Difference between revisions of "Van Aubel's Theorem"
(→Proof 1: Complex Numbers) |
(→Proof 1: Complex Numbers) |
||
Line 6: | Line 6: | ||
== Proof 1: Complex Numbers== | == Proof 1: Complex Numbers== | ||
− | + | <asy> | |
size(220); | size(220); | ||
import TrigMacros; | import TrigMacros; | ||
rr_cartesian_axes(-3,8,-2,8,complexplane=true,usegrid = false); | rr_cartesian_axes(-3,8,-2,8,complexplane=true,usegrid = false); | ||
− | pair | + | pair A, B, C, D, O, P, Q, R, SS; |
O = (0,0) ; | O = (0,0) ; | ||
A = (2,1.5); | A = (2,1.5); | ||
Line 30: | Line 30: | ||
//draw(WW--Y,red); | //draw(WW--Y,red); | ||
//draw(X--Z,blue); | //draw(X--Z,blue); | ||
− | dot(" | + | dot("$a$",U,SW); |
− | dot(" | + | dot("$b$",V,2*E); |
− | dot(" | + | dot("$c$",WW,E); |
− | dot(" | + | dot("$d$",Z,NNW); |
− | dot(" | + | dot("$p$",P,E); |
− | dot(" | + | dot("$q$",Q,S); |
− | dot(" | + | dot("$r$",R,N); |
− | dot(" | + | 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 | ||
Revision as of 13:56, 4 March 2023
Theorem
On each side of quadrilateral , construct an external square and its center:
,
,
,
; yielding centers
. Van Aubel's Theorem states that the two line segments connecting opposite centers are perpendicular and equal length:
, and
.
Proofs
Proof 1: Complex Numbers
Putting the diagram on the complex plane, let any point
be represented by the complex number
. Note that
and that
, and similarly for the other sides of the quadrilateral. Then we have
From this, we find that
Similarly,
Finally, we have , which implies
and
, as desired.