An IBL Introduction to Geometries

Let \(T\) be a rotation, and let \(C, P\text{,}\) and \(Q\) be points, so that \(P^\prime = T(P)\) and \(Q^\prime = T(Q)\text{.}\) By definition, \(\|\overline{CP}\| = \|\overline{CP^\prime}\|\) and \(\|\overline{CQ}\| = \|\overline{CQ^\prime}\|\) Note \(T\) is the same rotation for \(P\) and \(Q\text{,}\) so \(m\angle PCP^\prime = m\angle QCQ^\prime = \alpha\text{.}\) If we let \(\beta = m\angle PCQ\text{,}\) then \(m\angle P^\prime CQ^\prime = \beta + \alpha - \alpha = \beta\text{,}\) so \(\angle PCQ \cong \angle P^\prime CQ^\prime\text{.}\) Thus, we^\primeve found an SAS correspondence between the two triangles \(\triangle CPQ\) and \(\triangle CP^\prime Q^\prime\text{,}\) so \(\triangle CPQ \cong \triangle CP^\prime Q^\prime\text{.}\) By CPCTC, \(\|\overline{PQ}\| = \|\overline{P^\prime Q^\prime}\|\text{.}\) Therefore, T is an isometry.

Consider a transformation with points \(A, B,\) and \(C\text{.}\) We will show that \(||A - B|| = ||A^\prime - B^\prime||, ||A - C|| = ||A^\prime - C^\prime||,\) and \(||B - C|| = ||B^\prime - C^\prime||\) through direct proof.

\(||A - A^\prime|| = ||\vec{v}||, ||B - B^\prime|| = ||\vec{v}||,\) and \(||C - C^\prime|| = ||\vec{v}||\) by definition of a transformation. Construct quadrilateral \(A^\prime C^\prime CA\) by drawing sides \(\overline AA^\prime, \overline A^\prime C^\prime, \overline CC^\prime,\) and \(\overline AC.\) Note that \(A^\prime C^\prime CA\) has two sides \(\overline AA^\prime\) and \(\overline CC^\prime\) that are \(\vec{v},\) so these two sides are parallel and have the same length. This means that the other two sides \(\overline A^\prime C^\prime\) and \(\overline AC\) are equidistant where they meet both of \(\vec{v},\) the two sides have the same slope and are thus parallel (from ruler postulate and equidistance postulate). The quadrilateral is now known to be a parallelogram by definition of a parallelogram. \(\overline AC \cong \overline A^\prime C^\prime\) (also by definition of a parallelogram), so \(||A - C|| = |A^\prime - C^\prime||.\)

For transformations with only two original points, we are done. For larger transformations, we can similarly show that \(||A - B|| = ||A^\prime - B^\prime||\) and \(||B - C|| = ||B^\prime - C^\prime||\) (and so on if the transformation is larger than three points). From the last three equations, by definition of isometry, translations are isometries.

Let \(P\) and \(Q\) be points and let \(T\) be a reflection in the plane over the line \(\ell\text{.}\) Without loss of generality, let \(P\) be the point further from the line \(\ell\text{.}\) Let \(q = \|Q-\ell\|\) and \(p = \|P-\ell\|\text{.}\) Also, let \(A\) and \(B\) be the points where \(\|\overline{PT(P)}\|\) and \(\|\overline{QT(Q)}\|\) intersect \(\ell\) respectively. Since \(T\) is a reflection, \(q = \|T(Q)-\ell\|\) and \(p = \|T(P)-\ell\|\text{.}\) Then, let \(r = p - q\text{.}\) Place a point \(R\) on \(\|\overline{PT(P)}\|\) such that \(R\) is \(r\) away from \(P\text{.}\) Then, \(\triangle PQR\) is a right triangle. Furthermore, \(\overline{AB} \cong \overline{RQ}\text{.}\) All of these constructions can be performed similarly for the transformed points, and so we have that \(\triangle T(P)T(R)T(Q)\) is a right triangle, and \(\overline{AB} \cong \overline{T(R)T(Q)}\) . Also, \(\overline{PR} \cong \overline{T(P)T(R)}\) as points and their reflections are equidistant to \(\ell\) and they are all collinear. So, by SAS, \(\triangle PQR \cong \triangle T(P)T(R)T(Q)\) and therefore \(\|P-Q\| = \|T(P)-T(Q)\|\)