The Elements

...again, I will make the other extremity, namely point B, the center; and by the same postulate original: "petitionem," referring to Postulate 3: to describe a circle with any center and distance and according to the same quantity, I will draw the circle C.A.D.B. These circles will intersect each other at two points, which let be C and D. I will join one of the two intersections, such as intersection D, with both extremities of the given line by drawing lines D.A. and D.B. according to the first postulate. Since lines A.D. and A.B. are drawn from point A (which is the center of circle C.B.D.) to its circumference, they will be equal by the definition of a circle. Similarly, since lines B.A. and B.D. are drawn from point B (which is the center of circle C.A.D.) to its circumference, they will also be equal. Because each of the two lines A.D. and B.D. is equal to line A.B., as has been proven, they will be equal to each other by the first common notion Common Notion 1: Things which are equal to the same thing are also equal to one another.. Therefore, we have constructed an equilateral triangle equilateral: a triangle where all three sides are the same length upon the given line, which is what was proposed.

¶ If, however, one wishes to construct the remaining two species of triangles upon the same line—namely, the triangle of two equal sides an isosceles triangle and the triangle of three unequal sides a scalene triangle—let line A.B. be extended in both directions until it meets the circumferences of both circles at two points F and B The text likely refers to points created by the extension. With point A as the center of the line, circle E.B.G. is drawn according to the quantity of line A.B.; likewise, with point B as the center, circle E.F.G. is drawn according to the quantity of line B.F. These circles will intersect each other at two points, E and G. Therefore, let the extremities of the given line be joined with one of these intersections by two straight lines, A.G. and B.G. Because these lines A.B. and A.F. proceed from the center of circle C.D.F. to its circumference, they will be equal. Likewise, A.B. and B.B. likely a typo for another point on the circumference, because they proceed from the center of circle C.A.D.B. to its circumference, they will be equal. Since each of the two lines A.F. and B.B. is equal to line A.B., they will be equal to each other. Therefore, with A.B. being common, B.F. will be equal to A.B. But B.F. is equal to B.G. because both proceed from the center of circle E.F.G. to its circumference. Similarly, A.B. is equal to A.G., and each of them is greater than A.B. because each of the two lines B.F. and A.B. is greater than A.B. The text here is describing the construction of an isosceles triangle where two sides are longer than the base. Thus, we have constructed a triangle of two equal sides upon the given line.

¶ We shall also construct a triangle of three unequal sides upon the same line: if we take some point existing on the circumference of either of the two larger circles that is not in either of the two intersections, and which is not met by F.B. when extended continuously and directly in either direction, and join it by two straight lines with both extremities of the given line. Let point K be marked on the circumference of circle E.F.G., and let it not be in either intersection, nor let F.B. meet it when extended continuously to its circumference. I will therefore draw lines A.K. and B.K. Line A.K. will cut the circumference of circle E.B.G.; let it cut it at point L. B.K. will be equal to A.L. because B.K. is equal to B.G. and A.L. is equal to A.G. Therefore, A.K. is greater than B.K., but B.K. is also greater than A.B. Thus, triangle A.B.K. is one of three unequal sides. So, in this way, we have constructed all species of triangles upon the given line.

A geometric diagram shows the construction for three types of triangles. Two large circles overlap with centers 'a' and 'b'. A horizontal line passes through points 'f', 'a', and 'b'. Various lines connect these centers to points on the circumferences, including 'c', 'd', 'e', 'g', 'k', and 'l', demonstrating equilateral, isosceles, and scalene triangle constructions on the base segment 'ab'.

Proposition 2

A large ornate woodcut initial 'A' is decorated with white scrolling floral and vine patterns on a dark, dotted background. This marks the beginning of the second proof.

A Let A be the given point and B.C. the given line. I wish to draw from point A a single line equal to line B.C. in whichever direction it may fall. I will therefore join point A with whichever extremity of line B.C. I wish. I will join A with extremity C by the line A.C., upon which I will construct an equilateral triangle according to the teaching of the preceding [proposition], which shall be A.C.D. In that extremity of the given line with which I joined the given point—namely, in extremity C—I will place the stationary foot of the compass and describe upon it a circle according to the quantity of the given line itself, which shall be circle E.B....

A geometric diagram for Proposition 2 shows a circle with center 'c' and radius 'cb'. A point 'a' is connected to 'c', and the line is extended towards 'f'. This represents the initial steps for transferring the length of line 'bc' to start from point 'a'.

The mark "a 3" at the bottom of the page is a "signature," used by early printers to help binders assemble the pages of a book in the correct order.