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Difference between revisions of "Van Aubel's Theorem"

(New page: = Theorem = Construct squares <math>ABA'B'</math>, <math>BCB'C'</math>, <math>CDC'D'</math>, and <math>DAD'A'</math> externally on the sides of quadrilateral <math>ABCD</math>, and let the...)
 
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= Theorem =
 
= Theorem =
 
Construct squares <math>ABA'B'</math>, <math>BCB'C'</math>, <math>CDC'D'</math>, and <math>DAD'A'</math> externally on the sides of quadrilateral <math>ABCD</math>, and let the centroids of the four squares be <math>P, Q, R,</math> and <math>S</math>, respectively.  Then <math>PR = QS</math> and <math>PR \perp QS</math>.
 
Construct squares <math>ABA'B'</math>, <math>BCB'C'</math>, <math>CDC'D'</math>, and <math>DAD'A'</math> externally on the sides of quadrilateral <math>ABCD</math>, and let the centroids of the four squares be <math>P, Q, R,</math> and <math>S</math>, respectively.  Then <math>PR = QS</math> and <math>PR \perp QS</math>.
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  <geogebra>21cd94f930257bcbd188d1ed7139a9336b3eb9bc</geogebra>
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  <geogebra> 21cd94f930257bcbd188d1ed7139a9336b3eb9bc <geogebra>
  
 
= Proofs =
 
= Proofs =
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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.
 
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.
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==See Also==
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== Proof 2: Mean Geometry Theorem ==
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[[Category:Theorems]]

Revision as of 00:17, 20 July 2020

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.

See Also

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