The theorem is easily proved with complex numbers.
Assume the vertices A, B, and C are represented by the complex numbers α, β, and γ, respectively. Multiplication by the imaginary number i is equivalent to rotation through p/2 in the counterclockwise direction. Mulitplication by 1/i = -i does a similar job but in the clockwise direction. We then have:
(Ab + Ba)/2 = (α + β)/2 + (β - α)·i/2
independent of C = γ. The latter identity also shows that ΔAMB is right-angled and isosceles.
Here's an additional touch. Point M plays exactly the same role for triangle ΔABC as it does for ΔAcBcC. (Just turn the diagram upside down!) Therefore, ΔAcMBc is also right-angled and isosceles.
(Professor W. McWorter observed an additional property of Bottema's configuration and also that the squares can be replaced with arbitrary similar parallelograms.)
Bottema's theorem entered mathematical folklore in an adventurous guise. To make the connection clear, I shall add in parentheses points referring to the applet's diagram.
Someone found in his attic an old Description of a pirate who died long time ago. It read as follows: Go to the island X, start at the gallows (C), go to the elm tree (B) and count the steps. Then turn left by 90°, go the same number of steps and mark the point (Ab). Again, go to the gallows (C) and go from there to the fig tree (A) counting steps. Turn 90° right and proceed the same number of steps. Mark the point (Ba). A treasure is buried midway between the two marked points. The fellow travelled to the island and found the fig and the elm trees, but could not locate the gallows. Still, he was able to find and dig out the treasure. How did he do that?
Bottema's theorem helps solve a Japanese Sangaku.