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Difference between revisions of "Van Aubel's Theorem"

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

Revision as of 14:54, 21 February 2023

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.

See Also

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