Scroll down for a solution to this problem.
There is exactly one rotation, unless the corresponding sides are parallel, in which case there are none.
The first step is to find the location of the rotation centre by intersecting the perpendicular bisectors of AA’ and BB’. If AB and A’B’ are parallel, so will the bisectors be and they will not intersect. In this case there exists no rotation.
In all other cases there will be one possible rotation centre O. Now we have to prove that this O is also equidistant from C and C’. Looking at the figure above, we deduce from SSS that triangles ABO and A’B’O are congruent. Therefore angles OAC and OA’C’ are equal. From SAS it follows that triangles AOC and A’OC’ are congruent and hence OC and OC’ are congruent.
Finally, we must have that the rotation angles of the three vertices are equal. From the figure above we see that angle AOA’ equals α+β and so is angle BOB’. Therefore the are equal. Similarly they are equal to COC’.
Two twin triangles
Court hopefully each other
Every one does what must be done
And revolves around both
So that everyone follows
His own desire
A reciprocal attraction
A very strong alchemy
Like the two poles of a magnet
Two poles of love