o-f-n-t-b-m. f, being squarable, is equal to the n b, which is non-squarable. I add t n to f and make it squarable; and likewise I join m to the n b to make it squarable, that is, equal to f t n, and so I square b m, which have equal curvature. There remains n, non-squarable, which I set aside; and I make a square of f t n and another square of b m, which is smaller than the first by the whole part n. And so I take the smaller square from the greater, and the remainder of that greater is equal to n, whence n comes to be squared.

a - f - e d - c b. Divide for me the known lune a b c into equal parts by the line f e. Then square the quantity d b e c, which has equal sides. Then take up half the squaring of a b c, and what remains uncovered of that half will be equal to the sector f d e. I have a known surface enclosed between two curved boundaries and one straight one, and the curved ones are equal and alike; and I wish to make it equal to another known surface of two equal boundaries but not alike.

f e d - First. The first surface proposed is d e, which is known because its squaring is the rectilinear triangle e f, as is proved in the fifth of this; and it is enclosed between 2 curved side lines and one straight one for a base. And I wish to make it equal to another known surface, namely a b; and it is known because its squaring is b c by the sixth of this. And its curved boundaries are equal in quantity but not alike in curvature, because the shorter is the curvature of a circle double to the opposite, longer curvature.

d e a - b f c. If, within the hypotenuse of the right-angled triangle with 2 equal sides, a portion be removed, giving back a half-portion double the first outside the axis, you have restored the quantity of that right-angled figure. And it comes to the same to remove the half-portion inside the axis and put back a whole portion outside the hypotenuse, as to remove that whole portion inside the hypotenuse and give back the half outside the axis. No other difference results from the aforesaid operations, except that the triangle remaining within the right angle, between the half and the whole portion, sometimes has its axis curved and sometimes straight, and the hypotenuse does likewise.

e-L-d g-c b a-K f i [h]. If the lune a b c d is divided by the curve e f, it will be divided precisely into two equal parts, that is, a b will be equal to c d. And it is proved thus: h e f is equal to i e L K, because the one sector is an eighth of a circle double to another, of which the second sector is a quarter; and if you take g b from the eighth, a remains; and if you take b d from the quarter, there remains c equal to a, and b is equal to g and to d. Therefore the first lune, being composed of c and of d which are parts equal to a and to b, the lune a b is equal and alike to the lune c d. Therefore it is concluded that the great lune is divided into 2 equal parts by the line e f. Beside the left margin, lunes:

equal - equal.