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.

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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:

1. To draw a straight line from any point to any point. [We can draw a finite straight line connecting any two points.]
2. To produce a finite straight line continuously in a straight line. [We can extend a line segment.]
3. 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:

1. Things which are equal to the same thing are also equal to one another.
2. 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