If from the square a squarable curvature is removed, the remainder will be squarable. And if from the square a non-squarable curvature is removed, the remainder will again be non-squarable.

It is proved. Let there be removed from the square a squarable curvilinear surface, and let the square that results from it be taken out of that whole square; without doubt the remainder of that square will be squarable, because a rectilinear surface will remain.

Let a b be a rectilinear square, from which is taken the curvilinear square a, which is squarable. There will remain the square b, as is shown in c by drawing out d. But if the part that is separated is by this science non-squarable, as is the surface of a single curved side, you will not be able to take the lesser square from the greater; and consequently that greater square, not being able to be squared, will remain non-squarable.

If from a squarable curved figure you remove a squarable curvature, and let it be squared, the remainder will be squarable.

If to a squarable curved figure you add a squarable quantity, the sum will again be squarable.

First - a b c d.

The curvilinear triangles a b c d, being squarable, leave the remainder of the square squarable.

Second - f - b - c d - a n - e n - p o.

The concave triangle f equals the concave triangle c d. It is proved to be so, because the triangle c d is composed of two triangles similar to the greater one and similar to each other; but each of them is half the greater, and their concavities are of half-size circles. Furthermore, to add the greater portion b to the triangle f is as much as to add it to the triangle c d, since c d is equal in value to f.

a equals b; and because the curvilinear triangle b c d is composed of the two triangles c d and of the portion b, likewise the square d a e is composed of two similar triangles d e, equal to the c d triangles, and the a equal to the b. It follows that the curvilinear triangle b c d equals the square d a e. It follows that the whole greater square, which is set within the circle, equals the four curvilinear triangles made in the same circle.

Again: the triangle c d a equals the triangle c d b, and equals the square d a e, and equals the rectilinear triangle n m o p.

First - o n; Second - p q.

Again, if you look at the first and second figures set in the margin, you will see the first to be composed of 4 double portions, like n, and the second to be composed of four single portions of value equal to those double ones; because if the 2 are double in number, this one is double in quantity, of which p n is one. Again, if the first is finished with the 4 concave triangles, as is shown in o, the second is finished with 4 other triangles of similar value, of which q is one. And so it is concluded that the first and the second figure are equal to each other in quantity. And the one and the other are equal to the square marked in the second figure, and equal to the square marked inside the first.

Having proved the 4 portions of circle a to be of value similar to the 4 b, which, being taken from b and placed in a, fill the inner square in continuous quantity and make the inner square equal to the remainder of the outer square, restoring to the outside the portion a.

Squaring.

The first figure is squared by transmuting it into the second, joining the value of n with triangle o. And so it is shown:

o b m; a b c d; e f g - L h - m K i

First - o n m; Second - p c q.

These two surfaces are of equal quantity, being composed of equal parts; and each of them is squarable by itself and squarable with the help of the other; and the squaring is the square marked in the second. The second fills c with p on 4 sides and leaves a on 4 other sides where they are made; and the first fills m with o in 4 places and leaves n.

o r.

The first one above is squared by removing the triangles placed in the corners of the squares, like o, and its value is rendered similar and placed in r.

Lorenzo is to give 4 giuli. Hay was bought with it. For Christmas.