I've been asked to prove the following theorem, but I seem to be able to prove more than the theorem requires. I would like to check if what I proved is justified.
Let $F$ be an arbitrary isometry of the plane. If $F$ does not leave any point fixed, then F is either a translation, or the composite of a translation and a rotation, or the composite of a translation, a rotation, and a reflection through a line.
The following theorems have already been proven:
$(1.)$ Let $F$ be an isometry which leaves one point $O$ fixed. Then either F is a rotation, or F is a rotation composed with a reflection through a line.
$(2.)$ Let $P, Q$ be distinct points. Let $F$ be an isometry which leaves $P$ and $Q$ fixed. Then either $F$ is the identity, or F is a reflection through the line Lpq passing through $P$ and $Q$.
I approached the proof this way: I took three arbitrary points, $A, B$ and $C$. Since $F$ leaves no fixed points, $F(A)$ is not $A, F(B)$ is not $B$, and same for $C.$ If there is a T such that $T(A) = F(A), T(B) = F(B), \text{ and } T(C) = F(C),$ then $T-1F =$ Identity, since it is an isometry that maintains three fixed points. Hence, I proved the first part of the theorem. $T-1$ is inverse of $T.$
If there is a T such that $T(A) = F(A), T(B) = F(B),$ but $T(C)$ is not $F(C),$ then Theorem $2$ above applies, implying $T-1F =$ Identity (which it is not), or $T-1F =$ reflection, which means $F$ is a composite of a reflection and a translation (which makes sense when I visualize it).
If there is a $T$ such that only $T(A) = F(A),$ then via Theorem $1,$ $T-1F =$ Rotation or $T-1F =$ Rotation.Reflection. This would mean $F$ is a composite of translation and rotation, or F is a composite of a translation, rotation and reflection, which also makes sense to me visually.
However, the theorem does not account for the case where $F$ is a composite of translation and reflection only. Yet I (seem to) have proved it. Am I correct to make such an assertion?