24 - 1 - 4 - 9 - 16 - 25

1-1 biangle

2-2 biangles - 1 biangle

3 - a - b - 7 biangles - 7 biangles

c - 15 biangles - 3 c d - 15 biangles

f - d - c - 26 biangles - 26 biangles

5 - 40 biangles - 40 biangles

The six equilateral biangles, in their meeting at one and the same point, do not cut one another, being made from the circumference of one and the same circle, or of similar circles. But the circumference could be made of so great a circle, or divided into parts so minute, that the sides of the portions would have angles so acute that more than a hundred would meet at one and the same point without cutting one another. The hexagonal surface is composed of biangular figures, or, if you will, portions of a circle, because every equilateral biangle can be divided into two other biangles of unequal sides, called portions of a circle (three figures a, b, c follow), as is shown here in a, b, c: where a is the equilateral biangle, b is the biangle divided into two other biangles of different sides, that is one side straight and the other curved, and c is the third biangle, called a portion of a circle.

Second - f d e - i h - g. Here follows the equilateral triangle e f g, to which is added the portion e f d on the upper side, and below there is as much value in the four portions of the circle h i g. And this quantity taken together is called a sector of a circle, six of which, similar and equal, joined together make a circle divided into six sectors, as is shown here following.

First: c b a - e d - f

First. In this first figure it is shown that the two unequal biangles d f, e f are worth one equilateral biangle equal in value to the greatest portion a b c. And six of these are worth the following figure, which, if you remove from it the six portions, remains square; and the same will happen if you remove the six middle biangles, leaving the outer ones in place.

The surface of six equal sides composes of itself the hexagonal figure. Always multiply every even number of portions by six, and you will find it good when six is contained whole within, that is when 6 is an aliquot part of it, that is a multiplicative part without fractions.

f - b a - d c - n - e. Six of these sectors joined together within one and the same circle, in such a way that each is a sixth part of it, will make a surface of twenty-four biangles, and the four portions c d e are worth the greatest portion a b f. Now if you remove the greatest portion a b f from the equilateral triangle a b e, there will remain the rectilinear triangle a b e. And if you remove the 4 lesser portions, the remainder will be squarable, that is it will remain equal to the greatest rectilinear triangle. It is proved by the common notion, that is, if from equal things you take equal parts, the remainder stays equal. Therefore, by proving the greatest portion to be equal to the 4 lesser portions, we verify our promise. And what is said of the sixth part of the circle, I intend to have said of the whole circle ... and of all circles.