By this method it is manifest, that the base of a semiparabola will be divided into 3 and 5; and the base of the first three-sided figure of two means, into 4 and 6; and of the first three-sided figure of four means, into 6 and 8. The fractions, therefore, of the second row denote the proportions, into which the bases of the figures of the first row are divided by the diameters of equiponderation. But the first row begins one place higher than the second row. [Sidenote: The centre of equiponderation of the half of any of the figures of the second row of the same table may be found out by the numbers of the fourth row.] 12. The centre of equiponderation of the half of any of the figures in the second row of the same table of art. 3, chap. XVII, is in a strait line parallel to the axis, and dividing the base according to the numbers of the fraction in the fourth row, two places lower, so as that the numerator be answerable to that part which is next the axis. Let the half of the second three-sided figure of two means be taken; and let it be A B C D (in fig. 7); whose complement is B C D E, and the rectangle completed A B E D. Let this rectangle be divided by the two strait lines L K and N O, cutting one another in the centre M at right angles; and because A B E D is to A B C D as 5 to 3, let A B be divided in G, so that A G to B G be as 3 to 5; and let F G be drawn parallel to the base. Also because A B E D is (by art. 9) to B C D E as 5 to 2, let B E be divided in the point I, so that B I be to I E as 5 to 2; and let I H be drawn parallel to the axis, cutting L K and F G in P and Q. Let now P R be so taken, that it be to P Q as 3 to 2, and let R M be drawn and produced to F G in S. Seeing, therefore, R P is to P Q, that is, R M to M S, as A B C D is to its complement B C D E, and the centres of equiponderation of A B C D and B C D E are in the strait lines F G and H I, and the centre of equiponderation of them both together in the point M; R will be the centre of the complement B C D E, and S the centre of the figure A B C D. And seeing P M, that is I N, is to Q S, as R P is to R Q; and I N or P M is 3 of those parts, of which B E, that is F G, is 14; therefore Q S is 5 of the same parts; and E I, that is F G, 4; and F S, 9; and G S, 5. Wherefore the strait line S T being drawn parallel to the axis, will divide the base A D into 5 and 9. But the fraction 5⁄9 is found in the fourth row of the table, two places below the fraction ⅗ in the second row.