Revision as of 13:56, 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

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

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Difference between revisions of "Van Aubel's Theorem"

Revision as of 13:56, 4 March 2023

Contents

Theorem

Proofs

Proof 1: Complex Numbers

See Also

@@ Line 6: / Line 6: @@
 == Proof 1: Complex Numbers==
-[asy]
+<asy>
 size(220);
 import TrigMacros;
 rr_cartesian_axes(-3,8,-2,8,complexplane=true,usegrid = false);
-pair U, V, WW, Z, O, P, Q, R, SS;
+pair A, B, C, D, O, P, Q, R, SS;
 O = (0,0) ;
 A = (2,1.5);
@@ Line 30: / Line 30: @@
 //draw(WW--Y,red);
 //draw(X--Z,blue);
-dot("<math>a</math>",U,SW);
+dot("$a$",U,SW);
-dot("<math>b</math>",V,2*E);
+dot("$b$",V,2*E);
-dot("<math>c</math>",WW,E);
+dot("$c$",WW,E);
-dot("<math>d</math>",Z,NNW);
+dot("$d$",Z,NNW);
-dot("<math>p</math>",P,E);
+dot("$p$",P,E);
-dot("<math>q</math>",Q,S);
+dot("$q$",Q,S);
-dot("<math>r</math>",R,N);
+dot("$r$",R,N);
-dot("<math>s</math>",SS,S);
+dot("$s$",SS,S);
-[/asy]
+</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