Of every square of equidistant lines and equal angles, the center of its gravity is equally distant from its angles. Let the square a b c d be of equidistant 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 point K. Therefore K will be the center of gravity of such a square; and if this square is hung at point K it will come to weigh equally, because all its opposite parts are equally distant from the center K and weigh equally, whence they are equally distant from the center of the world, being suspended at the said center K.

Of every bracket-shaped space (mensola) the center of its gravity is on the line that divides it into two equal parts, when two of its sides are parallel. Let the said bracket be a b c d, which you divide equally by the line f g, on which I say the center of gravity of this bracket lies. It is proved by drawing the line from angle a to angle d, which divides the bracket into two triangles a c d and a d b; of which, by the 2nd, find the centers, and let that of triangle a c d be h and that of triangle a d b be L. Now draw a line from center h to center L, on which line you will find the center of the two triangles as was proved by the 4th. And the line f g divides the bracket into two equal parts a b f g and c d f g, and divides the line h L at point K, which is the center of the said bracket, because the line f g divides the bracket into equal parts, whence one weighs as much as the other: so that on this line lies the aforesaid center of gravity, that is on h L, which is divided by the line f g at point K. Whence the proportion from h K to K L is that from the triangle a c d to the triangle a d b, so that they stand in balance. And because the center of gravity is on the line h L and on the line f g, and K is on the one line and on the other, therefore K is the center of gravity of the said bracket, which is what was proposed. And if the bracket 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 point K, so that L K is the root of 5 and 11/25 and h K the root of 2 and 94/225.

If a b is parallel to c d, the proportion of the triangle n to the 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 high as the triangle c d b.

Of every two equal triangles the center of their gravity is at the middle of the line that departs from the center of the one triangle and ends at the center of the other triangle. Let there be the two triangles a b c and d e f; let the center of a b c be g and the center of d e f be h. Now draw a line from center g to center h, and let the line be g h, which you divide equally at point K. I say that, if they weigh equally, they will ponder equally, being equal; because the line g h that issues from their centers is divided equally at point K, in such a way that the proportion from g K to K h is that from one triangle to the other. Therefore K is the center 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 balance g h suspended at the pole K. From the first of the first of Archimedes On Weights.

If two unequal triangles weigh equally at unequal distances, the greater will be set at the lesser distance and the lesser at the greater distance. You have the two triangles a b c and d e f. The first is the greater, and its center is g, and the center of the smaller triangle is h. Now draw from g to h, and you will have on that line the center of gravity of the two triangles at the point K. And at its extremes the two triangles will be suspended, of which the proportion will be such as that of the opposite distances they have from the point K, their fulcrum. Because the triangle a b c is sesquitertian to the triangle d e f, and likewise the space K h is sesquitertian to the space g K; whence, these triangles not being equal, the opposite proportions they have with their distances from the pole of the balance make them ponder 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.