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 angle AOA' is twice as large as angle MON .

Now, A, E, and A' are all on the circle centered at O whose radius is |O, A|, 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, angle MOE is the reflection of angle AOM. Since reflections preserve angle size, we conclude that angles AOM and MOE have the same size. Similarly, A' is the reflection of E about ON, and a similar arguement will allow us to conclude that angles EON and NOA' have the same size. 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.

• The first case where B is inside angle MON.
• The second case will be where B is outside of angle MON but where its reflection about OM will be inside the angle.
• The third case is where B is outside of angle MON and so is its reflection about OM, but O is outside angle BB"B'.
• The fourth case is where B is outside of angle MON and so is its reflection about OM, and O is inside angle BB"B'.
• Finally we consider the case where angle BOM is obtuse.

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