In every square of parallel lines and equal angles, the centre of their gravity is equally distant from its angles. Let the square be a b c d, of parallel sides and equal angles. Draw a line from angle a to angle d; and from angle b to angle c: these lines will intersect in the middle of the square at the point K. Therefore K will be the centre of gravity of that square; and if that square is hung at the point K, it will come to weigh equally, because all its opposite parts are equally distant from the centre K, and weigh equally, and so are equally distant from the centre of the world, being suspended at the said centre K.

Of every trapezoidal space, the centre of its gravity is on the line that divides it into two equal parts, when two of its sides are parallel. Let the said trapezium be a b c d, which you divide into equal parts with the line f g, on which I say the centre of gravity of such a trapezium lies. It is proved by drawing the line from angle a to angle d, which will divide the trapezium, making of it two triangles a c d and a d b; of which, by the 2nd, find the centres, and let that of triangle a c d be h, and that of triangle a d b be L. Now draw a line from centre h to centre L: on which line you will find the centre of the two triangles, as was proved by the 4th. And the line f g divides that trapezium into two equal parts a b f g and c d f g, and divides the line h L at the point K, which is the centre of the said trapezium, because the line f g divides the trapezium into equal parts, so that one weighs as much as the other: so that on that line h L is the aforesaid centre of gravity, which is divided by the line f g at the point K. Whence the proportion from h K to K L is as that from triangle a c d to triangle a d b, so that they stand in balance. And because the centre of gravity is on the line h L and on the line f g, and K is on both lines, therefore K is the centre of gravity of the said trapezium, which is what was proposed. And if the trapezium weighs 60, the triangle a b d weighs 36, and the triangle a c d weighs 24, and the line h L is the root of 15 and 1/9, divided in the proportion of the triangles at the point K, so that L K is root of 5 and 11/25 and h K is root of 2 and 94/225.

If a b is parallel to c d, the proportion of triangle n to triangle m will be such as that of the line a b to the line c d.

It is proved, because the triangle a b c is as tall as the triangle c d b.

Of every two equal triangles, the centre of their gravity is at the middle of the line that goes from the centre of the one triangle and ends at the centre of the other triangle. Let the two triangles be a b c and d e f; let the centre of a b c be g; let the centre of d e f be h. Now draw a line from centre g to centre h; and let it be the line g h; which you divide into equal parts at the point K. I say that, if they weigh equally, they will balance equally, being equal; because the line g h that issues from their centres is divided into equal parts at the point K, in such a way that the proportion from g K to K h is as from the one triangle to the other. Therefore K is the centre of gravity of the two triangles, being between them in the same proportion as that of their distances from the point K, the fulcrum of their equilibrium g h suspended at the pole K. First [proposition] of the first [book] of Archimedes On Weights.

If two unequal triangles balance equally at unequal distances, the greater will be placed at the lesser and the lesser at the greater distance. You have the two triangles a b c and d e f. The first is greater, and its centre is g, and the centre of the smaller triangle is h. Now draw a line from g to h, and on that line you will have the centre of gravity of the two triangles at the point K. And at its ends the two triangles are suspended; of which their proportion will be as that of the opposite distances they have from the point K, their fulcrum. Because the triangle a b c is one-and-a-third to the triangle d e f, and likewise the space K h is one-and-a-third to the space g K; whence, the triangles not being equal, the opposite proportions they have with their distances from the pole of the equilibrium make them balance equally; because the space g K is 3, and the space K h is 4, and the triangle d e f is 4 and the triangle a b c is 3.