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}). 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_{CD}, and <math>P_{AB}P_{CD} \perp P_{BC}P_{CD}, | ||
= Proofs = | = Proofs = | ||
== Proof 1: Complex Numbers== | == Proof 1: Complex Numbers== | ||
| − | Putting the diagram on the complex plane, let any point <math>X< | + | 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 |
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\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
| − | Finally, we have <math>(p-r) = i(q-s) = e^{i \pi/2}(q-r)< | + | Finally, we have </math>(p-r) = i(q-s) = e^{i \pi/2}(q-r)<math>, which implies </math>PR = QS<math> and </math>PR \perp QS$, as desired. |
==See Also== | ==See Also== | ||
[[Category:Theorems]] | [[Category:Theorems]] | ||
Revision as of 14:45, 21 February 2023
Theorem
On each side of quadrilateral 
, construct an external square and its center: (
, 
, 
, 
; yielding centers 
P_{AB}P_{CD} = P_{BC}P_{CD}, and $P_{AB}P_{CD} \perp P_{BC}P_{CD},
= Proofs =
== Proof 1: Complex Numbers==
Putting the diagram on the complex plane, let any point$ (Error compiling LaTeX. Unknown error_msg)X
x
\angle PAB = \frac{\pi}{4}
PA = \frac{\sqrt{2}}{2}AB$, and similarly for the other sides of the quadrilateral. Then we have
<cmath>\begin{eqnarray*}
p &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a \\
q &=& \frac{\sqrt{2}}{2}(c-b)e^{i \frac{\pi}{4}}+b \\
r &=& \frac{\sqrt{2}}{2}(d-c)e^{i \frac{\pi}{4}}+c \\
s &=& \frac{\sqrt{2}}{2}(a-d)e^{i \frac{\pi}{4}}+d
\end{eqnarray*}</cmath>
From this, we find that <cmath>\begin{eqnarray*} p-r &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a - \frac{\sqrt{2}}{2}(d-c)e^{i \frac{\pi}{4}}-c \\ &=& \frac{1+i}{2}(b-d) + \frac{1-i}{2}(a-c). \end{eqnarray*}</cmath> Similarly, <cmath>\begin{eqnarray*} q-s &=& \frac{\sqrt{2}}{2}(c-b)e^{i \frac{\pi}{4}}+a - \frac{\sqrt{2}}{2}(a-d)e^{i \frac{\pi}{4}}-c \\ &=& \frac{1+i}{2}(c-a) + \frac{1-i}{2}(b-d). \end{eqnarray*}</cmath>
Finally, we have$ (Error compiling LaTeX. Unknown error_msg)(p-r) = i(q-s) = e^{i \pi/2}(q-r)
PR = QS
PR \perp QS$, as desired.
