Let there be given two rotations gP and gQ with centers P and Q. Assume gP is executed first, gQ second. gP leaves P in place. Denote gQ(P) = P'. gP carries some point Q' onto Q which is then left in place by gQ. The point O that lies at the intersection of perpendicular bisectors of QQ' and PP' is equidistant from P and P' and also from Q and Q'.
Now prove that
POP' =
Q'OQ =
PQP' +
Q'PQ. Finally, consider what happens with an arbitrary point (or two) under two successive rotations.
(The fact just established provides a powerful tool for solving geometric problems, see, for example, proofs of Bottem's theorem and an analogue.)
Copyright © 1996-2008 Alexander Bogomolny