Theorem 1. Let L be a line and O be a point not lying on L. Let R be the 180-degree

rotation around O. Then R maps L to a line parallel to L itself.

As I mentioned last week, my proof of the Line Parallel to Mirror Theorem is actually based on Wu's proof of Theorem 1. Now earlier I was considering repeating this same proof for Theorem 1 itself, but in some ways, this is a waste. Often, if the same technique used to prove one theorem is used to prove another theorem, it may be that the first theorem itself can be used to prove the theorem. And as it turns out, we can prove Wu's Theorem 1 using several other theorems -- including the Line Parallel to Mirror Theorem.

But let's consider the case where

*O*lies on

*L*first, then we can handle the related case where the point doesn't lie on the line.

Lemma. Let L be a line and O be a point lying on L. Let R be the 180-degree

rotation around O. Then R maps L to L itself.

Proof:

By the Two Reflection Theorem for Rotations, R is the composite of two reflections. The two mirrors must intersect at

*O*, and the angle between them must be half of 180, or 90 degrees. As it turns out, we can choose

*any*two lines through

*O*that we like, as long as the angle between them is a right angle. So which lines should we choose?

Well, we already have a line through

*O*-- namely

*L*itself! So let that be our first reflecting line. If we reflect line

*L*over line

*L*, we obviously obtain line

*L*again. This is because every point on the mirror is a fixed line for the reflection.

So where is our second line? It must be the line passing through

*O*perpendicular to

*L*-- a line which exists by the Angle Measure (Protractor) Postulate. If we reflect

*L*in this line, what do we get? By the Line Perpendicular to Mirror Theorem, the image of

*L*is again

*L*. So both of the reflections map

*L*to

*L*itself, and therefore its composite -- the rotation R -- maps

*L*to itself. So

*L*is an invariant line for any rotation of 180 degrees whose center lies on

*L*. QED

Now we can prove Theorem 1 in a similar fashion:

Proof:

Once again, we must choose two reflections whose composite is

*R*. Once again, we can choose any two perpendicular lines passing through

*O*that we like. But unfortunately, we can't choose

*L*itself, since

*L*doesn't pass through

*O*this time.

But let's choose a line perpendicular to

*L*again, since

*L*is an invariant line of that rotation. So let

*m*be the line passing through

*O*that is perpendicular to

*L*-- a line that exists by the Uniqueness of Perpendiculars Theorem. So far, the image of

*L*under the first reflection is

*L*itself.

Now let's try the second reflection. The second mirror must be the line passing through

*O*that is perpendicular to

*m*-- call this new line

*n*. So we have both

*L*and

*n*perpendicular to

*m*-- and so by the Two Perpendiculars Theorem,

*L*must be parallel to

*n*. And so we can now use that Line Parallel to Mirror Theorem mentioned earlier, to prove that the reflection image of

*L*over

*n*must be a another line parallel to

*L*. So the first reflection maps

*L*to

*L*, and the second maps

*L*to some other line parallel to

*L*. Therefore their composite, the rotation R, maps

*L*to a line parallel to

*L*. QED

Notice what happens when we rotate a line

*L*180 degrees. If the center is

*not*on the line, then the rotation maps

*L*to a line parallel to

*L*, and if the center

*is*on the line, then the rotation maps

*L*to

*L.*I must now remind you about the U of Chicago definition of parallel, which states that a line is parallel to itself. Therefore, a rotation of 180 degrees whose center can be

*any*point in the plane maps

*L*to a line parallel to

*L*.

Theorem 1 (U of Chicago def. of parallel). Let L be a line and O be a point. Let R be the 180-degree

rotation around O. Then R maps L to a line parallel to L itself.

I actually like this proof better than simply using Wu's proof. Our theorem is proved using other theorems that we (unlike Wu) previously proved -- the Two Reflection Theorem for Rotations, Protractor Postulate, Two Perpendiculars, and Line Perpendicular/Parallel to Mirror. Furthermore, what's nice about the Two Reflections Theorem for Rotations is that it can also help us with later theorems, such as rotational symmetry. As it turns out, a figure with two intersecting lines of (reflectional) symmetry

*must*have rotational symmetry, and the proof follows directly from the Two Reflections Theorem.

On the other hand, some readers might point out that this lesson singles out 180-degree rotations from rotations of other magnitudes. In fact, it might be possible to combine Wu's technique with the U of Chicago's -- begin by defining a special rotation that only works for 180 degrees, use it to prove that reflections are well-defined as in Wu, then define other rotations as the composite of two reflections as in U of Chicago. Indeed, some authors define a "point reflection":

The reflection of

*A*over a point

*C*is

*B*if and only if

*C*is the midpoint of

*AB*. (It goes without saying that the point reflection image of

*C*is

*C*itself.)

So a point reflection is equivalent to a 180-degree rotation. Even some texts that don't define point reflection still define "point symmetry" to mean 180-degree rotational symmetry. (So a student writing down the rotational symmetries of a regular hexagon are supposed to write "60, 120, point.")

This Theorem 1 is important enough to deserve its own section, since we are just about to use it to prove the Parallel Tests. I had to come up with the Exercises for this on my own, since neither the U of Chicago nor any other source devotes a full lesson to this theorem -- but many of my exercises are just earlier exercises are rotations, adopted for 180-degree rotations.

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