> [!NOTE] Lemma (Composition of isometries $\mathbb{R}^{2}$ is also an isometry on $\mathbb{R}^{2}$) > Let $f,g:\mathbb{R}^{2}\to \mathbb{R}^{2}$ be [[Isometry of The Plane|plane isometries]]. Then $g\circ f:\mathbb{R}^{2}\to \mathbb{R}^{2}$ is an isometry on $\mathbb{R}^{2}.$ ^1c0497 *Proof*. Let $a,b\in \mathbb{R}^{2}.$ Then since $f$ is an isometry $|f(a)-f(b)|=|a-b|$and since $g$ is an isometry $|g(f(a))-g(f(b))|=|f(a)-f(b)|=|a-b|$and so $g\circ f$ is an isometry.