a; b; d c. On portions. Of portions in sixteenfold proportion, the sixteenth part of the greater is worth the whole of the lesser. It follows that the sector d, which is the sixteenth part of the said greater portion, is worth the portion b, which is one-sixteenth of the greater portion. Therefore the portions cd are worth the two sectors ab.

c-ef-h-b da-g-S-f. All the cuts made from the arc to the chord that enclose within them the portions of the circles are to be directed toward the centre. The portion of a circle is divisible infinitely into similar and equal parts. Proof: let the portion of the circle be acb, which is to be divided into 4 similar and equal parts. First it is divided into two equal parts by the line cd, so that cdb remains equal to the other half cda. Then this half-portion cdb is divided into two other equal parts, first drawing the straight line cb, chord of the arc ceb, and you will have made the second portion cebh, which is divided into equals by the perpendicular eh. Besides this, the triangle cbd is divided into equals by the straight line hd, and this is proven to be divided into equals by one [proposition] of the Elements of geometry. And thus we have divided the half-portion into equals but not into similars; and to make them similar we draw a straight line equidistant [parallel] to the line chb, which will be the line dS, and we will have made the parallelogram cbdS, which we cut in half with the line ef, extending from the circumferential line to the centre of the circle. Then I will draw the lower angle of the triangle cdb to touch that line and I will make the triangle cbg equal to the triangle cdb. Such equality is proven by one [proposition] of the Elements of geometry, and thus I have divided the greatest portion into 4 parts similar in figure and equal in quantity.

Of all the triangles made on an equal base with equal heights, converging to one same point: they are equal among themselves.

First - Quarter. Second - Eighth. Third - Sixteenth.

bda-c-gcf. Given that the sector adbg is right-angled and that I wish to reduce it to equal angles, I will draw the straight line ab parallel to the line fg, and I will draw the angle g to the position c below the middle of the arc ab at the point d, and thus it will be of equal angles. (Two figures follow.)