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.

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 $w \in \C$.

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

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 $

$=$

$\ds \dfrac {2 \cmod a^2 - 2 \cmod b^2} {\left({a + b}\right) \left({\overline a + \overline b}\right)}$

$\ds $

$=$

$\ds 0$

as $\cmod a = \cmod b$

The result follows.

$\blacksquare$

1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Example $1$.