# Theorem

Construct squares $ABA'B'$, $BCB'C'$, $CDC'D'$, and $DAD'A'$ externally on the sides of quadrilateral $ABCD$, and let the centroids of the four squares be $P, Q, R,$ and $S$, respectively. Then $PR = QS$ and $PR \perp QS$.

<geogebra> 21cd94f930257bcbd188d1ed7139a9336b3eb9bc <geogebra>


# 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.