Equal quantities placed at equal distances weigh equally; and if equal quantities weigh equally, they are placed at equal distances. And if unequal quantities weigh equally, they are placed at unequal distances, the greater at the lesser distance and the lesser at the greater distance. And if equal quantities do not weigh equally, they are placed at unequal distances; the one that weighs more is at the greater distance. If two quantities at equal distances weigh equally, then any equal quantities placed at those distances will weigh equally. And every plane space has the centre of gravity on its surface within its bounds.

Every triangle has the centre of its gravity at the intersection of the lines that go from the angles and end at the middle of their opposite bases.

For example, let the triangle a b c be of equal sides, and b d the perpendicular, which holds within it the centre of gravity of such a triangle, and divides it into two equal parts, making of it two similar and equal triangles, so that they weigh equally. That this is so, I prove. a b, side of the triangle a b d, is equal to the side b c of the triangle b c d, and a d, side of the triangle a b d, is equal to c d, side of the triangle b c d. And thus b d is a side equal and common to both triangles. So that, by what has been said, it is proved that these two triangles are similar and equal to one another. And the same is concluded by the second [proposition] of the first [book] of Euclid. Now draw the line to the angle a, ending at the middle of the side b c at the point e. And then draw the line from the point b to the middle of the side a c at the point d, and a e at the point g. I say that the centre of gravity is on the line a e. Then draw the line from the angle c to the middle of the side a b at the point e, which will pass through g, at which I say the centre of gravity lies. And because each of these lines divides the whole triangle into two parts similar and equal in weight and in measure, and on each is the centre of gravity of the whole triangle, which, although it be on the two lines, cannot be more than one; whence it is concluded that such a centre must of necessity be at the intersection of the two aforesaid lines, that is at the point g. And if that triangle is 12 on each side, then d g will be the root of 12, which is 1/3 of b d, which is the root of 108; and b g is the root of 48, which is 2/3 of b d, by the 18th of the 14th [book] of Euclid; etc.

This same is concluded for triangles of different sides. It is proved. You have that the lines drawn from each of their angles to the middle of the bases divide the triangle in each direction into two equal parts, and on each line is the centre of all the gravity of the triangle; and the centre not being able to be more than one, it must necessarily be at the intersection of the said lines.

Of every equilateral pentagon the centre of its gravity is at the centre of the circle that circumscribes it. This is easily proved, because the equilateral pentagon touches the circle with all its angles, of which are made five similar and equal triangles, and their forces all terminate at the centre of the circle that circumscribes it. So that, suspending the pentagon at that centre, it will balance equally, because its angles are equally distant from the centre of such a pentagon, which is also the centre of the circle that circumscribes it.

a, o, r, p, m are the centres of the five triangles enclosed in the circle, and they make up the aforesaid pentagon. And the line a r and the line o p are the lines drawn from the centres of the four triangles; and the line c d issues from the midpoints of the said lines drawn from the aforesaid four centres, the middle of which is s. Now s is the centre of gravity of those four triangles. There remains the 5th triangle m, which must be counterbalanced against the other 4 triangles. Whence from the centre m draw the line to the middle of the line c d at the point s; and let this line m s be divided into five equal parts, and let it be balanced at the point n in such a way that the line is divided in the sesquiquartan proportion, that is 4 against one, and on the one part attach the 4 triangles, and on the 4 parts attach the one triangle, that is m; and thus you will find the line s m divided at the point n in that proportion, which is that of the weight of the 4 triangles against one.

When the pentagon is not of equal sides, it is divided into triangles; and of those the centres are found, and a line is drawn from one centre to the other, and according to the proportions of the weights of the triangles the distances of the centres are divided.

Let the pentagon a b c d e be proposed, which is divided into triangles; and with the given rule let their centres be found; and let the centre of the triangle a b d be o, and of the triangle a c d be n, and likewise let r be the centre of the triangle c d e; now draw a line from the centre of triangle a b d to the centre of triangle a c d, and you will have the line n o. On which, by the 4th, it is proved to be the centre of gravity of the two triangles. Now divide a b into equal parts at the point s, and divide c d into equal parts at the point p; then draw a line from the point s to the point p, and this line divides n o at the point m, which will be that proportion in the divided parts that is in the weight of the two opposite triangles. Therefore m is the centre of gravity of the two triangles, both for being in proportion of the distances and by the 5th of Archimedes, and because the centre of gravity is on the line n o and on the line s p, which divides it at the point m, their intersection. Now draw the line from the centre m to the centre r.