Theorem

On each side of quadrilateral $ABCD$ , construct an external square and its center: ( $ABA'B'$ , $BCB'C'$ , $CDC'D'$ , $DAD'A'$ ; yielding centers $P_{AB}, P_{BC}, P_{CD}, P_{DA}$ ). Van Aubel's Theorem states that the two line segments connecting opposite centers are perpendicular and equal length: $P_{AB}P_{CD} = P_{BC}P_{DA}$ , and $\overline{P_{AB}P_{CD}\}} \perp \overline{P_{BC}P_{DA}}$ .

Proofs

Proof 1: Complex Numbers

Putting the diagram on the complex plane, let any point $X$ be represented by the complex number $x$ . Note that $\angle PAB = \frac{\pi}{4}$ and that $PA = \frac{\sqrt{2}}{2}AB$ , and similarly for the other sides of the quadrilateral. Then we have

$\begin{eqnarray*} p &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a \\ q &=& \frac{\sqrt{2}}{2}(c-b)e^{i \frac{\pi}{4}}+b \\ r &=& \frac{\sqrt{2}}{2}(d-c)e^{i \frac{\pi}{4}}+c \\ s &=& \frac{\sqrt{2}}{2}(a-d)e^{i \frac{\pi}{4}}+d \end{eqnarray*}$

From this, we find that $\begin{eqnarray*} p-r &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a - \frac{\sqrt{2}}{2}(d-c)e^{i \frac{\pi}{4}}-c \\ &=& \frac{1+i}{2}(b-d) + \frac{1-i}{2}(a-c). \end{eqnarray*}$ Similarly, $\begin{eqnarray*} q-s &=& \frac{\sqrt{2}}{2}(c-b)e^{i \frac{\pi}{4}}+a - \frac{\sqrt{2}}{2}(a-d)e^{i \frac{\pi}{4}}-c \\ &=& \frac{1+i}{2}(c-a) + \frac{1-i}{2}(b-d). \end{eqnarray*}$

Finally, we have $(p-r) = i(q-s) = e^{i \pi/2}(q-r)$ , which implies $PR = QS$ and $PR \perp QS$ , as desired.

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Van Aubel's Theorem

Contents

Theorem

Proofs

Proof 1: Complex Numbers

See Also