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Book 1 – Proposition 9

Proposition 9

To bisect a given rectilinear angle.

This proposition is a construction that shows how to bisect an angle. We begin with the angle BAC.

Choose a point D at random on AB and cut off a segment AE from AC that is equal to AD. (Proposition 3)
Join DE.

Construct the equilateral triangle DEF be constructed using a point F on the opposite side of DE from point A . (Proposition 1)

Let AF be joined.

Now we will show that AF bisects the angle BAC by using the triangles DAF and EAF.

AD is equal to AE, since AE was cut off AC to equal AD.
AF is common to both triangles.
DF is equal to EF, since they are two of the sides of an equilateral triangle.

The conditions for Proposition 8 have been met, so angles DAF and EAF are equal to each other. Therefore the angle BAC has been bisected.

Now that we have learned how to bisect a rectilinear angle, we will learn how to bisect a given straight line (segment) in the next proposition.
As always, questions and feedback is welcome – George

Book 1 – Proposition 8

Proposition 8

If two triangles have the two sides equal to two sides respectively, and also have the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.

[Desmos graphs can be found here.]

Let ABC and DEF be two triangles such that AB is equal to DE and AC is equal to DF, and that the base BC is equal to the base EF.

We will prove that angle BAC is equal to angle EDF.

If triangle ABC is applied to triangle DEF with the point B placed on the point E and the straight line BC on the straight line EF, then the point C will also coincide with the point F because BC is equal to EF.
BA and AC will also coincide with ED and DF; if they did not then they would fall beside them and coincide with EG and GF for some other point G.

If that were true then given two straight lines (ED and DF) constructed on a straight line EF [from its extremities] and meeting at a point, there has been constructed on the same straight line [from its extremities] and on the same side of it two other straight lines (EG and GF) meeting at another point and equal to ED and DF. Proposition 7 tells us that this construction is not possible.

Therefore BA and AC must coincide with ED and DF. So, the angle BAC will coincide with the angle EDF. Thus the two angles BAC and EDF are equal to one another.

Q.E.D.

In this proof if triangle ABC coincided with triangle DEF, then the two angles BAV and EDF would be equal. So instead we assumed that the triangle ABC did not coincide with triangle DEF and showed that this was not possible.

The next four propositions will consist of constructions. I’d love to hear your feedback on Book 1 Proposition 8, or on anything about this blog – George

Book 1 – Proposition 7

Proposition 7

Given two straight lines constructed on a straight line [from its extremities] and meeting in a point, there cannot be constructed on the same line [from its extremities], and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

Desmos.com link to graphs

The strategy for this proof is to perform a proof by contradiction. We will assume that two other straight lines can be constructed in this fashion and will reach a contradiction.

Suppose two straight lines AC and CB are constructed on the straight line AB and meeting at the same point C as shown.

Basically, we have started with a line segment AB and added a point C to complete a triangle. Proposition states that we cannot add another point D on the same side of AB such that AC is equal to AD and BC is equal to BD. Let’s add a point D on the same side of AB as C is.

This will be a proof by contradiction. Assume that AC is equal to AD and BC is equal to BD.

Since AC is equal to AD, the angle ACD is equal to the angle ADC by Proposition 5.

Since angle DCB is smaller than angle ACD (because it is contained in that angle), we know that angle ADC is greater than angle DCB.

The following show that angle CDB is greater than angle DCB:

Angle CDB is greater than angle CDA (because angle CDA is contained in angle CDB).
Angle CDA is equal to angle ACD as stated previously.
Angle ACD is greater than angle DCB (because angle DCB is contained in angle ACD).

[CDB > CDA = ACD > DCB]

Now we shift our attention to triangle BCD. Since BC is equal to BD, Proposition 5 tells us that angle CDB is equal to angle DCB.

But we just showed that if our assumptions were true that angle CDB is greater than angle DCB, which is a contradiction. Therefore, there cannot be constructed on the same line [from its extremities], and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

Q.E.D.

That wraps up one week’s worth of these propositions. I hope you are finding it rewarding – George

Book 1 – Proposition 6

Proposition 6

If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

(Note: If you’d like to click through the images in this proof, they are available in this graph on Desmos.com.)

In Proposition 5 we started with two equal sides and proved that the angles that were opposite of those sides were equal. In this proposition we will begin with two equal angles and prove that the two sides that are opposite those angles are equal. We will begin with the triangle ABC, and let angle ABC be equal to angle ACB.

This proof is an example of a proof by contradiction. We will assume that sides AB and AC are not equal to one another, and as we proceed we will run into a contradiction that will tell us that our initial assumption is false and that sides AB and AC must be equal to one another.

Assume that sides AB and AC are not equal to one another. In that case one side must be longer than the other. Suppose that side AB is longer than side AC. From the longer side AB let DB be cut off equal to AC. (Proposition 3)
Join the points C and D as shown.

We will now be comparing the triangles DCB and ACB.
We know that DB is equal to AC, and the two triangles have the side BC in common. Also, the angle DBC is equal to the angle ACB as those are the two equal angles from triangle ABC.

Since the two triangles have two pairs of equal sides with an equal angle between them, Proposition 4 tells us that triangle DCB is equal to triangle ACB. However triangle DCB is clearly less than triangle ACB, so we have reached a contradiction. Our assumption that AB is not equal to AC is absurd, and therefore the two sides AB and AC must be equal to each other.

Q.E.D.

Proof by contradiction is an often used form for proving propositions or theorems in mathematics. The U.S. court system uses this principle in its trials – we assume that the defendant in innocent and then search for evidence that contradicts this assumption. A statistical hypothesis test uses the same format – assuming that the null hypothesis is true prior to searching for evidence that it is false.

Note: I have decided to stop previewing propositions. I will cover one proposition per post – George

Book 1 – Proposition 5

Proposition 5

In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.

An isosceles triangle has two sides that are equal. In the picture below the sides AB and AC are equal.

Now we produce the straight lines BD and CE further in a straight line with AB and AC. (Postulate 2)

We are trying to show that the two angles ABC and ACB are equal. Those are the two angles that are opposite the equal sides. We are also trying to prove that angles DBC and ECB are equal. (We will change the labels for those in a moment.)

Place a point F at random on BD, and from AE cut off a straight line AG that is equal to AF (by Proposition 3).

Let the lines FC and GB be joined.

We begin by focusing on triangles ACF and ABG, trying to prove that they are equal.

Sides AF and AG are equal by our choice of where we placed point G. Side AC is equal to side AB because they are the two equal sides from the isosceles triangle. The pairs of equal sides contain the common angle FAG. So, side CF must be equal to side BG by Proposition 4. Also, the triangle ACF is equal to the triangle ABG, and the angles ACF and ABG are equal, as are angles AFC and AGB.

We now turn our attention to triangles CBF and BCG.

Since AF is equal to AG and AB is equal to AC, the remainder BF is equal to the remainder CG. We proved above that the side CF is equal to the side BG.
So, the two sides BF, FC are equal to the two sides CG, GB respectively, and the angle BFC is equal to the angle CGB. The base BC is common to both triangles. Therefore the triangle CBF is equal to the triangle BCG, and the corresponding angles FCB and GBC are equal. The angles FBC and GCB are also equal to one another.

Now we have all of the tools needed to prove that angles ABC and ACB are equal.

We know that angle ABG is equal to angle ACF, and we know that angle CBG is equal to angle BCF. Therefore, the remaining angle ABC (cutting angle CBG out of angle ABG) and the remaining angle ACB (cutting angle BCF out of angle ACF) must be equal to one another.

We have also proved that the angles under the base will be equal to one another.

When we proved that triangle CBF was equal to the triangle BCG, we showed that the angles FBC and GCB are equal and those are the angles under the base.

Q.E.D.

Here’s a summary of the steps in this proof that angles ABC and ACB are equal.

Show that triangle ACF is equal to triangle ABG.
This gives us that angle ACF is equal to angle ABG.
Show that triangle CBF is equal to triangle BCG.
This gives us that angle BCF is equal to angle CBG.
(It also proves that the angles under the base – FBC and GCB – are equal.
Use the fact that angle ACB remains when angle BCF is cut from angle ACF and that angle ABC remains when angle CBG is cut from angle ABG.

If you want to step through the graphs I used in this proof, you can find them here on Desmos.com.

Proposition 6

If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

This proof will be interesting. We will assume that the two sides are not equal and reach a contradiction that proves our assumption must be false.

If you have any questions or feedback on the proof of Proposition 5, please leave a comment – George

Book 1 – Proposition 4

Proposition 4
If two triangles have the two sides equal to two sides, respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.

Suppose you have two triangles ABC and DEF with side AB equal to side DE and side AC equal to DF, and angle BAC equal to angle EDF, as shown below.

We are trying to prove the following four statements:

Side BC is equal to side EF
Triangle ABC is equal to triangle DEF
Angle ABC is equal to angle DEF
Angle ACB is equal to angle DFE

If we apply triangle ABC to triangle DEF by placing point A on point D and the straight line AB on the straight line DE, then point B will also coincide with point E because AB is equal to DE.

The straight line AC will also coincide with the straight line DF because AB coincides with DE and angle BAC is equal to angle EDF.

Thus, point C coincides with the point F because AC is equal to DF.

So, base BC coincides with the base EF and is equal to it because the point B coincides with the point E and the point C coincides with the point F.
(Common Notion 4: Things which coincide with one another are equal to one another.)

Thus the triangle ABC coincides with the triangle DEF, and the two triangles are equal by Common Notion 4.

Now we know that angle ABC coincides with angle DEF, so those two angles are equal. By similar reasoning angles ACB and DFE are equal as well.

Q.E.D.

(Note: Q.E.D. stands for the Latin “quod erat demonstradum” and means that which was to have been demonstrated. It is common to write Q.E.D. to signify the end of a proof.)

Proposition 5

In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.

The first part of the proof is to prove that the angles that are opposite of the equal sides in an isosceles triangle are equal to one another. This proof relies on Proposition 3 (If two line segments have different lengths, a segment of the longer segment can be cut off that is equal to the length of the shorter segment) and Proposition 4 (If two triangles have two equal sides and an equal angle between the two sides, then the two triangles are equal.)

Book 1 – Proposition 3

Proposition 3
Given two unequal straight lines, to cut off from the greater a straight line equal to the less.

(Note: You can click through the corresponding graphs on the Desmos website.)

We begin with two line segments. I have labeled the shorter line segment as C, and the endpoints of the longer line segment are A and B.

The goal is to cut off from AB a line that is the same length as C. Again, you might think we should just trace C on top of AB and cut off the overlap, but we must do this construction using the previous propositions, postulates, definitions, and common notions. Using Proposition 2, we can draw a line segment from point A to a point D in such a way that its length is equal to that of C.

Consider a circle centered at A whose radius is equal to the length of AD.

Since the length of AB is greater than the length of AD, the circle must intersect AB at some point. Let’s call it E.

The length of AE is equal to the length of AD, and the length of C is also equal to the length of AD.

Since things that are equal to the same thing are also equal to one another (common notion 1), we now know that the length of AE is equal to the length of C.

So, from the longer line segment (AB) we have cut off a line segment (AE) that has the same length as the shorter line segment (C).

Being what it was required to do.

One key idea that we have seen in each of the first three propositions is that the introduction of a circle can be used to show that two line segments are of equal length. Keep that in mind as we progress.

If you have two triangles ABC and DEF with side AB equal to side DE and side AC equal to DF, and angle A equal to angle D, can you prove that sides BC and EF are equal? Can you also prove that angle B is equal to angle E and that angle C is equal to angle F?

This proof is the first proof we will undertake, and depends primarily on common notion 4: Things that coincide with one another are equal to one another.

The goal of this proof is to show that all of the sides and all of the angles coincide and are therefore equal. See if you can do it before reading the next post. – George

Book 1 – Proposition 2

(Note: If you want to recreate these graphs as you make your way through the construction, check out this graph on Desmos.com. Click the folder icons for Steps 1 through 4 to make the graphs appear.)

Proposition 2

To place at a given point a straight line equal to a given straight line.

Here is a picture of a point A and a line segment BC.

We are being asked to draw a line segment starting at point A that has the same length as line segment BC. I drew line segment BC to be vertical, but that choice is arbitrary and will not affect the proof. You may be thinking “Why not just measure BC and then draw a line segment starting at A of the same length?” The concept of this construction is that it must be done only using the previous definitions, postulates, common notions, and propositions – without actually measuring.

According to Proposition 1, we know that we can construct an equilateral triangle (ABD) on the line segment AB.

At first glance that may not appear to help, but this is where you must really appreciate the abilities of these ancient mathematicians. Constructing the equilateral triangle will provide us with two points (B and D) that can be the centers of two circles that will allow us to relate the length BC to another line segment starting at A.

According to Postulate 2 we can extend the line segment DA beyond A to another point E and we can extend the line segment DB beyond B to another endpoint F, as shown.

We know that the length BC has to come into play at some point. We can draw a circle, centered at B, that has a radius (or, as Euclid would call it, a distance) equal to BC.

We label the point where the circle intersects line segment DF as G.
The segment DG can be thought of as the sum of DB and BG.

We know that the segment DA is the same as DB (they are two sides in an equilateral triangle), and if we can find a point L on the line segment DE such that the length DL is the same as the length DG, then will show that AL is the same as BG, which we have already shown as being the same as BC, and we will have the line segment we were looking for.

To make this work we will need a circle centered at D, and we will make the radius equal to DG.

The point where this circle intersects the line segment DE is labeled as L.
We know that DL is equal to DG, because L and G are both on the same circle that is centered as D. We also know that DA is the same as DB because those are two sides of an equilateral triangle.

Now we use Common Notion 3. If equals (DA and DB) be subtracted from equals (DL and DG respectively), the remainders (AL and BG) are equal.

Put another way, we had two line segments of equal length (DL and DG) and we cut equal pieces (DA and DB) off of those segments. The remaining segments (AL and BG) must also be equal to each other.

So, AL is equal to BG. But we also know that BG is equal to BC because they were on the same circle centered at B. By Common Notion 1 (Things which are equal to the same thing are also equal to each other), we therefore know that AL is equal to BC.

We have found a line segment beginning at point A that is equal to the given line segment BC.

Being what it was required to do.

As you make your way towards being a mathematician you begin with the final picture and see if you can prove that AL is equal to BC. Once you are able to do that, the next step for you is to see if you can determine the next step if the circle centered at D was not provided – in other words, can you reverse engineer this process. This will make you aware why the circle centered at D was important. You can then restart without either circle being provided, and see if you can justify those two circles, and so on.

This construction is nowhere near obvious, but understanding it will be useful in later proofs as well as in later courses.

Proposition 3
This is another construction.

Given two unequal straight lines, to cut off from the greater a straight line equal to the less.

If you are given two line segments of different lengths, can you cut the segment with greater length so that it is equal to the segment of shorter length? This could be within your reach.

Hint 1: Use Postulate 2 to draw a segment equal to the shorter segment that shares an endpoint with the longer segment.

Hint 2: Draw a circle.

Any questions or comments? Please leave feedback below. – George

Book 1 – Proposition 1

Proposition 1

On a given finite straight line to construct an equilateral triangle.

We begin with a line segment AB.

I have drawn a horizontal line segment, but you can draw the segment in any way you wish, with any length that you wish. We say that the selection of the line segment is arbitrary, and that means that we will not use any of the properties of this particular segment in the proof.

Now we will draw a circle centered at A, whose radius is equal to the length of the line segment AB. That means that the circle passes through point B. I have also labeled two other points on the circle: C and D. (Euclid says “With center A and distance AB let the circle BCD be described.“) Make note that the distance between point A and any other point on that circle is equal to the distance AB.

Now we will draw a second circle, centered at B, whose radius has the same length as AB. In addition to points A and C, I have also labeled point E on the circle.

Point C is the point where the two circles intersect. Now we can draw the triangle ABC.

The length of side AC is the same as the length of side AB, because points B and C are both on the circle whose center is A.

The length of side BC is the same as the length of side AB, because points A and C are both on the circle whose center is B.

So we have AC equal to AB and BC equal to AB. Since “things that are equal to the same thing are equal to each other“, therefore AC is also equal to BC. We have shown that the three lengths AB, AC, and BC are all equal to one another, and since points A, B, and C form the vertices of a triangle, triangle ABC is an equilateral triangle.

“Being what it was required to do.“

Blogger’s Note: I used the website Desmos.com to create the graphs used in this post.You can click your way through the graphs and experiment with them using this link.

Proposition 2

To place at a given point a straight line equal to a given straight line.

In other words, if you are given a particular line segment BC, can you draw a line segment of equal length that starts at a point A that is not on the line segment BC?

The proof makes use of Proposition 1 (proved above).

It also makes use of three postulates:

To draw a straight line from any point to any point. [We can draw a finite straight line connecting any two points.]
To produce a finite straight line continuously in a straight line. [We can extend a line segment.]
To describe a circle with any center and distance. [We can draw a circle provided that we know its center and its distance (or radius).]

It also makes use of two of Euclid’s common notions:

Things which are equal to the same thing are also equal to one another.
If equals be subtracted from equals, the remainders are equals.

Here is a picture you may start with.

If you have any feedback on the construction in Proposition 1, or if you have another way to do it, please leave a comment below. Thanks – George

Before We Begin … Definitions

The building blocks for classical mathematics are definitions and theorems (or propositions). The goal is to prove new theorems based solely on theorems that have already been proved and definitions. Examples and pictures can be used to gain an idea whether a statement is true, but are not sufficient when it comes to proving that statement.

Definitions – Book 1

The definitions for Book 1 can be found on pages 1 and 2 of the book by Green Lion’s Press (Euclid’s Elements). You can also find a free digital version here.

To work through Proposition 1 you should be familiar with the following definitions:

Finite straight line: A straight line is a line which lies evenly with the points on itself. A finite straight line can be thought of as a line segment.
Circle: A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another; and the point is called the center of the circle.
Equilateral triangle: Rectilinear figures are those which are contained by straight lines, trilateral figures being those contained by three. An equilateral triangle is that which has its three sides equal.

You should also be familiar with these two postulates. Let the following be postulated:

To draw a straight line from any point to any point. [We can draw a finite straight line connecting any two points.\
To describe a circle with any center and distance. [We can draw a circle provided that we know its center and its distance (or radius).]

Finally, there is a common notion that we will use: Things which are equal to the same thing are also equal to one another.

Proposition 1

On a given finite straight line to construct an equilateral triangle.

In other words, if you are given a line segment, can you explain how to construct an equilateral triangle from it? All of the tools you will need are listed above. See if you can develop a plan. A construction will be presented in the next post.

If you have questions, or would like to share your construction, please use the comments section. Thanks – George