf a + \mathbf b$.

By Geometrical Interpretation of Complex Subtraction, the diagonal $BD$ is represented by the complex number $a - b$, expressed as a vector $\mathbf a - \mathbf b$.

Let $a - b = w \paren {a + b}$ for some complex number $w \in \C$.

For $A \perp BD$, it is necessary and sufficient from Complex Multiplication as Geometrical Transformation for the argument of $w$ to be $\dfrac \pi 2$.

That is, for $w$ to be wholly imaginary.

From Complex Number equals Negative of Conjugate iff Wholly Imaginary, that is for $w + \overline w = 0$.

It remains to demonstrate this.

So:

 $\ds w + \overline w$ $=$ $\ds \dfrac {a - b} {a + b} + \dfrac {\overline a - \overline b} {\overline a + \overline b}$ $\ds$ $=$ $\ds \dfrac {2 a \overline a - 2 b \overline b} {\left({a + b}\right) \left({\overline a + \overline b}\right)}$ $\ds$ $=$ $\ds \dfrac {2 \cmod a^2 - 2 \cmod b^2} {\left({a + b}\right) \left({\overline a + \overline b}\right)}$ Modulus in Terms of Conjugate $\ds$ $=$ $\ds 0$ as $\cmod a = \cmod b$

The result follows.

$\blacksquare$