Версія теореми Наполеона з квадратами

van Aubel's Theorem

van Aubel's Theorem states that if you are given a quadrilateral and construct squares on each side, the line segments connecting the centers of these squares are perpendicular and have equal lengths.

Proof

To prove this, we take a similar approach as we did for Napoleon's Theorem.

Let $j$ be a $45^\circ$ rotation in the complex plane ($j=e^{\pi i/2}$). Since $ABFG$, $BCHG$, $CDJI$, and $DALK$ are squares, we note that if we map $B$ about the spiral symmetry $S(A,\frac{\sqrt 2}2,45^\circ)$, we get $W$. Thus, we can apply similar transformations for the rest of the vertices to get: $$w = \dfrac{\sqrt 2}2 (b-a)e^{\pi i/4} + a$$ $$x = \dfrac{\sqrt 2}2 (c-b)e^{\pi i/4} + b$$ $$y = \dfrac{\sqrt 2}2 (d-c)e^{\pi i/4} + c$$ $$z = \dfrac{\sqrt 2}2 (a-d)e^{\pi i/4} + d$$ Subtracting these equations, we get: $$\begin{align} w - y &= \dfrac{\sqrt 2}2 (b-a)e^{\pi i/4} + a - \dfrac{\sqrt 2}2 (d-c)e^{\pi i/4} - c \\ &= \dfrac{(1+i)(b-d)}2 + \dfrac{(1-i)(a-c)}2 \end{align}$$ $$\begin{align} x - z &= \dfrac{\sqrt 2}2 (c-b)e^{\pi i/4} + b - \dfrac{\sqrt 2}2 (a-d)e^{\pi i/4} - d \\ &= \dfrac{(1+i)(c-a)}2 + \dfrac{(1-i)(b-d)}2 \end{align}$$

It's not immediately clear what we can do with this. However, if we note that $w-y = i(x-z)$, then $WY=XZ$ and $WY \perp XZ$, so we are done. $\Box$