Any Composition of Rigid Motions Is a Rigid Motion

The fact that any composition of isometries is also an isometry is true and quite obvious.

The proof is a trivial application of the fact that equality and congruence are transitive.

Suppose that we have a segment $AB$ which is mapped by an isometry $I_1$ to $CD$, which is mapped by another isometry $I_2$ to $EF$.

Thus, we have the following: $$I_1(AB)=CD$$ $$I_2(CD)=EF$$ $$I_2\circ I_1(AB)=EF$$ Since $I_1$ is an isometry, $AB=CD$.

Since $I_2$ is an isometry, $CD=EF$.

By the Transitive Property of Equality, $AB=EF$.

Since $AB$ was an arbitrary segment, we can see that the following argument holds for all segments.

Thus, the composition $I_2\circ I_1$ preserves distances, so it is an isometry.