Analytic Proof of Bottema's Theorem

Consider a configuration of two squares ACBcBa and BCAcAb with a common vertex C.

Bottema's theorem

Bottema's theorem claims that the midpoint M of the segment AbBa is independent of C.

Proof

A short analytic proof by Hubert Shutrick demonstrates how simple the theorem actually is.

If \(A = (-1,0)\), \(B = (1,0)\) and \(C = (x,y)\), then \(A_b = (y+1, 1-x)\), \(B_a = (-1-y, 1+x)\) and their midpoint is \((A_{b} + B_{a})/2 = (0,1)\).

Moreover, the proof makes it rather obvious that the angle \(AMB\) is right and, therefore, is the center of the square erected on \(AB\) on the side of \(C\).

As a practice in analytic geometry we may elaborate on the above diagram:

Bottema's theorem

As Hubert Shutrick has observed, \(\triangle A_{c}MB_{c}\) is right isosceles. He also noticed that \(OECM\), where \(E\) is the midpoint of \(A_{c}B_{c}\), is a parallelogram. Analytic geometry helps confirm another fact, the one noticed by William McWorter, viz., \(\triangle A_{b}B_{a}D\) is also right isosceles.

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