Upon the strait line F Z, the fourth part of the radius A B, let the equilateral triangle _a_ Z F be constituted; and upon the centre _a_, with the radius _a_ Z, let the arch Z F be drawn; which arch Z F will therefore be equal to the arch Q F, the half of the arch B F. Again, let the strait line Z O be cut in the midst in _b_, and the strait line _b_ O in the midst in _c_; and let the bisection be continued in this manner till the last part O _c_ be the least that can possibly be taken; and upon it, and all the rest of the parts equal to it into which the strait line O F may be cut, let so many equilateral triangles be understood to be constituted; of which let the last be _d_ O _c_. If, therefore, upon the centre _d_, with the radius _d_ O, be drawn the arch O _c_, and upon the rest of the equal parts of the strait line O F be drawn in like manner so many equal arches, all those arches together taken will be equal to the whole arch B F, and the half of them, namely, those that are comprehended between O and Z, or between Z and F, will be equal to the arch B Q or Q F, and in sum, what part soever the strait line O _c_ be of the strait line O F, the same part will the arch O _c_ be of the arch B F, though both the arch and the chord be infinitely bisected. Now seeing the arch O _c_ is more crooked than that part of the arch B F which is equal to it; and seeing also that the more the strait line X _c_ is produced, the more it diverges from the strait line X O, if the points O and _c_ be understood to be moved forwards with strait motion in X O and X _c_, the arch O _c_ will thereby be extended by little and little, till at the last it come somewhere to have the same crookedness with that part of the arch B F which is equal to it. In like manner, if the strait line X _b_ be drawn, and the point _b_ be understood to be moved forwards at the same time, the arch _c b_ will also by little and little be extended, till its crookedness come to be equal to the crookedness of that part of the arch B F which is equal to it. And the same will happen in all those small equal arches which are described upon so many equal parts of the strait line O F. It is also manifest, that by strait motion in X O and X Z all those small arches will lie in the arch B F, in the points B, Q and F. And though the same small equal arches should not be coincident with the equal parts of the arch B F in all the other points thereof, yet certainly they will constitute two crooked lines, not only equal to the two arches B Q and Q F, and equally crooked, but also having their cavity towards the same parts; which how it should be, unless all those small arches should be coincident with the arch B F in all its points, is not imaginable. They are therefore coincident, and all the strait lines drawn from X, and passing through the points of division of the strait line O F, will also divide the arch B F into the same proportions into which O F is divided.