Difference between revisions of "Van Aubel's Theorem"
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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_{ | + | <math>P_{AB}P_{CD} = P_{BC}P_{DA}</math>, and $\overline{P_{AB}P_{CD}} \perp \overline{P_{BC}P_{DA}}, |
= Proofs = | = Proofs = |
Revision as of 15:52, 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 $\overline{P_{AB}P_{CD}} \perp \overline{P_{BC}P_{DA}},
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.