a b c d e f g h i K m L n o p q – r S t u x y 3 7

Here the 4 portions of the larger circle are restored from the 8 portions of the smaller circle. There remain to us another 4 portions of the larger circle which, having no restoration, are not to be removed.

In this experiment there remain to me below 8 half-portions, equal to the 8 portions of the whole smaller circle. Therefore I shall remove the 8 half-portions below from the half of the larger circle, and there will remain to me the square part that was joined to those 8 half-portions, that is the hatched or cross-hatched part; and I shall leave the 8 whole portions to their smaller circle, joined with its remaining square, throwing away the 8 half-portions that were above the larger circle, that is a b c d n o p q. Then I shall draw the hatched part of the square part joined to the smaller circle with its 8 whole portions, and I shall draw that hatched part from the outer side, that is from where the 8 half-portions above were detached. Then I shall remove the 8 lesser portions from their conjoined part, which will remain within, curvilinear, of the whole smaller circle.
* But it is better, before those 8 portions are removed from the smaller circle, that you square all of its squarable part, leaving those lesser portions within it.
It is better that to the hatched part, together with its 8 half-portions, there be added so much of the rectilinear part joined with the 8 portions of the smaller circle, that the said part joined with the 8 half-portions below be made equal to all the part that joins with the 8 portions of the smaller circle together joined with that smaller circle; and it will remain thus (Figures m, n). That is, the hatched part of m is that which was removed from the hatched part of n, and there remained the unhatched part of n together with its 8 small whole portions, equal to all of m, hatched and unhatched, together with its 8 half-portions. This done, I shall remove the whole and half portions from both surfaces, and there will remain — the hatched part above with its rectilinear addition — there will remain all squarable, and around the smaller circle there will remain the rectilinear surface on the outer side, and within, all curvilinear of the whole curvature of the smaller circle. And thus a square is given equal to a surface of a single curved side, and this was our intention. It remains for us to find a way of making the part that remains below, that is of the 8 half-portions with its hatched part, equal to its remainder above, by removing from that remainder such a rectilinear part that, joined to the rectilinear part below, makes the whole below equal to the remainder above; and thus far, at first sight, it seems an easy thing, having the portions equal in value in both surfaces.

In this case one must first square both surfaces, lending to or taking from each a part that is squarable, if it can be done; or, exchanging, again remove from the smaller circle its 8 whole portions and give back on the outside the larger circle with 8 half-portions, and you will have made the surface of the required nature in part of the surface below. But this is difficult to take away, unless you were to take from the rectilinear fronts (Figure); yet here, because the two surfaces that are worked are not equal or similar, one cannot proceed with true knowledge of the parts removed or added. But I shall take this one (Figure) which, by taking above the 4 half-portions and giving back below the 4 whole portions of the same value, remains squarable; and I shall do the same below, and there will remain the squarable part in this form (Figure), and the hatched remainder stands thus (Figure) on one side and likewise on the other. Whence, knowing the value of its 2 portions, that is (Figure a b) — that is, a b being equal to 4 of the smaller portions — I shall take half of the smaller circle with 4 of its portions, that is (Figure), and I shall remove so much from the inner rectilinear part, that is within, between the arc and the chord, that the remainder will be equal, as you see (Figures a, b), that is a to b. But by this I do not have knowledge of what part b is of its circle, and a remains in itself unsquarable, as b was at first. After we have found (Figure n) n to be squarable, I shall remove the squarable part that stands joined with it, as is shown above, which has portions equal to this, and it will be placed here (Figure m) in m, and I shall square n, and it will be this o (Figure o). Then I shall draw o, the smaller, from m, the larger, and I shall divide the remainder of m in half and distribute it to each surface. But it is better to throw it away, or else you add it to the hatched part below, and you will have made the hatched part equal to the figure o. Therefore that hatched part (Figure), remaining in itself unsquarable, is made squarable by the example equal to it and squarable.