Given any triangle $\triangle ABC$ and any congruent triangle $\triangle A'B'C'$, there exists a unique isometry $T: \mathbb{R}^2 \longrightarrow \mathbb{R}^2$ such that...

We begin with a triangle, $\triangle ABC$ and suppose $T$ is an isometry. Click the buttons below to see the steps. You can full-screen the dynamic "Try it" part by clicking the full-screen icon on the bottom-right.

We begin with an arbitrary point $P$. Because $P'$, the image of $P$, must be the same distance from $A'$, $B'$, and $C'$ as $P$ is to $A$, $B$, and $C$ (respectively), this forces $P$ to exactly one point. See the diagrams below.