As I wrote in my September 12th post, there are three periods with an aide (one Study Skills and the two eighth grade English classes) and two with a co-teacher (both seventh grade English). Thus, as usual, there is no "A Day in the Life" today.
Of course, today is Halloween. The eighth graders had a special Halloween assignment where they read an article about spiders, answer questions, and then color a spider where the correct colors are determined by their answers.
In my September 12th post, I write that the seventh graders were reading some short story I'd never heard of, titled "Lambert." (But it had nothing to do with quadrilaterals with three right angles.) In today's seventh grade classes, they take a quiz on another short story. I don't remember anything about it.
The eighth graders also watch the following video about a ghost ship:
I continue to hand out candy and pencils for the holidays. I don't always sing music for classes when there is an aide or co-teacher, but I can't resist singing "Ghost of a Chance" today. I start singing in the Study Skills class, then continue again in the English classes when students from the Study Skills class requests the song again.
Some students are dressed in pajamas. This is part of Red-Ribbon "Week." Actually, Red-Ribbon Week is supposed to be the last full week in October -- but some interpret this to mean that the entire week from Monday-Friday must be in October, while others interpret this to mean that Halloween must be included in the week. Still others take advantage and consider the last eleven days of October this year, from the 21st to today, to be Red-Ribbon "Week." (A week "with" eleven days -- hmm, they must be using the Eleven Calendar!)
Chapter 12 of Ian Stewart's The Story of Mathematics is called "Impossible Triangles: Is Euclid's Geometry the Only One?" Here's how it begins:
"Calculus was based on geometric principles, but the geometry was reduced to symbolic calculations, which were then formalized as analysis."
This chapter is all about one of my favorite topics -- different geometries, including both spherical and projective geometry. Stewart writes:
"The discovery that Euclid was not alone, that there can exist logically consistent types of geometry in which many of Euclid's theorems fail to hold, emerged from a renewed interest in the logical foundations of geometry, debated and developed from the middle of the 18th century to the middle of the 19th."
And so we begin with a little history:
"As far as Europe was concerned, geometry was becalmed in the doldrums between the years 300 and 1600."
But one application of different geometries was in Renaissance art. Stewart continues:
"Ordinary Euclidean geometry is about features that remain unchanged by rigid motions -- lengths, angles."
And of course, by "rigid motions" the author means "isometries" -- Common Core transformations such as reflections, rotations, and translations. (He reminds us that Euclid himself used congruent triangles -- and it's still a debate today whether to use isometries or congruent triangles.) On the other hand, projective geometry is different. Parallel lines appear to intersect at the horizon:
"They behave like this on an ideal infinite plane, not just on a slightly rounded Earth. In fact, they only behave exactly like this on a plane."
Stewart mentions a projective theorem, proved by Desargues in 1648. (I blogged about this theorem more than two years ago.) It can be proved by considering projections of 3D figures onto the plane:
"So we can use Euclidean methods to prove projective theorems. Projective geometry differs from Euclidean geometry as far as its viewpoint goes (pun intended)."
On the other hand, non-Euclidean geometry is different. One mathematician who worked with Euclid's axioms should be familiar to us -- Adrien Legendre. I spent two summers on the blog focusing on Legendre's work. He proved that Euclid's Fifth Postulate was equivalent to the existence of similar triangles:
"But he, and most other mathematicians, wanted something even more intuitive. In fact, there was a feeling that the Fifth Postulate was simply superfluous -- a consequence of the other axioms."
By the way, it's only fitting that I'm writing about Euclidean and non-Euclidean geometry on the day that I'm posting Lesson 5-5. During the first two years of this blog, I considered Lesson 5-5 to be the first lesson that required the Fifth Postulate, since it's all about trapezoids, which, of course, have a pair of parallel sides. (I had intentionally delayed Lesson 3-5, on parallel lines, until covering Lesson 5-5 back then.) Two lessons later is a brief mention of non-Euclidean geometry in our text.
(And so I'm also adding the "Spherical Geometry" label to this post.)
Indeed, Lesson 5-7 is on Triangle-Sum, and Legendre was the one who proved that the defect of a triangle (180 minus the angle sum) is proportional to its area:
"This seemed promising: if he could construct a triangle whose sides were twice those of a given triangle, but with the same angles, then he would obtain a contradiction, because the larger triangle would not have the same area as the smaller one."
But of course, he failed to prove the Fifth Postulate. We move on to another familiar name, Gerolamo Saccheri, who tried to use the quadrilaterals named after him to prove the Fifth Postulate. (Recall that a Saccheri quadrilateral is defined as one with two base angles that are right, and two congruent sides that are opposite each other and each adjacent to one base angle.) Indeed, he considered Saccheri quads whose summit angles are acute:
"Eventually he proved a rather complicated theorem about a family of lines all passing through one point, which implied that two of these lines would have a common perpendicular at infinity."
As it turned out, this wasn't really a contradiction. Well, since we already mentioned Saccheri and his quadrilaterals, of course we're going to mention Lambert and his quadrilaterals. (We know Lambert quadrilaterals to be those with at least three right angles.) He also proved that if the fourth angle of a Lambert quad is acute, something interesting happens with the angles of any polygon:
"Add all the angles, and subtract this from 2n - 4 right angles: the result is proportional to the polygon's area."
This is, of course, the defect of the polygon. The author also mentions a result of spherical geometry that was known to Lambert -- the excess, rather than the defect, of a spherical triangle or other polygon is proportional to the area. (On the sphere, the summit angles of a Saccheri quad and the fourth angle of a Lambert quad are obtuse, but we consider the acute case as well.)
By now, mathematicians were starting to suspect that Euclid's Fifth Postulate is independent of the other four postulates. We consider Nikolai Lobachevsky:
"By 1840 Lobachevsky was publishing a book on the topic, in which he complained about the lack of interest."
Even though Stewart doesn't give all the details, we already know them. Eventually, two new types of non-Euclidean geometry were discovered -- spherical (elliptic) and hyperbolic geometry. Indeed, hyperbolic geometry is the one where the summit angles of a Saccheri quad, and the fourth angle of a Lambert quad, are acute:
"It has numerous intriguing features, which distinguish it from Euclidean geometry. If the surface has zero curvature, like a Euclidean plane, then it is the Euclidean plane, and we get Euclidean geometry."
Non-Euclidean geometry led scientists to wonder, what is the shape of the universe? Before other geometries were discovered, it was naturally assumed that space was Euclidean:
"What else could it be? This question ceased to be rhetorical when logically consistent alternatives to Euclid's geometry began to appear."
And of course, the answer to that question is still unknown:
"Thanks to some imaginative and unorthodox thinking, often viciously contested by a less imaginative majority, it is now understood -- by mathematicians and physicists, at least -- that there are many alternatives to Euclid's geometry, and that the nature of physical space is a question for observation, not thought alone."
On that note, Stewart concludes the chapter as follows:
"For that matter, much of mathematics bears no obvious relation to reality at all -- but is useful, all the same."
Unlike previous chapters, this one has no biographical sidebars -- only what non-Euclidean geometry did for them and what it does for us.
I found a link to a puzzle about Halloween and spherical geometry (on a pumpkin, of course):
https://fivethirtyeight.com/features/can-you-carve-the-perfect-pumpkin/
I want to carve the perfect eye for my pumpkin this Halloween, but I can’t seem to make it the right size. Since symmetry is the key to beauty, the triangular eye should be equilateral and equiangular, which is easier said than done on the surface of a spherical pumpkin!
Anyway, the triangular eye that I made is way too big. Its sides all meet at right angles, and the resulting eye takes up a whole eighth of the pumpkin’s surface.
Instead, I want an eye that’s precisely half that size, or one-sixteenth of the pumpkin’s surface. For such an ideal pumpkin eye, at what angle should each of the sides meet?
Lambert's theorem about the spherical excess is the key to solving this puzzle. Try it!
By the way, there's been another recent Numberphile video on non-Euclidean geometry. But first, let's look at the definition of rectangle:
- A rectangle is a four-sided polygon, all of whose angles are right.
We know that rectangles, as defined above, exist only in Euclidean geometry. But there are several generalizations that do exist in non-Euclidean (either spherical or hyperbolic) geometry:
- Eliminate right: A rectangle is a four-sided polygon, all of whose angles are congruent.
- Eliminate all: A rectangle is a four-sided polygon, most of whose angles are right.
Choice 1) refers to equiangular quads, while choice 2) leads to Lambert quads. It can be proved that a Saccheri quad is a compromise between 1) and 2) -- it is a four-sided polygon with two right base angles and two congruent (not necessarily right) summit angles.
But there is a fourth possible definition as well:
- Eliminate four: A rectangle is a polygon, all of whose angles are right.
This is the definition implied by the following Numberphile video. (Actually it discusses squares, but we can define these as equilateral rectangles, whatever those are.)
Notice that by this fourth definition, a rectangular n-gon has n right angles. We already stated above that the angles of a Euclidean n-gon has 2n - 4 right angles, and the difference between 2n - 4 and the actual angle sum of a non-Euclidean n-gon is the defect (or excess) of the n-gon.
It's easy to solve 2n - 4 > n (defect), 2n - 4 = n (Euclidean), and 2n - 4 < n (excess) to show that it's a defect (hyperbolic) when n > 4, Euclidean when n = 4, and an excess (spherical) when n < 2. This shows us how many sides a "rectangle" can have in each of the three geometries.
We already know that a Euclidean rectangle can only have four sides. In the spherical case, at n = 3, there is a three-sided rectangle -- this is our well-known triangle with three right angles (the much too big pumpkin eye in the puzzle above). Each side of this triangle is a quadrant of the sphere. Notice that n = 2 case also makes sense on the sphere -- it is a 90-degree lune. Two of the n = 3 triangles can be glued together to create this lune.
The video shows us n = 5 for hyperbolic geometry. Not all five-sided hyperbolic rectangles are squares, though. We can glue two of these square pentagons to form n = 6, a six-sided rectangle. Of course, the resulting hexagon won't be square, since two of the sides would be twice as long as the other four. But I believe we can adjust the sides to form a square hexagon. All rectangular hexagons (square or not) have the same area, since they all have the same defect (two right angles).
I believe it's possible to glue together two square pentagons to form a hexagon, two hexagons to form an octagon, and two octagons to form a dodecagon (12-gon), in such a way that the resulting dodecagon is square (with all sides twice as long as those of the original pentagon).
The Numberphile video is the first time I've seen the square/rectangle definition loosened so that they no longer have to be quadrilaterals. The "equiangular" definition 1) is the only other loosening I've ever seen in a text. We don't need to loosen "rectangle" to mean a Lambert or Saccheri quad because they already have the established names, Lambert or Saccheri quad.
The Numberphile video is the first time I've seen the square/rectangle definition loosened so that they no longer have to be quadrilaterals. The "equiangular" definition 1) is the only other loosening I've ever seen in a text. We don't need to loosen "rectangle" to mean a Lambert or Saccheri quad because they already have the established names, Lambert or Saccheri quad.
Lesson 5-5 of the U of Chicago text is called "Properties of Trapezoids." In the modern Third Edition of the text, trapezoids appear in Lesson 6-6.
This is what I wrote last year in describing this worksheet:
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.
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 ofID
Prove: m is the perpendicular bisector ofZO
m is a symmetry line for ZOID
Proof:
Statements Reasons
1. m is the perp. bis. ofID 1. Given
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.ZO | | DI 7. Definition of trapezoid
8.ZO perpendicular to 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).
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.
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 therotations and reflections that carry it onto itself.
(emphasis mine)
The rotational symmetry of the parallelogram appears in the Third Edition. Therefore, the many of the changes I made three years ago can also be viewed as "fixing" my Second Edition to include material from the Third Edition. The tricky part is that in adhering to the Second Edition order, the students don't actually see rotations until Chapter 6. So my old worksheet mentions rotations, but that means there's Chapter 6 material on a Chapter 5 worksheet. I'll keep this discussion of rotations here on the blog, but teachers may want to save rotations for Chapter 6.
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 diagonalAC and we will use Theorem 1 -- Wu's First Theorem (that a 180-degree rotation maps a line to a parallel line, already proved on this blog) to explore the implications of the 180-degree rotation R around 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 segmentBC to the segment joining 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 DA, by the Unique Line Assumption. Thus by the Figure Reflection Theorem, R maps ABCD to CDAB, so ABCD has rotational symmetry. Since rotations preserve distance, BC = AD, as desired. QED
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.
This is what I wrote last year in describing this worksheet:
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.
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
Prove: m is the perpendicular bisector of
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.
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).
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.
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 therotations and reflections that carry it onto itself.
(emphasis mine)
The rotational symmetry of the parallelogram appears in the Third Edition. Therefore, the many of the changes I made three years ago can also be viewed as "fixing" my Second Edition to include material from the Third Edition. The tricky part is that in adhering to the Second Edition order, the students don't actually see rotations until Chapter 6. So my old worksheet mentions rotations, but that means there's Chapter 6 material on a Chapter 5 worksheet. I'll keep this discussion of rotations here on the blog, but teachers may want to save rotations for Chapter 6.
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
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
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.
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