Van Aubel's Theorem
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
[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; 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("",U,SW);
dot("
",V,2*E);
dot("
",WW,E);
dot("
",Z,NNW);
dot("",P,E);
dot("
",Q,S);
dot("
",R,N);
dot("
",SS,S);
[/asy]
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.