আইসোমেট্রির শ্রেণিবিন্যাস

Let's say that we have a congruence transformation $\alpha$ in the plane. Then there is a sequence of steps that will show how to represent $\alpha$ as a combination of shifts, flips, and rotations. For convenience, we will be considering three arbitrary non-collinear points $A$, $B$, and $C$ in the original plane.

Step 1: The first transformation in representing $\alpha$ is the shift which takes the origin $A$ to $\alpha(A)$ (the image of $A$ in $\alpha$). We'll call this transformation $T_1$.

Step 2: The next transformation is the rotation centered at $T_1(A)$ that takes $T_1(B)$ to $\alpha(B)$. We'll call this transformation $T_2$.

Step 3: If $T_2(T_1(C)) = \alpha(C)$, then we let $T_3$ be a trivial transformation -- that is, a transformation that takes every point to itself; otherwise, we let $T_3$ be a flip about line $AB$.

We can show that this process will always successfully create $\alpha$ as $T_3 \circ T_2 \circ T_1$ by using the fact that it's a congruence transformation. Consider an arbitrary point $D$, not necessarily distinct from $A$, $B$, and $C$. Notice that $D$ is the intersection of three circles: the circle centered at $A$ with radius $AD$, the circle centered at $B$ with radius $BD$, and the circle centered at $C$ with radius $CD$. Since the centers of these three circles are not collinear, the three circles can only intersect at at most one point (specifically, their radical center). Now, the image of these three circles are also circles with the same radii, and whose centers are also not collinear, so they too intersect at at most one point. However, by the properties of congruence transformations, this point must be the image of $D$! Thus since $\alpha$ and $T_3 \circ T_2 \circ T_1$ take $A,B,C$ to the same points, they must take $D$ to the same point, so since $D$ was arbitrary, they must be the same transformation.