**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*.

- The first case will be where
*B*is outside of__/__*MO**N*, but where its reflection about*OM*will be inside the angle. - The second case where
*B*is inside__/__*MON*. - The third case is where
*B*is outside of__/__*MON*and so is its reflection about*OM*, but*B*"*O*which is perpendicular to*OM*. - The fourth case is where
*B*is outside of__/__*MON*and so is its reflection about*OM*, and*B*" is on the other side of the line through*O*which is perpendicular to*OM*.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*.Analytic Foundations of Geometry