Difference between revisions of "Van Aubel's Theorem"
(→Proof 1: Complex Numbers) |
(→Proof 1: Complex Numbers) |
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draw(B--(B + rotate(-90)*(C-B))--(C + rotate(90)*(B-C))--C); | 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(C--(C + rotate(-90)*(D-C))--(D + rotate(90)*(C-D))--D); | ||
− | draw(D--(D + rotate(-90)*(A-D))--(A + rotate(90)*(D-A))-- | + | draw(D--(D + rotate(-90)*(A-D))--(A + rotate(90)*(D-A))--A); |
P = (B + (A + rotate(-90)*(B-A)))/2; | P = (B + (A + rotate(-90)*(B-A)))/2; | ||
Line 30: | Line 30: | ||
//draw(WW--Y,red); | //draw(WW--Y,red); | ||
//draw(X--Z,blue); | //draw(X--Z,blue); | ||
− | dot("$a$", | + | dot("$a$",A,SW); |
− | dot("$b$", | + | dot("$b$",B,2*E); |
− | dot("$c$", | + | dot("$c$",C,E); |
− | dot("$d$", | + | dot("$d$",D,NNW); |
dot("$p$",P,E); | dot("$p$",P,E); |
Revision as of 13:58, 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.