By the same method, if in the same second row there be taken the second three-sided figure of three means, the centre of equiponderation of the half of it will be found to be in a strait line parallel to the axis, dividing the base according to the numbers of the fraction 6⁄10, two places below in the fourth row. And the same way serves for all the rest of the figures in the second row. In like manner, the centre of equiponderation of the third three-sided figure of three means will be found to be in a strait line parallel to the axis, dividing the base, so that the part next the axis be to the other part as 7 to 13, &c. Coroll. The centres of equiponderation of the halves of the said figures are known, seeing they are in the intersection of the strait lines S T and F G, which are both known. [Sidenote: The centre of equiponderation of the half of any of the figures in the same table being known, the centre of the excess of the same figure above a triangle of the same altitude and base is also known.] 13. The centre of equiponderation of the half of any of the figures, which (in the table of art. 3, chap. XVII) are compared with their parallelograms, being known; the centre of equiponderation of the excess of the same figure above its triangle is also known. For example, let the semiparabola A B C D (in fig. 8) be taken, whose axis is A B; whose complete figure is A B E D; and whose excess above its triangle is B C D B. Its centre of equiponderation may be found out in this manner. Let F G be drawn parallel to the base, so that A F be a third part of the axis; and let H I be drawn parallel to the axis, so that A H be a third part of the base. This being done, the centre of equiponderation of the triangle A B D will be I. Again, let K L be drawn parallel to the base, so that A K be to A B as 2 to 5; and M N parallel to the axis, so that A M be to A D as 3 to 8; and let M N terminate in the strait line K L. The centre, therefore, of equiponderation of the parabola A B C D is N; and therefore we have the centres of equiponderation of the semiparabola A B C D, and of its part the triangle A B D. That we may now find the centre of equiponderation of the remaining part B C D B, let I N be drawn and produced to O, so that N O be triple to I N; and O will be the centre sought for. For seeing the weight of A B D to the weight of B C D B is in proportion reciprocal to that of the strait line N O to the strait line I N; and N is the centre of the whole, and I the centre of the triangle A B D; O will be the centre of the remaining part, namely, of the figure B D C B; which was to be found.