Four Books on Conics (Commandino 1566)

the circumference e f; and d e is common and at right angles, the base d f will be smaller than d a. By the same reasoning, d a is also smaller than d b. Since d e has been shown to be smaller than d f; and likewise d f smaller than d a, and d a smaller than d b; it follows that d e is the minimum and d b the maximum; and that which is closer to d e is always smaller than that which is farther from it. But let the perpendicular d e fall outside the circle a b c g f, as in the second figure; again let the center of the circle k be taken; and having joined e h k, let it be produced to b; and let d b and d h be joined. Moreover, let two equal circumferences be taken on either side of point h, which are f h and h g; and on either side of b, let two others be taken, a b and b c. Finally, let e f, e g, f k, k g, d f, d g, a b, b c, a k, k c, d a, and d c be joined. Thus, since the circumference b f is equal to h g, the angle h k f will be equal to the angle h k g. But the straight line f k is equal to k g, for they are drawn from the center to the circumference; and k e is common. Therefore, the base f e is equal to the base g e. Furthermore, d e is common and at right angles. Therefore, the base d f is equal to the base d g. Again, since the circumference b a is equal to the circumference b c, and the angle a k b is equal to c k b, the remaining part of the two right angles a k e will be equal to the remaining part c k e. Also, the lines a k and k c are equal to each other, for they are from the center; and k e is common. Therefore, the base a e is equal to the base c e. Again, since d e is common and at right angles, the base d a will be equal to the base d c. Similarly, all others that are equally distant from d b or d h will be demonstrated to be equal to each other. And since e h is smaller than e f, and e d is common and at right angles, the base d h will be smaller than the base d f. Again, since the line drawn from point e touching the circle is greater than all those falling from the same point onto the convex circumference, and the rectangle a e l is equal to the square of e f when e f touches the circle, as shown in the third book of the Elements, it will be that as a e is to e f, so is e f to e l. But e f is greater than e l, for that which is closer to the minimum is always smaller than that which is further away. Therefore, a e is also greater than e f. But e d is common and at right angles. Therefore, the base d f is smaller than the base d a. Again, since a k is equal to k b, and k e is common, the two lines a k and k e will be equal to the two e k and k b, that is, the whole e b. But the two a k and k e are greater than e a. Therefore, b e is also greater than a e. Moreover, d e is common and at right angles. Wherefore, the base d a is smaller than the base d b. Thus, since d h is smaller than d f, and d f is smaller than d a, and d a is smaller than d b, d h will be the minimum and d b the maximum; and that closer to d h will always be smaller than that which is further away.

27. of the third.

8. of the first

33. of the first.

Geometrical diagram showing a circle with center k and various lines (d, a, b, c, f, g, h, e) extending from a point d outside the circle to points on and within the circle. A decorative floral sprig is placed to the left of the diagram.
8. of the third

36. of the third

14. of the sixth

8. of the third

14. of the fifth

Geometrical diagram similar to the one above, showing a circle in perspective with lines radiating from an external point d. A decorative scroll ornament is placed to the left.
20. of the first

Finally, let the perpendicular d e fall within the circle a b c g f, as in the third figure; having taken the center of the circle k, let the joined e k be produced in both directions to the points b h. And let d h and d b be joined. Moreover, let equal circumferences f h and h g be taken on either side of point h; and on either side of b, let a b and b c be taken; finally, let e f, e g, f k, k g, d f, d g, k a, k c, e a, e c, d a, d c, a b, and b c be joined. Since, therefore, the circumference h f is equal to the circumference h g, and the angle h k f is equal to the angle h k g; indeed the line k f is equal to k g; and k e is common. Therefore, the base f e will be equal to the base g e. But d e is common, and the right angle f e d is equal to the right angle g e d. Therefore, the base d f is equal to the base d g. Again, since the circumference a b is equal to the circumference b c, the angle a k b will be equal to the angle c k b. Therefore, the remaining part of the two right angles a k e is equal to the remaining part c k e. Also, the line a k is equal to k c and k e is common. Therefore, the base a e is equal to the base c e. But since d e is common and the angle a e d is equal to the angle c e d, because both are right angles, the base d a will be equal to the base d c.

27. of the third

4. of the first