b, n, d, c a o e, f

I wish with certainty to divide every kind of angle into 2 equal parts.

c d, e f, a b, 2 4

Let the angle that is to be divided into 2 equal parts be the angle a d b e. Of which I have to consider and find the curvatures of its sides, and from those find the circles and proportion them together by the given rules. And because I find the curvature a b d to be of a circle which stands to the circle of the second curvature b n e as 2 stands to 8, that is, quadruple, now I have to seek a circle that is the mean proportional between 2 and 8, which is 4, which is double to 2 just as 8 is to 4.

I have a circle of 2 and a circle of 8 and I want to know the geometric mean proportional between them. And I will say: "2 times 8 makes 16," and the root of 16 is 4; therefore 4 will be the sought answer.

Let the greater circle be a b. And the smaller one, subquadruple to it, be c d. Then the mean proportional will be e f, which will be as much greater than c d as it is smaller than a b. And likewise in this pyramid you will find the mean proportional between every variety of extremes, so triples as doubles and quadruples, and so on to infinity.

h o r, f, i, e, a b c d

If you want the mean proportional between 2 numbers, multiply the first by the last, and the root of the resulting sum will be that mean.

The right-angled triangle a h i is right-angled, and the supplement of the falcata r will always be equal to the half-portion a h d.

4 6, a b, a is equal to b, 6 2 1/4 4, 24

Such proportion will be from circle to circle as that from diameter to diameter multiplied by itself. Whence you will say for the smaller circle: "4 times 4 makes 16." And for the greater say: "6 times 6 makes 36." Now divide 36 by 16, and there comes 2 and 1/4, that is, a double-sesquiquarter proportion.