A plane figure F is a reflection-symmetric figure if and only if there is a line

*m*such that r(F)=F. The line

*m*is a symmetry line for the figure.

In other words, it's what one usually means when one uses the word "symmetry." Some geometry texts use the term "line-symmetric" instead of "reflection-symmetric." Some geometry and algebra texts use the term "axis of symmetry" instead of "symmetry line" -- especially Algebra I texts referring to the axis of symmetry of a parabola. Some biology texts use the term "bilateral symmetry" instead of "reflection (or line) symmetry" - in particular, when referring to symmetry in animals. As animals are three-dimensional, instead of a symmetry line there's a sagittal

*plane*.

Indeed, it is this last topic that makes symmetry most relevant and interesting. Most animals -- including humans -- have bilateral symmetry. I once read of a teacher who came up with an activity where the students look for the most symmetrical human face. The teacher blogged about how students who are normally indifferent to geometry suddenly came fascinated and engaged to learn about the relationship between symmetry and human beauty. Unfortunately, this was more than a year ago, and I can't remember or find what teacher did this activity -- otherwise I'd be posting a link to that teacher's blog right here!

In the Common Core Standards, symmetry is first introduced as a fourth grade topic:

CCSS.MATH.CONTENT.4.G.A.3

Recognize a line of symmetry for a two-dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts. Identify line-symmetric figures and draw lines of symmetry.

Later on, symmetry appears in the high school geometry standards:

CCSS.MATH.CONTENT.HSG.CO.A.3

Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.

Notice that if a reflection over a line carries a polygon to itself, then that line is a symmetry line. But symmetry lines for polygons formally appears in Chapter 5 of the U of Chicago text. Right here in Chapter 4, we only cover symmetry lines for simpler shapes -- segments and angles. The text reads:

"In the next chapter, certain polygons are examined for symmetry. All of their symmetries can be traced back to symmetries of angles or segments."

For segments, the text presents the Segment Symmetry Theorem:

A segment has exactly two symmetry lines:

1. its perpendicular bisector, and

2. the line containing the segment.

The text gives an informal proof of this -- as the mirror image of an endpoint, there can only be two possible reflections mapping a segment

*A*to

*B*and

*B*to

*A*-- and that mirror must be the perpendicular bisector of

*A*to

*A*and

*B*to

*B*-- which means that both

*A*and

*B*must lie on the mirror, since the image of each is itself. No other symmetry is possible. QED

But we also want to work with angles. The first theorem given is the Side-Switching Theorem:

If one side of an angle is reflected over the line containing the angle bisector, its image is the other side of the angle.

An informal proof: the angle bisector divides an angle into two angles of equal measure. The picture in the U of Chicago text divides angle

*ABC*into smaller angles 1 and 2. Now the reflection must map ray

*AB*onto a ray that's on the other side of the angle bisector

*BD*, but forms the same angle with

*BD*that

*AB*does with

*BD*. And there's already such a ray in the correct place -- ray

*BC*. Notice that part b of the Angle Measure Postulate from Chapter 3 already hints at this -- the "Two sides of line assumption" gives two angles of the same measure, one on each side of a given ray. QED

The other theorem, the Angle Symmetry Theorem, follows from the Side-Switching Theorem:

Angle Symmetry Theorem:

The line containing the bisector of an angle is a symmetry line of the angle.

Earlier this week, I wrote that we'd be able to prove the Converse of the Perpendicular Bisector Theorem after this section. As it turns out, the Side-Switching Theorem is the theorem we need.

Converse of the Perpendicular Bisector Theorem:

If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

Given:

*PA*=

*PB*

Prove:

*P*is on the perpendicular bisector

*m*of segment

*.*~~AB~~

Now I was considering giving a two-column proof of this, but it ended up being a bit harder than I would like for the students. But as it turns out, even though the U of Chicago text doesn't prove this converse, in Section 5-1 it gives a paragraph proof of what it calls the "Isosceles Triangle Symmetry Theorem," and the proof of this one and the Converse of the Perpendicular Bisector Theorem are extremely similar. After all, we're given that

*PA*=

*PB*-- so

*PAB*is in fact an isosceles triangle!

Proof:

Let

*m*be the line containing the angle bisector of angle

*APB*. First, since

*m*is an angle bisector, because of the Side-Switching Theorem, when ray

*PA*is reflected over

*m*, its image is

*PB*. Thus

*A'*, the reflection image of

*A*, is on ray

*PB*. Second,

*P*is on the reflecting line

*m*, so

*P' = P*. Hence, since reflections preserve distance,

*PA'*=

*PA*. Third, it is given that

*PA*=

*PB*. Now put all of these conclusions together. By the Transitive Property of Equality,

*PA'*=

*PB*. So

*A'*and

*B*are points on ray

*PB*at the same distance from

*P*, and so

*A'*=

*B*. That is, the reflection image of

*A*over

*m*is

*B*.

But, by definition of reflection, that makes

*m*the perpendicular bisector of

*AB*-- and we already know that

*P*is on it. Therefore

*P*is on the perpendicular bisector

*m*of segment

Let's think about what we're trying to prove here. We want the Converse of the Perpendicular Bisector Theorem -- and consider what I wrote earlier about the proof of converses. The proof of the converse of a statement often involves the forward direction of the theorem and a uniqueness statement -- and even though we didn't use the forward direction of the theorem here, we did use a uniqueness statement here. As it turns out, given two distinct points

*A*and

*B*, there exists only one line

*m*such that the mirror image of

*A*over

*m*is

*B*-- and that line is the perpendicular bisector of the segment

*A*over

*m*is

*B*, we'll have proved that

*m*is the perpendicular bisector of

*. So that's exactly what we did above -- we proved that a certain line (the*~~AB~~

*angle bisector*of

*APB*) is the perpendicular bisector of

In this section, we found symmetry lines for simple figures such as segments and angles. But can we find symmetry lines for the simplest figures? As it turns out, a point has infinitely many lines of symmetry -- any line passing through the point is a symmetry line. But a ray has only one line of symmetry -- the line containing the ray.

Finally, does a

*line*have a line of symmetry? This is exactly the answer to Question 25 of this section, in the Exploration/Bonus Section. A line -- considered as a straight angle -- contains more than one symmetry line. This is because

*any*point on the line can be taken as the vertex of that straight angle. Since straight angles measure 180, their angle bisectors must divide them into pairs of 90-degree angles. Therefore,

*any*line perpendicular to a line (straight angle) is a symmetry line of the given line. This is what I called the Line Perpendicular to Mirror Theorem. It implies that a line (straight angle) has infinitely many symmetry lines. (Of course, the line has one more symmetry line that I didn't mention -- namely the line itself.)

I included Question 24, even though it appears to mention corresponding and same-side interior angles formed by two lines and a transversal. But nowhere in the question does it mention anything about the two lines being

*parallel*.

I left out Questions 16 and 17, which give the construction of an angle bisector. I finally plan on going to constructions sometime next week. But here's another video from Square One TV, where doctors have to perform a "bisectomy" on an angle. (Unfortunately, only the entire 30-minute show is available on YouTube -- the "bisectomy" doesn't begin until the 11-and-a-half-minute mark.)

## No comments:

## Post a Comment