View source for Van Aubel's Theorem

= Theorem =
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>.

= 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 <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


<cmath>\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*}</cmath>

From this, we find that 
<cmath>\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*}</cmath>
Similarly,
<cmath>\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*}</cmath>

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]]