Rigid Motions Fixing 3 Noncollinear Points Fix All Points

Suppose that we have an isometry $F$ that fixes three noncollinear points $A$, $B$, and $C$.

Now, let us take an arbitrary point $D$ in the plane of $ABC$.

Draw a line through $D$ that intersects the triangle in two points; call these points $E$ and $F$.

From a theorem on an earlier page, any isometry that fixes two points fixes all points on the line containing these two points. Thus, we may conclude that any point on the line $AB$, $BC$, or $CA$ is fixed by the isometry.

Notice that these points include our points $E$ and $F$.

Since $D$ lies on the line $EF$, we may use the above theorem again. Since $E$ and $F$ are fixed by the isometry, $D$ is fixed as well.

Since $D$ was an arbitrary point in the plane, we could have used any point in the plane to attain the same result. Thus, any isometry that fixes three noncollinear points fixes all points in the plane containing those three points.

However, this brings up an important point: the title should not say "all points," but rather "all points in the plane." To prove that it doesn't apply to a point not in the plane, consider reflection across a given plane. It would be easy to find three fixed points. However, points on the plane are fixed, but points not on the plane are not fixed.

This may have been obvious, and perhaps we only care about the planar case anyway, since that is what we usually deal with. However, it is still worth mentioning, since nothing was said about the number of dimensions of space.