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Difference between revisions of "Van Aubel's Theorem"

(→‎Proof 2: Mean Geometry Theorem)
(→‎Proof 1: Complex Numbers)
 
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= Theorem =
 
= Theorem =
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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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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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<geogebra> 21cd94f930257bcbd188d1ed7139a9336b3eb9bc <geogebra>
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<math>P_{AB}P_{CD} = P_{BC}P_{DA}</math>, and <math>\overline{P_{AB}P_{CD}} \perp \overline{P_{BC}P_{DA}}</math>.
  
 
= Proofs =
 
= Proofs =
  
 
== Proof 1: Complex Numbers==
 
== Proof 1: Complex Numbers==
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<asy>
 +
size(220);
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import TrigMacros;
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rr_cartesian_axes(-3,8,-2,8,complexplane=true,usegrid = false);
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pair A, B, C, D, O, P, Q, R, SS;
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O = (0,0) ;
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A = (2,1.5);
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B= (4,1.8);
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C = (5.3,3);
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D= (3,5.3);
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draw(A--B--C--D--cycle);
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draw(A--(A + rotate(-90)*(B-A))--(B + rotate(90)*(A-B))--B);
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draw(B--(B + rotate(-90)*(C-B))--(C + rotate(90)*(B-C))--C);
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draw(C--(C + rotate(-90)*(D-C))--(D + rotate(90)*(C-D))--D);
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draw(D--(D + rotate(-90)*(A-D))--(A + rotate(90)*(D-A))--A);
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P = (B + (A + rotate(-90)*(B-A)))/2;
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Q = (C + (B + rotate(-90)*(C-B)))/2;
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R = (D + (C + rotate(-90)*(D-C)))/2;
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SS = (A + (D + rotate(-90)*(A-D)))/2;
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//draw(WW--Y,red);
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//draw(X--Z,blue);
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dot("$a$",A,SW);
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dot("$b$",B,2*E);
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dot("$c$",C,E);
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dot("$d$",D,NNW);
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dot("$p$",P,E);
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dot("$q$",Q,S);
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dot("$r$",R,N);
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dot("$s$",SS,S);
 +
</asy>
 
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
 
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>
  
−
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-s)</math>, which implies <math>PR = QS</math> and <math>PR \perp QS</math>, as desired.
  
 +
==See Also==
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Latest revision as of 13:01, 4 March 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

[asy] size(220); import TrigMacros; rr_cartesian_axes(-3,8,-2,8,complexplane=true,usegrid = false); pair A, B, C, D, O, P, Q, R, SS; O = (0,0) ; A = (2,1.5); B= (4,1.8); C = (5.3,3); D= (3,5.3); draw(A--B--C--D--cycle); draw(A--(A + rotate(-90)*(B-A))--(B + rotate(90)*(A-B))--B); draw(B--(B + rotate(-90)*(C-B))--(C + rotate(90)*(B-C))--C); draw(C--(C + rotate(-90)*(D-C))--(D + rotate(90)*(C-D))--D); draw(D--(D + rotate(-90)*(A-D))--(A + rotate(90)*(D-A))--A); P = (B + (A + rotate(-90)*(B-A)))/2; Q = (C + (B + rotate(-90)*(C-B)))/2; R = (D + (C + rotate(-90)*(D-C)))/2; SS = (A + (D + rotate(-90)*(A-D)))/2; //draw(WW--Y,red); //draw(X--Z,blue); dot("$a$",A,SW); dot("$b$",B,2*E); dot("$c$",C,E); dot("$d$",D,NNW); dot("$p$",P,E); dot("$q$",Q,S); dot("$r$",R,N); dot("$s$",SS,S); [/asy] 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 $(p-r) = i(q-s) = e^{i \pi/2}(q-s)$, which implies $PR = QS$ and $PR \perp QS$, as desired.

See Also

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