[p] – [b] – m c a n – d – t S – o. a b c d are among themselves of equal value, as is proved. And further, because the circular parallel n p m o is worth the circle it encloses, that is a b c d, and removing the half of the circle a b c d, that is a c, there remains the aforesaid circular parallel with the half of the circle a b c d, that is there remains b d. Therefore the greatest circle keeps 3/4 of its value when the quarter a c is removed, which is worth the circles b d. Whence we shall remove the three quarters of the 4 greatest portions, that is n o m, and the remainder is a square, by removing the 4 portions of a circle worth 3/4 of the greatest circle n p m o.

n – b – c a – d – m. a b c d are among themselves of equal value, because n m are worth the 4 greatest portions.

i – K – e a – f b – h g d c. The two circles a b c d, e f g h are worth the semicircle i K a c. And removing i from the semicircle e c d, and g h from the two circles, one has removed parts equal to the semicircle and to the two circles, and the remainder K of the semicircle is worth the remainders b f of the two circles.

e – m L – b – f – c K i a – g – p d o – h. a b c d are of equal value and are worth the 4 greatest [portions] of the greatest circle. The remainder of the whole is divided into equal parts with various figures, that is the surfaces e f g h, i K, L m, o p, joined together as S shows here.

S 2 – 2 – 4. These 3 surfaces are squarable and are of equal value.

a – f e d – c b – r o. Portion a is worth the portions b c; f is common to the 2 equal surfaces placed one upon the other. It follows that the rectilinear triangles o r are worth the curvilinear ones d e. a b c is not to be understood except as parts removed from the two quantities f b c o r, a d f e.

f – b d – o – a – n – c e – r. a is worth b c. It is proved because the two circles are worth the semicircle, and the parts removed from the semicircle, that is d c, are the same as those removed from the two circles, that is they are common, and the portions lacking to the two semicircles in their junctions are worth f o n r. Therefore the remainder, or residue, of the two surfaces are equal, that is a is worth the two triangles b c.

e – a – d c – b – f. The circle a b c d is worth the two circles e a, b f. And if two quantities are in part superimposed one upon the other, whatever of them touches will be equal in quantity and similar in figure, and what of them does not touch will be equal in quantity and not in figure. Therefore a b will be similar in quantity and in figure; c d e f will be equal in quantity and not in figure, et cetera.

a – c b – d. The parallels a d are worth the lunes b c. And if a d is in itself squarable, and the two lunes are squarable, then the union of the lunes with the 2 parallels is also squarable, which square is worth the greatest square of the greatest circle.

d c b a. a b c d is worth double its antecedent, that is double its square.

b a. a b with its adjuncts is squarable, being itself the field of the antecedent.

b a – d c. a b c d is worth its antecedent.