...in weight this square which is on a x to that which is on a b d. Now the line x is indeed greater than x p, but less by that excess by which the axis is greater than the sesquialteral (half again as much) of the part reaching to the axis. Let there be fitted in between a certain one of the conic sections a p o l, a z d, namely n o equal to x, and let it cut the remaining section of the cone at one point, and the straight line r s at b. It will be demonstrated that it is so, just as it was demonstrated that p ω is double of s x. From o let there be drawn o s tangent to the section a p o l, and a line perpendicular upon b d; and let a be joined to n. Then a n and q n will be equal to one another. For since in the similar segments a p o l, a x d, lines drawn from the bases to the segments a n, a q make equal angles at the bases, q a and a n will have the same ratio to l a, a d, on account of the second figure of the aforesaid propositions. Therefore a n is equal to q n and parallel to o s. It must be demonstrated that, when the segment is let down into the fluid so that its base does not touch the surface at a single point, the axis makes with the surface of the fluid an acute angle greater than the excess. Let it be demonstrated: let it be set so that its base touches the surface of the fluid at one point. Let the segment be cut through the axis by a plane at right angles to the surface of the fluid; let the section of the segment's surface be a p o l, the section of a right-angled cone, that of the fluid's surface o a, and the axis of the section and diameter b d; and let b d be cut at k and r, as has been said. Let there be drawn also p g, a straight line parallel to a o, tangent to the section a p o l at p, and p t parallel to b d, and p s perpendicular upon b d. Since therefore the segment immersed in the fluid has in weight the ratio of the square on a x to that on a b d, and whatever ratio the segment has to the fluid, this the submerged part of the segment has to the whole, and whatever the submerged part has to the whole, that has the square on a t p to the square on a d b, the line x will be equal to t p, and therefore n o too is equal to t p. Wherefore the segments a p g, a p s are equal to one another. And since in the equal and similar segments a p o l, a b l k z b, the lines t a, a q are drawn from the extremities of the bases, and the segments cut off make equal angles to the diameters, on account of the aforesaid...