Van Aubel's Theorem

Theorem

On each side of quadrilateral , construct an external square and its center: (, , , ; yielding centers P_{AB}P_{CD} = P_{BC}P_{CD}, and $P_{AB}P_{CD} \perp P_{BC}P_{CD},

= Proofs =

== Proof 1: Complex Numbers== Putting the diagram on the complex plane, let any point$ (Error compiling LaTeX. Unknown error_msg)Xx\angle PAB = \frac{\pi}{4}PA = \frac{\sqrt{2}}{2}AB$, and similarly for the other sides of the quadrilateral. Then we have

<cmath>\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*}</cmath>

From this, we find that <cmath>\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*}</cmath> Similarly, <cmath>\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*}</cmath>

Finally, we have$ (Error compiling LaTeX. Unknown error_msg)(p-r) = i(q-s) = e^{i \pi/2}(q-r)PR = QSPR \perp QS$, as desired.