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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}).  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:
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</math>P_{AB}P_{CD} = P_{BC}P_{CD}, and  <math>P_{AB}P_{CD} \perp P_{BC}P_{CD},
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<math>P_{AB}P_{CD} = P_{BC}P_{CD}, and  </math>P_{AB}P_{CD} \perp P_{BC}P_{CD},
  
 
= Proofs =
 
= Proofs =
  
 
== Proof 1: Complex Numbers==
 
== Proof 1: Complex Numbers==
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Putting the diagram on the complex plane, let any point </math>X<math> be represented by the complex number </math>x<math>.  Note that </math>\angle PAB = \frac{\pi}{4}<math> and that </math>PA = \frac{\sqrt{2}}{2}AB<math>, and similarly for the other sides of the quadrilateral.  Then we have
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Putting the diagram on the complex plane, let any point <math>X</math> be represented by the complex number <math>x</math>.  Note that <math>\angle PAB = \frac{\pi}{4}</math> and that <math>PA = \frac{\sqrt{2}}{2}AB</math>, and similarly for the other sides of the quadrilateral.  Then we have
  
  
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\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
  
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Finally, we have </math>(p-r) = i(q-s) = e^{i \pi/2}(q-r)<math>, which implies </math>PR = QS<math> and </math>PR \perp QS$, as desired.
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Finally, we have <math>(p-r) = i(q-s) = e^{i \pi/2}(q-r)</math>, which implies <math>PR = QS</math> and <math>PR \perp QS</math>, as desired.
 
==See Also==
 
==See Also==
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Revision as of 14:46, 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: $P_{AB}P_{CD} = P_{BC}P_{CD}, and$P_{AB}P_{CD} \perp P_{BC}P_{CD},

Proofs

Proof 1: Complex Numbers

Putting the diagram on the complex plane, let any point $X$ 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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