Difference between revisions of "Van Aubel's Theorem"
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= 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: | 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>P_{AB}P_{CD} = P_{BC}P_{DA}</math>, and <math>\overline{P_{AB}P_{CD}\}} \perp \overline{P_{BC}P_{DA}}</math>. |
= Proofs = | = Proofs = | ||
Revision as of 14:53, 21 February 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.
