# On Bottema's Shoulders IIWhat Is This About? A Mathematical Droodle

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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 that the result still holds if the squares are replaced with similar parallelograms so that the angles BaAC and BCAc are equal. The proof is actually the same as that for Bottema's original theorem. One just has to replace the factor i with an arbitrary complex number. The proof can be slightly simplified by placing A at the origin, such that a = 0.

Let c be a complex number. Define

Ba = cg,
Ac = g + c(b - g), and
Ab = Ac + (b - g).

Then

Ab = b + c(b - g).

For the midpoint M of AbBa, we get

 M = (Ab + Ba)/2 = (cg + b + cb - cg)/2, = b(1 + c)/2,

which is independent of g, i.e. C.