math – uCLID

Book 1 – Proposition 39

Equal triangles which are on the same base and on the same side are also in the same parallels.
[Desmos graphs can be found here]

Let triangles ABC and DBC have the same base BC and both be on the same side of it.

Join AD. We will prove that AD is parallel to BC.

If AD is not parallel to BC, let A be joined to a point E on BD such that AE is parallel to BC. (Proposition 31 covers drawing a line through a given point that is parallel to a given line.) Join EC.

By Proposition 37 (triangles on the same base and in the same parallels are equal) triangle ABC is equal to triangle EBC as they share the same base (BC) and are in the same parallels.

Since triangle ABC is equal to triangle DBC, then triangle EBC is also equal to triangle DBC which is impossible.
Therefore AE cannot be parallel to BC.
The same is true for any other straight line except AD.
Therefore AD is parallel to BC.

Q.E.D.

In the next proposition we will extend “the same base” to “equal bases” – George

Book 1 – Proposition 30

Straight lines parallel to the same straight line are also parallel to one another.
[Desmos graphs can be found here]

If the two lines AB and CD are both parallel to another line EF, we will prove that AB and CD are parallel to each other.

Let the straight line GK fall upon the three lines as shown.

By Proposition 29 (alternate angles are equal), since AB and EF are parallel angle AGK is equal to the angle GHF.

Since EF is parallel to CD, Proposition 29 tells us that angle GHF is equal to angle GKD. (Exterior angle is equal to the interior and opposite angle.)

Therefore angle AGK is equal to angle GKD. (AGK = GHF = GKD)

Since angles AGK and GKD are alternate, Proposition 27 tells us that the lines AB and CD are parallel to one another. (If a straight line falls on two straight lines and the alternate angles are equal, then the two straight lines are parallel.)

Q.E.D.

The next proposition involves the construction of a line through a given point that is parallel to a given line – George

Book 1 – Proposition 28

If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

Let the straight line EF fall on the two straight lines AB and CD, intersecting AB at a point G and intersecting CD at a point H, is such a way that exterior angle EGB is equal to the interior and opposite angle GHD or the interior angles on the same side (BGH and GHD) are equal to two right angles.

Starting with angle EGB being equal to angle GHD we will prove that AB is parallel to CD.

Angle EGB is equal to angle GHD by hypothesis.
Angle EGB is also equal to angle AGH by Proposition 15 (vertical angles are equal)

So, angle GHD is equal to angle AGH. Since these two angles are alternate angles, AB and CD are parallel by Proposition 27. (If a line falling on two straight lines produces equal alternate angles then the two lines are parallel.)

Now we will start with angles BGH and GHD being equal to two right angles, and will once again prove that AB and CD are parallel.

Angles AGH and BGH are also equal to two right angles by Proposition 13. (A straight line set up on a straight line will make angles equal to two right angles.)

So, the angles AGH and BGH together are equal to the angles BGH and GHD together. Subtracting angle BGH from each, the remaining angle AGH is equal to the remaining angle GHD.

Since angles AGH and GHD are alternate angles, Proposition 27 again tells us that AB and CD are parallel.

Q.E.D.

Book 1 – Proposition 16

In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior or opposite angles.
[Desmos graphs can be found here]

Start with triangle ABC, and let one side of BC be produced to point D.

We will prove that the exterior angle ACD is greater than either of the interior and opposite angles BAC and CBA.

By Proposition 10 (that allows us to bisect a line segment) let AC be bisected at E.
Let BE be joined and produced to a point F so that EF is equal to BE. (Proposition 3 allows us to construct a line segment equal to a given segment.)
Let FC be joined.

Finally, let the side AC be drawn through to a point G as shown.

We will now examine the two triangles ABE and CFE.

Sides AE and EC are equal, since E bisects AC.
Sides BE and EF are equal according to the way we selected point F.
Angles AEB and CEF are equal to each other because they are vertical angles.
(Proposition 15 states that vertical angles are equal.)

So, by Proposition 4 (“side-angle-side”), the side AB is equal to the side CF and the remaining corresponding angles are equal. In particular, angle BAE is equal to angle ECF.

Angle ECD is greater than angle ECF, and that implies that angle ACD is greater than angle BAE. (Angle ACD is equal to angle ECD, and angle BAE is equal to angle ECF.)

In a similar fashion, we can prove that angle ACD is greater than angle ABC.
(Bisect BC … angle BCG equals angle ACD (vertical angles) … construct triangles and show that angle BCG is greater than angle ABC.)

Q.E.D.

Later we will see that the three interior angles ABC, BCA, and ACB are equal to two right angles.
The adjacent angles ACB and ACD are also equal to two right angles.
That shows that the two angles ABC and BCA are equal to the angle ACD, and angle ACD must be greater than either of those two angles.

If you have feedback, I’d love to hear it – George

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.

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.

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