Theorem 5.12: A rotation is a composition of two reflections, and hence is an invertible isometry.
Proof: Let O be the fixed point of the rotation. Let A be any point and let A' be the point to which it is rotated.
Let E be any point on the arc between A and A' and consider the ray OE .
Let M be the midpoint of the line segment AE, and let N be the midpoint of the line segment EA'.
The claim is that the rotation can be accomplished by reflecting first about OM and then about ON.
First we will show that / AOA' is twice as large as / MON .
Now, A, E, and A' are all on the circle centered at O whose radius is |OA|, and since M is taken to be the midpoint of the line segment AE, by Theorem 3.10, OM is perpendicular to AE. By the definition of a reflection, then, E is the reflection of A about OM. Since O and M are fixed by this reflection, / MOE is the reflection of / AOM. Since reflections preserve angle size, we conclude that m/ AOM = m/ MOE. Similarly, A' is the reflection of E about ON, and a similar arguement will allow us to conclude that m/ EON = m/ NOA'. So
Now let B be any point in the plane. Let B" be the reflection of B about OM, and let B' be the reflection of B" about ON. Since O is fixed by both of these reflections, and since reflections preserve distances, it follows that
So we need only show that
There are several cases to consider. First we will consider all of the cases where B is on the same side of the line through O which is perpendicular to OM as A.
For the cases where B is on the other side of the line through O, which is perpendicular to OM as A, we apply the cases above to the vertical angle to / MON, which is obtained from taking the rays opposite to the rays from O through M and N.