n-c a-b - o; f d - e. Here I have a which, being of unequal sides, is non-squarable. So I lend it the known triangle b c and compose the single surface a b c, and within that triangle I complete the lune a b, which quantity, being made of equal and alike sides, is made squarable; and thus rectilinear I set it aside in d e, and I join to it, as I did before, the triangle e f, and I have made d e f equal to a b c.

a; c b a. I have a, and to make it square ... and of it I use only b, and I have made the lune a b, which is squarable, by removing a from its place and putting it back at c; and by this I have consumed all of the first a and consumed the lent part b c ... to the portion ... it concludes that of the quantity a b c ... the portion a to make its squaring; of the b c of the other figure that follows, half of that portion is consumed.

b e a - d f c. The lune is divided into 2 equal parts along its height in circles that are double one to the other. Such a proportion holds from circle to circle as from square to square made of their diameters.

To the ...

a b; f o m n. a b are equal. I take from m n, which is like b; I take away n and give it back o, which is half of n. There remains m with half of n, o having been given back to it, which o is also taken from f. Therefore m o remain equal to f.

a-d b e f; o a r-c b; a-d b. I have a b equal to c d, and I have c d e f double a b. I take b d from a c and take f from c d, and there remains a c equal to c d e. Then I take a b, which in itself is squarable, and add to it the sector d, which is non-squarable, whose quantity I join indeterminately with b and make a single surface, which in itself, having straight sides without any division, is squared. This done, I square the lune a b, which stands thus (figures) and before (figure with t o r p) stood thus. And because, in squaring the lune a r which we know with certainty to be equal to a rectilinear triangle, its quantity is known, we shall set such knowledge aside thus (figure). Then we shall take the triangle b d and add it incorporated in this manner (figure with a n o), and it will be equal, being all single, to the quantity a b d single, because the whole a n is equal to the whole a b, since it is its true squaring by the fifth of this; and so the whole n o joined together is equal to the whole b d joined together. And this saying 'joined together' is necessary because as yet there is no knowledge of the parts, that is, how much b is smaller than d; but we shall leave that division unknown and square the whole of the triangle b d, which is known. This done, the whole is reduced to a square.

b a - c; b a-e d-c-f; a-e d; e-c-f; a-e d; e-c-f; e a d; f c. I find a b known, that is, squarable, and I find b c known and equal to a b, and they are two squarable and equal surfaces. And by the fifth of this I lay one over the other so that b remains double and a c remains single. And by the sixth of this, what of them touches is equal in quantity and equal in figure, and what of them does not touch remains equal in quantity and unequal in figure. Therefore a is equal to e in quantity and not in figure. It follows by the seventh that if to equal things equal quantities be added, the sums will be equal: d e equal to f added to a b, and f added to e c, the sums remain equal — a d e and e c f — separately and divided equally among themselves, as shown in the following figures. a d e, therefore, being separately figured, is equal to e c f, as proved above. I say that in a d e a known triangle in the shape of a lune is included, which is a d, and its remainder is a part of the triangle d e; and although that whole triangle d e is known, the part e, having a curved side, remains unknown, and our whole demonstration extends to making it known, that is, squared.

a-e d; n-o r. I lent to a the triangle d e, because, not having equal sides, it was not squarable. Therefore a by itself, then added with the triangle d e, makes a quantity in which a, without other help from d e, can be squared. a d e is not in such a state, but in another like it; as, making apart the triangle n r equal to a d, then with the addition you will make the triangle r o equal to the triangle d e, and there remains n equal to a. I have two equal surfaces, namely a d e and n r o; and they are equal because n r is the squaring of a d by the sixth; and for such squaring I lent d e to a, because in itself a, not having equal sides, was not squarable. Then I have r o equal to d e, and each is single, that is, a simple surface. Therefore they are equal.

Conclusion. In a b d a known triangle in the form of a lune is included, which is a b, and its remainder is a part of the triangle b d; and the whole quantity a b d joined together is known, but not each one separately by itself — that is, a b is not known in itself, nor b d in itself separated from a b, nor b, nor d in itself; but everything joined together composes, as I said before, a rectilinear triangle drawn from a b d, in that a b is a known rectilinear triangle, which is b d. And if you wished to separate the known b d from a, a remains unknown and non-squarable; and if you wished to draw the known a b out of the known b d, d would remain unknown and non-squarable.

a b is equal to b c, but a b is not equal in itself, nor is it necessary that it be so, nor is it possible to make it so; but since it does no harm, it does not matter, provided that a is equal to c.

This is proved by division, as shown in the following figures.

I lend to a the triangle d e, and I complete in d e the whole known triangle a d, and I make it rectilinear.

... is of one and the same quantity ... the unknown part made known, and ... triangle e.

a - e d a; b ... remains equal.