The areas of circles are to each other as the squares of their
diameters. If you have a circle 2 in. in diameter and another 4 in. in
diameter, then one circle will be four times as great in area as the
other, because the square of 4 is four times as great as the square of
2. Now, if we refer to Diagram 1, we see how two equal squares may be
cut into four pieces that will form one larger square; from which it is
self-evident that any square has just half the area of the square of its
diagonal. In Diagram 2 I have introduced a square as it often occurs in
ancient drawings of the Monad; which was my reason for believing that
the symbol had mathematical meanings, since it will be found to
demonstrate the fact that the area of the outer ring or annulus is
exactly equal to the area of the inner circle. Compare Diagram 2 with
Diagram 1, and you will see that as the square of the diameter CD is
double the square of the diameter of the inner circle, or CE, therefore
the area of the larger circle is double the area of the smaller one, and
consequently the area of the annulus is exactly equal to that of the
inner circle. This answers our first question.
[Illustration: 1 2 3 4]
In Diagram 3 I show the simple solution to the second question. It is
obviously correct, and may be proved by the cutting and superposition of
parts. The dotted lines will also serve to make it evident. The third
question is solved by the cut CD in Diagram 2, but it remains to be
proved that the piece F is really one-half of the Yin or the Yan. This
we will do in Diagram 4. The circle K has one-quarter the area of the
circle containing Yin and Yan, because its diameter is just one-half the
length. Also L in Diagram 3 is, we know, one-quarter the area. It is
therefore evident that G is exactly equal to H, and therefore half G is
equal to half H. So that what F loses from L it gains from K, and F must
be half of Yin or Yan.