1/8 subduple 4, a b, c, d, 1/16 double 8

I have here 2 sectors of a circle which are of equal quantity between them, that is, a b is 1/8 of a circle of 4, and d is 1/16 of a circle of 8, which are double one another. Hence the eighth of the smaller is worth as much as the quarter of the greater. So, by this, removing the portion c from each, the remainder of each stays equal between them. Hence a b is equal to d. Beyond this, I shall remove the part a and remove as much from d; the remainder will still stay equal. Of these 2 parts of the remainder, one, of d, remains squarable, and the other triangular, of 2 curved lines of unequal curvature. And thus that triangle remains squarable, being of sides of unequal curvature, and so one can make infinitely many.

m c f, d, e, n a b g

Let there be given a triangle of 2 curved sides, equal to a triangle of any variety of straight sides that can be given. In this case one must draw 2 equidistant (parallel) lines, of which one is joined to the base of the given triangle and the other touches its apex. Then, between the said lines, make a second triangle upon the same base as the first triangle, and let its apex touch exactly the upper line where the other triangle is. Then remove a portion of a circle from one of the sides of that triangle, and replace it on the opposite side. And thus we shall have made a curvilinear triangle equal to a given irregular triangle. As, let the given triangle be a b f upon the base a b. Then draw the 2 parallel lines m f n g, and let the line n g pass through the base of the triangle, and the line m f touch the apex. Then make the rectilinear triangle a b c that touches the said parallels, and from one side remove the portion b e c, and replace it in a d c.