The Classification Theorem: Every isometry is one of the following: the identity, a translation, a rotation, a reflection, or a glide reflection.

For background, see the page on the theory of isometries.

Proof:

Let T be an isometry. Let ABC be a triangle [A, B, C noncollinear], and let A'B'C' be the image of ABC by the transformation T. [That is, T(A)= A', T(B) = B', and T(C) = C'.] To show that T is one of the types identity, translation, rotation, reflection, or glide reflection, we just have to show that there's an isometry F of one of these types such that F also takes the triangle ABC to A'B'C'. [That is, F(A) = A', F(B) = B', and F(C) = C'.] Then, by the uniqueness part of the triangle theorem, T = F. [In other words, for all points X we have T(X) = F(X).]

The proof is organized into four cases:

Case 1. T fixes A, B, and C; i.e. A'= A', B = B', and C = C'. Then T is the identity.

For if I is the identity, then I(A) = A, I(B) = B, and I(C) = C, so T = I by uniqueness.

Case 2. T fixes exactly two of the points A, B, C. Then T is a reflection.

Suppose that A = A' and B = B', but C is not C'. Since T is an isometry, AC = AC' and BC = BC'. In other words, C' lies on the circle c1 with center A through the point C, and C' lies on the circle c2 with center B through the point C.

The intersection of two different circles must be empty, one point, or two points. The point C lies on both circles c1 and c2. If C were the only point common to c1 and c2, then we would have AC + BC = AB, so C would lie between A and B, and C would be collinear with A and B, contrary to hypothesis. Therefore the intersection of c1 and c2 is two points, C and C'. Since A is equidistant from C and C', A lies on the perpendicular bisector of CC'. Since B is equidistant from C and C', B also lies on the perpendicular bisector of CC'. Thus AB is the perpendicular bisector of CC'. Thus if F is the reflection with mirror AB, we have F(A) = A, F(B) = B, and F(C) = C'. Thus T = F by uniqueness.

Case 3. T fixes exactly one of the points A, B, C.

3(a) If T is orientation reversing, then T is a reflection.

Suppose A = A', but B is not B', and C is not C' Since T is orientation reversing, the proper angle measures of the angles CAB and C'AB' have opposite signs. Since T is an isometry, CA = C'A and BA = B'A.

Let D be the midpoint of the segment BB'. Since the triangle BAB' is isosceles, AD is the perpendicular bisector of BB', and AD is the bisector of the angle BAB'. Since the angles CAB and B'AC have equal proper angle measures, AD is also the bisector of the angle CAC'. Since the triangle CAC' is isosceles, AD is the perpendicular bisector of CC'. Thus if F is the reflection with mirror AD, we have F(A) = A, F(B) = B', and F(C) = C'. Therefore T = F by uniqueness.

3(b) If T is orientation preserving, then T is a rotation.

Again suppose A = A', but B is not B', and C is not C'. Since T is orientation preserving, the proper angle measures of the angles CAB and C'AB' are equal. Since T is an isometry, CA = C'A and BA = B'A. Thus B and B' lie on the same circle c1 with center A, and C and C' lie on the same circle c2 with center A.

The angles CAB and C'AB' have equal proper angle measure: a'(CAB) = a'(C'AB'). Now a'(CAC') = a'(CAB) + a'(BAC') and a'(BAB') = a'(BAC') + a'(C'AB'). So a'(CAC') = a'(BAB'). Thus if R is the rotation with center A and angle CAC', we have R(A) = A, R(B) = B', and R(C) = C'. Therefore T = R by uniqueness.

Case 4. T fixes none of the points A, B, C.

4(a) If T is orientation preserving, then T is a translation or a rotation.

Since A, B, C are noncollinear, one of the points B, C does not lie on the line AA'. Suppose B does not lie on AA'. Thus the lines AB and A'B' are different.

(i) Suppose the lines AB and A'B' are parallel. Then the vectors AB and A'B' are either equal (i.e. ABB'A' is a parallelogram) or opposite (i.e. ABA'B' is a parallelogram). If ABB'A is a parallelogram then AA' is parallel to BB'. Since T is orientation preserving, if ABB'A' is a parallelogram, then so is ACC'A'. [The point C cannot lie on both AA' and BB'. We may assume that C does not lie on AA'.] So if S is the translation with vector AA', we have S(A) = A', S(B) = B', and S(C) = C'. Therefore T = S by uniqueness.

On the other hand, if ABA'B' is a parallelogram [and C does not lie on BB'] then CBC'B' is a parallelogram. In this case, the segments AA' and BB' bisect each other, since they are the diagonals of the parallelogram ABA'B'. For the same reason the segments BB' and CC' bisect each other. Let O be the common midpoint of the three segments AA', BB', CC'. [Special cases: If C lies on BB' but not on AA' then CAC'A' is a parallelogram, and the three segments AA', BB'. CC' still have a common midpoint O. If C lies on both AA' and BB' then C = C' and we let O = C.] If R is the rotation with center O and angle 180 degrees, then R(A) = A', R(B) = B', and R(C) = C', so T = R by uniqueness.

(ii) Suppose the lines AB and A'B' are not parallel.

Let L be the perpendicular bisector of AA', and let M be the perpendicular bisector of BB'. Let O be the intersection of L and M.

[If L and M are parallel, then AA' and BB' are parallel, so ABB'A' is an isosceles trapezoid, and hence L = M. In this case we let O be the intersection of AB and A'B'.] The (yellow) triangles AOB and A'OB' are congruent, by the side-side-side congruence theorem. Therefore the angles AOB and A'OB' have the same proper angle measure. Thus the angles AOA' and BOB' have the same proper angle measure. Let R be the rotation with center O and angle AOA'. Then R(A) = A' and R(B) = B'. Since T and R are orientation preserving, R(C) = C'. Therefore T = R by uniqueness.

4(b) If T is orientation reversing, then T is a reflection or a glide reflection.

Note that if two angles have corresponding sides parallel, and their proper angle measures are negatives of each other, then both angles must be right angles. Thus, since T is orientation reversing, all three sides of the triangle ABC cannot be parallel to the corresponding sides of triangle A'B'C'. Suppose that AB is not parallel to A'B'. Let P be the midpoint of segment AA' and let Q be the midpoint of segment BB'. Let L be the line PQ. [The points P and Q are not equal, for if they were equal then the triangles APB and A'PB' would be congruent, and AB would be parallel to A'B'.] Let X be the foot of the perpendicular from A to L, let X' be the foot of the perpendicular from A' to L, let Y be the foot of the perpendicular from B to L, and let Y' be the foot of the perpendicular from B' to L. Triangles AXP and A'X'P are congruent, so AX = A'X'. (If X = P then X' = P, and again AX = AX'.) Similarly, BY = BY'. Since AB = A'B', it follows that the (yellow) trapezoids AXYB and A'X'Y'B' are congruent. Therefore XY = X'Y', so XX' = YY'.

Let G be the glide reflection with mirror L and vector XX'. (If X = X' then G is the reflection with mirror L.) Then G(A) = A' and G(B) = B' since XX' = YY'. Since T and G are orientation reversing, G(C) = C'. Therefore T = G by uniqueness.

This completes the proof of the classification theorem!