Here it is shown how the proportion of the beam with its hypotenuse is the same as that which the weight of the beam has with the weight that this beam transmits to its two hypotenuses.

I say that the beam h c is 3/4 of one of its hypotenuses c f. Therefore the weight m is 3/4 of the weight that the two beams feel.

What is said is proved. Because from the line b c a quarter is cut off in b h, there remain the 3/4 in h c. It follows that these 3/4 are in such proportion with their whole b c as this beam was with its hypotenuse. Therefore the hypotenuse is similar to the whole perpendicular b c. But what has been said up to here does not conclude. It is proved by the balance f h and f e, which has 2 of counter-lever in f h and 3 of lever in f e: it follows that 3 at h can prevail against the 2 at e, and thus the weights are in the same proportion as the arms of the balance were among themselves, that is 2/3. Now because at m is 4, and in this second demonstration we have set it 3 against the 2 of the appendage c i; now because at m is 3 and at i is 2, that 2 being 2/3 of 3, and it answers to the proportion of the arms of the said balance; therefore at i is 2/3 of the weight m which is 3. In this proof the weight that touches a single hypotenuse is made 2, and as much is loaded on the opposite hypotenuse, which makes 4; and thus we have made the weight m, which is 3, be 3/4 of the weight that the two hypotenuses feel; and it meets with the first, that the beam was 3/4 of one of its hypotenuses.

The semicircle f h c is made upon the hypotenuse f c only to find within it the right angle f e c, which is the right-angled junction that the lever f e makes with its appendage e i.

I suspend the weight b, which is 2, attached to the cord h K. What counterweight must there be at a, c to sustain it in the situation of equality, the balances being equal and their arms equal among themselves?

The second figure here is of the same substance as the first above.

Here is sought the weight 4, in what way it varies on its supports f, g, h placed in various situations, and their junctions with the shaft being always at right angles.

I wish to move a heavy body suspended in the air along different lines, and I seek to know what power its mover has to use for each of the given lines.

As many are the varieties of weights that one and the same heavy body gives of itself to its mover, as are the varieties of the obliquities whence the mover moves that weight.

This suspended heavy body is conditioned in such a way that it is wholly at b by likeness of power, whence the side of the rectangle f a feels no resistance at the beginning of the motion which it makes to move a from its place. The mover e feels such proportion with its remainder as the lever g b has with the counter-lever b a; it is 13/14, that is, because the lesser lever is 13 and the greater 14, a being 13 pounds, and g would want 14 of power. There follows the mover d n, at right angle to the lever n b, which the lever has, being 10/14 of the counter-lever b a, whence 14 at d resists 10 at a; then the mover c m of the lever m b which has 5/14 of the counter-lever b c, whence 14 at c is against 5 at b.

This calculated figure is like heavy bodies dragged over different obliquities, and nothing else varies except the power of the friction of the weight, because in the air there is no friction given.