With numbers you have, for the first, b d which is 6 and b K which is 3 3/7, and the triangle a b c is 36, and the remainder of the portion is 20 and 4/7; for that proportion which b d (⁶ has to b K (3 3/⁷ is the same as the triangle a b c (³⁶ has to the remainder of the portion (20 4/⁷. For the second, you have the centre of the triangle a b c, which is e, and the centre of the two portions, which is r; and from r to e is 1 4/7, divided at point K, and e K is 4/7 and r K is 1; for that proportion which K r (¹ has to r e (4/⁷ is the same as the triangle a b c (³⁶ has to the remainder of the portion (20 4/⁷, which makes the proposition clear, etc.

Of every pyramid, round, triangular or square or of however many sides, the centre of its gravity is at the fourth part of its axis nearest the base.

Let the pyramid be a b c d with base b c d and let the apex be a. Find the centre of the base b c d, which will be f; then find the centre of the face a b c, which will be e, as was proved by the first. Now draw the line a f, on which lies the centre of gravity of the pyramid, because f is the centre of the base b c d, and the apex a is perpendicularly above f, and the angles b, c and d are equally distant from f and weigh equally, so that the centre lies on the line a f. Now draw from the angle d to the centre e of the face a b c a line cutting a f at point g; I say, for the reasons stated above, that the centre of gravity lies on the line d e. Therefore, the centre being on each and being able to be no more than one, it must lie at the intersection of those lines, which is the point g, because the angles a, b, c and d are equally distant from that g.

The lower axes of the faceted pyramids, which arise at the third of the axis of their bases, will intersect at the fourth of their length toward the base.

See here in the first figure, going toward the left hand, that which is promised above; and in the middle figure it is shown how the lines a f and d e arise at their angles and terminate at the third of the height of the axis of their faces. And in the third figure one sees how, the lines being drawn from the third of the height of the axis of the aforesaid sides and terminated at the opposite angles, these lines intersect at the fourth of their length, as is shown by the lines p t and n r intersecting at the point s.

Let the portion of a half-circle be divided into 8 similar and equal triangles, of which find the centres by the first rule. Then draw a line from centre to centre and you will have four lines, each of which terminates at the centres of two triangles, which triangles are a, b, c, d, e, f, g, h. Now take the middle of the line c d and of the line e f, which will be r and s, and draw the line from r to s; then draw the line x u from the middle of the line a b and of the line g h; then draw the line n t from the middle of the line r s and from the middle of the line x u. Now it is concluded that the centre of gravity of the portion is at the middle of the line n t, at the point m.

For every segment of a rectangle the centre of gravity is found on its diameter, dividing it in sesquitertian proportion and residing in the smaller part toward the base.

Let the segment be a b c and a c the base, which is divided in half at point d, and above d the perpendicular b d, which is the diameter of the segment, within which the triangle a b c is described. Now divide b d, the cathetus of the triangle a b c, at point K, so that the proportion of b K to K d is as that of the triangle to the remainder of the portion; and it will follow that, just as the portion consists of the triangle and the residue, so on the cathetus consist the triangle and the remainder of the portion, that is b K and K d. So that, these quantities being in one proportion, as has been proved, I say K is the centre of the segment a b c, and b K is sesquitertian to K d. Furthermore, divide the diameter of the two portions, called the residue, in sesquitertian proportion, which diameters are L m and o p. Now divide L m at point n so that L n is sesquitertian to n m, and likewise let o p be divided at point q so that o q is to q p in sesquitertian proportion. Then draw a line from q to n and you will cut b d at point K, which K is the centre of the two portions, because it is at the intersection of the two lines, that is b d and q n, at which intersection is the common centre of the two portions. Then find the centre of gravity of the triangle by the given rule, which will be at e; now at e is suspended the triangle, and at K are suspended the two portions.

R divides K e in sesquitertian proportion; the triangle is at the lesser distance, the residue at the greater.