b a; c; d; f–h e; g–i. Here are two portions or semicircles, of which the greater is double the lesser. It follows that, the lesser being taken from the greater, the remainder of that greater is equal to the whole lesser; therefore the crescent c is equal to the portion d. Now I will lend to c and to d two rectilinear surfaces that will be equal to each other; therefore their sums will likewise become equal. I will do it thus: I will cut the middle of the greater arc of the crescent c and open it out, as is shown below in e f, and I will lend it the triangle h, and I will have made the surface e f h. And then to g, which is like d, I will lend another triangle like h, and it will be i; and doing so I will have made two equal surfaces, that is e f and g i. Then I will subtract g from h and the remainder of e f h will be equal to i. But because h cannot receive g, I will take half of the crescent b and half of the semicircle a and...

b–c a; a–b a c e–i b g f. But it is better to follow the manner set out below, taking half of the outer crescent and the whole of the inner part, which is equivalent to the outer arc; and then I will open it out, as is shown in the last figure at the foot, and I will give it the triangle c a, and the like I will do to the portion a, giving it the triangle b. Then I will subtract the portion a from both surfaces and there will remain a triangle equal to the remainder of the second surface c d e f g h i. Then I will subtract e, a portion double the two portions below it at f g; there will remain d, a greater portion equal to the two lesser portions h i, and in this way I will have removed e f g, a squarable part, which, once squared, I will subtract from triangle b, and the remainder d e h i opposite will remain squared; I will remove them and square them as I did with f g, and this second squaring I will subtract from b, and the remainder of that b will be equal to the triangle c, of two straight sides and one curved. Therefore, there remaining in b a part equal to c, I will have found and squared one of the four corners of the square, since from it is taken the greatest circle of which such a square is capable.

a–n; b–o–p; c. a b are equal, c is half of each; n does the like, and so o p.

b a. I will take the arc a from the arc of b, and the remainder of a will be equal to the remainder of b, and that which is of the upper arc of b will be equal to its lower arc, so that the lower is capable of the upper. The following texts refer to the figure in the lower corner; with: By lending — k i h — e d q — p n — r b c o — S. I will square r p and r K, and this squaring I will subtract from the parallelogram c b, and the remainder of that parallelogram will be equal to the remainder of the lune, that is p k h n o. Therefore the squaring of that remainder of the lune is known, being equal to the remainder of the parallelogram o p r s. Again I will remove an equal amount of arc below and above, being k q above and p o below, and there will remain to me the convex arc q n, equal to the concave arc n o, of which the first, q n, will fill the second concave arc n o. Behold how the law of nature, in cases of quadrature, has done its due by filling the hollows with the reliefs of the arcs. And it will always be found that every curvilinear figure for which a square equal to it can be given can also be squared in itself by means of its parts, by taking and returning parts of its arc. And those curvilinear surfaces to which a square equal to them cannot be given cannot be squared by themselves by means of their parts either.

By lending — k i h — e d q — p n — r b c o — S. I will square r p and r K, and this squaring I will subtract from the parallelogram c b, and the remainder of that parallelogram will be equal to the remainder of the lune, that is p k h n o. Therefore the squaring of that remainder of the lune is known, being equal to the remainder of the parallelogram o p r s. Again I will remove an equal amount of arc below and above, being k q above and p o below, and there will remain to me the convex arc q n, equal to the concave arc n o, of which the first, q n, will fill the second concave arc n o. Behold how the law of nature, in cases of quadrature, has done its due by filling the hollows with the reliefs of the arcs. And it will always be found that every curvilinear figure for which a square equal to it can be given can also be squared in itself by means of its parts, by taking and returning parts of its arc. And those curvilinear surfaces to which a square equal to them cannot be given cannot be squared by themselves by means of their parts either.

Here the whole lune is equal to the whole parallelogram below. I will remove from the lune the part o r, which is worth the two portions o p and r s, and thus, removing these two portions o p and r s, there remains to me the rectilinear parallelogram a b c t equal to the lune less n o. * I will remove from the lune the wholly curvilinear parallelogram, and take the portion of its front and put it back into the concavity of the opposite front, and I will have... * But because the biangular figure of two lines of equal curvature is worth the two portions o p and r s, I will leave that biangular figure in its place and remove an equal value with two portions h i and i k, and there will remain a b c t rectilinear, equal to the remainder of that lune. Therefore I will remove only one portion of the lune at n o, and the like at o p, and the parallelogram will remain equal to the lune. Then, by superimposing, I will take away the portion of the rectilinear front that remained under the portion n o, and there will remain the wholly rectilinear parallelogram equal to all the remainder of the lune. Then I will remove two equal arcs from the lune at n h and at o p, or else from the opposite side.

Elements of curvilinear geometry. Workings-out of various curvilinear figures.

First. To take from one side and return the same on the opposite side.

b–c a. a b is worth a c. Therefore I will remove c and return b to the said a. To take from one side and return the value on another side, but let it not be of the same figure as the one that was taken.

I take the two portions from the lune and give it one portion equal to those two. To take on several sides of a curvilinear surface and return the value on a single side. To lend two equal and similar parts to two equal and similar surfaces. To lend two equal and unlike parts to two equal and unlike surfaces. To lend two equal and similar surfaces to two equal and unlike surfaces. To lend two equal and unlike parts to two equal and similar surfaces. To make one same curvilinear surface into various figures. To give portions of circles in whatever proportion, one to another, that may be asked.

To give whole portions equal to whatever part may be asked of another portion.

b; a. To give crescents equal to other crescents in various figures. The crescent a is equal to half the crescent b.

To divide a crescent into equal parts.

c b–d a. The parallelogram a b c d is a third of the semicircle and a sixth of the whole circle. And thus, always throwing away four portions, you will do the like in every problem, or say four sectors. To give a known part of a circle or semicircle between two parallel lines. To vary to infinity the equality of two surfaces.

o c b a; f n. To give a crescent equal to a portion, or if you will, a sector.

r–t. To make one same value from two or more crescents of different arcs.

a–r–b–o n m–s t. To vary infinitely the angle of a sector without any loss of its original value.