f c b a. a and b are set equal.

n o p q - e b d. Having set above a equal to b, and c to f, look below and you will find that be is equal to bd; and taking away b from each of the said surfaces, there remains the square d equal to the curvilinear figure e. Therefore, since bd is equal to c and the portion b is removed, the square d remains equal to the aforesaid lune e. Take the parallel a and with it remove as much of f as was before removed from c, which remained in the squarable lune e, and this will be o p q below. And to prove that this quantity o p is equal to the portion b below, you will take the parallel o p q drawn from such a circle that the portion o will be equal to all the little portions of b. But first remove the little portions from b, and the remainder will be squared, that is, sided (rectilinear). Then square that sided portion, and you will then make the square p equal to the said sided portion b. And this you will do by raising and lowering the portion o p q within the semicircle f - although such a procedure would not be geometrical - but first make the quadrature of the sided portion, then square half of the parallel above without the portion o, and you will have made a square greater than the quadrature of the sided portion. Therefore you will subtract the smaller square from the greater; then return the smaller square p to its first width of the first parallel, which you will do by the seventh proposition of my first book, and it will be the part p, to which you will add the portion o, and o p will be equal to the portion b at the side. And thus you will have made 4 equal surfaces, and of these c is equal to n and d is equal to q. Now subtract the lower portion from q; there will remain a square of q smaller than d, which you will subtract from that d. There will remain a little particle of the squared d, equal to the aforesaid portion of q.

a - b - c. Draw the parallel b from such a circle that its removed portion is equal to all the removed portions of the sector c; and let all of b first be equal to all of c, and let c be equal to all of a.

a - b. It is therefore necessary that the parallel b, being composed of 2 portions, as is shown below in a b, should be such that you find 2 sectors equal to the portion a or c. This will be found in this way: you have that the sector c is 1/4 of a circle, and from it a sector has been taken that is 1/6 of that sector c. It will have one of the 6 portions that you see removed, and such a removed sector will be 1/24 of its whole circle. Now, to find another sector whose portion is equal to the 6 portions of c, it is necessary that such a circle be 24 times greater than the whole circle of c.

The quadrature of the circle by Archimedes is well said and badly given. It is well said where he stated the circle to be equal to a right-angled triangle made from the circumferential line and the semidiameter of a given circle. And it is badly given where he squares a sided figure of 96 sides, from which 96 portions detached from those 96 sides come to be missing. And this in no way is to be called a quadrature of the circle; but in truth, by such rules, it is impossible to do otherwise.