OF PAPPUS OF ALEXANDRIA

LEMMAS ON THE FIRST BOOK OF APOLLONIUS'S CONICS.

WITH THE COMMENTARIES OF FEDERICO COMMANDINO OF URBINO.

FIRST LEMMA.

An ornamental drop cap "S" with floral and scrollwork background.Let there be a cone whose base is the circle a b and whose vertex is the point c. If, therefore, the cone is isosceles, it is manifest that all lines drawn from the point c to the circumference of the circle a b are equal to each other. But if it is scalene, it is necessary to find which is the greatest and which is the least.

A geometrical diagram shows a scalene cone with vertex c and circular base a b. A perpendicular line c d is dropped to the plane of the base. Points a, b, e, f, and g are marked on the base.

Let a perpendicular line be drawn from point c to the plane of the circle a b, which let us first assume falls within the circle; let it be c d. Let the center of the circle be taken, which is e, and let the joined line d e be produced in both directions to the points a and b; then let a c and c b be joined. I say that b c is the greatest and a c is the least of all the lines that belong to point c and the circle a b. For let another line c f be drawn, and let f d be joined. Therefore, b d is greater than d f, and c d is common, and the angles at d are right angles. Therefore, b c is greater than c f. In the same way, c f will be shown to be greater than c a. From this, it appears that the line c b is the greatest of all, and a c is the least. marginal note: 7th of the third [book], 2nd definition, 11th, 47th of the first [book].

Again, let the perpendicular line drawn from point c fall onto the circumference of the circle a b; let it be c a. And with the circle's center d, let a d be joined and produced to b, and let b c be joined. I say that b c is the greatest and a c is the least. It is clear that the line c b is greater than c a. However, let another line c e be drawn, and let a e be joined. Therefore, since a b is the diameter, it will necessarily be greater than a e, and a c contains a right angle with a b and a e. marginal note: 18th of the first [book]. Therefore, b c will be greater than c e, and similarly greater than all others. marginal note: 15th of the first [book], 2nd definition of the eleventh [book]. In the same way, e c will be shown to be greater than c a. Wherefore, it follows that b c is the greatest and a c is the least of all the lines that belong to point c and the circle a b.

With the same things posited, let the perpendicular c d fall outside the circle, and let d e be produced to the center e of the circle; let a c and b c be joined. I say that b c is the greatest and a c is the least of all that are drawn from point c to the circle a b. For it is constant that b c is greater than c a. But it will also be greater than all that fall from point c to the circumference of the circle a b. For let another line c f be drawn, and let d f be joined. Since therefore b d passes through the center, it is greater than d f. But c d is perpendicular to the lines d b and d f, since it is also perpendicular to the plane itself. marginal note: 8th of the third [book], 2nd definition of the eleventh [book]. Therefore b c will be greater than c f, and similarly greater than all others. It is clear, therefore, that c b is the greatest. But we shall show that a c is the least in this way. For since a d is smaller than d f, and d c is perpendicular to them, a c will be smaller than c f, and thus smaller than the others. marginal note: 8th of the third [book].