Transformations and Isometries Definition: A transformation in absolute geometry is a function f that associates with each point P in the plane some other point P in the plane such that (1) f is one-to-one (that is, if for any two points P and Q, then P = Q). and (2) f is onto (that is, if Q is any point in the plane, then there is a point P such that ). We will often denote f(p) by P Definition: A transformation of the plane is said to have a point A as a fixed point iff f(a) = A. Definition: A transformation of the plane is said to be the identity mapping if every point of the plane is a fixed point. The identity mapping is denoted by e. Thus e(p) = P for all points P of the plane. Definition: An isometry is a transformation of the plane that preserves distances; that is, if P and Q are two points, then. Theorem: An isometry preserves collinearity, betweenness, and angles. That is, if A, B, and C are three points in the plane and their images under an isometry are A, B, and C, then: 1. If A, B, and C are collinear, then A, B, and C are also collinear. 2. If A*B*C, then A *B *C. 3. If A, B, and C are non-collinear, m ABC = m A B C For (1), suppose A, B, and C are collinear. One of the points must be between the others; WLOG, suppose A*B*C. Then AB + BC = AC, so because we have an isometry A B + B C = A C. By the triangle inequality, the only way A, B and C could fail to be collinear is if A B + B C > A C, which we have shown is not true. In addition, we also conclude A *B *C, which proves (2). For (3), note that AB = A B and AC = A C and BC = B C so by SSS ABC is congruent to A B C. By CPCF, the corresponding angles have equal measure.
Theorem: The inverse of an isometry is an isometry. Let f be an isometry and. Let R and S be points of the plane. Let, so. Then: thus preserves distances. Theorem: The composition of two isometries is an isometry. Choose points A and B, and isometries f and g. Then and. Theorem: In addition to preserving lines, angles, betweenness, angles, and distances, an isometry f also preserves: a. Segments: for points A and B, (and ) b. Betweenness of rays: If, then. c. Triangles: for noncollinear A, B, and C, is a triangle and. d. Circles: If is a circle with radius r and center O, then is a circle with radius r and center. For (a), if X is a point such that A*X*B, then A *X *B so X is on 1 segment. Now f is an isometry as well, so if A *Y*B,, so, so for every Y on segment there is an X (namely ) on such that. Statement (b) follows from preservation of angle measures. Statement (c) follows from SSS. Statement (d) is almost immediate from the definitions of isometry and circle.
Definition: Let l be a line. Define the reflection over line l as a function s that assigns to each point P a point P defined as follows: l 1. If P is on l, then s l(p) = P. 2. Otherwise, drop a perpendicular from P to l, with foot F. Find a point P on the other side of l from P that lies on the perpendicular and such that PF = FP. Let s l(p) = P. l Note that s has the property that l is the perpendicular bisector of every segment formed by a point and its image under the function.
Theorem: A reflection is an isometry. We show that for any two points P and Q, PQ = P Q. We consider cases: Case 1: If P and Q are both on l they are fixed so P = P and Q = Q and the result is trivial. Case 2: P is on l and Q is not. P is fixed so P = P. Then QFP Q FP by SAS, and CPCF gives QP = Q P = Q P. Case 3: Neither P nor Q are on l, but they are on the same side of l. Let F and G be the feet of the perpendiculars to l from Q and P, respectively. By SASAS, FQPG FQ P G. By CPCF, PQ = P Q. Case 4: Neither P nor Q are on l, and are on opposite sides of l. As in Case 3, let F and G be the feet of the perpendiculars to l from Q and P, respectively. Again by SASAS, FQP G FQ PG. By CPCF, PQ = P Q and QP P Q PP. Then SAS gives us Q PP QP P and CPCF gives us PQ = P Q.
We are now going to prove an important result about isometries: Every isometry is defined by what it does to three noncollinear points. We make use of the ideas of fixed points and the identity transformation. We need a couple of nice lemmas first: Lemma: If P and Q are points on a line such that AP = AQ and BP = BQ, then P = Q. Use the ruler postulate to establish a coordinate system. Using the notation P[x} to denote that the coordinate of point P is x. Now let A[0], B[b] with b > 0, P[p], and Q[q]. Then, so p = ±q. If p = q, then P = Q by the ruler postulate and we are done. Otherwise, suppose q = -p. Now so (p - b) = ± (q - b). If (p - b) = (q - b), then p = q, and this combined with q = -p gives us p = -p so p = 0, and P = A = Q. If (p - b) = -(q - b), then p - b = - q + b. This, together with q = -p, gives us -b = b, or b = 0, a contradiction. Thus in all possible cases, P = Q. Lemma: If an isometry f fixes two points A and B, it also fixes every point on. Let P be on. Because f is an isometry it will map to another line, and since A and B are fixed it will in fact map unique line containing A and B, namely, itself. Now if P is a point of, we need to show that. P is on to the and in fact AP = A P = AP and BP =B P = BP since f is an isometry. By the lemma above, P = P, so P is a fixed point for f.
Theorem: The only isometry to have three noncollinear fixed points is the identity. Let A, B, and C be noncollinear, and suppose f has A, B, and C as fixed points. That is,. We will show that for any other point P of the plane P is also a fixed point. We note first that by the previous lemma, every point on the triangle ABC is fixed. Now suppose P is a point not on ABC. By picking a point D in the interior of any side of ABC, we can guarantee that intersects the triangle exactly twice (either it contains a vertex, or PASCH guarantees that is crosses another side); call the other intersection point E. Since f fixes every point on the triangle, in particular points E and D, it fixes all of, including point P. Thus f fixes every point, and must be the identity. Theorem: If f and g are isometries and A, B, and C are noncollinear points such that, then f = g. If, then. In other words, is an isometry that fixes three noncollinear points. By the previous theorem, must be the identity, so for all points P. But this implies that for all P,.
Theorem: Given two congruent triangles a unique isometry that maps ABC onto DEF., there is First, note that it is sufficient to show that there is a unique isometry that maps A to D, B to E, and C to F, because segments map to segments. We build the isometry from reflections. We proceed in stages: Stage 1: Let l be the perpendicular bisector of and let s l be the reflection across line l. This maps A onto D. We now consider, where.
Stage 2: If s l mapped B onto E, proceed to Stage 3. Otherwise, let m be the perpendicular bisector of. We know that ED = BA = B A = B D, so B and E are equidistant from D. Thus m will contain the point D (=A ) and so D will be a fixed point for the reflection s m. Reflect over line m. This will map B onto E and leave A mapped onto D. Stage 3. If s mapped C onto F, we are done. The mapping s s is the m desired mapping. Otherwise. A = D and B = E, so has been mapped onto. Let n be the line. We know that a reflection through n will leave fixed. Moreover, we know that EF = BC = B C = EC and DF = AC = A C = DC so C and F are equidistant from both D and E. Thus is the perpendicular bisector of. Thus s n will map C onto F, while leaving A = D and B = E fixed. The mapping snss m l(the composition of s lfollowed by s m followed by s n) is thus the desired mapping that takes onto. m l
Theorem: Every isometry is the product of at most three reflections. Let f be an isometry and A, B, and C be noncollinear points. Let. Because f is an isometry and preserves angles and distances,. By the previous theorem, we know that is also mapped to by a product of at most three reflections. Because this product of reflections maps the three noncollinear points A, B, and C to the same points as f, f is equivalent to the product of reflections.