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 .

/ AOA' = / AOE + / EOA'

= / AOM + / MOE + / EON + / NOA '

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

m/ AOA' = 2m/ MOE + 2m/ EON = 2m/ MON

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

|OB| = |OB"| = |OB'|

So we need only show that

/ BOB' = / AOA '

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.