** To prove in class:**

On a given finite straight line construct an equilateral triangle.

Definitions:

A point is that of which there is no part.

A line is a widthless length.

A straight line is one which lies evenly with the points on itself.

A circle is a plane figure contained by one line [which is called the circumference], and all the straight lines coming from one point of those lying within the figure and falling [upon the circumference of the circle] are equal to one another.

An equilateral triangle is that which has its three sides equal.

Postulates:

From any point to any point a straight line may be drawn.

With any center and any radius a circle may be drawn.

A Common Notion:

Things equal to the same thing are also equal to one another.

** To discuss:**

Is this a proof?

How do we know that the perimeters meet each other?

Killing's postulate: "If a line [in this case the circumference of one of the triangles] belongs entirely to a figure [i.e. the same plane] which is divided into two parts [the part inside the circumference of the other circle and the part outside] and if the line has at least one point common with each part, it must also meet the boundary between the parts."

What about the triangle on the other side?

Which "givens" were useful? Which were not?

Note: no numbers.

Note: no notion of equal lengths: saved for propositions 2 and 3.

Euclid's words in my translation:

Let there be the given, straight, bounded line, AB. It is necessary, then, upon the straight line, AB, to construct an equilateral triangle. With center A, then, and with radius AB, let a circle be drawn, the circle BCD. And again, with center B, then, and radius BA, let a circle be drawn, the circle ACE. And from point C, through which the circles cut one another, to the points A, B let there be joined straight lines, the lines CA, CB. And since the point A is the center of the circle CDB, the line AC is equal to the line AB. Again, since the point B is the center of Circle CAE, the line BC is equal to BA. It has been shown that CA is also equal to AB. For indeed, both of the lines CA and CB are equal to the line AB. Things, then, that are equal to the same thing are also equal to one another. And the line CA is surely equal to the line CB. The three lines, then, CA, AB, BC are equal to one another. Equilateral, then, is the triangle ABC. And it has been constructed upon the given, straight, bounded line, AB. [Upon the given, straight, bounded line, then, an equilateral triangle has been constructed.] The very thing it was necessary to make.

** Assignment: Complete Euclid's Second Theorem.**

Upon a given point place a straight line equal to a given straight line

Let there be the given point A, and then the given straight line BC. It is necessary, then, upon the point A to place a straight line equal to the straight line BC.

Then let there be joined from the point A to the point B the straight line AB, and let there be constructed upon it the equilateral triangle DAB, and at the straight lines DA and DB, let the straight lines AE and BF be extended; and with the center B, and then with the distance BC, let there be drawn the circle CGH, and again with the center D and the distance DG, let a circle be drawn, GKL.

Use this additional "common notion" in your proof:"If from equal things equal things should be taken away, the remaining things are equal."