h - b a - m n o - e - g d c f

These are 2 semicircles, double one to the other, and of the greater there remains a semi-annulus equal to the whole semicircle of the lesser circle. And from the semi-annulus I take away the half-portions a c f and b d g, and to the remainder I give its squaring. Then I take away the squared lune; there remains ...

d a - L K - o e c n - h m f 4 3 2 1 b i g

In this experiment the semi-annulus is placed of the same value as the semicircle circumscribing that annulus, and the 4 portions of the semicircle are worth the 2, a d, of the semi-annulus. Whence I will take the 2 portions a d from that semi-annulus and give back to it the same value with the 4 portions b c e f of the semicircle, and with this exchange I shall have made the surface n b - K c - L e - o f, which is rectilinear, worth the curvilinear surface n - a K - d L - o. Then I will take away - there remaining to me the 4 triangles 1 2 3 4, that is, the remainder of the semicircle - taking from it its 4 portions b c e f.
There remains to me the remainder of the semicircle in the figure of the 4 triangles 1 2 3 4, which is greater than the squared part of the semi-annulus by as much as is g i m h; whence, to reduce every part to the entire squaring of its whole, I will return the semicircle to its whole (Fig. a), and the like I will do with the part of the semi-annulus (Fig. b); and the like in quantity will be placed in the squaring of that part of such semi-annulus (Fig. e); and so I shall have b c equal among themselves and each less than the semicircle by as much as is the value of g i - m h. And for this I will take the 4 portions from the semicircle, and there will remain the 4 triangles 1 2 3 4. And the like in value I will take from g i - m h, and there will remain two square surfaces, and they will be (Fig. c a b) a b c. Then I will draw off that i m, or rather here I will draw off c d from the a; there will remain a (Fig.) in this figure; and it will be equal to the part of the semi-annulus. And so I shall have 3 equal surfaces in 3 various figures, which are a b c (Fig. a b e).
Therefore we have above the 4 triangles 1 2 3 4, equal to the squaring of the whole semi-annulus in b - K c - L e - o f m. But all this squaring of the whole semi-annulus is not equal to the whole curvilinear semi-annulus, but less by the g h portions, which do not have their value within, because the 4 portions of the semicircle are the value of 2 only of the semi-annulus. Therefore the 2 portions g h stay outside such working, and yet still the 4 portions are removed from the semicircle, because they are equivalent to those 2 ...

The case is to find in the portion h the value of the portion f, in the figure of another portion, without changing the periphery h.

e c - h m f D b i g

The squaring of the part of the annulus in n - a K - d L - o m will be n b - K c - L e - o f m. Or you may take the 4 triangles 1 2 3 4, which are of the same value, and the square parts are a b (Fig. a b) and are each equal to the c (Fig. c).
And this happens because the 4 portions of the b are worth the 2 of the c, the whole semicircle b being first equal in value to the whole semi-annulus.
After that, the sector 4 f is worth as much as the part of the annulus o m h, and since I take from the sector its portion f and lend it to the semi-annulus o m h, that semi-annulus has grown by all of f and the sector has diminished by all of it. I will take away the crescent of h, equal to the portion f in value, and still the semi-annulus stays greater than the remainder of the sector by a portion similar and equal to the f. Whence, f staying joined to its sector, the o m h would lack the value of the crescent equal to the f; whence I again take f from its sector and lend it no more to the semi-annulus, and the surfaces will stay equal, that is, the semi-annulus and the remainder of the sector, which is a rectilinear triangle, which, put back in the place f and taken from outside, there stays f o m, equal to the triangle 4. And here I have not found anything new.