For example, let the first figure of three means be taken, whose half is A B C D (in fig. 6), and let the rectangle A B E D be completed. The complement therefore will be B C D E. And seeing A B E D is to the figure A B C D (by the table) as 5 to 4, the same A B E D will be to the complement B C D E as 5 to 1. Wherefore, if F G be drawn parallel to the base D A, cutting the axis so that A G be to G B as 4 to 5, the centre of equiponderation of the figure A B C D will, by the precedent article, be somewhere in the same F G. Again, seeing, by the same article, the complete figure A B E D, is to the complement B C D E as 5 to 1, therefore if B E and A D be divided in I and H as 5 to 1 the centre of equiponderation of the complement B C D E will be somewhere in the strait line which connects H and I. Let now the strait line L K be drawn through M the centre of the complete figure, parallel to the base; and the strait line N O through the same centre M, perpendicular to it; and let the strait lines L K and F G cut the strait line H I in P and Q. Let P R be taken quadruple to P Q; and let R M be drawn and produced to F G in S. R M therefore will be to M S as 4 to 1, that is, as the figure A B C D to its complement B C D E. Wherefore, seeing M is the centre of the complete figure A B E D, and the distances of R and S from the centre M be in proportion reciprocal to that of the weight of the complement B C D E to the weight of the figure A B C D, R and S will either be the centres of equiponderation of their own figures, or those centres will be in some other points of the diameters of equiponderation H I and F G. But this last is impossible. For no other strait line can be drawn through the point M terminating in the strait lines H I and F G, and retaining the proportion of M R to M S, that is, of the figure A B C D to its complement B C D E. The centre, therefore, of equiponderation of the figure A B C D is in the point S. Now, seeing P M hath the same proportion to Q S which R P hath to R Q, Q S will be 5 of those parts of which P M is four, that is, of which I N is four. But I N or P M is 2 of those parts of which E B or F G is 6; and, therefore, if it be as 4 to 5, so 2 to a fourth, that fourth will be 2½. Wherefore Q S is 2½ of those parts of which F G is 6. But F Q is 1; and, therefore, F S is 3½. Wherefore the remaining part G S is 2½. So that F G is so divided in S, that the part towards the axis is in proportion to the other part, as 2½ to 3½, that is as 5 to 7; which answereth to the fraction 5⁄7 in the second row, next under the fraction 4⁄5 in the first row. Wherefore drawing S T parallel to the axis, the base will be divided in like manner.