First - t r - f n - S; Second o - b a - p -; Third - g - h. Let a circle be made of the second figure a b, from which are missing the 8 hatched bi-angles that represent the 4 greatest [portions], and to that round let there be added, or else let there be missing, the two quadrangles o p; and then let the quadrangles o p be added around that circle, so that the circle is left squared. So go to the first figure and make the circle S equal to the circle o of the second figure, and clothe this with the addition that squares it, which will be similar to the quadrangle o. Once you have done this, double that n f by the rule given for circles which requires that the square outside the circle be double the square inside the circle; and by this the second square r t is worth twice the square n f. Therefore such a double square receives within itself a circle double the circle S, and consequently such a circle is worth the two circles of the second figure; which two circles, without the 2 quadrangles, are worth the two lunes that clothe those circles. And so the 2 quadrangles joined together are worth the 4 angles of the greatest square of the first figure, which are placed in the third figure at g, where the angle g is similar to the three others of its acute quadrangle, and the same angle g is similar to the other angles that surround the circle h. Behold, then, that we have reduced the figure a b o p into the square and circle h, and thus our purpose is concluded, and this same square is worth the greatest square of the greatest circle of the second figure. The larger circle of the first figure ought not to be there. So make it below and remove the third figure.

First - t - v e o r - v i p-S h - m n - g. This is the first and the third above, and is made to reduce the two empty circles into a single circle, and likewise to reduce the a b lunes into a circle worth the two circles written above; and besides this, to spread the 4 angular supplements o e i v around the circle p i n m. The acute quadrangle r t v S is proved to be worth the angles that surround the circle p i n m, of which i is common to the acute quadrangle and is common to the angles i p n m.

Sixth. This sixth lacks the 4 greatest [portions] and lacks 4 lunules. Therefore it is worth half the greatest square of the greatest circle.

b a. This is worth the second above and worth the first placed here opposite, but it was born of a circle quadruple to this one opposite, and the angle b is worth the 4 angles o e v i, and it is squared as shown below, that is, by removing the greatest portion a b and giving back its value with c d.

a c - e - d - b. This is worth the second placed at the front of this page.

Fourth - e a - c - b d. c is worth b; and if you remove c d and give back d b, b is double. So halve it and give the value back into c, which had been made empty by the separation of the c d portion.

To give the quadrature of the lune a b, I lend it the portion c d and make the surface a b c d, from which I remove the lent c d with the same value, taking away from it d b, and there remains a c rectilinear equal to the curvilinear a b. It is proved thus: I set the greatest portion a e c b d as double the least portion c d, and by this it follows that the half of the greatest portion a e c b d, that is d b, is worth the c d portion which is sub-double to that greatest portion. Then I remove away the e portion and there remains a b, which in itself is squarable and is joined in company with the portion c d. Now I will remove c d from a b and give it back as much as I took with the half greatest portion d b, and so I will have the rectilinear triangle a e equal to the curvilinear surface a b. Brevity. The lune a b is worth the triangle a c. It is proved, because a c b d is a compound of two unequal surfaces, since c d is greater than a b, and again there is another compound equal and similar in proportion to the aforesaid, that is a c d b. So, being thus, it follows that c d, a whole portion of the sub-double circle, is worth as much as half the portion d b of the double circle; it follows that the triangle a c is worth the lune a b. On the equality of the parts. c is worth b. It is proved by saying: if two equal surfaces are partly superposed one on the other, etcetera. It follows that, c d being a sub-double portion equal to d b the half portion, and their common contact being the part d, it is necessary that c b be equal to one another.

a - b - c. This portion is divisible to infinity into other similar portions and lunes, as is shown above in the fifth, where the portion o is worth the lune d, because the parts o n p are worth o n d and their contact is equal and equal is that which is not touched. It is true that here, in the diminution of the parts, the sagittae of the arcs, in the diminutions of the portions, always diminish in proportion to the antecedent.

Fifth - a - m - S b o - c c n - d - p. Here is given the quadrature of the lune a b c d with the rectilinear surface a b c o. It is proved, because the half of the greatest portion o n p m b S c d is divided into two equals, that is the lune m b S c d and the portion o n p, and it is divided in half in another way, that is m o b S c and n d p. Now I will remove n d p, which is worth the portion o n p, and there will remain the lune a b c o squared, joined in company with n d p, as it was with o n p, which made it curved and now is made rectilinear; and this cannot be squared by any other way than this, and the same happens with the right-hand quadrature.

The arcs made on a straight line are divisible to infinity into portions and lunes in equal number; and every portion divides into 4 equals, that is two lunes and 2 portions.

This division does not generate lunes, but always generates portions, and diminishes the length of the sagittae more than their chords and their arcs.

Second - m n r - e - f d b - o g a - h - c - S. The line n o can be moved from the top as much as one wishes, keeping it fixed below at o, and thus the d g can be widened as much as one wishes. To square the lune d g h I join it with the portion b a c and form the surface a b d g h c; then I remove the portion b a c and give back as much to the lune, giving it the surface a h c. But because h is double, that is the lune that was doubled in laying upon it the surface a c h, you halve it and give the value back to the empty b, which then remains the square b d g.

Conceptions. First. - Of surfaces of double proportion, the half of the greater is worth all the lesser. Second. - Of surfaces of double proportion entirely superposed one on the other, the excess of the greater is worth all the lesser. By the first, the portion b a c is worth the lune e d f g h, because all this surface is double the portion b a c, and consequently the lune remaining from the whole is worth that portion. All this surface is divisible into 4 equal parts in two ways, of which one is the portion a b c which is worth the lune e d f g h, the second way is b d e f g, the other is a b c. It is proved, because each of them is composed of equal parts, that is of half the triangle r m S, divided in half by the line r h, and is composed of half the portion m o S, divided in half at o g.

First - d a - c b. This half portion can be the half of the narrowest portions that I can make of the circle. The portion a b, being the half of a d b c, is worth the lune d c, and again a d is worth a b and is worth b c. But our purpose is to prove a to be of the value of c; and it is proved by the ninth [proposition] of this, which says: if there be two equal surfaces from which equal parts are removed, the remainder will be equal. Therefore I will remove b from a b and there will remain a; and I will remove b from b c and there will remain c. Therefore it is concluded that the portion a is worth the lune c, and ...

Third - m n - f c b - g h a - r d c. Here the lune e h d is to be squared, which joins with the semicircle b a c and makes the compound b a c e h d. Now I will disjoin the semicircle from the lune, and give to that lune as much value as the semicircle, and it will be the eighth part of a circle eight times the circle of which the semicircle b a c is the half. And this eighth of a circle will be a c d. Now we have made a second compound of the sector of the eighth part of a circle a c d with the semicircle b a c, of which the excesses that exceed their common contact a c are b d, equal in value. But because in such transmutation of the compound, b has remained empty and the lune d has been doubled, we will halve that lune in the mind, imagining the portion b to have its deficiency restored to it, and so we will have the rectilinear triangle b e h equal to the lune e h d; and such a triangle can be increased and diminished by the motion of the upper end of the line n r, and so the triangle will increase and diminish together with the lune e h d.

Fourth - d a f - e - b - c g. Quadrature of the imperfect lune a b c. The deficiency of its perfection is the half portion d e, and to square it I first remove d e and reintegrate it by giving it the f g portions that are worth d e; and so f a b c g are worth a d e b c. Then I remove away g c which is worth b e, and there remains to me the rectilinear triangle f a b which is worth the imperfect lune a b c. And to make a shorter quadrature I will remove from f a b c the lune c and place it in f, because e is worth as much as f, as is proved, and I will likewise have f a b equal to a b c. But in the third below, the space e h is free and can be large or small, and in the fourth not, because the triangle f a b is always orthogonal, that is with one right angle; and in the third such a triangle is always free and extends toward infinity by extending the line n m.

Here are 4 equal parts, that is the semicircle is equal to the lune, and the two sectors of the eighth part, each of them, is worth the lune and is worth the semicircle. And the motion of the line n r is of marvelous virtue, because the motion of its upper part increases the space of the lune together with its quadrature, made by the increase of the triangle together with the lune e h d. It matters as much to e h, the third surface, to be joined with the lune d as with the portion b, because b is worth as much as d.

Measurement. San Paolo of Rome has 5 naves and 8 columns, and is wide, within the width of its naves, 130 braccia, and from the steps of the high altar to the door 155 braccia, and from those steps to the last wall behind the high altar 70 braccia; the portico is 130 braccia long and 17 braccia wide. Done on the ... of August 1516.

Fifth - a - c - b. a is worth b. It is proved by the fifth which says: if 2 equal things are superposed one on the other, their common contact will be equal among itself and equal will be their remainder. So, the sector b c being equal to the semicircle a c, c being their common contact, a b the residue, that is the lune b, is equal to the portion a. The sector is the eighth part of a circle of quadruple proportion to the circle of the semicircle a c.