Geometry.

a — d e — o — b c

Make them with narrow bases.

Let there be given a crescent surface equal to a triangle. This crescent that is sought is a d o b e c, equal to the triangle a b c. It is proved because the two semicircles a d b and a e c are similar and equal to one another, and if from equal things equal parts are taken, the remainder is equal. It follows that, removing the part a d o from the semicircle a e c and removing the same a d o from the semicircle a d b, there remains a b o equal to a d e o c. And if to equal things equal things are added, the joined wholes will be equal; and for this reason I add the triangle o c b to the aforesaid crescent, and I add the like triangle to the triangle a b o, and I shall have the triangle a b c equal to the crescent a d e b c, and thus we have proved what was needed, etc.

f — a n h — o — d c e

n b c is worth b o c e.

The contained semicircle is worth half the circle that contains it, their positions of curvature being opposite.

a — d — c e — f — b

Here are two circles, one double the other, made upon one same centre, and for this the similar parts of the greater are double the similar parts of the lesser. It follows that, removing the two portions a b from the greater and removing the four portions c d e f from the lesser, the remainder of the greater is worth the remaining square of the lesser.

The animals, movers of carts, have most simply demonstrated to us the squaring of the circle that the wheels of such carts have, by means of the track of the circumference of such wheels which becomes rectilinear.