This theorem is first mentioned in Lesson 6-6, where it's called the Glide Reflection Theorem. The text states that its proof will be given in a later lesson -- that is, Lesson 13-6. But in that lesson, the theorem is unnamed. Still, we might as well call it the Glide Reflection Theorem anyway for lack of a better name. Here it is:

Glide Reflection Theorem:

In a glide reflection, the midpoint of the segment connecting a point to its image lies on the glide-reflection line.

Let's repeat the proof that I mentioned last week:

Given: C = T o r_

*l*(*A*), where T is a translation parallel to*l*.*N*is the midpoint of

Prove:

*l*contains*N*.
Proof:

Let ~~AB~~. From the Midpoint Connector Theorem, line ~~BC~~. But, from the definition of reflection, ~~BC~~. Thus both ~~BC~~ through

The other theorems that we mentioned in the last few days -- the parallelogram theorems and the Midpoint Connector Theorem -- lead up to this conclusion, which is why we spent the last few days discussing them. But what is the significance of the Glide Reflection Theorem itself? That is, when exactly would we have a glide reflection and need to find its mirror?

The following Common Core Standard seems relevant here:

CCSS.MATH.CONTENT.8.G.A.2

Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations;

[emphasis mine]

Theorems like the Glide Reflection Theorem help us do exactly that -- let's say we have two triangles,

Let's say we happen to know that~~AA'~~ and ~~BB'~~. Then the mirror for the glide reflection is exactly line

Notice that ordinarily, we don't even need point

It could be that reflecting point

Of course, there exist infinitely many reflection-translation pairs whose composite is a glide reflection, but there is only one such pair that is

The Glide Reflection Theorem only works when the preimage and the image have opposite orientation, not the same orientation. If a figure and its image have the same orientation, then we know that the isometry mapping one to the other is either a translation or a rotation. This case may be a bit tricky -- it could be that the easiest way is simply to translate

We know that the center of rotation is equidistant from~~AA'~~. For the same reason, the center lies on the perpendicular bisector of ~~BB'~~. So where these two points intersect is the center of rotation. Notice that if these two perpendicular bisectors are parallel, then the above reasoning constitutes an indirect proof that there is no rotation mapping one to the other -- that is, there is a translation map instead.

Today's worksheet covers all of Lesson 13-6. This means that not only are there questions about the Glide Reflection Theorem, but also about uniqueness in general. A modern form of Euclid's original five postulates are given.

*B*= r_*l*(*A*) and let*M*be the midpoint of*MN*is parallel to*l*contains*M*, and because the translation T is parallel to*l*,*l*is parallel to*l*and line*MN*are parallel to*M*. By the Uniqueness of Parallels Theorem,*l*and line*MN*must be the same line. Thus*l*contains*N*. QEDThe other theorems that we mentioned in the last few days -- the parallelogram theorems and the Midpoint Connector Theorem -- lead up to this conclusion, which is why we spent the last few days discussing them. But what is the significance of the Glide Reflection Theorem itself? That is, when exactly would we have a glide reflection and need to find its mirror?

The following Common Core Standard seems relevant here:

CCSS.MATH.CONTENT.8.G.A.2

Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations;

**given two congruent figures, describe a sequence that exhibits the congruence between them**.[emphasis mine]

Theorems like the Glide Reflection Theorem help us do exactly that -- let's say we have two triangles,

*ABC*and*A'B'C'*-- that we know to be congruent. So, by the definition of congruent (meaning), there exists an isometry mapping*ABC*to*A'B'C'*. So how exactly do we find this isometry?Let's say we happen to know that

*ABC*and*A'B'C'*have opposite orientation. Thus the isometry mapping one to the other must be either a reflection or a glide reflection. The Glide Reflection Theorem tells us how to find the mirror for this isometry -- we let*P*be the midpoint of*Q*be the midpoint of*PQ*. This works because of uniqueness (which is part of the reason why this is in the uniqueness lesson) -- through any two points there is a unique line, so as soon as we know two points on the line, such as*P*and*Q*, that's all we need to know to determine the line.Notice that ordinarily, we don't even need point

*C*-- just*A*,*B*, and their respective images are sufficient to produce a mirror. It is possible to prove that if*AA'*,*BB'*, and*CC'*are concurrent -- that is, that they intersect in one point (which isn't enough to establish a line), then the original points*A*,*B*, and*C*must be collinear rather than the vertices of a triangle.It could be that reflecting point

*A*over line*PQ*already gives point*A'*. In this case, a mere reflection is sufficient rather than a glide reflection. Otherwise, the image of*A*must be translated to give*A'*, and that would be the translation part of the glide reflection.Of course, there exist infinitely many reflection-translation pairs whose composite is a glide reflection, but there is only one such pair that is

*canonical*-- that is, where the translation is actually parallel to the mirror. This is the mirror found by applying the Glide Reflection Theorem.The Glide Reflection Theorem only works when the preimage and the image have opposite orientation, not the same orientation. If a figure and its image have the same orientation, then we know that the isometry mapping one to the other is either a translation or a rotation. This case may be a bit tricky -- it could be that the easiest way is simply to translate

*A*to*A'*and see whether this translation maps*B*to*B*' -- if not, then a rotation is necessary. But how do we find the center?We know that the center of rotation is equidistant from

*A*and*A'*. Thus it lies on the perpendicular bisector ofToday's worksheet covers all of Lesson 13-6. This means that not only are there questions about the Glide Reflection Theorem, but also about uniqueness in general. A modern form of Euclid's original five postulates are given.

## No comments:

## Post a Comment