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On Bottema's Shoulders: What Is This About?
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Copyright © 1996-2008 Alexander Bogomolny
On Bottema's Shoulders
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Consider a configuration of two squares ACBcBa and BCAcAb with a common vertex C. Bottema's theorem claims that the midpoint M of the segment AbBa is independent of C. Professor W. McWorter has observed another property of the configuration: if D is the point that completes the parallelogram ADBC, then 
BaDAb is isosceles and angle D is right: 
BaDAb = 90o.
The theorem is easily proved with complex numbers but also has a simple synthetic proof.
Continuing with the proof of Bottema's theorem, we assume that the vertices A, B, and C are represented by the complex numbers a, b, and g, respectively. We found there that
| Ba = a + (g - a)·i, | |
| Ab = b + (g - b)·(-i), | |
| M = (a + b)/2 + (b - a)·i/2, |
where i2 = -1, as usual. We then obviously have
| i(Ab - M) = -i(Ba - M). |
Denoting the point M + i(Ab - M) by D,
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The latter tells us that ADBC is a parallelogram.
The synthetic proofs starts with the observation that in triangles ABaD and BDAb,
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ABa = BD and AD = BAb. |
The angles between the equal sides are also equal, since both are found to be
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360o - 90o - A - B,
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the latter being the angles of ABC at vertices A and B, respectively.
By Bottema's theorem, the clockwise rotation around M through 90o, maps B on A. Since BAb is perpendicular to BC which, in turn, is parallel to AD, the same rotation maps BAb on AD and, similarly, BD on ABa. It follows that the third sides of the triangles ABaD and BDAb are also mapped on each other. Thus the angle between them is 900, as required.
Copyright © 1996-2008 Alexander Bogomolny
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