a c - b d

I have two known quantities, such as 6 and 6; I take from them two known parts, such as 2 and 2, which leaves 4.
c a, being of like and equal sides, is squarable, and remains thus (rectangle figure). Then the half goes as you see here (rectangle divided at a b): a b divided into equal parts. Having ... the true half of c a equal to the half of b, and it being unknown in itself, I will cut from the body a part between straight lines equal to the half of the known c a, and it will stand thus (figure with e b o b – n e), and the whole o n will be equal to the half of c a. There remains to me b b e e equal to the other half of c a. Now, because the curve of b e is like but not equal to the opposite curve of the other b e, I will take from the greater so much of its convex curve as suffices to fill the concavity of the shorter concave curve, and the remainder of the greater curve, which is b b, will be set apart and stand thus (figure with b b). Then, the square of their remainder e e being made, I will draw it from the half of the known c a, and the remainder of that known c a will be equal to b b, that is (figure with b b); and from this I will draw the portion, and the rest lent is squared, which, added to the other square and placed upon the squared half of c a, does not fill it all, but a known part will remain, equal to the aforesaid portion.

a c - m n - b d - f e

I find myself with 2 known quantities, that is c a, d b, a surface equal to the surface a b e f when they are separated; and I take from c a d b the whole known d b, and from a b e f I take f equal to d b. There remains c a, a known part, equal to b e, an unknown part, and it stands as the following figure (in the margin four figures with: a - b - e; a c; b d; f). There remains, then, c a equal to a b e, and the equal remainders d b f are set apart. Then I make the squaring of c a, as is shown below, and I make the squaring of a b e, which is in quantity like a c. Then, because a is double and set one upon the other, I take it from the one surface and from the other, and there remains c equal to b e. And the half that I took by removing a - a is not a true half, but I take the imagined and true half of the true squaring of the whole c a, which is squarable, and of the whole a b e — because if the whole a c is equal to the whole a b e, separated from it, the half of the squaring of c a is equal to the half of the squaring of a b e. This done, and there remaining to me c equal to b e squared, and although b e is in itself unsquarable, I have squared it because it is equal to the half of c a squared, that is c. Now there remains to me b e equal to the square c; and because b c, being whole, is unsquarable — and this happens because its curved sides, convex and concave, although they are alike, are not equal, for which reason the squaring is impossible — I will therefore take from it e, the squarable part, and square it and draw it from c, and the remainder of that e will remain squared, and will then be equal to the b that before was never in itself squarable.

e b a; a c; e b; c; e b; c; e; b – e c