[First].

The 12 portions of the enclosed half-sized circle are worth the 6 portions of the circle that encloses it; and this derives because the two circles are one double the other, and the number of these portions grows as much as their figures diminish, which, being diminished by half of their figures, have grown by half of their number, that is: the double portions are 6 and the subduple ones are made 12 and are of the value of those six. By what is said here above, make it that the number of the 6 greatest portions is always an aliquot part of the number of the portions into which they must be divided. As if the six greatest portions above had to divide the number of the 12 inner portions, since the divisor enters twice into 12; therefore six, which enters 2 times into 12, is an aliquot part of that twelve, because it divides it into two parts like itself; and 18 is divided by that 6 into 3 equal parts, and 24 into 4, and so it does infinitely.

Second.

Here the 6 greatest portions are an aliquot part, or you may say a multiplicative part, of the 18 portions enclosed by them. And this arises because the 2 circles are in triple proportion one to the other. Therefore the numbers of the portions of the circles have among them such proportion as the enclosing circles have with the enclosed circles.

Third.
In this third demonstration there are 24 portions of a circle against six greatest portions of the circle that encloses them; and the 2 circles enclosed one in the other are in quadruple proportion, and the same are the numbers of the portions, that is 6 greatest portions against 24 lesser portions; and it does as much to take the 6 greatest portions from the greatest circle as to take the 24 lesser portions from the lesser circle; that is, if you take the 6 greater portions leaving the lesser portions annulled, it is as much as to take the lesser portions and leave the greater portions standing.

False.
Here are four equal portions made upon the four portions of two circles and upon the two semidiameters of the same circles. Therefore each of these portions occupies a place that has two names, that is semidiameter and circumference. It follows that one same portion is upon the semidiameter of the one circle and upon the circumference of the other circle.

a-cb-d
Here are four equal portions, and each of them is made upon the sixth part of the circumference and upon the semidiameter of two circles equal to one another; and from this it follows that such portions are equal to one another, and each of them is upon the semidiameter of one circle and upon the circumference of the other.

* Those two circles will be equal to one another whose circumferences pass over their centres. It is proved to be so because the semidiameter of the one, which extends from the one centre to the other, is common to the two circles and is equal to all the other semidiameters of the two same circles.

* The two circles intersected one with the other in such a way that reciprocally the circumference of the one circle passes over the centre of the other ... Those two circles will be equal to one another when the circumference of the one passes over the centre of the other, just as the other passes over the centre of the one.