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Geometry for Elementary School

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[edit] Further Reading

Geometry http://en.wikibooks.org/wiki/Geometry

[edit] External Links

Euclid's Elements

Introduction

Geometry for Elementary School
	Print version	Our tools: Ruler and compass

[edit] Why geometry?

Geometry is one of the most elegant fields in mathematics. It deals with visual shapes that we know from everyday life.

Learning geometry does not require previous skills like basic arithmetic. Hence, geometry is suitable as an introduction to mathematics for elementary school.

[edit] Who should use this book?

This book is intended for use by a parent (or a teacher) and a child. It is recommended that the parent have some familiarity with geometry, but this is not necessary. The parent can simply read the chapter before teaching the child and then learn it together.

[edit] Book guidelines

The classic book about geometry is Euclid's Elements. This book helped teach geometry for hundreds of years, so we feel that writing this book based on the Elements is a correct step.

We will adapt parts of the book for children and modify the order of some topics in order to make the book clearer.

The learning will be based on constructions and proofs. A construction is a method of creating a geometric object (such as a triangle) using a set of tools. In the case of this book, the tools we will be using are a compass and a ruler. A proof is a logical trail where we can prove one fact by starting with some given information and make a series of conclusions based on that information. Often times it is more difficult to prove a result than to simply find the result.

The constructions are useful for letting the child experience geometric ideas and get visual results.

The proofs are a good way to understand geometry and are a good basis for future study of logic.

Since the book is for children, we omit some of the proof details and use intuition instead of precise definition. On the other hand, we insist on correct and elegant proofs. Precise definitions and exact proofs can be found in regular geometry books and can be used to extend to material to some of the children.

[edit] Notation

The notation that is used in the book is defined the first time it is used. However, in order to simplify its use, it is also summarized in the "Notation" chapter at the end of the book.

[edit] Euclid's Elements online

There is a wonderful online version of Euclid's Elements at this web site. The site was created by David E. Joyce, a Professor of Mathematics and Computer Science at Clark University. This site includes all the text of the Elements, applets that display the constructions, and many insightful comments. We give reference in this book to the original sources and encourage the reader to read those materials on his own.

Ruler and Compass

Geometry for Elementary School
Introduction	Our Tools: Ruler and compass	Points

[edit] Introduction

This is a picture of a compass and a ruler.

[edit] How to draw a line?

Using a ruler (based on Book I, Postulate 1), we can draw a straight line.

We will use the notation $\overline{AB}$ for the line segment the starts at A and ends at B. Note that we don't care about the segment direction and therefore $\overline{AB}$ is the same as $\overline{BA}$ .

Line $\overline{AB}$

Line $\overline{BA}$

[edit] How to draw a circle?

Using the compass (based on Book I, Postulate 3).

We use the notation $\circ A,\overline{AB}$ for the circle whose center is the point A and its radius length equals that of the segment $\overline{AB}$ .

Note that in other sources, such as Euclid's Elements, a circle is described by any 3 points on its circumference, ABC.

The center-radius notation was chosen because of its suitability for constructing circles with a ruler and a compass.

Points

Geometry for Elementary School
Our tools: Ruler and compass	Points	Lines

A Point is the limit of a circle whose size is decreasing.

You might be surprised to know, but these are not points:
	The reason that this shape is not a point is that it is too large, it has area. This is a 'ball'.
	Even when taking a ball of half that size we don't get a point.
	And that is too large as well...

We can consider that the "ball", circles, and the other non-points heve and represent points at their centers. A point can also be considered to exist where two lines cross, (Such as +, X, and * ).

A point is so small that even if we divide the size of these balls by 100, 1,000 or 1,000,000 it would still be much larger than a point. A point is considered as infinitely small. In order to get to the size of a point we should keep dividing the ball size by two - forever. Don't try it at home.

A point seems to be too small to be useful. Luckily, as we will see when discussing lines we have plenty of them.

Why define a point as an infinitely small dot? For one thing it has a very precise location, not just the center of a rough dot, but the point itself. Another reason is that if the drawing is made much bigger or smaller the point stays the same size. A point which is an infinitely small dot would be too small to see, so we must use a big old visible normal dot, or where two lines cross to represent it and its approximate location on paper.

Lines

Geometry for Elementary School
Points	Lines	Constructing equilateral triangle

[edit] Lines

A line is as wide as a point, infinitely thin, having an infinite number of points, (in a straight row), extending forever in both the directions. Any two lines can intersect at only a single point.

[edit] Line segments

A line segment is a part of a line, which has two endpoints. The endpoints give the line segment a fixed, or finite length.

Line segments A B, and C D, can be written as $\overline{AB}$ , and $\overline{CD}$

[edit] Rays

A ray is a line that has only one endpoint. A ray is infinite in one direction.

[edit] Axiom: there is only a single straight line between two points

Axiom: there is only a single straight line between two points.

show by halving that there are infinite number of points in a line
show that the number of points in a long line and a short line is equal

Constructing an Equilateral Triangle

Geometry for Elementary School
Lines	Constructing equilateral triangle	Copying a line segment

[edit] Introduction

Equilateral triangle

In this chapter, we will show you how to draw an equilateral triangle. What does "equilateral" mean? It simply means that all three sides of the triangle are the same length.

Any triangle whose vertices (points) are A, B and C is written like this: $\triangle ABC$ .

And if it's equilateral, it will look like the one in the picture.

[edit] The construction

The construction (method we use to draw $\triangle ABC$ ) is based on Book I, proposition 1.

Using your ruler, Draw a line whatever length you want the sides of your triangle to be.
Call one end of the line A and the other end B.
Now you have a line segment called $\overline{AB}$ .
It should look something like the drawing below.
Using your compass, Draw the circle $\circ A,$ whose center is A and radius is $\overline{AB}$ .
Again using your compass Draw the circle $\circ B,\overline{AB}$ , whose center is B and radius is $\overline{AB}$ .
Can you see how the circles intersect (cross over each other) at two points?
The points are shown in red on the picture below.
Choose one of these points and call it C.
We chose the upper point, but you can choose the lower point if you like. If you choose the lower point, your triangle will look "upside-down", but it will still be an equilateral triangle.
Draw a line between A and C and get segment $\overline{AC}$ .
Draw a line between B and C and get segment $\overline{BC}$ .
Construction of $\triangle ABC$ is completed.

[edit] Claim

The triangle $\triangle ABC$ is an equilateral triangle.

[edit] Proof

The points B and C are both on the circumference of the circle $\circ A,\overline{AB}$ and point A is at the center.
So the line $\overline{AB}$ is the same length as the line $\overline{AC}$ . Each is a radius of circle $\circ A$ , or more simply $\overline{AB}=\overline{AC}$ .
We do the same for the other circle:
The points A and C are both on the circumference of the circle $\circ B,\overline{AB}$ and point B is at the center.
So we can say that $\overline{AB}=\overline{BC}$ .
We've already shown that $\overline{AB}=\overline{AC}$ and $\overline{AB}=\overline{BC}$ . Since $\overline{AC}$ and $\overline{BC}$ are both equal in length to $\overline{AB}$ , they must also be equal in length to each other. This can be shown by substitution. So we can say $\overline{AC}=\overline{BC}$
Therefore, the lines $\overline{AB}$ , $\overline{AC}$ , and $\overline{BC}$ are all equal. $\overline{AB}=\overline{AC}=\overline{BC}$
We proved that all sides of $\triangle ABC$ are equal, so this triangle is an equilateral triangle by definition.

[edit] Problems with the proof

The construction above is simple and elegant. One can imagine how children, using their legs as compass, accidentally find it.

However, Euclid’s proof was wrong.

In mathematical logic, we assume some postulates. We construct proofs by advancing step by step. A proof should be made only of postulates and claims that can be deduced from the postulates. Some useful claims are given names and called theorems in order to enable their use in future proofs.

There are some steps in its proof that cannot be deduced from the postulates. For example, according to the postulates he used, the circles $\circ A,\overline{AB}$ and $\circ B,\overline{AB}$ do not have to intersect.

Although the proof was wrong, the construction is not necessarily wrong. One can make the construction valid, by extending the set of postulates. Indeed, in later years, different sets of postulates were proposed in order to make the proof valid. Using these sets, the construction that works so well using pencil and paper is also logically sound.

This error of Euclid, the gifted mathematician, should serve as an excellent example of the difficulty in mathematical proof and also the difference between proof and our intuition.

Copying a Line Segment

Geometry for Elementary School
Constructing equilateral triangle	Copying a line segment	Copying a triangle

This construction copies a line segment $\overline{AB}$ to a target point T. The construction is based on Book I, prop 2.

[edit] The construction

Let A be one of the end points of $\overline{AB}$ . Note that we are just giving it a name here. (We could replace A with the other end point B).
Draw a line $\overline{AT}$
Construct an equilateral triangle $\triangle ATD$ (a triangle that has $\overline{AT}$ as one of its sides).
Draw the circle $\circ A,\overline{AB}$ , whose center is A and radius is $\overline{AB}$ .
Draw a line starting from D going through A until it intersects $\circ A,\overline{AB}$ and let the intersection point be E . Get segments $\overline{AE}$ and $\overline{DE}$ .
Draw the circle $\circ D,\overline{DE}$ , whose center is D and radius is $\overline{DE}$ .
Draw a line starting from D going through T until it intersects $\circ D,\overline{DE}$ and let the intersection point be F. Get segments $\overline{TF}$ and $\overline{DF}$ .

[edit] Claim

The segment $\overline{TF}$ is equal to $\overline{AB}$ and starts at T.

[edit] Proof

Segments $\overline{AB}$ and $\overline{AE}$ are both from the center of $\circ A,\overline{AB}$ to its circumference. Therefore they equal to the circle radius and to each other.
Segments $\overline{DE}$ and $\overline{DF}$ are both from the center of $\circ D,\overline{DE}$ to its circumference. Therefore they equal to the circle radius and to each other.
$\overline{DE}$ equals to the sum of its parts $\overline{DA}$ and $\overline{AE}$ .
$\overline{DF}$ equals to the sum of its parts $\overline{DT}$ and $\overline{TF}$ .
The segment $\overline{DA}$ is equal to $\overline{DT}$ since they are the sides of the equilateral triangle $\triangle ATD$ .
Since the sum of segments is equal and two of the summands are equal so are the two other summands $\overline{AE}$ and $\overline{TF}$ .
Therefore $\overline{AB}$ equals $\overline{TF}$ .

Constructing a Triangle

Geometry for Elementary School
Copying a line segment	Constructing a triangle	Copying an angle

In this chapter, we will show how to construct a triangle from three segments. The construction is based on Book I, proposition 22

[edit] The construction

Given three line segments $\overline{AB}$ , $\overline{CD}$ and $\overline{EF}$ we build a triangle whose sides equal the segments.

Copy the line $\overline{CD}$ to point A.

If you have forgotten how to do this, follow the instructions in the previous section. Your construction should look like the grey lines in the picture below. Call the new line $\overline{AG}$

It's a good idea to erase your construction lines now, so all that's left are the four line segments shown below.
Copy the line $\overline{EF}$ to point B

Your construction should look like the grey lines in the picture below. Call the new line $\overline{BH}$
Draw the circle $\circ A,\overline{AG}$ , whose center is A and radius is $\overline{AG}$ .
Draw the circle $\circ B,\overline{BH}$ , whose center is B and radius is $\overline{BH}$ .
Let J be an intersection point of $\circ A,\overline{AG}$ and $\circ B,\overline{BH}$ .
Draw a line $\overline{AJ}$ .
Draw a line $\overline{BJ}$ .

[edit] Claim

The sides of the triangle $\triangle ABJ$ equal to $\overline{AB}$ , $\overline{CD}$ and $\overline{EF}$ .

[edit] Proof

The segment $\overline{AB}$ is a side of the triangle and equal to itself.
The segment $\overline{AJ}$ is equal to $\overline{AG}$ because they are both radii of circle $\circ A,\overline{AG}$ . And because it was copied, $\overline{AG}$ = $\overline{CD}$ . Therefore $\overline{AJ}$ is also equal to $\overline{CD}$
The segment $\overline{BJ}$ is equal to $\overline{BH}$ because they are both radii of circle $\circ B,\overline{BH}$ . And because it was copied, $\overline{BH}$ = $\overline{EF}$ . Therefore $\overline{BJ}$ is also equal to $\overline{EF}$
Hence the sides of the triangle $\triangle ABJ$ are equal to $\overline{AB}$ , $\overline{CD}$ and $\overline{EF}$ .

[edit] Testing the procedure

Draw a line $\overline{AB}$ of some length.
Copy the line $\overline{AB}$ to an arbitrary point C and get $\overline{CD}$ .
Draw a line $\overline{EF}$ such that it length is three times the length of $\overline{AB}$ . (We didn't specify how to construct such a segment and we give it as an exercise. Use chapter Copy the line as a guide for the solution.
Construct a triangle from $\overline{AB}$ , $\overline{CD}$ and $\overline{EF}$ .

[edit] Why you couldn't construct the triangle in the test?

The reason we couldn’t build the triangle in the test is that the circles we constructed did not intersect. One cannot use any three segment to construct a triangle. The length of the segments must obey a condition called “The triangle inequality”. The triangle inequality states that any of the segments should be shorter that the sum of the length of the other two segments. If one of the segments is longer the circles do not interest. If one segment equals to the sum of the other two, we get a line instead of a triangle.

Therefore, the construction is correct but one should condition the segments on which it can be applied. Note that the original construction was conditioned by Euclid, hence there is no error in the construction or in its proof.

Why the Constructions are not Correct

Geometry for Elementary School
Copying an angle	Why the constructions are not correct?	The Side-Angle-Side congruence theorem

In the previous chapters, we introduced constructions and proved their correctness. Therefore, these constructions should work flawlessly. In this chapter, we will check whether the construction are indeed flawless.

[edit] Testing a construction

Draw a line of $\overline{AB}$ of length 10cm.
Copy the line segment to a different point T.
Measure the length of the segment you constructed.

[edit] Explanation

I must admit that I never could copy the segment accurately. Some times the segment I constructed was of the length 10.5cm, I did even worse. A more talented person might get better results, but probably not exact.

How come the construction didn't work, at least in my case?

Our proof of the construction is correct. However, the construction is done in an ideal world. In this world, the lines and circles drawn are also ideal. They match the mathematical definition perfectly.

The circle I draw doesn't match the mathematical definition. Actually, many say that they don't match any definition of circle. When I try to use the construction, I'm using the wrong building blocks.

However, the construction are not useless in our far from ideal world. If we use approximation of a circle in the construction, we are getting and approximation of the segment copy. After all, even my copy is not too far from the original.

[edit] Note

In the Euclidian geometry developed by the Greek the rule is used only to draw lines. One cannot measure the length of segments using the rulers as we did in this test. Therefore our test should be viewed as a criticism of the use of Euclidian geometry in the real world and not as part of that geometry.

Side-Side-Side Congruence Theorem

In this chapter, we will start the discussion of congruence and congruence theorems. We say the two triangles are congruent if they have the same shape. The triangles $\triangle ABC$ and $\triangle DEF$ are congruent if and only if all the following conditions hold:

The side $\overline {AB}$ equals $\overline {DE}$ .
The side $\overline {BC}$ equals $\overline {EF}$ .
The side $\overline {AC}$ equals $\overline {DF}$ .
The angle $\angle ABC$ equals $\angle DEF$ .
The angle $\angle BCA$ equals $\angle EFD$ .
The angle $\angle CAB$ equals $\angle FDE$ .

Note that the order of vertices is important. It is possible that $\triangle ABC$ and $\triangle ACB$ are not congruent even though both refer to the same triangle.

Congruence theorems give a set of the fewest conditions that are sufficient in order to show that two triangles are congruent.

The first congruence theorem we will discuss is the Side-Side-Side theorem.

[edit] The Side-Side-Side congruence theorem

Given two triangles $\triangle ABC$ and $\triangle DEF$ such that their sides are equal, hence:

The side $\overline {AB}$ equals $\overline {DE}$ .
The side $\overline {BC}$ equals $\overline {EF}$ .
The side $\overline {AC}$ equals $\overline {DF}$ .

Then the triangles are congruent and their angles are equal too.

[edit] Method of Proof

In order to prove the theorem we need a new postulate. The postulate is that one can move or flip any shape in the plane without changing it. In particular, one can move a triangle without changing its sides or angles. Note that this postulate is true in plane geometry but not in general. If one considers geometry over a ball, the postulate is no longer true.

Given the postulate, we will show how can we move one triangle to the other triangle location and show that they coincide. Due to that, the triangles are equal.

[edit] The construction

Copy The line Segment side $\overline {AB}$ to the point D.
Draw the circle $\circ D,\overline{AB}$ .
The circle $\circ D,\overline{AB}$ and the segment $\overline{DE}$ intersect at the point E hence we have a copy of $\overline {AB}$ such that it coincides with $\overline{DE}$ .
Construct a triangle with $\overline{DE}$ as its base, $\overline{BC}$ , $\overline{AC}$ as the sides and the vertex at the side of the vertex F. Call this triangle triangles $\triangle DEG$

[edit] The claim

The triangles $\triangle DEF$ and $\triangle ABC$ congruent.

[edit] The proof

The points A and D coincide.
The points B and E coincide.
The vertex F is an intersection point of $\circ D,\overline{DF}$ and $\circ E,\overline{EF}$ .
The vertex G is an intersection point of $\circ D,\overline{AC}$ and $\circ E,\overline{BC}$ .
It is given that $\overline{DF}$ equals $\overline{AC}$ .
It is given that $\overline{EF}$ equals $\overline{BC}$ .
Therefore, $\circ D,\overline{DF}$ equals $\circ D,\overline{AC}$ and $\circ E,\overline{EF}$ equals $\circ E,\overline{BC}$ .
However, circles of different centers have at most one intersection point in one side of the segment that joins their centers.
Hence, the points G and F coincide.
There is only a single straight line between two points, therefore $\overline {EG}$ coincides with $\overline {EF}$ and $\overline {GD}$ coincides with $\overline {DF}$ .
Therefore, the $\triangle DEG$ coincides with $\triangle DEF$ and the two are congruent.
Due to the postulate $\triangle DEG$ and $\triangle ABC$ are equal and therefore congruent.
Hence, $\triangle DEF$ and $\triangle ABC$ are congruent.
Hence, $\angle ABC$ equals $\angle DEF$ , $\angle BCA$ equals $\angle EFD$ and $\angle CAB$ equals $\angle FDE$ .

[edit] Note

The Side-Side-Side congruence theorem appears as Book I, prop 8 at the Elements. The proof here is in the spirit of the original proof. In the original proof Euclid claims that the vertices F and G must coincide but doesn’t show why. We used the assumption that “circles of different centers have at most one intersection point in one side of a segment that joins their centers”. This assumption is true in plane geometry but doesn’t follows from Euclid’s original postulates. Since Euclid himself had to use such an assumption, we preferred to give a more detailed proof, though the extra assumption.

Copying a Triangle

Geometry for Elementary School
Copying a line segment	Copying a triangle	Copying an angle

In this chapter, we will show how to copy a triangle $\triangle ABC$ to other triangle $\triangle CDE$ . The construction is a excellent example of the reduction technique – solving a problem by solution to a previously solved problem.

[edit] The construction

Construct a triangle from the sides of $\triangle ABC$ : $\overline{AB}$ , $\overline{AC}$ , $\overline{BC}$ and get $\triangle CDE$ .

[edit] Claim

The triangles $\triangle ABC$ and $\triangle CDE$ are congruent.

[edit] Proof

$\overline{AB}$ , $\overline{AC}$ , $\overline{BC}$ are sides of the triangle $\triangle ABC$ and therefore obey the triangle inequality.
Therefore one can build a triangle whose sides equal these segments.
The sides of the triangle $\triangle ABC$ and $\triangle CDE$ are equal.
Due to the The Side-Side-Side congruence theorem the triangles $\triangle ABC$ and $\triangle CDE$ congruence.

Copying an Angle

Geometry for Elementary School
Copying a triangle	Copying an angle	Why the constructions are not correct?

In this chapter, we will show how to copy an angle $\angle ABC$ to other angle $\angle CDE$ . The construction is based on Book I, proposition 23.

[edit] The construction

Draw a line between A and B and get $\triangle ABC$ .
Copy the triangle $\triangle ABC$ and get $\triangle CDE$

[edit] Claim

The angles $\angle ABC$ and $\angle CDE$ are equal.

[edit] Proof

The triangles $\triangle ABC$ and $\triangle CDE$ congruence.
Therefore the angles of the triangles equal.
Hence, $\angle ABC$ and $\angle CDE$ are equal.

Note that any two points on the rays can be used to create a triangle.

Bisecting an Angle

[edit] Introduction

BISECT ANGLE $\angle ABC$

Use a compass to find points D and E, equidistant from the vertex, point B.
Draw the line $\overline{DE}$ .
Construct an equilateral triangle on $\overline{DE}$ with third vertex F and get $\triangle DEF$ . (Lines DF and EF are equal in length).
Draw the line $\overline{BF}$ .

[edit] Claim

The angles $\angle ABF$ , $\angle FBC$ equal to half of $\angle ABC$ .

[edit] The proof

$\overline{DE}$ is a segment from the center to the circumference of $\circ B,\overline{BD}$ and therefore equals its radius.
Hence, $\overline{BE}$ equals $\overline{BD}$ .
$\overline{DF}$ and $\overline{EF}$ are sides of the equilateral triangle $\triangle DEF$ .
Hence, $\overline{DF}$ equals $\overline{EF}$ .
The segment $\overline{BF}$ equals to itself
Due to the Side-Side-Side congruence theorem the triangles $\triangle ABF$ and $\triangle FBC$ congruent.
Hence, the angles $\angle ABF$ , $\angle FBC$ equal to half of $\angle ABC$ .

[edit] Note

We showed a simple method to divide an angle to two. A natural question that rises is how to divide an angle into other numbers. Since Euclid’s days, mathematicians looked for a method for trisecting an angle, dividing it into 3. Only after years of trials it was proven that no such method exists since such a construction is impossible, using only ruler and compass.

[edit] Exercise

Find a construction for dividing an angle to 4.
Find a construction for dividing an angle to 8.
For which other number you can find such constructions?

Side-Angle-Side Congruence Theorem

Geometry for Elementary School
Why the constructions are not correct?	The Side-Angle-Side congruence theorem	Some impossible constructions

In this chapter, we will discuss another congruence theorem, this time the Side-Angle-Side theorem. The theorem appears as Based on Book I, prop 4 at the Elements.

[edit] The Side-Angle-Side congruence theorem

Given two triangles $\triangle ABC$ and $\triangle DEF$ such that their sides are equal, hence:

The side $\overline {AB}$ equals $\overline {DE}$ .
The side $\overline {CA}$ equals $\overline {DF}$ .
The angle $\angle CAB$ equals $\angle FDE$ (These are the angles between the sides).

Then the triangles congruent and their other angles and side are equal too. Success!

[edit] Proof

We will use the method of superposition – we will move one triangle to the other one and we will show that they coincide. We won’t use the construction we learned to copy a line or a segment but we will move the triangle as whole.

Superpose $\triangle ABC$ on $\triangle DEF$ such that A is place on D and $\overline {AB}$ is placed on $\overline {DE}$ .
It is given that $\overline {AB}$ equals $\overline {DE}$ .
Hence, B coincides with E.
It is given that the angle $\angle CAB$ equals $\angle FDE$ .
Hence, $\overline {CA}$ is placed on $\overline {DF}$ .
it is given that $\overline {CA}$ equals $\overline {DF}$ .
Hence, C coincides with F.
Therefore, $\overline {CB}$ coincides with $\overline {EF}$ .
The triangles $\triangle ABC$ and $\triangle DEF$ coincide.
The triangles $\triangle ABC$ and $\triangle DEF$ congruent.

Bisecting a Segment

[edit] Introduction

In this chapter, we will learn how to bisect a segment. Given a segment $\overline{AB}$ , we will divide it to two equal segments $\overline{AC}$ and $\overline{CB}$ . The construction is based on book I, proposition 10.

[edit] The construction

Construct the equilateral triangle $\triangle ABD$ on $\overline{AB}$ .
Bisect an angle on $\angle ADB$ using the segment $\overline{DE}$ .
Let C be the intersection point of $\overline{DE}$ and $\overline{AB}$ .

[edit] Claim

Both $\overline{AC}$ and $\overline{CB}$ are equal to half of $\overline{AB}$ .

[edit] The proof

$\overline{AD}$ and $\overline{BD}$ are sides of the equilateral triangle $\triangle ABD$ .
Hence, $\overline{AD}$ equals $\overline{BD}$ .
The segment $\overline{DC}$ equals to itself.
Due to the construction $\angle ADE$ and $\angle EDB$ are equal.
The segments $\overline{DE}$ and $\overline{DC}$ lie on each other.
Hence, $\angle ADE$ equals to $\angle ADC$ and $\angle EDB$ equals to $\angle CDB$ .
Due to the Side-Angle-Side congruence theorem the triangles $\triangle ADC$ and $\triangle CDB$ congruent.
Hence, $\overline{AC}$ and $\overline{CB}$ are equal.
Since $\overline{AB}$ is the sum of $\overline{AC}$ and $\overline{CB}$ , each of them equals to its half.

Some Impossible Constructions

Geometry for Elementary School
The Side-Angle-Side congruence theorem	Some impossible constructions	Parallel lines

In the previous chapters, we discussed several construction procedures. In this chapter, we will number some problems for which there is no construction using only ruler and compass.

The problems were introduced by the Greek and since then mathematicians tried to find constructions for them. Only in 1882, it was proven that there is no construction for the problems.

Note that the problems have no construction when we restrict ourself to constructions using ruler and compass. The problems can be solved when allowing the use of other tools or operations, for example, if we use Origami.

The mathematics involved in proving that the constructions are impossible are too advanced for this book. Therefore, we only name the problems and give reference to the proof of their impossibility at the further reading section.

[edit] Impossible constructions

[edit] Squaring the circle

The problem is to find a construction procedure that in a finite number of steps, to make a square with the same area as a given circle.

[edit] Doubling the cube

To "double the cube" means to be given a cube of some side length s and volume V, and to construct a new cube, larger than the first, with volume 2V and therefore side length ³√2s.

[edit] Trisecting the angle

The problem is to find a construction procedure that in a finite number of steps, constructs an angle that is one-third of a given arbitrary angle.

[edit] Further reading

A Wikibookian has nominated this page for cleanup.
You can help make it better. Please review any relevant discussion.

Proving that the constructions are impossible involve mathematics that is not in the scope of this book.

The interested reader can use these links to learn why the constructions are impossible.

The Four Problems Of Antiquity has no solution since their solution involves constructing a number that is not a constructible number. The numbers that should have being constructed in the problems are defined by these cubic Equations.

It is recommended to read the references in this order:

Pythagorean Theorem

In this chapter, we will discuss the Pythagorean theorem. It is used the find the side lengths of right triangles. It says:

In any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of areas of the squares whose sides are the two legs (i.e. the two sides other than the hypotenuse).

This means that if $\triangle ABC$ is a right triangle, the length of the hypotenuse, c, squared eqauls the sum of a squared plus b squared. Or:

$a^2 + b^2 = c^2 \,$

Here's an example:

In a right-angled triangle, a=5cm and b=12cm, so what is c?

$a^2 + b^2 = c^2 \,$

$c = \sqrt{ 5^2 + 12^2 }\,$

$c = \sqrt{ 25 + 144 }\,$

$c = \sqrt{ 169 }\,$

$c = 13 \,$

If c is not larger than a or b, your answer is incorrect. There may be a number of reasons that your answer is incorrect. The first is that you have calculated the sums wrong, the second is that the triangle you are trying to find the hypotenuse of is not a right angled triangle or the third is you have mixed up the measurements. There may be more finer points to having a wrong answer but the three stated are the most common

[edit] Exercise

Write about Pythagorean Theorum and its use to prove that $\sqrt{0.5}$ is an irrational number.

Parallel Lines

Geometry for Elementary School
Some impossible constructions	Parallel lines	Squares

[edit] Definition

The definition of parallel lines is based on Book I, definition 23.

Parallel lines are straight lines that never intersect, which means that they never cross. Notice that when we look at parallel parts of shapes there is no place where they intersect even if we extend the lines.

[edit] The parallel lines postulate

The postulate appears in Euclid’s elements as the fifth postulate.

Let there be two lines. If there is a third line that intersects them such that the sum of the interior angles on one side is smaller than two right angles then the two lines intersect.

This postulate was suspect as redundant. Mathematicians though that instead of assuming it, the postulate can be deduced from other postulates. However, the attempts to deduce this postulate failed. The reason to this failure is that the indeed, the parallel postulate doesn’t follow from the other ones. While we assume it in plane geometry, one can define different geometries (e.g., on a ball) for which this postulate is not valid.

Squares

Geometry for Elementary School
Parallel lines	Squares	Notation

A square is a geometric figure comprised of four lines of equal length, which are connected at right angles.

Prove properties of different squares using parallel lines theorems.
Show that the more properties we have the more we can prove but on fewer shapes.

A Proof of Irrationality

In mathematics, a rational number is a real number that can be written as the ratio of two integers, i.e., it is of the form

a/b where a and b are integers and b is not zero. An irrational number is a number that cannot be written as a ratio of two integers, i.e., it is not of the form

a/b .

[edit] History of the theory of irrational numbers

The discovery of irrational numbers is usually attributed to Pythagoras, more specifically to the Pythagorean Hippasus of Metapontum, who produced a proof of the irrationality of the $\sqrt{2}$ . The story goes that Hippasus discovered irrational numbers when trying to represent the square root of 2 as a fraction (proof below). However, Pythagoras believed in the absoluteness of numbers, and could not accept the existence of irrational numbers. He could not disprove their existence through logic, but his beliefs would not accept the existence of irrational numbers and so he sentenced Hippasus to death by drowning. As you see, mathematics might be dangerous.

[edit] Irrationality of the square root of 2

One proof of the irrationality of the square root of 2 is the following proof by contradiction. The proposition is proved by assuming the negation and showing that that leads to a contradiction, which means that the proposition must be true.

The term coprime is used in the proof. Two integers are coprimes if neither of them divides the other.

Assume that $\sqrt{2}$ is a rational number. This would mean that there exist integers a and b such that a / b = $\sqrt{2}$ .
Then $\sqrt{2}$ can be written as an irreducible fraction (the fraction is shortened as much as possible) a / b such that a and b are coprime integers and (a / b)² = 2.
It follows that a² / b² = 2 and a² = 2 b².
Therefore a² is even because it is equal to 2 b² which is obviously even.
It follows that a must be even. (Odd numbers have odd squares and even numbers have even squares.)
Because a is even, there exists a k that fulfills: a = 2k.
We insert equation of (6) into the last equation of (3): (2k)² = 2b² is equivalent to 2b² = 4k² is equivalent to b² = 2k².
Because 2k² is even it follows that b² is also even which means that b is even because only even numbers have even squares.
By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).

Since we have found a contradiction the assumption (1) that $\sqrt{2}$ is a rational number must be false. The opposite is proven. $\sqrt{2}$ is irrational.

Fractals

All the previous constructions we considered had one thing in common. The constructions were ended after a final number of steps. When one recalls that mathematicians actually used a ruler and compass in order to execute the constructions, this requirement seems to be in place. However, when we remove this requirement we can construct new interesting geometric shapes. In this chapter we will introduce two of them. Note that these shapes are not part of Euclidian geometry and were considered only years after its development.

[edit] Cantor Set

For a full overview of Cantor set see the article at wikipedia on which this section is based. The Cantor set was introduced by German mathematician Georg Cantor.

The Cantor set is defined by repeatedly removing the middle thirds of line segments. One starts by removing the middle third from the unit interval [0, 1], leaving [1, 1/3] ∪ [2/3, 1]. Next, the "middle thirds" of all of the remaining intervals are removed. This process is continued for ever. The Cantor set consists of all points in the interval [0, 1] that are not removed at any step in this infinite process.

[edit] What's in the Cantor set?

Since the Cantor set is defined as the set of points not excluded, the proportion of the unit interval remaining can be found by total length removed. This total is the geometric series

$\frac{1}{3} + \frac{2}{9} + \frac{4}{27} + \frac{8}{81} + \cdots = \sum_{n=0}^\infty \frac{2^n}{3^{n+1}} = \frac{1}{3}\left(\frac{1}{1-\frac{2}{3}}\right) = 1.$

So that the proportion left is 1 – 1 = 0. Alternatively, it can be observed that each step leaves 2/3 of the length in the previous stage, so that the amount remaining is 2/3 × 2/3 × 2/3 × ..., an infinite product which equals 0 in the limit.

From the calculation, it may seem surprising that there would be anything left — after all, the sum of the lengths of the removed intervals is equal to the length of the original interval. However a closer look at the process reveals that there must be something left, since removing the "middle third" of each interval involved removing open sets (sets that do not include their endpoints). So removing the line segment (1/3, 2/3) from the original interval [0, 1] leaves behind the points 1/3 and 2/3. Subsequent steps do not remove these (or other) endpoints, since the intervals removed are always internal to the intervals remaining. So we know for certain that the Cantor set is not empty.

[edit] The Cantor set is a fractal

The Cantor set is the prototype of a fractal. It is self-similar, because it is equal to two copies of itself, if each copy is shrunk by a factor of 1/3 and translated.

[edit] Koch curve

For a full overview of Koch curve see the article at wikipedia on which this section is based.

The Koch curve is a one of the earliest fractal curves to have been described. It was published during 1904 by the Swedish mathematician Helge von Koch. The better known Koch snowflake (or Koch star) is the same as the curve, except it starts with an equilateral triangle instead of a line segment.

One can imagine that it was created by starting with a line segment, then recursively altering each line segment as follows:

divide the line segment into three segments of equal length.
draw an equilateral triangle that has the middle segment from step one as its base.
remove the line segment that is the base of the triangle from step 2.

After doing this once the result should be a shape similar to the Star of David.

The Koch curve is in the limit approached as the above steps are followed over and over again.

The Koch curve has infinite length because each time the steps above are performed on each line segment of the figure its length increases by one third. The length at step n will therefore be (4/3)ⁿ.

The area of the Koch snowflake is 8/5 that of the initial triangle, so an infinite perimeter encloses a finite area.

What's Next?

Geometry for Elementary School/What next?

Notation

Geometry for Elementary School
Squares	Print version

This chapter summaries the notation used in the book. An effort was made to use common conventions in the notation. However, since many conventions exist the reader might see a different notation used in other books.

One who is interested seeing an example of past notation might be interested in Byrne's edition of Euclid's Elements. See for example the equilateral triangle construction.

[edit] Point

A point will be named by a bolded English letter, as in the point A.

[edit] Line segment

We will use the notation $\overline{AB}$ for the line segment that starts at A and ends at B. Note that we don't care about the segment direction and therefore $\overline{AB}$ is similar to $\overline{BA}$ .

[edit] Angle

We will use the notation $\angle{ABC}$ for the angle going from the point B, the intersection point of the segments $\overline{BA}$ and $\overline{BC}$ .

[edit] Triangle

A triangle whose vertices are A, B and C will be noted as $\triangle ABC$ . Note that for the purpose of triangles' congruence, the order of vertices is important and $\triangle ABC$ and $\triangle BCA$ are not necessarily congruent.

[edit] Circle

We use the notation $\circ A,\overline{BC}$ for the circle whose center is the point A and its radius length equals that of the segment $\overline{BC}$ .

Note that in other sources, such as Euclid's Elements, a circle is describe by any 3 points on its circumference, ABC.

The center, radius notation was chosen since it seems to be more suitable for constructions.

Geometry for Elementary School/Print version

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Introduction

[edit] Why geometry?

[edit] Who should use this book?

[edit] Book guidelines

[edit] Notation

[edit] Euclid's Elements online

Ruler and Compass

[edit] Introduction

[edit] How to draw a line?

[edit] How to draw a circle?

Points

Lines

[edit] Lines

[edit] Line segments

[edit] Rays

[edit] Axiom: there is only a single straight line between two points

Constructing an Equilateral Triangle

[edit] Introduction

[edit] The construction

[edit] Claim

[edit] Proof

[edit] Problems with the proof

Copying a Line Segment

[edit] The construction

[edit] Claim

[edit] Proof

Constructing a Triangle

[edit] The construction

[edit] Claim

[edit] Proof

[edit] Testing the procedure

[edit] Why you couldn't construct the triangle in the test?

Why the Constructions are not Correct

[edit] Testing a construction

[edit] Explanation

[edit] Note

Side-Side-Side Congruence Theorem

[edit] The Side-Side-Side congruence theorem

[edit] Method of Proof

[edit] The construction

[edit] The claim

[edit] The proof

[edit] Note

Copying a Triangle

[edit] The construction

[edit] Claim

[edit] Proof

Copying an Angle

[edit] The construction

[edit] Claim

[edit] Proof

Bisecting an Angle

[edit] Introduction

[edit] Claim

[edit] The proof

[edit] Note

[edit] Exercise

Side-Angle-Side Congruence Theorem

[edit] The Side-Angle-Side congruence theorem

[edit] Proof

Bisecting a Segment

[edit] Introduction

[edit] The construction

[edit] Claim

[edit] The proof

Some Impossible Constructions

[edit] Impossible constructions

[edit] Squaring the circle

[edit] Doubling the cube

[edit] Trisecting the angle

[edit] Further reading