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>$$\begin{array}{rcl}p& =& \frac{\sqrt{2}}{2}(b-a)e^{i\frac{\pi }{4}}+a\\ q& =& \frac{\sqrt{2}}{2}(c-b)e^{i\frac{\pi }{4}}+b\\ r& =& \frac{\sqrt{2}}{2}(d-c)e^{i\frac{\pi }{4}}+c\\ s& =& \frac{\sqrt{2}}{2}(a-d)e^{i\frac{\pi }{4}}+d\end{array}$$</cmath>

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