The first theorem that we have is the Isosceles Trapezoid Symmetry Theorem:

The perpendicular bisector of one base of an isosceles trapezoid is the perpendicular bisector of the other base and is a symmetry line for the trapezoid.

Notice that this theorem is similar to the Kite Symmetry Theorem. In many ways, there is a sort of dualism between the kite and the isosceles trapezoid. A kite is defined by having consecutive equal

*sides*, while an isosceles trapezoid has consecutive equal

*angles*. The symmetry line for the kite bisects two of the

*angles*, and the symmetry line for the isosceles trapezoid bisects two of the

*sides*.

In fact, this dualism in some ways justifies our definition of

*isosceles trapezoid*. Defining an isosceles trapezoid to be a trapezoid having a pair of equal opposite sides -- which works for the exclusive definition of

*trapezoid*but not the inclusive definition -- would be just like defining a kite in terms of having a pair of equal opposite angles.

Here is the proof of the Isosceles Trapezoid Symmetry Theorem as given by the U of Chicago. This time, since Section 5-5 gives a paragraph proof, let me post a two-column proof here on the blog.

Given:

*ZOID*is an isosceles trapezoid with angles

*I*and

*D*equal in measure.

*m*is the perpendicular bisector of

~~ID~~

Prove:

*m*is the perpendicular bisector of

~~ZO~~

*m*is a symmetry line for

*ZOID*

*Proof:*

Statements Reasons

1.

*m*is the perp. bis. of

2.

*D'*=

*I*,

*I'*=

*D*2. Definition of reflection

3. angle

*I*= angle

*D*3. Given

4. ray

*DZ*reflected is

*IO*4. Reflections preserve angle measure.

5.

*Z'*lies on ray

*IO*5. Figure Reflection Theorem

6.

*ZOID*is a trapezoid 6. Given

7.

8.

*m*8. Fifth Postulate

9.

*Z'*lies on line

*ZO*9. Definition of reflection

10.

*Z'*=

*O*10. Line Intersection Theorem

11.

*O'*=

*Z*11. Flip-Flop Theorem

12.

*ZOID*reflected is

*OZDI*12. Figure Reflection Theorem

13.

*m*symm. line of

*ZOID*13. Definition of symmetry line

14.

*m*is the perp. bis. of

*OZ*14. Definition of reflection

This proof is quite long and can be intimidating for students. A teacher can break it down by actually folding the isosceles trapezoid along line

*m*. The teacher can ask, "Where does

*D*fold to?" The students will probably answer

*I*, only to have the teacher ask "Why?"

Then the tougher part is to show that

*Z*folds to

*O*. We do it one step at a time -- first we show that since reflections preserve angle measure,

*Z'*must be somewhere on ray

*OI*-- although not necessarily

*O*(but of course students will want to jump to that conclusion). Notice that we then have to the Fifth Postulate to show that side

*ZO*is also perpendicular to

*m*, so that

*Z'*also lies on

*ZO*-- and this is why I stated our Fifth Postulate yesterday. We didn't use Playfair or any other Parallel Postulate, but we used the Perpendicular to Parallels Postulate directly. This is one reason why I let Perpendicular to Parallels be our Fifth Postulate.

Now we have another theorem, the Isosceles Trapezoid Theorem:

In an isosceles trapezoid, the non-base sides are equal in measure.

In other words, our definition of isosceles trapezoid implies what we usually think of when we hear the word

*isosceles*. The text states that this is merely a corollary of the previous theorem -- since the reflection of one of the non-base sides is the other.

For readers of this blog, let's go back to what I wrote about non-Euclidean geometry. I stated that in hyperbolic geometry, there's something called a Saccheri quadrilateral -- defined as a quadrilateral with the base angles both

*right angles*and the non-base sides equal. As it turns out, a Saccheri quadrilateral is also an (inclusive) isosceles trapezoid and it also has a line of symmetry. Cutting the Saccheri quadrilateral along this line of symmetry gives two new figures called Lambert quadrilaterals. But the above Isosceles Trapezoid Symmetry Theorem, as stated above, can't be used to prove that a Saccheri quadrilateral is symmetrical, because its proof uses the Fifth Postulate.

Instead, we can prove that a Saccheri quadrilateral is symmetrical, as follows: the first five steps can remain as they are. But we can easily get that

*Z'*is

*O*, because the fact that

*DZ*=

*IO*is now part of the

*given*information (rather than proved as a corollary later). So

*Z'*is the point on ray

*IO*that is the correct distance from

*I*-- and that point is exactly

*O*. Then we pick up the proof from step 11 of the proof above. (Now we have a new corollary -- namely that

*and*~~ZO~~

*are parallel. This follows from the fact that*~~ID~~

*m*is the perpendicular bisector of both

*ZO*and

Returning as a text, another corollary given is the Rectangle Symmetry Theorem:

Every rectangle has two symmetry lines, the perpendicular bisectors of its sides.

This corollary follows from the Isosceles Trapezoid Symmetry Theorem in the same way that the Rhombus Symmetry Theorem (which shows that a rhombus also has two symmetry lines) follows from the Kite Symmetry Theorem. It shows again the beauty of using inclusive definitions.

So far in this section, we've mentioned the symmetries of the kite, rhombus, isosceles trapezoid, as well as the rectangle. A general trapezoid, as we've mentioned before, has no symmetry. But there's one special quadrilateral that's missing -- the parallelogram. So does a parallelogram have symmetry?

Also, so far in this section we've mentioned

*reflections*and reflection symmetry, but there's another transformation that we've learned about --

*rotations*and rotational symmetry. So do any of our special quadrilaterals have rotational symmetry.

We can answer both of these questions at once: A parallelogram has rotational symmetry!

Notice that the U of Chicago text doesn't discuss rotational symmetry of any figure, much less that of the parallelogram. Yet we see in the Common Core 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.

(emphasis mine)

And so we should discuss rotational symmetry in general and that of the parallelogram in particular.

But let's begin with the rhombus and rectangle before going on to the general parallelogram. We recall that our definition of rotation is the composite of reflections in intersecting lines -- and if those lines are symmetry lines of a figure, each one maps the figure to itself. Therefore, the entire rotation must map the figure to itself. It follows that any figure that contains two lines of symmetry must automatically have rotational symmetry as well! And the Two Reflection Theorem for Rotations tells us exactly what the center and magnitude of the rotation are. The center is where the two symmetry lines intersect, and the magnitude is double the angle between the intersecting lines.

Corollary to Two Reflection Theorem for Rotations:

If

*l*and

*m*are intersecting symmetry lines of a figure, then it also has rotational symmetry, where the magnitude is twice the non-obtuse angle between

*l*and

*m*and the center is the point of intersection of

*l*and

*m.*

*Applying this to the rhombus, we see that both its diagonals are symmetry lines. Because a rhombus is a kite, its diagonals are perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.*

And now let's look at the rectangle. Its symmetry lines are the perpendicular bisectors of its sides, and we can use the Two Perpendiculars Theorem and the Fifth Postulate to show that these symmetry lines are also perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.

But as it turns out, all parallelograms -- not just the rhombuses and rectangles -- can be shown to have 180-degree rotational symmetry. The U of Chicago doesn't prove this, but Hung-Hsi Wu does. In his Theorem 4, he proves that the opposite sides of a parallelogram are equal, by showing that the parallelogram has rotational symmetry.

Here is Wu's proof of his Theorem 4, in paragraph form (as Wu himself gives it):

Given:

*ABCD*is a parallelogram

Prove:

*AD*=

*BC*

*Proof:*

Let

*M*be the midpoint of the diagonal

*M*.

Because

*MA*=

*MC*and rotations preserve distance, we have

*C"*=

*A*, so that R maps line

*BC*to a line passing through

*A*and parallel to line

*BC*. Since the line

*AD*has exactly the same two properties by assumption, Playfair implies that R maps line

*BC*to line

*AD*. Similarly, R maps line

*AB*to

*CD*. Thus since

*B*lies on both

*AD*and

*CD*, its image

*B"*is a point that lies on both

*AD*and

*CD*. By the Line Intersection Theorem, this point is exactly

*D*.

Recall we also have

*C"*=

*A*. Therefore R maps the segment

*D*(which is

*B"*) to

*A*(which is

*C"*) by the property that a rotation (like a reflection) maps segments to segments. The latter segment has to be the segment

*ABCD*to

*CDAB*, so

*ABCD*has rotational symmetry. Since rotations preserve distance,

*BC*=

*AD*, as desired. QED

Notice that Playfair is used in the proof. A Saccheri quadrilateral, despite being a parallelogram, doesn't have rotational symmetry.

This proof is very similar to the Isosceles Trapezoid Symmetry Theorem. Both theorems depend on what Wu calls "Lemma 6" -- identifying the reflection or rotation image of a point by finding two lines that intersect at the preimage point and noting that the images of these two lines must intersect at the image of the point.

Like the Isosceles Trapezoid Symmetry Theorem and its corollary the Isosceles Trapezoid Theorem, we should call part of the proof the Parallelogram (Rotational) Symmetry Theorem and then derive that opposite sites are equal as a corollary. Wu also derives as a corollary that opposite angles are equal -- but we can also derive this by applying the Trapezoid Angle Theorem twice -- since a parallelogram is a trapezoid. Even the third major property of parallelograms -- that their diagonals bisect each other -- can be derived as a corollary (since rotations preserve distance, we must also have

*MB*=

*MD*in the above proof).

So we can prove all the major properties of parallelograms using only rotations and not using triangle congruence at all. The U of Chicago derives the properties of other figures using reflection symmetry but reverts to triangle congruence for the parallelogram properties (in Chapter 7).

Finally, notice that since rhombuses and rectangles are parallelograms, we can show that they have 180-degree rotational symmetry without looking at their lines of symmetry. But I still like using the Corollary to the Two Reflection Theorem for Rotations because it can be used to determine the rotational symmetry for figures other than quadrilaterals. We earlier proved that an equilateral triangle has three lines of symmetry -- so it also has rotational symmetry.

Tomorrow I will be tutoring my geometry student, so I'll have another chance to test out some of my geometry material posted to this blog.

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